This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
1. 9 We remark that the estimates in the proposition are crude. Additionally, one can show that if G has a NTRM, it has one in which the boxes form a m a x i m u m clique, that G has at most n 2 / 4 maximum cliques, and that these cliques can be obtained efficiently. See [Ma93] for the details. From now on we assume that the vertices of G are already partitioned for us into boxes B, segments S and points P, with B forming a maximal clique, the vertices that become isolated in G - B forming the set P, and the remaining vertices forming the set S and inducing a bipartite graph. We ask whether we can position these boxes, segments and points to form a NTRM so that their intersections represent the adjacencies of G. We may now assume that no two boxes are twins, namely have the same closed neighborhood, for otherwise we can delete one of them without affecting the answer (if we find a N T R M without this box, we can then add it as an exact or approximate copy of its twin box). Similarly we may assume that no two points are twins, namely have the same open neighborhood. As for the segments, we assume for the moment that the bipartite graph that they induce is connected, and therefore the segments partition uniquely into two sets that we may assign arbitrarily as horizontal and vertical segments. We may then assume that no two horizontal and no two vertical segments are twins, for similar reasons. The search for a NTRM is done in two stages. In the first stage we ignore the tips of the segments and try to place the other key-points (corners of boxes, points, and bases of segments) in the plane so as to reflect the adjacencies of the corresponding vertices, except for adjacencies between segments. In the second stage we arrange the segments, without disturbing their intersections with the boxes, to reflect their adjacencies with each other. Let us define a binary relation ~ on Ii{2 as follows" ( x l , y e ) __ ( x 2 , y 2 ) , when X 1 ~ X 2 and y~ _< y2. We say that ( x i , y l ) d o m i n a t e s (x~,y2), written as (xl,yl) -~ (x2, y2), when (xl, yl) ~___ (x2, Y2)and (xl,Yl) -r (x2,y2). Two different points that do not dominate each other are said to be incomparable. Clearly ~ is a partial order on Ii{2, and it is 2-dimensional, since it is the
230
2-Threshold Graphs
intersection of the linear orders of the separate z and y coordinates. The strict partial order -~ is also 2-dimensional. We are interested in finding domination relations between some of the key-points in every NTRM of the given graph G, namely between the corners b of the boxes, the bases s of the segments and the points p (since we are now in the first stage, we ignore the tips of the segments). It turns out that to a large extent these can be deduced directly from the neighborhoods of G. Every NTRM of G must satisfy the following, where we use the letters b, s, p to denote both the key-points of the geometrical objects and the corresponding vertices of G" 9 For a box corner b, a base s and a point p we have s-.< b < > s c:_ N(b)
(8.11)
p--< b ~
peN(b).
(8.12)
N[bl] C N[b2]
(8.13)
9 For box corners bl a n d b2 we have b1 ~ b2 ~
(the ~ the ~
part follows from b~ C b~ and from the absence of box twins; part follows from Step 5 of the normalization of the TRM).
9 For a point p and a box b we have b-.~ p ~
Yb' E N(p)
(8.14)
b-~ b'
follows from Step 3 of the normalization of the TRM). Hence (the < by (8.13) (8.15) b -~ p r Vb' e N(p) N[b] C N[b']. 9 For points pl and p2 we have
Pl ~ P2 ~
(8.16)
N(pa) D N(p2)
(the ~ follows from the absence of point twins; the < Step 3 of the normalization of the TRM).
follows from
9 For a point p and a segment s we have s -.~ p ~
(the ~
N(p) C_ N ( s ) N B
follows from Step 3 of the normalization of the TRM).
(8.17)
8.7
Intersection Threshold Dimension 2
231
9 For segment bases 81,82 E ~x or 81,82 E ~y we have the implication 81 "~ 82 ,r
N ( S l ) N B D N(s2) N B.
(8.18)
Let Kb, Kp, Kx, Ky be the sets of box corners, points and bases of the vertical and horizontal segments, respectively. Let R be the binary relation on K = Kb O I(p O K~ O Ky, R such that (a, b) C R if and only if a -~ b according to (8.11)-(8.18). The relation R can be computed from the neighborhoods of G. If G and the partition of its vertices into B, P, S~, Sy have a NTRM, then P = (K, R) is a strict poset of dimension 2, which we know how to check in polynomial time. If it turns out that P has a realizer, we can use the positions of the elements of K in the two linear extensions of P as coordinates of the corresponding key-points in R 2. However, we want to ensure that the bases of the vertical segments are lower than any other key-point and that the bases of the horizontal segments are to the left of any other key-point. Therefore we solve the restricted poset dimension 2 problem given by P and the following two sets of incomparable pairs of P:
{(a,b) : a C Kx, b ~ Kx,(a,b) ~ R} R2 = {(a,b) : a C Ky,b ~ Ky,(a,b) ~ R}
1~ 1
:
(i.e., we ask whether R is the intersection of two linear extensions containing RU R1 and Rt.J R2, respectively; this can be determined in polynomial time by Theorem 8.6.14). If the answer is negative, the current B is discarded. If the answer is positive, the two linear extensions are used to place the box corners, the points and the segment bases in the plane. Since the bases of the vertical segments are lower than any other key-point, we can lower them down further to the x axis without changing their intersections with the boxes or any domination relations, and similarly for the bases of the horizontal segments. We then obtain a NTRM for G except for the intersections between segments, since we have not yet placed the segment tips. Now we enter the second stage and try to arrange the segments bases and tips without changing the intersections between segments and boxes, so as to reflect the intersections between segments. Let us ignore the intersections between segments and boxes for the moment. We say that a bipartite graph has a segment intersection model when it is the intersection graph of vertical and horizontal segments in the first quadrant whose bases are on the coordinate axes. The question is then whether the bipartite subgraph of G induced by the segment vertices has a segment intersection model.
232
2-Threshold Graphs
P r o p o s i t i o n 8.7.2 A bipartite graph has a segment intersection model if and only if it is the intersection of two difference graphs with the same color classes. Proof. " O n l y if"" Consider a segment intersection model and its bipartite intersection graph Q. Consider the grid defined by the lines containing the segments. The horizontal and vertical lines of the grid correspond to the rows and columns of the adjacency matrix A of Q, and A has a 1 in a given position if and only if the corresponding horizontal and vertical segments intersect. Thus A is the intersection (component-wise product) of two adjacency matrices A1 and A2:A1 has l's in the positions corresponding to the grid points on the horizontal segments, and similarly for A2 and the vertical segments. Therefore Q is the intersection of the the bipartite graphs/-/1 and /-/2 whose adjacency matrices are A1 and A2, respectively. Now //1 is a difference graph, since every two rows of A1 are comparable by inclusion according to the lengths of the corresponding segments. Similarly//2 is a difference graph, since every two columns of A2 are comparable. Figure 8.26 illustrates this argument. "If": We assume that the bipartite graph Q is the intersection of difference graphs //1 and //2 having the same color classes. Let the color classes be X : { X l , . . . , x k} and Y : {yl,. 99, Yl}, where N , l (xl) D " ' " D N , 1 (Xk)
(8.19)
N , 2 ( Y l ) _~ " ' " __~ NH~(y,).
(8.20)
Consider the adjacency matrices A of Q, A1 of HI and A2 of//2 as having columns corresponding to Xl,..., xk from left to right and rows corresponding to y l , . . . , y t from bottom to top. By (8.19) and (8.20), the l's in every row of A1 are consecutive from the first column on, and the l's in every column of A2 are consecutive from the first row on. We construct horizontal segments corresponding to the l's in the rows of A1 and vertical segments corresponding to the l's in the columns of A2. The i-th vertical segment intersects the j-th horizontal segment if and only if AI(j, i) = A2(j, i) = 1, which is equivalent to A ( j , i ) = 1 since A is the component-wise product of A1 and A2. In other words, the i-th vertical segment intersects the j-th horizontal segment if and only if xi and yj are adjacent in Q. Therefore we have constructed a segment intersection model for Q. ,, Proposition 8.7.2 reduces the recognition of bipartite graphs having a segment intersection model to the recognition of intersection difference di-
8.7
233
Intersection Threshold Dimension 2
Figure 8.26: A segment intersection model and its intersection graph Q. The adjacency matrix A of Q is the intersection of adjacency matrices A1 and A2 of difference graphs.
Y4
Yl
Ys
Y2
Y3
y
Y2
Y4
Yl X1
Xl
X2
X3
X2
X3
y4
0
Y4
1
1
0
y4
0
1
0
ys
0
Y3
1
0
0
y3
1
1
0
y2
1
Y2
1
1
1
y2
1
1
1
yl
0
yl
1
1
0
yl
1
1
1
Xl
X2
X3
Xl
X2
X3
Xl
X2
A
X3
A1
A2
2-Threshold Graphs
234
mension 2 (or equivalently to the recognition of union difference dimension 2 of the bipartite complement). In turn, this problem is reduced by Theorem 8.6.12 to recognizing poset dimension 2. However, we must also take into account the domination relations between segment bases imposed in the first stage by the intersections of segments with boxes. We partition the sets I(= and h'y of segment bases on the x and y axes into equivalence classes called windows. Two elements of I(~ or two elements of I(y are in the same window when they intersect the same boxes. After the first stage we are free to exchange the positions of segment bases only if they are in the same window; if 81,82 E I(= are in different windows, then the first stage has imposed a definite order relation between them, say sl -4 s: (because N(Sl)M B D N(s2) M B). As the proof of Proposition 8.7.2 shows, the segments are induced by the ordering of the roots, whereas the roots on each axis are ordered by the neighborhood containment relations in one of the two difference graphs. We can therefore formulate the problem of finding a segment intersection model for the bipartite subgraph Q = (S=, Sy; E) of G induced by the segment vertices as follows. The color classses S= and Sy are partitioned into windows X0,... ,Xp and Y0,..., Yq, respectively (corresponding to the partitions of the segment bases into windows). We seek difference graphs HI a n d / / 2 with color classes S=, 5'y such that H1 ('1/-/2 = Q and such that
Va C Xi, b E Xi+l,
NH, (a) D___NH, (b)
Va C Y/, b C Y/+I, NH2(a)D___NH2(b). Equivalently we can consider the bipartite complement Q - (S=, Sy; E) of Q and seek difference graphs H1 and H2 with color classes S=, Sy such that H1 U/-/2 - Q and such that
Va e X,, b e X,+,,
Va C
b C Y,+,,
NT, ' (a) C_ NT, ' (l,)
(a) C_
(t,).
Construct a new bipartite graph Q' - (S= u U, S v u V; E U Eu U Ev), where U = {ua,...,uq} and V = {v~,...,vp} are sets of new vertices and
Eu = { u ~ y : y C Y j , i < _ j } Ev
=
{vix:xCXj,i a l > ' ' ' > am > 0 such that x i x j is an edge of G1 if and only if ai q- aj > S. We may assume that S - al >_ 2 (otherwise multiply all these integers by 2), hence we can insert an integer T such that S > T > a l > ' " > am > O. We now continue the construction to obtain the bj's. Let G2 be the graph induced by Y. G2 also is a threshold graph, hence there exists a permutation r of { 1 , . . . , n} such that each yr(j) is either an isolated vertex or a dominating vertex in the subgraph induced by y~(~),..., yr(j). The initial step of this part of the construction is as follows. If y~(1) is not adjacent to any xi, set S " - S + 2,
T " - T + 2,
ai " - ai q- 1 for all i,
br(1) "- 1,
and if yr(1) is adjacent to X l , . . . ,Xj and non-adjacent to x j + l , . . . ,x,., then do not change S, T and the ai, and put br T - aj. We continue by constructing br b,(~). Note that we may assume for each j that v ( 1 ) , . . . , r ( j ) are consecutive indices among 1 , . . . , n. Therefore at the general step of this part of the construction, we have constructed thresholds S, T and weights a l , . . . , a m , bk+l,...,bs-1, and must now construct either bk, where the next vertex yk to be introduced is adjacent to each of y k + l , . . . , y~-l, or b~, where the next vertex y~ to be introduced is adjacent to none of them. C a s e 1: yk is introduced. It is adjacent to X l , . . . , x i and non-adjacent to Z i + l , . . . , Zm for some i, 0 _ T
(9.6)
4(bk+l - ai) - 1 < ;- S - T + 1
>bk+l-a~.
(9.22)
250
The Dilworth Number
Inequality (9.22) is a relaxation of (9.21), and by repeating these steps as necessary, we eventually find a suitable value for b~. In the extreme case i = m, the argument is the same with am+~ = 0. In the extreme case i = 0, the argument is even easier, since inequality (9.16) drops and there is no lower bound for b~ except for the strict lower bound 0. Therefore we change the existing weights and threshold as follows: aj bj b~ S T
"""-
aj + 1 bj -]- 1 1 S+2 T+2.
j- 1,...,m j-k+l,...,s-1
This completes the construction in all cases. 9 The next Theorem follows directly from Lemma 9.2.3 and Lemma 9.2.4. T h e o r e m 9.2.5 A graph is a TS graph if and only if its Dilworth number is at most 2. C o r o l l a r y 9.2.6 Every TS graph is a comparability graph. P r o o f . Let G - (V, E) be a graph of Dilworth number at most 2 with its vertex set partitioned into two sets X - { x ~ , . . . , x m } and Y - { y ~ , . . . , y ~ } such that x l ~- " . ~ xm and yl ~ "'" ~- y~. Orient each edge ab as (a, b) if a - xi, b - yj for s o m e i , j , o r a - x i , b - xj w i t h i < j , o r a yi, b - yj with i < j. The resulting orientation is transitive. 9 9.2.2
Forbidden
Subgraphs
By the Dilworth Theorem, a graph has Dilworth number at most 2 if and only if it does not contain an antichain of size 3. Thus by obtaining all the minimal graphs with an antichain of size 3, we get a characterization of TS graphs by forbidden subgraphs as stated below. A proof of the next theorem can be found in [BHd85b]. T h e o r e m 9.2.7 A graph G is a TS graph if and only if it does not contain any of the 21 graphs listed in Figure 9.2 as induced subgraphs. Using Theorem 9.2.7, one can characterize the split graphs of Dilworth number 2 as follows.
9.2
251
G r a p h s of D i l w o r t h N u m b e r 2
Figure 9.2: The forbidden subgraphs for TS graphs; G2i+l - G2i, i - 1,..., 10.
_
i I
G1
G2
G4
G6
I I I
X>
alo
G8
a12
. k/Z/_ v
v
w
Q14
v
w
G18
G16
w
v
w
w
G2o
v
252
The Dilworth Number
T h e o r e m 9.2.8 ([FH77a, B H d 8 5 b ] ) For each graph G, the following ditions
con-
are equivalent:
1. G and G are interval graphs; 2. G is a split TS graph; 3. G contains as induced subgraphs neither 2K2, C4, C5, nor Gs, Gg, G14, G15 of Figure 9.2. Proof. 1) =~ 3)" As we have mentioned after Definition 1.4.1, G is an interval graph if and only if G is triangulated and G is a comparability graph. Since G and G are interval graphs, G contains neither C4, C5 (being a triangulated graph) nor Gs, G14 (being a comparability graph) nor their complements 2K2, Gg,
G15. 3) =~ 2)" Since G does not contain 2K2, C4, C5, it is a split graph by Theorem 5.2.1. For the same reason, the only induced subgraphs of Figure 9.2 that G might possibly contain are Gs, Gg, G14, G15. But since by assumption G does not contain them, it is a TS graph by Theorem 9.2.7. 2) =~ 1)" Since G is a split TS graph, it is both triangulated and a comparability graph (by Corollary 9.2.6). But G is also split, and it is a TS graph by Theorem 9.2.5. So G is also triangulated and a comparability graph. Thus G and G are triangulated and cocomparability graphs, and therefore both are interval graphs by the above-mentioned characterization of interval graphs.
9.3 T h e D i l w o r t h N u m b e r and P e r f e c t G r a p h s In this section we examine the relationship between the perfectness of a graph and its Dilworth number. A graph is called minimally imperfect if it is not perfect, but every proper induced subgraph is perfect. L e m m a 9.3.1 If G is a minimally imperfect graph, then
D(G) -IV(G)I. Proof. The proof amounts to showing that V(G) is an antichain. Assume that, if possible, V(G) is not an antichain, and let x ~ Y- If x and y are
9.3 T h e D i l w o r t h N u m b e r and Perfect Graphs
253
not adjacent, then color the perfect subgraph G - {x} with w(G) colors, and then assign to x the color of y, proving that x(G) - c0(G), a contradiction to minimal imperfectness. If x and y are adjacent, then they are not adjacent in G, which is also minimally imperfect [Lov72], and for which y ~ x, so the previous argument applies. Therefore V(G) is an antichain, as required. 9 C o r o l l a r y 9.3.2 ([Pay83]) If D(G) deg(E). By the incomparability, there exist vertices B and D such that A B and E D are edges and AD and E B are nonedges. Since deg(A) > deg(E), there e x i s t s a vertex C such t h a t AC is an edge and E C is a nonedge. Thus G contains the configuration .T'(A,B, C , D , E ) and is not matrogenic. An example of a non-matrogenic BT graph is illustrated in Figure 10.2. 9 Figure 10.2: A non-matrogenic BT graph.
I I I I The following two results for BT graphs generalize similar results for threshold graphs.
Proposition 10.2.3 The complement of a B T graph is BT. Proof. Complementation does not alter vertex incomparabilities or equality of degrees. 9
Theorem 10.2.4 B T is a forcible property of a degree sequence. In other words, if G and H are graphs with the same degree sequence and G is BT, then H is also BT. We may thus unambiguously define a degree sequence to be B T if (all) its realizations are B T graphs.
Box-Threshold Graphs
260
P r o o f . Since there is a degree-preserving bijection between the vertices of G and H, we may assume that G and H have the same vertices and that each vertex has equal degrees in G and H. Then by Corollary 3.1.11, H can be obtained from G by a sequence of transfers, where a transfer involves dropping the two edges xu and zy and adding the two nonedges x z and yu of an alternating 4-cycle. We may therefore assume that H is obtained from G by a single transfer as above. Moreover, by the symmetry of the four vertices of the alternating 4-cycle, it is enough to show that if x II P in H, where p is a fifth vertex, then deg(x) = deg(p). Since x II p in H, x and p are opposite corners of an alternating 4-cycle A in H. If neither x u nor x z is a side of A, then A is also an alternating 4-cycle in G and the conclusion follows since G is BT. If both o f x u and x z are sides of A, then x II Y II p i n G, and hence deg(x) = deg(y) = deg(p). If xu is a side of A and x z is not, then the configuration in G and H is as illustrated in Figure 10.3, where the vertices of A are x, u, p, q. Figure 10.3" Configurations in the proof of Theorem 10.2.4. q
x
z
q
x
z
T
9
9
9
i I I
I I I
I I I
I I I
I ,k v
v
p
u G
y
I
I
i
9
~k
.L
v
v
v
p
u
y
H
If pz is an edge, then x I] P in G; otherwise x II Y It P in G. Therefore the conclusion follows in both cases as before. Finally, if xz is a side of A and x u is not, we repeat the argument with pu replacing pz. 9 The next theorem gives an easy test of the BT property. To state it, we use the following definitions. D e f i n i t i o n 10.2.5 A bipartite graph with bipartition ( X , Y ) is said to be c o l o r - r e g u l a r ( C R ) if all the vertices of X have the same degree and all the vertices of Y have the same degree. Two disjoint sets X and Y of vertices in a graph G are said to be c o m p l e t e l y c o n n e c t e d , C R c o n n e c t e d , or
10.2
Elementary Properties
261
c o m p l e t e l y d i s c o n n e c t e d when the edges joining them in G form a com-
plete, a CR, or an edgeless bipartite graph, respectively. A similar definition applies in the case of a connection of X to itself, where the subgraph induced by X should be complete, regular, or edgeless, respectively. Note that complete connection and complete disconnection are special cases of CR connection. D e f i n i t i o n 10.2.6 Assume that a graph G has the degree sequence
d? i.e., exactly m i vertices have degree di, with d l > of degree di constitute the i - t h b o x Bi of G.
"'" > d~. The rni vertices
is B T if and only if for each subscript i there is some k(i) such that Bi is completely connected to B I , . . . ,Bk(i)_l, Ct~ connected to Bk(i), and completely disconnected from
Theorem
10.2.7 A graph with boxes B I , . . . , B ~
Bk(0+l,. 9 9 B~. P r o o f . " O n l y if"" Recall that N(x) denotes the set of neighbors of a vertex x. For every vertex x C B~, let k(z) be the largest k such that N ( z ) N Bk r ~. Then by the BT property (10.2), {x} is completely connected to B 1 , . . . , Bk(x)-I and completely disconnected from Bk(x)+l,..., B~. Moreover, if y is any other vertex of B~, then since deg(z) = deg(y), we must have k(x) = k(y) and IN(x) N Bk(~)l = IN(y)N Bk(y)l. This means, first, that k is a function of i alone and not of the particular x in Bi, and may be denoted by k(i); and, second, that for j = k(i) and trivially for each j 5r k(i) all the vertices x E Bi have the same IN(x)N Bj]. In particular, all the vertices x C Bk(~) have the same IN(z)N Bil. Therefore Bi and Bk(~) are CR connected. " I f " : Let x II Y for some x C B~, y E Bj. We have to show that i = j. First of all, k(i) = k(j), for if, say, k(i) > k(j), then N(y) C_ N(z), contradicting non-comparability. Let us denote by p the common value of k(i) and k(j). The only way that x and y can be non-comparable is for each of them to have a neighbor in Bp that is not a neighbor of the other. Therefore Bi and Bj are neither completely connected to Bp nor completely disconnected from Bp. This implies that k(p) is equal to both i and j. " Note that a BT graph is a threshold graph if and only if every two boxes are either completely connected or completely disconnected. This means that
Box-Threshold Graphs
262
the adjacency matrix is equal to the corrected Ferrers diagram of the degree sequence. Deleting a vertex of degree 1 from the graph of Figure 10.2 results in a non-BT graph. This shows that BT is not hereditary, and hence cannot be characterized by forbidden configurations. However, the next theorem shows that the BT property is preserved when an entire box is deleted from the graph. The results of Section 10.4 can be interpreted as a characterization of the BT property in terms of forbidden configurations of boxes. T h e o r e m 10.2.8 If G is BT and H is obtained from G by deleting an entire
box, then H is also BT. P r o o f . let the vertices of the deleted box have degree d in G. Assume that, if possible, H is not BT. Then it has vertices x, y, z such that degu(x ) > degH(y), yet z is adjacent to y but not to x. Since G has the same configuration and is BT, dega(x ) _< dega(y). Hence d e g a ( y ) - degH(y) > dega(x ) - d e g u ( x ) , which means that in G, Y has more neighbors of degree d than x has. By Theorem 10.2.7, y is adjacent in G and H at least to all vertices u such that dega(u) > d, whereas x is adjacent in H at most to these vertices. Hence degu(y ) > degH(x), which is a contradiction. ..
10.3
A Transportation M o d e l
Theorem 10.2.4 raises the question of characterizing BT sequences without constructing an arbitrary realization and testing it for the BT property. Here we give such a characterization, which does not even assume that the sequence is realizable. A transportation network consists of suppliers A1,...,A~ with supplies al,. 9 a~ of some commodity, consumers B 1 , . . . , B~ with demands b l , . . . , b~, and capacities c~j of the route from A~ to Bj. A flow is a matrix (f~j) such that 0 deg a y and show that z ~ y in G. Some third vertex z is adjacent to z but not to y. We assert that for each vertex w of G, i f t ~ C B, then wz C B. Indeed, 22 C B U R and 2~ ~ B. If zx C B, then t~2 C B by Condition 2 as asserted. If 22 C R, then 29 ~ R by Condition 3, i.e., 2~ is not an edge of F, hence xw C B U R by Condition 1. But t~ 7~ 2 because t ~ is an edge of F and 2~ is not, hence xw ~ R by Condition 3, and therefore 2t~ C B as asserted. Returning to the proof that x ~- y, we assume that, if possible, w r x, y, wy is an edge of G and wx is not. Then t~2 ~ B, so t ~ ~ R by the assertion, hence wy C R. Therefore t~2 ~ R by Condition 3, i.e., t~2 is not an edge of F. Since zx is an edge of F, we have 2 :/- t~, and since wy C R, we have zy ~ R by Condition 3. Hence zy C B by Condition 1, and this contradicts the fact that z is not adjacent to y in G. " T h e o r e m 10.4.4 Let F be a graph with loops whose edges are colored black or red. Then F with its colors is the frame of some B T graph if and only if 1. F is a threshold graph with loops; 2. the black edges of F form a threshold graph with loops; 3. the red edges of F form a matching with loops; ~. every two vertices of F differ in their number of incident black edges or in their number of incident red edges. We say that the vertices differ in their black degree or their red degree, where a loop contributes 1 to the degree. Moreover, in that case no three vertices of F can have the same black degree.
268
Box-Threshold Graphs
P r o o f . " O n l y if"" Assume that F is the frame of a BT graph G. By Theorem 10.4.3, Conditions 1 through 3 hold. To prove Condition 4, assume that ~ and ~ are distinct vertices of F, say deg a z > deg a y, and that ~ and have the same black degree. By Condition 2, for every z, z z E B if and only if ~2 E B. But some z is adjacent to x and not to y in G, hence zy ~ B and 22 E R. By Condition 3, ~,~ ~ R, i.e., 29 is not an edge of F. We assert that no red edge is incident with 9, which shows that 2 and 9 differ in their red degrees. Indeed, if @~ E R, then w x ~ R by Condition 3 and @2 ~ B by the above, and this contradicts Condition 1, thus proving t h e assertion and Condition 4. Moreover, if 2 and r have the same black degree, then by Conditions 3 and 4, their red degrees differ by exactly 1, and so no three vertices can have the same black degree. "If"" We construct a BT graph G whose frame is F. For each vertex v of F, let G have four vertices v0, vl, v2, v3. For each black edge v w E t3 of F with v 7~ w, join each vi to each wj. For each red edge v w E R of F with v -~ w, join each vi to wi and Wi+l (indices modulo 4). For each black loop vv E B of F, join each vi to each vj, i 7~ j . For each red loop vv E R of F, join each Vi to Vi+I and v i - , . This construction is illustrated in Figure 10.5. Note that each red edge or loop of F contributes 2 to the degrees of the vertices of G corresponding to its ends, a black loop contributes 3, and a black edge that is not a loop contributes 4. It is sufficient to show that F is the frame of G, for this implies that G is BT by Theorem 10.4.3. Clearly for each vertex v of F, all the vi have the same degree in G. Therefore it only remains to show that if v and w are distinct vertices of F, then deg a v0 =fi dega w0. If v and w have the same black degree, then by Condition 2, either both have a black loop or both do not, so the black edges incident to v and w contribute equally to the degrees of v0 and w0. Also by Condition 4, v and w have different red degrees, hence deg a v0 r deg a w0 as required. Assume now that the black degree of v is larger than the black degree of w. Then by Condition 2, for each vertex x of F, z w E B implies x v E B , but not conversely. Therefore the black edges contribute more to deg a v0 than to dega w0. We now prove that for each vertex y of F, y w E R implies yv E R U B , which shows that deg C v0 > dega w0 as required. Assume that, if possible, y w E R and yv ~ R U B. Let x be a vertex of F such that x v E B and x w ~ B. Then x-~ y (sinceyv ~ B ) , hence by Condition 3, x w ~ R, i.e., x w is not an edge of F. This contradicts Condition 1. .. The construction used in the "if" part of the proof of Theorem 10.4.4 can be generalized to yield all the degree sequences of BT graphs with a given frame.
10.4
Frames of BT Graphs
269
Figure 10.5: The construction of a BT graph with a given frame.
vo . . . . ~
wo
V0
W0
Vl
Wl
-
Wl
V2 ~
W2
.,.
W2
V3 -
. W3
?-)3
.,.
vwEB
W3
vwER
V2
"
-
131
V2
"
V3
...
-
VO
Y3
.
vv E B
~
Vl
-
vv E R
VO
Chapter 11 Matroidal and Matrogenic Graphs 11.1
Introduction
We have seen in Theorem 1.2.4 that a graph G - (V, E) is threshold if and only if G does not contain any alternating 4-cycle. Therefore, if we call a set of edges I C_ E "independent" when no two edges of I induce an alternating 4-cycle, then a spanning subgraph (V, I) is a threshold graph if and only if I is independent. Obviously the independent subsets of E form an independence system, i.e., a subset of an independent set must be independent. The circuits of this independence system, namely the minimal dependent sets of edges, are the sets of two edges whose endpoints induce an alternating 4-cycle in G. We call these sets couples. We are interested here in those graphs for which the above independence system is a matroid. As is well-known, one definition of a matroid is an independence system such that if C and C ~ are distinct circuits and e C C N C', then C U C ' - { e } is dependent. In the present situation, where every circuit has cardinality 2, this condition simply reads as follows" if a, b, c are distinct edges and {a, b}, {b, c} are couples, then {a, c} is a couple.
(11.1)
An equivalent condition is the following. Define a reflexive and symmetric binary relation R on E by having aRb if and only if a = b or {a,b} is a couple in G. Then our independence system is a matroid if and only if R is transitive. In that case we say that G is a matroidal graph. Although the 271
272
Matroidal and Matrogenic Graphs
corresponding matroids are uninteresting, the matroidal graphs themselves have a nice structure, which permits us to find their threshold dimension and show that they are perfect graphs. Peled [Pe177] introduced these graphs, characterized them by two forbidden configurations, the chordless pentagon and configuration ~ of Definition 11.2.1, and found their structure. We present his results in the next section. FSldes and Hammer [FH78b] introduced a variation of the above concept. Instead of an independence system on the edge set E, they considered an independence system on the vertex set V whose circuits, i.e., minimal dependent sets, are the sets of four vertices that induce an alternating 4-cycle of G. If this independence system is a matroid, the graph G is said to be matrogenic. The matrogenic graphs are characterized by forbidding configuration 9r alone. Furthermore, a matrogenic graph can have at most one chordless pentagon, and hence the structure of matrogenic graphs is almost the same as that of matroidal graphs, even though the two are defined in an apparently different way. We characterize the matrogenic graphs and present their structure Section 11.3. In Section 11.4 we present the work of Marchioro et al. [MMPS84] that characterizes the degree sequences of matrogenic graphs. These results were also obtained independently by Tyshkevich [Tys84].
11..2
Matroidal Graphs
11.2.1
Forbidden Configurations
We begin our study of the matroidal graphs by characterizing them by forbidden configurations. Definition 11.2.1 A configuration ~ consists of vertices A, B, C, D, E such that A is adjacent to B, C but not to D, whereas E is adjacent to D but not to B, C. As with all configurations, unspecified edges (in this case among B, C, D and between A and E) may or may not exists. See Figure 11.1. We denote the fact that A, B, C, D, E generate a configuration jz in this order by 5(A,B,C,D,E).
T h e o r e m 11.2.2 A graph is matroidal if and only if it contains neither the configuration ~" nor a chordless pentagon.
11.2
M a t r o i d a l Graphs
273
Figure 11.1: The forbidden configuration $'(A, B, C, D, E). A
B
E
P r o o f . The "only if" part is immediate: if .T(A, B, C, D, E), then the edge D E forms a couple with both A B and AC, which do not form a couple with each other, contradicting transitivity of the binary relation R; similarly each edge of a chordless pentagon forms a couple with two adjacent edges. To prove the "if" part, assume that R is not transitive, so that there exist edges a, b, c satisfying aRb, bRc, but not aRc. It follows that a, b, c are distinct edges having a total of five or six endpoints, according as a and c have a common endpoint or not.
Five endpoints. Put a = AB, b = CD, c = B E with AC and B D nonedges (Figure ll.2(a)). If E C is a nonedge, we have ~ ( B , A, E, D, C), so assume that E C is an edge. Since b, c form a couple, B C and D E are nonedges (Figure ll.2(b)). Now if AD is a nonedge, then JZ(B, A , E , C, D), and if AD is an edge, then A, B, E, C, D induce a chordless pentagon if A E is a nonedge and $'(A, B, E, C, D) otherwise. Six endpoints. Put a = AB, b = CD, c = E F with AC, BD, CE, D F nonedges (Figure 11.3). Since a r c does not hold, some endpoint of a is adjacent to some endpoint of c. By symmetry we may assume that B E is an edge, and then we have .T(B, A, E, D, C). 9
C o r o l l a r y 11.2.3 The complement of a matroidal graph is matroidal.
P r o o f . This follows immediately from the fact that 5 and the chordless pentagon are self-complementary. 9
274
Matroidal and Matrogenic Graphs
Figure 11.2" Configurations in the proof of Theorem 11.2.2.
c
A a
AT
b
B
a
D
/
B
---
c
b D
,,~,
E
E
(~)
(b)
Figure 11.3" A configuration in the proof of Theorem 11.2.2.
j
A
C
//
B
/%
/
D
/
/
/
/
/
/
c
11.2.2
F
P4-vertices
In this subsection we describe the s t r u c t u r e of those m a t r o i d M graphs for which every vertex belongs to an induced P4 (a p a t h on 4 vertices). 1 1 . 2 . 4 Let vertices A1, A2, B1, B2 induce a P4 with edges Al131, BIB2, B2A2 (Figure 11.~). We then say that A1,A2, B1,B2 i n d u c e a P4 in t h i s o r d e r and denote this fact by 7 9 4 ( A ~ , A 2 , B~,B2). We call the A~ the e x t e r n a l v e r t i c e s and the Bi the i n t e r n a l v e r t i c e s of this P4, and similarly call the edges AiBi the e x t e r n a l e d g e s and the edge B1B2 the i n t e r n a l e d g e of the P4. Definition
1 1 . 2 . 5 Assume that ~P4(A1,A2, B1,B2) in a matroidal graph G, and let X be a fifth vertex of G. If X is adjacent to A1, then X is adjacent to A2, B1 and B2. If X is adjacent to B1, then X is adjacent to B2.
Lemma
11.2
Matroidal Graphs
275
Figure 11.4: P4(A1, As, B1, B2). B1 "
i
"" B2
\
/ \
/ x
/ /
"
\ \ \
A1
A2
P r o o f . If X is adjacent to A1 but not to A2, then we have the configuration S'(A1, B I , X , B2, A2). If X is adjacent to A1 and A2 but not to B1, then in case X is not adjacent to B2 the five vertices induce a chordless pentagon, and in case it is we have ~ ( X , A2, B2, B1,A1). Finally, if X is adjacent to B1 but not to A1, A2, B2, then we have JZ(B1,A1,X, A2, B2). " 1 1 . 2 . 6 In a matroidal graph, if edges a and b form a couple of the form P4 and edges b and c form a couple, then the latter couple is also of the Lemma
P r o o f . We may obviously assume that a, b and c are distinct edges. Assume that P4(A1,A2, B1, B2) with a = AIBI and b = A2B2. Since c forms a couple with each of a and b, there are a fifth vertex X and a sixth vertex Y with c = X Y . If the couple bc is of the form 2K2, then by L e m m a 11.2.5 X and Y are not adjacent to B1, and hence X Y forms a couple with BIB2, contradicting transitivity. If the couple bc is of the form C4, then by L e m m a 11.2.5 X and Y are both adjacent to A2 and B2, a contradiction. " C o r o l l a r y 1 1 . 2 . 7 Let R' be the binary relation defined on the edges of a
graph G so that e R ' f if and only if e = f or e and f form a couple of the form P4. If G is matroidal, then R' is an equivalence relation. The next l e m m a says that if a vertex in a matroidal graph belongs to some P4's, then it is either external in all of them or internal in all of them, and we may call it an external vertex or an internal vertex unambiguously. The same applies to edges too. Lemma
1 1 . 2 . 8 A vertex or an edge in a matroidal graph cannot be internal
in one P4 and external in another P4.
276
Matroidal and Matrogenic Graphs
The result for edges follows from the one for vertices, which we prove below. Assume that P4(A1,A2, B1,B2), and assume that, if possible, A1 is an internal vertex in another P4. Then A1 must have neighbors in the new P4, which by Lemma 11.2.5 are also adjacent to B1 and A2. It follows that B1 cannot belong to the new P4, so A1, having degree 2 in the new P4, must have two new neighbors X and Y in it. But one of X and Y is external in the new P4, and since it is adjacent to A2, A2 is also adjacent to the internal vertex A1 of the new P4 by Lemma 11.2.5 applied to the new P4. This is a contradiction.
L e m m a 11.2.9 In a matroidal graph, all the internal vertices form a clique and all the external vertices form a stable set. P r o o f . We prove the result for the internal vertices, and the result for the external vertices then follows by Corollary 11.2.3. If BIB2 and B2B3 are two internal edges, then B1 is adjacent to B3 by Lemma 11.2.5. Now assume that B1B2 and B3B4 are two internal edges and show that the B i form a clique. If some of B1, B2 are adjacent to some of B3, B4, the result follows from Lemma 11.2.5. If not, let A1 be an external vertex adjacent to B1. By Lemma 11.2.5 A1 is not adjacent to B3 and B4, and therefore B3B4 forms couple with each of B1A1 and BIB2, contradicting the transitivity of R. 9 In order to describe the structure of the equivalence classes of the binary relation R ~ of Corollary 11.2.7, we introduce the following terminology. Definition 11.2.10 Let A 1 , . . . , A p , B 1 , . . . , B p be 2p distinct vertices for some p > 2. We say that these vertices induce (in this order) a net (resp. a n e t - c o m p l e m e n t ) of o r d e r p and denote this fact by A f ( A ~ , . . . , Ap, B 1 , . . . , Bp) (resp. A T ( A ~ , . . . , Ap, B 1 , . . . , B p ) )
if all the Ai form a stable set, all the Bi form a clique, and each A~ is adjacent to Bi and to no other Bj (resp. each Ai is adjacent to all the Bj except for Bi). Figure 11.5 displays a net and a net complement of order 3. If N ' ( A I , . . . , A;, B I , . . . , Bp) holds in G, then ~?(B1,...,/3p, A1, . . . . Ap) holds in G, and conversely; and if 794(A1, A2,/~1~/~2), then Af(A1, A2,/31,/32) and A?(A1, A2,/~2,/~1 ). L e m m a 11.2.11 In a matroidal graph, two or more edges forming an equivalence class under the binary relation R' of Corollary 11.2.7 induce a net.
11.2
Matroidal Graphs
277
Figure 11.5: A net and a net-complement of order 3. A1
A1
B
2
_
A2
A3 (a) a net
A2
B1
A3
(b) a net-complement
P r o o f . Let A 1 B 1 , . . . , A p B ; be the distinct edges of the equivalence class with the Ai external and the Bi internal vertices. Clearly if Ai = Aj, then Bi = Bj and as the p edges are distinct, we have i = j. Similarly Bi = Bj implies i = j. Also by Lemma 11.2.8 A~ r Bj for all i,j. Thus we have 2p distinct vertices. By Lemma 11.2.9 the Ai form a stable set and the Bi form a clique. Finally for i -r j, A~ is not adjacent to Bj since we have 794( Ai, Aj, Bi , Bj ). .. Although the equivalence classes of R' are edge-disjoint, they may share vertices. The next lemma is crucial in determining the situation in these cases. L e m m a 11.2.12 In a matroidal graph G, let E and E ~ be distinct equivalence classes under the binary relation R' of Corollary 11.2.7, each having two or more edges. Assume that some vertex is an endpoint of an edge of E and of an edge of E'. Then each of E and E' has exactly two edges and these four edges induce in G a net-complement of order 3. P r o o f . Let us first consider the case that the common vertex is internal. Let 7)4(A1,A2, B1,B2) where AIB1 and A2B2 belong to E, and let the edge XB1 belong to E'. Then X must be external, since it follows from Lemma 11.2.5 that internal edges do not belong to couples, and so X -r B2 by Lemma 11.2.8. Also X r A1 since E -r E', and X :/- A2 since A2 is not adjacent to B1. Therefore X is a fifth vertex, which is adjacent to B2 by Lemma 11.2.5, but not to A1 or A2 by Lemma 11.2.9. By assumption ~P4(X, Y, B~, Z) for some
278
Matroidal and Matrogenic Graphs
vertices Y, Z. By Lemma 11.2.8 Z -~ X, A1, A2 and by definition of a couple Z -~ B1, B2. Thus Z is a sixth vertex, which is adjacent to B1 and B2 by Lemma 11.2.9 but not to X by definition of a P4 (see Figure 11.6). Figure 11.6" A configuration in the proof of Lemma 11.2.12.
al
B1
A2
B2
A
A /
/ /
\ /
u
We assert that Y = A2. Clearly Y r Z, B1, B2 by Lemma 11.2.8; Y -~ A1 since A1 is adjacent to B1 and Y is not; and Y -J= X. Therefore if Y -r A2, then Y is a seventh vertex. In the latter case, since YZ forms a couple with XB1, it cannot form a couple with XB2 by transitivity, and therefore Y must be adjacent to B2. But then Y is adjacent to B1 by Lemma 11.2.5, contrary to the assumption that YZ forms a couple with XB1. This proves the assertion Y = A2, from which it follows that Z is adjacent to A2 and thus also to A1 by Lemma 11.2.5. We have therefore shown that ~(A1,A2, X, B2,B1,Z). Moreover, since A2 was an arbitrary external vertex other than A1 belonging to an edge of E, it follows from the assertion that E has only two edges A1B1 and A2B2. By symmetry E' also has only two edges XB1 and A2Z, and the lemma is proved in this case. We now examine the other possibility, that the edges of E and E' share an external vertex but not an internal one. We show that this is impossible. Let 7)4(A1, A2, B1, B2) where A1B1 and A2B2 belong to E, and let the edge XA1 belong to E'. Then X must be internal, and therefore by assumption X -r A2, B1,B2. By Lemma 11.2.5 X is adjacent to A2, B1,B2. By assumption E' contains another edge that forms a P4 with XA1. This edge cannot contain B1 or B2 by assumption and it cannot be of the form A2Y, since Lemma 11.2.5 would imply that Y is adjacent to A1, contrary to the definition of P4. Thus this edge is of the form YZ, where Y is an internal sixth
11.2
M a t r o i d a l Graphs
279
vertex and Z is an external seventh vertex (see Figure 11.7). We now apply Lemma 11.2.5 several times to the two P4's. Since B1 is adjacent to A1, it is adjacent to Z. Since Z is adjacent to BI, it is adjacent to B2. Since B2 is adjacent to Z, it is adjacent to A1. But this contradicts the definition of P4.
Figure 11.7: A configuration in the proof of Lemma 11.2.12. z Y x v
A1
B1
w
B2
v
A2
In order to generalize Lemma 11.2.12 to more than two equivalence classes, let us consider a new abstract graph G whose vertices are the equivalence classes of two or more edges under R'. By Lemma 11.2.11, the edges of G that belong to a single vertex of G induce a net in G. Two vertices of G shall be adjacent when the corresponding two nets of G have vertices in common. L e m m a 11.2.13 For a matroidal graph G, let X l , . . . , 2 q be a sequence of
vertices of G such that X~ is adjacent to some of X 1 , . . . , X~-I for each i 2 , . . . , q, and q >_ 2. Then each )[i contains exactly two edges of G and these 2q edges induce a net-complement in G. P r o o f . The proof is by induction on q. The basis q - 2 is the content of Lemma 11.2.12. Assume that q > 2 and that the lemma holds for q - 1 . Then each of J ( 1 , . . . , Xi-1 contains exactly two edges of G and these 2q - 2 edges induce in G a net-complement N ' ( A ~ , . . . , Ap, B 1 , . . . , Bp). By assumption Xq is adjacent to 2 i for some i - 1 , . . . , q - 1. Hence by Lemma 11.2.12 J~q contains exactly two edges of G, which induce in G a net N(Q, T, R, S), and Xi contains exactly two edges of G, which induce in G a net N(A~, A~, B~, B~). It further follows from Lemma 11.2.12 that these two nets share exactly one internal vertex and exactly one external vertex, and their six distinct vertices induce in G a net-complement of order 3. Hence by renaming Q, T, R, S if ,v
280
Matroidal and Matrogenic Graphs
necessary we may assume that Q = A~ and S = B~, as illustrated in Figure 11.8. Figure 11.8: A net-complement in the proof of Lemma 11.2.13.
Q=Ar
R
w
As
8
v
S = Br
v
T
Clearly T is distinct from B I , . . . , Bp and R is distinct from A I , . . . , Ap. Assume that T is one of A I , . . . ,Ap, say T = At. Then R -r Bs because B~ is adjacent to T and R is not. If R r Bt, then the fact that R is adjacent to A~ but not to At contradicts Lemma 11.2.5 as applied to 7)4(As, At, Bt, B~). Therefore R - Bt and the vertices of G in ) ( 1 , . . . , 32q are just A I , . . . , Ap, B 1 , . . . , Bp, which induce a net-complement of order p in G, as was to be proved. Now assume that T is distinct from A 1 , . . . , Ap. Then R is distinct from A I , . . . , Ap, because by Lemma 11.2.5 R = Bt would imply that R is adjacent to T. Then for each t -r s we can apply Lemma 11.2.5 to P4(At, A~, B~, Bt) to conclude that T is adjacent to Bt but not to At and R is adjacent to both At and Bt. Thus we have a net-complement of order p + 1, ~?'(A1,..., Ap, T, B 1 , . . . , Bp, R), as was to be proved, tt C o r o l l a r y 11.2.14 For a matroidal graph G, the vertices belonging to a single connected component of G induce in G a net or a net-complement. These nets and net-complements are vertex-disjoint.
P r o o f . If a connected component C of G contains exactly one vertex of G, then the corresponding vertices of G induce in G a net by Lemma 11.2.11. If C contains two or more vertices of G, then these vertices can be arranged in a sequence satisfying the conditions of Lemma 11.2.13, and so the corresponding vertices of G induce a net-complement. Assume that C1 and C2
11.2
M a t r o i d a l Graphs
281
are connected components of G such that the corresponding nets or netcomplements N1 and N2 of G share a vertex X. Then X belongs to some P4 contained in N1, hence X is an endpoint of an edge of G belonging to a vertex X1 of G belonging to C1. Similarly X is an endpoint of an edge of G belonging to a vertex X2 of G belonging to C2. By definition of G, X1 is adjacent to 22, hence C1 - C2, hence N1 - N2. 9 Thus we have shown that in a matroidal graph G, the vertices that belong to P4's partition into disjoint sets, which we now call cells, such that each cell induces in G a net or a net-complement, and each P4 in G is induced by the vertices of a single cell. We number the cells as 1 , . . . , k and denote by E i (resp. P) the set of external (resp. internal) vertices of the i-th cell. We say that cell i dominates cell j, where i =/= j, when each vertex of E i is adjacent to each vertex of I j, and no vertex of E j is adjacent to any vertex of I i. L e m m a 11.2.15 The domination relation between the cells of a matroidal
graph is a linear order. P r o o f . By definition the domination relation is irreflexive and antisymmetric. We show that it is transitive and total. Totality. Consider any two cells, say cell 1 and cell 2, and let AiBi be any external edge of cell i, with Ai external and Bi internal (i = 1, 2). We assert that either A1 is adjacent to B2, or else A2 is adjacent to B1, but not both. If neither holds, then by Lemma 11.2.9 we have JP4(A1, A2, B1, B2), contrary to the fact that the vertices of each P4 are contained in a single cell. If both hold, then by Lemma 11.2.5 each vertex of E 1 is adjacent to each vertex of 12 and each vertex of E 2 is adjacent to each vertex of I 1. Let "]')4(Ai, A~, Bi, B~) in c e l l / - 1,2. Then we have 7'4(A1,A2, B;,B~), again contrary to the fact that the vertices of each P4 are contained in a single ceil. This proves the assertion, say A1 is adjacent to B2 and A2 is not adjacent to B1. Then it follows from Lemma 11.2.5 that cell 1 dominates cell 2. Transitivity. In view of the totality, it is enough to show that it is impossible for cell 1 to dominate cell 2, cell 2 to dominate cell 3, and cell 3 to dominate cell 1. Assume that, if possible, this is so, and let A1 E E 1, B2 C 12 , Aa E E 3, and B3 E 13, with A3 adjacent to B3. By Lemma 11.2.9 B2 is adjacent to B3 and A1 is not adjacent to A3. By our assumption we have 7?4(A1, A3, B2, B3), contrary to the fact that the vertices of each P4 are contained in a single cell. 9
282
Matroidal and Matrogenic Graphs
Lemma 11.2.15 allows us to renumber the cells so that cell i dominates cell j if and only if i < j. This determines completely the structure of the subgraph induced by the edges of couples of the form P4 in a matroidal graph. We summarize the results of this subsection by the following theorem. T h e o r e m 11.2.16 In a matroidal graph G, the vertices belonging to couples of the form P4 partition into disjoint sets called cells. Each cell induces in G a net or a net-complement. All the internal vertices form a clique and all the external vertices form a stable set in G. The cells can be indexed so that the external vertices of cell i are adjacent to the internal vertices of cell j if and only if i < j. We call a matroidal graph all of whose vertices belong to couples of the form P4 a cell graph.
11.2.3
2I(2-vertices and C4-vertices
In this subsection we complete the description of those matroidal graphs for which every vertex belongs to some edge in some couple. In particular, we show that every vertex of such a graph belongs either only to couples inducing a P4, or only to couples inducing a 2K2, or only to couples inducing a 6'4.. Moreover, a matroidal graph cannot contain couples of the form 2K2 and 6'4 at the same time. As usual, by a perfect matching inK2 we mean a graph with 2m vertices, all of degree 1. Figure 11.9 displays a perfect matching 3K2 and its complement 3K2. Figure 11.9: A perfect matching and its complement.
3K2
3K2
11.2
M a t r o i d a l Graphs
283
T h e o r e m 11.2.17 Let G be a matroidal graph in which the set M of vertices belonging to couples of the form 2K2 is not empty. Then M induces a perfect matching in G and the vertices of G outside M partition into a clique C whose vertices are adjacent to every vertex of M and a stable set I whose vertices are adjacent to no vertex of M . P r o o f . Let edges A B and C D form a couple of the form 2K2, and let X be a fifth vertex. We observe that X is adjacent to all of A, B, C, D or to none of them. Indeed, if X is adjacent to A, then by the transitivity of R, X A cannot form a couple with CD, and therefore X is adjacent to both C and D (Figure 11.10). Figure 11.10: A configuration in the proof of Theorem 11.2.17. A
C \
/ \
X
/ x
/
\
/
B
D
A repetition of this argument shows that X is also adjacent to B. From this observation it follows that the binary relation "equals to or forms a couple of the form 2K2 with" is transitive, and thus an equivalence relation. By definition each equivalence class induces in G a perfect matching, and from the assumption of the theorem there is at least one equivalence class g with at least two edges A B and CD. Let M' be the set of endpoints of the edges of g. By the above observation the vertices of G outside M ~ partition into a set C of vertices adjacent to every vertex of M ~ and a set I of vertices adjacent to no vertex of M ~. Any two vertices X, Y of I are not adjacent, otherwise the edge X Y would belong to g. Every two vertices X, Y of C are adjacent, otherwise X A would form a couple with both Y C and Y D , contradicting transitivity. Thus I is a stable set and C is a clique. It follows that the vertices of M ~ belong only to couples formed from two edges of g, and that C U I induces a split graph, which has only couples of the form P4, if any. Therefore M ~ = M. 9
Matroidal and Matrogenic Graphs
284
C o r o l l a r y 11.2.18 Under the conditions of Theorem 11.2.17, G has no couples of the form C4, C contains all the internal vertices and I contains all the external vertices of G. P r o o f . We have seen that G has no couples of the form C4. Let B1 be any internal vertex of G. Then we have 794(A1,A2, B1, B2) for some A1,A2, B2, and It follows from Lemma 11.2.5 that the edge B I ~ 2 does not belong to any couple of G. Therefore at least one of B1, B2 must be in C. If B1 is in I and B2 is in C, then A1 cannot be in C since it is non-adjacent to B2, and cannot be in I U M since it is adjacent to B1, a contradiction. Therefore B 1 must be in C and C contains all the internal vertices. It follows from this that A1 and A2 must be in I, so I contains all the external vertices. .. By taking the complement of G we obtain from Theorem 11.2.17 and Corollary 11.2.18 the following analogous results. T h e o r e m 11.2.19 Let G be a matroidal graph in which the set M of vertices belonging to couples of the form C4 is not empty. Then M induces a complement of a perfect matching in G and the vertices of G outside M partition into a clique C whose vertices are adjacent to every vertex of M and a stable set I whose vertices are adjacent to no vertex of M. C o r o l l a r y 11.2.20 Under the conditions of Theorem 11.2.19, G has no couples of the form 2K2, C contains all the internal vertices and I contains all the external vertices of G. The following summarizes the results of this subsection. D e f i n i t i o n 11.2.21 A matroidal graph G all of whose vertices belong to couples is called a s t r i c t m a t r o i d a l g r a p h . T h e o r e m 11.2.22 A graph G is strict matroidal if and only if the vertices of G partition into subsets M and S, M induces a perfect matching or a complement of a perfect matching, S induces a cell graph, and the vertices of M are adjacent to every internal vertex and no external vertex of S. 11.2.4
Threshold
Vertices
We now have to consider those vertices of a graph G that do n o t b e l o n g to any couples. We call these vertices threshold vertices, since they induce
11.2
M a t r o i d a l Graphs
285
a threshold subgraph of G. Recall the definition of the vicinal preorder on the vertices of G. We extend its definition to subsets of vertices so that X~Yifandonlyifx~yforallxCX, yEY. If G is matroidal, then as seen in the previous subsection, its vertices belonging to couples partition into sets M, I 1, E l , . . . , [k E k, where M is the set of vertices that belong to couples of the form 2K2 or C4, I t is the set of internal vertices of cell l, and E l is the set of external vertices of cell l. These vertices induce in G a strict matroidal graph S, relative to which
[k ~ ... ~_ I 1 ~ M ~ E 1 ~ "'" ~- E k.
(11.2)
The threshold vertices of G induce a threshold graph T, and they can be labeled as X 1 , . . . , Xt so that relative to T
X , ~ ... ~ Xt.
(11.3)
Moreover, (11.2) and (11.3) remain true relative to G, because they could only be violated if a vertex of T belonged to a couple of G (there is an exception in the case that Xi ~ Xj relative to T but Xi ~- Xj relative to G; in that case we adopt the convention that i < j). Thus the sets M , I ~ , E 1 , . . . , I k , E k, { X 1 } , . . . , {Xt} are totally ordered by ~ relative to G in a way that is a common refinement of both (11.2) and (11.3). Conversely, assume that S is a strict matroidal graph satisfying (11.2) and T is a threshold graph disjoint from S and satisfying (11.3). If S and T are connected by edges so that relative to the resulting graph G the sets M, I 1, E l , . . . , I k, E k, {X1},..., {Xt} are totally ordered by ~- in a way that is a common refinement of both (11.2) and (11.3), then G is matroidal and its threshold vertices are precisely the vertices of T. Indeed, the Xi are comparable with every vertex of G and hence do not belong to couples of G; the other vertices of G do belong to couples since S is strict; further, R is transitive on S by assumption and hence remains transitive on G. It remains to see how S and T are to be connected by edges so as to satisfy the above necessary and sufficient requirements. By Theorems 11.2.17 and 11.2.19, if a vertex of T is adjacent to some vertices of M, it is adjacent to all vertices of M. By Lemma 11.2.5, if a vertex of T is adjacent to some vertices of E t, it is adjacent to all vertices of E t t_J I t, and if it is adjacent to some vertices of I l, it is adjacent to all vertices of I t. It therefore follows that a vertex of T cannot be equivalent to a vertex of S' under the vicinal preorder. The remaining question is to which sets M, I 1, E l , . . . , I k, F_,k a given vertex of T should be adjacent.
Matroidal and Matrogenic Graphs
286
We begin to consider this problem by assuming that S is a single cell, i.e.,M-Oandk-1. T h e o r e m 11.2.23 Let S be a cell with external vertices E and internal vertices I. Let T be a threshold graph disjoint from S, whose vertices X 1 , . . . , Xt satisfy (11.3) relative to T. If S and T are joined by edges so that the resulting graph G is matroidal and its threshold vertices are precisely X 1 , . . . , Xt, then the following conditions hold: 1. there is an index r - O , . . . , t such that { X r + l , . . . , X t } is a stable set, or equivalently such that X~+I is not adjacent to X~+2 (provided r + 2 - X j , where j is the smallest index other than r such that X j and X~ are not adjacent, and i is the largest index such that Xi and X~+I are adjacent (if i or j is not defined, the corresponding condition on s is omitted; Xo :'- XN is considered to be true); 3. each vertex of I is adjacent to X1, . . . , X~, but not to X r + l , . . . , X t / ~. each vertex of E is adjacent to X1, . . . , X~, but not to X s + l , . . . , X t. Conversely, r and s can always be chosen to satisfy 1 and 2, and if S and T are connected according to 3 and ~, then the resulting graph G is matroidal and its threshold vertices are precisely X 1 , . . . , Xt. P r o o f . Assume that G is matroidal and its threshold vertices are X1, . . . , Xt. Then there are indices 0 _< s - A, X 1 , . . . , X ~ are adjacent to B and hence to every vertex of I. Since A >-- X r + l , . . . , Xt, X r + l , . . . , Xt are not adjacent to B' and hence not to any vertex of I. Condition 4 has a similar proof.Assume that 1 fails and there is an edge connecting two vertices among X ~ + I , . . . , Xt. Then by 3 this edge forms a couple with any internal edge of S, contradicting the assumption that X 1 , . . . , Xt are threshold vertices of G. It remains to prove 2. We assert that I ~ Xj. Indeed, otherwise we would have
11.2
M a t r o i d a l Graphs
287
Xj ;,-- I, and since the vertices of I are adjacent to X~, Xj would be adjacent to X~, contradicting the definition of j. Similarly we see that Xi ~ I. Since Xi ~- I ;,- Xj and s is the largest index such that X~ ~ I, we have i _ r + 1 would contradict the definition of i. Thus it is always possible to choose r and s to satisfy 1 and 2, often in different ways. Now assume that this was done and S and T were joined by edges according to 3 and 4. We complete the proof by showing that (11.4) holds relative to G. First we show that if u _< v, then X~ ~ X~ relative to G. Indeed, let B be an internal vertex adjacent to X~. Then by 3 v _< r, hence u _< r and by 3 B is adjacent to X~. A similar use of 4 results in the same conclusion for external vertices. It remains to show that X~ ~- E ~- X~+I and X~ ~- I ~- X~+I. X~ ~ E: Let Z be any vertex adjacent to some vertex of E. Then Z is in I o r T. In the first case Z is adjacent to X~ by 3 and we are done. In the second case we have Z - Xq for some q and q _< s by 4. If j is defined, then by 2 Xq ~ Xj relative to T, hence q < j, and by the definition of j we have that Z coincides with or is adjacent to X~. If j is undefined, then X~ is adjacent to all other vertices in T, so again Z coincides with or is adjacent to XT. E ~- X~+I" Let Z be any vertex adjacent to X~+I. Since s < r by 2, and by 3 and 4, Z must be in T, and so Z - Xq for some q. Then i is defined and satisfies q < i, and by 2 i < s. Therefore q < s and so by 4 Z is adjacent to the vertices of E. Xs ~ I: Let Z be any vertex adjacent to some vertex of I. If Z is in S, then by 3 and 4 and the fact that s < r, Z is adjacent to X~ and we are done. If Z is in T,we have Z - Xq for some q and q < r by 3. If j is defined, we distinguish the cases s 7~ r and s - r. In the first case, since by 2 X~ ~- Xj relative to T and therefore s < j, Xs is adjacent to XT by the definition of j, and therefore X~ coincides with or is adjacent to Xq. In the second case X~ ~- Xj by 2 so that r < j, by the definition of j { X 1 , . . . , X~} is a clique, and in particular since q, s _< r we have that Xq coincides with or is adjacent to X~. Again, if j is undefined, then X~, and hence X~, is adjacent to all
288
Matroidal and Matrogenic Graphs
other vertices in T, so t h a t Xq coincides with or is adjacent to X~. I ~- Xs+l" Let Z be any vertex adjacent to Xs+l. If Z is in S', then by 4 Z is internal, and therefore Z is adjacent to all other vertices of I and we are done. I f Z i s i n T , then Z - X q for s o m e q . Since b y 2 i_< s < s + l , and by the definition of i, X~+I and X~+I are not adjacent. This means t h a t either s + 1 - r + 1, in which case q _< r by 1, or else X~+I and XT+I are distinct non-adjacent vertices, in which case Xq ~- XT+I relative to T, hence again q _< r. Thus in both cases Xq is adjacent to the vertices of I by 3. .. We illustrate T h e o r e m 11.2.23 with the following example. Let T be the threshold graph illustrated in Figure 11.11, relative to which one has X1 ~- X2 "~ X3 ~- X4. Table 11.1 summarizes all the possibilities of m a t r o i d a l graph whose threshold vertices are X1, . . . . X4 and whose other vertices induce a cell. These graphs are illustrated in Figure 11.12 for the case t h a t the cell is a P4. Figure 11.11" A threshold graph.
X3
X1
X2
X4
Table 11.1" A summary of the possibilities for the graph of Figure 11.11. Case
r
i
j
s
(a) (b)
1 2
1 1
4 3
1 1
(c)
3
2
0
(d) (e)
3 4
2 1
1 0
Order of vertices X1 X1 I X1 I
~ ~ ~ ~~-
I I X1 I X1
~ ~ ~ ~~-
E X2 22 X2 X2
}.- 2 2 }.- f_, ,.-, 2 3 ,.~ X 3 ,~ X 3
~ ~ ~.~:,-
X3 X3 ~, E X4
~ ~~~~-
24 24 24 X4 E
Having e x a m i n e d the case where S' is a cell, we now generalize to the case t h a t S is a cell graph. The next theorem describes the situation in this case. Theorem
1 1 . 2 . 2 4 Let S be a cell graph with cells 1 , . . . , k satisfying Ik;, - . . . ~ - I 1 ~- E 1~- . . . ~ - E k
(11.5)
relative to S. Let T be a threshold graph disjoint from S with vertices X 1 , . . . , X t satisfying (11.3) relative to T. If S and T are joined by edges
tl.2
Matroidal Graphs
289
Figure 11.12: Matroidal graphs corresponding to Table 11.1.
X3
Xl
X2
X4
X3
X1
(~)
X3
X1
X2
X4
X2
X4
(b)
X2
X4
X3
X1
(c)
(d)
X3
X1
X2
X4
~v
v
v
v
(e)
Matroidal and Matrogenic Graphs
290
so that the resulting graph G is matroidal and its threshold vertices are precisely X 1 , . . . , Xt, then for each cell l there exist indices rl, st satisfying 1 of Theorem 11.2.23 with respect to I t and E l, and furthermore Sk ~
"'"
< 81
~
rl ~
...
~ rk.
(11.6)
Conversely, r~ and st can always be chosen to satisfy 1 and 2 as well as (11.6), and if the connecting edges satisfy 3 and ~ with respect to I t and E ~, then the resulting graph is matroidal and its threshold vertices are precisely X1, . . . , Xt.
P r o o f . If G is matroidal with threshold vertices X 1 , . . . , X t , then the sets 11, E l , . . . , I k, E k, {X1 } , . . . , {Xt } are totally ordered by ~ relative to G in a way that is a common refinement of both (11.5) and (11.3). For each cell l, I l U E t U { X 1 , . . . , Xt } induces a matroidal graph whose threshold vertices are X 1 , . . . , Xt and whose other vertices induce a cell. By Theorem 11.2.23 there are indices rl and st satisfying 1-4 of that theorem with respect to I t and E I. Using (11.4) applied to rt, st, I t and E t and also (11.5), we obtain (11.6). Conversely, assume that S and T are given as above. Then by Theorem 11.2.23, for each cell 1 it is possible to choose indices rt, st satisfying 1-2 of that theorem. Furthermore, it is also possible to satisfy (11.6), for example by choosing rl . . . . . rk and Sl . . . . = sk. We assert that if the vertices of I t are joined to X 1 , . . . , Xr~ and the vertices of E l are joined to X 1 , . . . , X~ so as to satisfy 3-4 of the theorem, then the resulting graph G is matroidal and its threshold vertices are X 1 , . . . , X t . To show this, it is sufficient to demonstrate that (11.3) and (11.5) continue to hold relative to G and that (11.4) holds relative to G with rl, sl, I l and E t It is immediate to obtain (11.3) from 3 and 4 of Theorem 11.2.23. As for (11.5), 11 ~- E 1 follows from Theorem 11.2.23. To show that 12 ~ 11, assume that Xq is adjacent to the vertices of I ~. Then q _< rl by (3), hence q _< r2 by (11.6), and Xq is adjacent to the vertices of 12 by 3. Similarly all the relations of (11.5) can be proved. It remains to prove the relations Xr z > E t > X~z+1 and Xs~ ~- I l ~ Xs~+1. Let us prove X~.t ~ E l , the other relations being similar. Let Z be any vertex adjacent to the vertices of E z. If Z is in T, then Z coincides with or is adjacent to X~ z by Theorem 11.2.23. If Z is in S, then by definition of a cell graph and by (11.5), Z must be in 11 U ... U I k. By 3 Z is adjacent at least to X 1 , . . . , X m , where m - m i n ( r l , . . . , r k ) . But m - rt by (11.6), and therefore Z is adjacent to X~. 9
Corollary 11.2.25 Let G be a matroidal graph all couples of which are of the f o r m P4. Then G is a split graph.
2
M a t r o i d a l Graphs
291
roof. We use the results and notations of Theorem 11.2.24. Let h be the rgest index such that Sl _< h _< rl and { X ~ , . . . , Xh} is a clique. Then if < rl, Xh is not adjacent to Xh+x. Therefore in all cases { X h + l , . . . , X~ } is a able set. We assert that in fact { X I , . . . , Xh} is a clique and { X h + l , . . . , Xt } a stable set. Indeed, if sl < h, the first statement is obvious; if Sl - h, follows from the facts that X1 >- I x, . . . , X~ >- I ~ and that X~ ~ are adjacent to all the vertices of 11. If h < rl, the second statement obvious; if h - rl, it follows from the fact that X~+I and X~+2 are not tjacent. Since X 1 , . . . , X~ 1 are adjacent to the vertices of 11 U . . . U I k, u U 11 U ... U I k is a clique. Since X~ l + l , . . . , X t are not tjacent to any vertex of E 1 tJ . . . t.J E k, { X h + l , . . . , X t } U E 1 U " " [_J E k a stable set. .. We conclude the description of the matroidal graphs by answering the ~estion which vertices of T are adjacent to M. h e o r e m 11.2.26 Let S be a cell graph and T a threshold graph as in Theo:m 11.2.2~. Let M be a perfect matching inK2 or a complement of a perfect ~atching inK2 disjoint from S and T, with rn > 2. If the edges joining M S and T are such that the resulting graph G is matroidal with threshold ~rtices X1, . 9 Xt , then 1. the vertices of M are adjacent to all the internal vertices and to none of the external vertices of S; 2. if h is the largest index such that Xh is adjacent to the vertices of M (h - 0 if none), then Sl < h < rl, { X l , . . . , X h } U 11 U . . . U I k is a clique and { X h + l , . . . , X t } t2 E 1 tJ . . . U E k is a stable set. ~onversely, it is possible to choose an index h such that S 1 < h < r l , X I , . . . , X h } is a clique and { X h + l , . . . , X t } is a stable set. If the vertices f M are joined to the vertices of { X 1 , . . . , X h } U 11 U . . . U I k and to no ther vertices, then the resulting graph G is matroidal with threshold vertices ~'1,''', Xt.
'roof. Condition 1 has been proved in Subsection 11.2.3, together with the mt that the vertices of S partition into a clique whose vertices are adjacent ) M and a stable set whose vertices are not. If h - 0, then the vertices of { X 1 , . . . , X t } U .~1 I,_J . . . I,.J E k are not djacent to M, hence they form a stable set, Sl - 0, and 2 follows. The
292
Matroidal and Matrogenic Graphs
case h - t is similarly easy, and therefore we now assume that 0 < h < t. Since (11.3) holds relative to G, the vertices of M are adjacent to X 1 , . . . ,Xh. It follows that { X 1 , . . . , X h } t0 I 1 tO ... tO I kis a cliqueand { X h + l , . . . , X t } tO E 1 U ... U E k is a stable set. Since M is comparable with every vertex of S and T relative to G, it satisfies in particular 11 P- M >- E 1 and Xh >- M >Xh+l relative to G. By transitivity of >- we have Xh >- E 1 and 11 >- Xh+l, from which it follows that Sl _< h _< rl and 2 is proved. Conversely, by Corollary 11.2.25 there is an index hi _< h _< rl such that { X 1 , . . . , X h } tO I1 U ... tO I k is a clique and {Xh+~,...,Xt} U E 1 U ..- tO E k is a stable set. Assume that M is joined to the vertices of { X 1 , . . . , X h } U I 1 U " " U I k and to no other vertices of S U T. In order to prove that the resulting graph G is matroidal with threshold vertices X 1 , . . . , Xt, it is sufficient to show that M is comparable with every vertex of S U T relative to G. We show below that each cell l satisfies I 1 >- M >- E t and t h a t X 1 > - M f o r q _ < h a n d M > - X q forq_>h+l. I t >- M: Let Z be any vertex adjacent to some vertex o f M . I f Z i s i n M or in S, then Z is adjacent to all vertices of I t. If Z is in T, say Z - Xq, then q h + 1" This is proved similarly. 9
11.2.5
Properties of Matroidal Graphs
The structure of matroidal graphs has been completely described in the previous three subsections. Each matroidal graph can be decomposed into a cell graph S, induced by the vertices that belong to P4's and described by Theorem 11.2.16, a perfect matching or a complement of a perfect matching M, induced by the vertices that belong to 2K2's or C4's, and a threshold graph T induced by the threshold vertices. The vertices of M are adjacent to all the internal and to none of the external vertices of S'. The edges joining T to S' are described by Theorems 11.2.23 and 11.2.24, and the edges joining T to M are described by Theorem 11.2.26. We can now determine the threshold dimension t(G) of a matroidal graph G, which is the original
11.2
M a t r o i d a l Graphs
293
reason of interest in these graphs. T h e o r e m 11.2.27 Let G be a matroidal graph and let E l , . . . ,Eq be the
equivalence classes of the relation R of page 271. Then
t(G) = max(IE, I,..., levi). P r o o f . Put t = m a x ( l E l l , . . . , IEql). If fewer than t subgraphs of G cover all its edges, then one of these subgraphs H has two edges from the same equivalence class Ei. These two edges form a couple in G and therefore in H too, and so H cannot be a threshold graph. This shows that t(G) >_ t. To prove the reverse inequality, we exhibit t threshold subgraphs of G covering its edges. We construct one of these subgraphs, called H, as follows. Since G is matroidal, it decomposes into S, M and T as above. Consider a cell of S. If it is a net of order p, J~f(AI,...,Ap, BI,...,Bp), then its p external edges AiBi form one equivalence class, and each internal edge BjBk forms an equivalence class by itself. There are p ways to select one edge from each of these equivalence classes, giving p threshold subgraphs of the net covering its edges. We select one of these p threshold subgraphs and put its edges in H. If the cell is a net-complement of order p, A/'(A1,..., dr, B 1 , . . . , Bp)), then each pair AjBk, AkBj, j ~ k, of external edges forms an equivalence class, and each internal edge BjBk forms an equivalence class by itself. The collection of the external edges AjBk with j < k and all the internal edges is a threshold subgraph of the net-complement containing one edge of each of these equivalence classes. This collection and a similar one obtained with j > k cover all the edges of the net-complement. We select one of these two and enter its edges in H. Now consider M, and assume it is a perfect matching inK2. Then each of its m edges forms an equivalence class by itself and is a threshold subgraph of M, and these m subgraphs obviously cover the edges of M. We select one of these edges and put it in H. If M is a complement inK2 of a perfect matching induced by the vertices A 1 , . . . , Am, A 1 , . . . , A" with all edges present except for AiA~, i - 1 , . . . , m, then each edge X Y forms an equivalence class with ZU, where Z is the unique vertex of M not adjacent to X and U is the unique vertex of M not adjacent to Y. The collection of edges AjA'k with j < k and all edges AjAk forms a threshold subgraph of M having one edge of each of these equivalence classes. This collection and its complement, which has a similar form, cover all the edges of M. We select one of these two collections and put its edges in H.
294
Matroidal and Matrogenic Graphs
Each of the remaining edges of G (namely the edges of T and the edges joining S, M, T and the different cells of S) forms an equivalence class by itself. We put all of them in H. The H constructed in this way has exactly one edge of each equivalence class Ei, and it is clear that we can construct t H's of this form that will cover all the edges of G. Moreover, we assert that each H constructed in this way is a threshold subgraph of G, in other words, no two edges a and b of H form a couple in H. Indeed, if both a and b are in the same cell of S, or if both are in M, then the assertion follows by construction. Otherwise a and b do not form a couple in G, because G contains enough edges so that the endpoints of a and b induce a triangle in G, namely the edges of T and the edges joining S, M, T and the different cells of S. But these edges are in H as well, so the endpoints of a and b induce a triangle in H too, and so a and b do not form a couple in H. This proves that t(G) 0, then b/rn _ D(G[X]) and in particular
D(G) >_ max(D(G[C]), D(G[Q])). On the other hand, no antichain of G can meet both C and Q, since every vertex of C is comparable to every vertex of Q. Hence
D(G) < max(D(G[C]), D(G[Q])). If T is a spanning threshold subgraph of G and G[X] is an induced subgraph of G, then T[X] is a spanning threshold subgraph of G[X]. Therefore t(G) >_ t(G[X]), in particular t(G) >_ On the other hand, if Tc is a spanning threshold subgraph of G[C] and T o is a spanning
11.4
Matrogenic Sequences
303
threshold subgraph of G[Q], then the spanning subgraph of G consisting of the edges of Tc U TQ together with all the edges of G[K] and all the edges of G joining C to K is a threshold subgraph of G. This can be easily verified by noting that no alternating 4-cycle of this subgraph can meet both C and Q. If we apply this fact to a minimum collection of spanning threshold subgraphs of G[C] covering its edges and similarly for G[Q], we see that
t(G) 1 a n d d~ _> 1. F r o m u - 1 - A1 - d~ - dl - 1 we o b t a i n dl - u a n d t h e r e f o r e :r
~
du+ 1 -
.
.. = dn_ 1 - 0.
3. B y 1 above dm - rn, a n d t h e r e f o r e d~n_ 1 >_ rn. H e n c e by 2, r n - 1 _< u. B u t m c a n n o t be u + 1, since by 1 we have Am - - 1 5r u - 1 - A , + I . T h e r e f o r e m < u.
11.4
Matrogenic Sequences
309
4. F r o m 3 we h a v e - 1 -
A.+I
-
d; - d,+l
-1
-
A,+2
-
d~,+l-du+2
-1
-
A~
-
d;_~-dn.
T h e n f r o m 2 we h a v e du+2 = ".. = dn - 1 a n d 1 _< d , + l - d; - u + 1. F r o m t h e l a t t e r e q u a t i o n we d e d u c e t h a t d; _> u a n d t h e r e f o r e d~ _> u. B u t f r o m 2 we h a v e u - dl >_ d,. T h e r e f o r e d. - u. 5. Since d~ - u, we have m >_ u by definition of m , b u t by 3 we h a v e m _< u. T h e r e f o r e m - u. H e n c e by definition of m d , + l < L,, f r o m w h i c h it follows t h a t d; _ u, a n d so d; - u. T h e first e q u a t i o n of 4 now i m p l i e s t h a t d,+l - 1. 6. T h u s d has r e m 5.4.4, it u v e r t i c e s of a s t a b l e set. u, G is a net
b e e n c o m p l e t e l y d e t e r m i n e d as d - (u ", 1~). B y T h e o follows now t h a t G is a split g r a p h . Since m - u, t h e first G f o r m a m a x i m a l clique, a n d so t h e last u v e r t i c e s f o r m Since t h e last u degrees are 1 a n d t h e first u degrees are of o r d e r u.
Condition a: We h a v e n - 2 - / ~ 1 - d~ - d 1 ~ (?% - 1) - 1. It follows t h a t d~ - n - 1 a n d dl - 1, a n d t h e r e f o r e d - (1 ~) a n d G is a p e r f e c t m a t c h i n g . Condition 5" W e h a v e 2 AI - d ~ - dl. If d 1 - 3 a n d dl - 1, t h e n d - (14, 0), w h i c h gives t h e w r o n g A. T h e only o t h e r p o s s i b i l i t y is d~ - 4, dl - 2. F u r t h e r m o r e , 2 - /k2 d ~ - d2. If d~ - 3 a n d d2 - 1, t h e n we would h a v e t h e a b s u r d d3 > d2. T h e only o t h e r p o s s i b i l i t y is d~ - 4, d2 - 2. T h i s d e t e r m i n e s d as d - (2s), a n d t h e r e f o r e G is a p e n t a g o n . 9 1 1 . 4 . 6 Let G ( K , S ) be a split graph with a clique K , a stable set S, and a proper degree sequence da. Let H be a graph with a proper degree sequence dH. Then the A sequence of G (9 H is given by
Lemma
A - (AK, AH, As), wh e re
A.As
-
k-tKI (/Nk+l(dG),.
. ., /Xk+s(dG)),
-
I 1"
Matroidal and Matrogenic Graphs
310
P r o o f . Put d = dH, e = da, f = da| diagram of e is given by
Since G is split, the corrected Ferrers
-IkQ 0s) where Jk is the all-1 matrix and Ik is the identity matrix of order k, O~ is the all-O matrix of order s, P is k x s, and Q is s x k. On the other hand, the corrected Ferrers diagram of f is given by
C(f) =
(
Jk--Ik
Jkx~
P
)
C(d)
Q
0~•
,
0~
where J, I and 0 are all-l, identity and all-O matrices of the indicated orders, respectively, and n = Iv(g)l. From the structure of C ( f ) , it is apparent that !
!
9 for i - 1 , . . . , k, f[ - fi - ei + n - ei - n - e i - ei; 9 fori-k+l
,..., k+n,
f" . fi . n + d.~ _ k -.n
9 fori-k+n+l,...,k+n+s,f[-fi-ei_
di-k
' k - di-k ; di_
-ei-~.
If d is a proper sequence with boxes B I , . . . , Br and a A sequence A = A(d), we denote by ABk the subsequence of A corresponding to the box Bk. We call a p r i m i t i v e sequence of length p (not necessarily a prime) a sequence of length p of the following types: (p-1,-1,...,-1),
(1,...,1, I-p),
(0,...,0).
We are now ready for the characterization of matrogenic split sequences. T h e o r e m 11.4.7 A proper sequence d is the degree sequence of a m a t r o g e n i c split graph if and only if all its ABk are p r i m i t i v e sequences and 1. the n u m b e r q of non-zero p r i m i t i v e sequences is even; 2. f o r k = 1 , . . . , q/2, the k-th and ( q - k ) - t h are equal.
non-zero p r i m i t i v e sequences
11.4
Matrogenic Sequences
311
P r o o f . " I f " : Let G be a realization of d = (dl,. 9 dn) with boxes B 1 , . 99, B r and A = Aa. The graph G is BT by Theorem 11.4.4, since the sum of the elements of every primitive sequence is 0. We use induction on r, the number of boxes of G. If r = 1, then A = 0 by Condition 1. Hence G is threshold and thus matrogenic split. For the induction step, we distinguish three cases: C a s e 1: G has dominating vertices (dl = n and S = 0. C a s e 2: G has isolated vertices (d~ = 0).
1). In this case, let K = B1
In this case, let K = O and
S=B~. C a s e 3: G has no dominating and no isolated vertices. K = B1 and S = B~.
In this case, let
In Case 3, B~ must be stable, otherwise by the BT property the vertices of B1 would be dominating in G. Similarly, each vertex of B~ has some neighbors in B1, otherwise the vertices of B~ would be isolated in G. Therefore B1 is a clique. Moreover, each vertex of B2 U ... U B~-I is adjacent to each vertex of B1 and to no vertex of B~. In all cases, the subgraph induced by K U S is a split graph G'(K, S), and if H denotes the subgraph induced by v(a)-v(a'), then G = G'(I'(,S)QH. Hence A = (AK, AH,/kS)
by Lemma 11.4.6. In Case 1, AK = 0 since the vertices of K are dominating, and As is empty, hence AH satisfies the conditions of the theorem. Similarly in Case 2, AK is empty and As = 0, and AH satisfies the conditions of the theorem. In Case 3, AK and As are not equal to the zero sequence as B1 and B~ are neither completely connected nor completely disconnected. Hence by Condition 2, Air = As, and again AH satisfies the conditions of the theorem. Since in Case 3 AK and As are equal primitive sequences, G' is a net or a net-complement (or a graph on two vertices) by Lemma 11.4.5. In all cases, H is a split matrogenic graph by the induction hypothesis, and hence G is split matrogenic by Theorem 11.4.3. " O n l y if"" Let G be matrogenic split with boxes B , , . . . , B~. We use induction on r. If r = 1, G must be complete or edgeless, hence A is the zero sequence. For the induction step, we have by Theorem 11.4.3 that G = G' @ H, where H is again a matrogenic split graph and G'(K, S) is a
312
Matroidal and Matrogenic Graphs
split graph which is (a) a complete graph or ( b ) a n edgeless graph or (c) a net or (d) a net-complement. By Lemma 11.4.6, A = (A/,-, AH, As), where in Case (a) As is empty and A/s- = 0, and in Case (b) A/,- is empty and As = 0. By Lemma 11.4.5, both A/s- and As are primitive sequences and if both are non-zero, they are equal. Moreover, in Case (a) in Case (b) B1 U B~, in Case (c)or (d).
B1, v(a')
-
B,,
The result then follows by induction.
Chapter 12 Domishold Graphs 12.1
Introduction
Recall from Definition 9.1.3 that a dominating set of a graph G = (V, E) is a subset S of V such that every vertex not in S has a neighbor in S, and that the domination number 7(G) of G is the size of a smallest dominating set of G. We have seen in that section that the domination number is bounded above by the Dilworth number. Also recall that the threshold graphs are defined as the graphs for which the characteristic vectors of the stable sets are separated from the characteristic vectors of the nonstable sets by a hyperplane. In a similar spirit Benzaken and Hammer [BH78] introduced the domishold graphs as the graphs for which the characteristic vectors of the dominating sets are separated from the characteristic vectors of the nondominating sets by a hyperplane. They characterized the domishold graphs and showed that the threshold graphs form a proper subclass of domishold graphs. They also introduced the concept of domistable graphs. Later, Payan [PayS0] introduced a class of equidominating graphs, and Cherynak and Chernyak [ccg0], introduced a class of graphs called pseudodomishold graphs. We study these results in this chapter. In addition, some enumeration problems and other characterizations of domishold graphs are discussed in Chapter 13.
12.2
Notation and Main Results
We begin with the definitions of domishold graphs and domistable graphs. 313
314
Domishold Graphs
1. A graph G - (V, E) is called a d o m i s h o l d g r a p h if there exist a positive real-valued function w on V and a real 0 such that for all S C V,
D e f i n i t i o n 12.2.1
S is dominating if and only if
>_ o. xES
2. A d o m i s t a b l e g r a p h is a graph all of whose minimal dominating sets
are stable sets. 3. A graph G - (V, E) is called e q u i d o m i n a t i n g if there exist a positive integer-valued function w on V and a positive integer 0 such that for all Q c_ V, Q is a minimal dominating set if
only if
w(x) -- 0 xEQ
Clearly, in the definition of domishold graphs one can assume without loss of generality that w and 0 are positive integers. We refer to w(x) as the weight of x and to 0 as the threshold value. The domishold graphs and the domistable graphs were defined in [BH78] and the equidominating graphs were defined in [PayS0]. The results and concepts in this section are from [BH78]. As usual, N(x) denotes the neighborhood of vertex x and we put M(x) = V - ( N ( x ) U {x}). a stable dominating pair is a set of two nonadjacent vertices each of which is adjacent to every other vertex. Let Sk denote an edgeless graph on k vertices, Kk denote a complete graph on k vertices, and J2k denote the complement of a perfect matching on 2k vertices (kK2). The graph J2k is called a cocktail-party graph. We write Q d o m G to indicates that Q is a dominating set of G. Let us now define a binary relation ~d on the vertex set V of G as follows" D e f i n i t i o n 12.2.2 For x, y E V, x Ld Y if and only if for all Q c V - { x , y}, (Q u {y}) dom G ~ (Q u {x}) dom G. L e m m a 12.2.3 The relation ~d is reflexive and transitive, i.e., it is a preorder.
P r o o f . The reflexivity is obvious. To prove the transitivity, we assume that i, j, k are distinct, i ~e j and j ~d k, and show that i Ld k. Let Q c_ V - { i , k}
12.2
315
Notations and Main Results
be such that ( Q U { k } ) d o m G . I f j ~ Q, t h e n Q U { j } ) d o m G a n d h e n c e (Q u {i}) dom G. If j C Q, put Q' = Q - j. Then Q u {k} = ( ( Q ' u {k}) u {j }) dom G, and hence ((Q'u { k}) U {i}) dom G, and the latter set contains k but not j. So ((Q'U {j}) u {i}) = Q u {i} dom G. Hence i ~d k. 9 In view of Lemma 12.2.3, the relation ~d is called the dominal preorder of G. We now state the main result of this section. T h e o r e m 12.2.4 For every graph G equivalent:
(V, E), the following conditions are
1. G is domishold; 2. the dominal preorder ~d is linear; 3. G can be built from the empty graph by repeated application of the operation G' ~ G", where G" = (G' U S;) | Kq | J2~ and p + q + r > O; in other words, G can be obtained from an empty graph by repeatedly adding an isolated vertex, a dominating vertex, or a stable dominating pair; ~,. V can be partitioned into sets 1/1, 1/2 and 1/3 inducing the graphs SIVll, KIV21, Jig31 respectively, with the following properties: (a) each vertex in 1/2 is adjacent to each vertex in 173; (b) for each i E Vx, g ( i ) N V3 induces a cocktail-party subgraph of G; (c) the neighborhoods of the vertices of V1 are nested; 5. G can be obtained from some threshold graph with split partition (K, S) by substituting cocktail-party graphs for some vertices of K; 6. the graphs of Figure 12.1 are not induced subgraphs of G. Proof. 1) =v 2): It is enough to show that for every two vertices x, y with w(x) > w(y), we have x ~d Y. This is indeed the c~se, since for every
QcV-{x,y}, Q u {y} dom G ==~w(Q) -4- w(y) >_ 0
w(Q) + w(x) > 0 :=~ Q u {x} dom G, where w(Q) - EzeQ w(z). 2) =V 3)" First we show that if x is any maximal vertex in the linear order
Domishold Graphs
316
Figure 12.1" The forbidden induced subgraphs of domishold graphs.
H1
Ha
H2
H4
Hs
~d, Q is any maximal stable set, and x ~ Q, then x is adjacent to every vertex in Q. Since Q is a maximal stable set, Q dom G. Now for every y C Q, Q = ((Q - y ) u {y}) dom G and hence ((Q - y ) u {x}) dom G, since x ~d Y. But the only neighbor of y in ( Q - y ) u {x} is x. Thus x is adjacent to y, as required. We now show that G contains an isolated vertex, a dominating vertex, or a stable dominating pair. The result then follows by induction since removing isolated vertices or a dominating set preserves Condition 2, as is easy to verify. Assume that G contains no isolated vertex or dominating vertex. Let x be a maximal vertex in ~d and y C M(x) (y exists since x is not dominating). We show that X(y) = N(x). Suppose that, if possible, zx C E and zy q~ E. Extend {y, z} to a maximal stable set Q, which will not contain x. Then, as shown above, x is adjacent to every vertex in Q, contradicting xy ~ E and y C Q. This shows that N(x) C_ N(y). It follows that y is also a maximal vertex in Le and hence, by symmetry, N(y) C_ N(x). It follows that every vertex in M(x) has the same set of neighbors as x. Thus if z C N(x) (z exists since x is not isolated), then {z, y} dom G and hence {x, y} dom G as x is a maximal vertex. Thus G has a stable dominating pair {x, y}, as required. 3) =~ 4): Let Go,..., Gt be a sequence of graphs with Go = 0, Gt = G and
G~+I - (G~ u Sp,) | Kq, @ &~,,
i-O,...,t-
i.
12.2
Notations
and
Main
317
Results
Putting i
i
i
and using an easy induction, we see that this partition has the desired property. 4) =~ 5)" Identify each vertex of V3 with its unique nonneighbor in V3 to obtain a clique V~. The resulting graph is clearly a threshold graph with split partition (1/2 U V3', 1/1), 5) =~ 6)" The graphs of Figure 12.1 are not threshold graphs, nor do they contain a stable dominating pair. Hence they cannot be obtained from a threshold graph by substituting cocktail-party graphs for some vertices in the clique. The result now follows since if G satisfies Condition 5, so does every induced subgraph of G. 6) => 3)" It is enough to show that G contains an isolated vertex, a dominating vertex, or a stable dominating pair. If G is not connected, then G contains isolated vertex since H1 is forbidden. So assume that G is connected. Then by a well-known result about P4-free graphs (Lemma 14.3.6), G is not connected since H2 is forbidden. If a connected component of G has one vertex, then it is a dominating vertex in G, and if it has two vertices, then they form a stable dominating pair in G. Otherwise, G contains two components, each with at least three vertices. Let xl, x2, x3 be vertices of a component such that x2 is adjacent to the other two. Similarly, let yl, y2, y3 be vertices of another component with y2 adjacent to the other two. Then, depending on the adjacency between xl and x3 and between yl and y3, G contains Ha,/-/4 or Hs as an induced subgraph, a contradiction. 3) =~ 1)" The empty graph is obviously domishold. Assuming that G' = (G u Sp) | Kq @ J2r and that G is domishold, we show that G' is domishold too. Let wl represent the weight of each 1 E V(G), and 0 the threshold for G. Let w* - mint wt/2. We may assume that 2w* < 0, since otherwise G is a clique, and we can change all weights wt for 1 C V(G) as well as 0 into 1 to achieve2w* < 0. Let us also put W - 1 + ~ t e y ( a ) wl and let us define 0 ~ 0 + p W and
i , wi -
W 0 + pW O+pW-w*
v(G),
iESp, i E Kq, i G J2~.
318
Domishold Graphs
The weights w~ and the threshold O' characterize the dominating sets of G'. Indeed, every minimal dominating set D' of G' is of one of the following three forms: 1. D ' = {k}, k e Kq; 2. D' = { x , y } , x C J2r, x r y, y C J2,. U Sp t2 V(G);
3. D' = D tO Sp, where D is a minimal dominating set of G. It is straightforward to verify that these are also precisely the minimal sets Q' of G' satisfying w'(Q') > 0'. tt Since each of Ha,//4, Hs of Figure 12.1 contains an induced 6'4, it follows that all threshold graphs are domishold graphs. The next theorem gives other characterizations of threshold graphs among the domishold graphs. T h e o r e m 12.2.5 For a domishold graph G, the following conditions are equivalent: 1. G is a threshold graph; 2. G is a C4-free graph; 3. G is a split graph; .~. G is an interval graph; 5. G is a domistable graph.
P r o o f . 1) =~ 2,3,4,5): This follows from Theorem 1.2.4 characterizing threshold graphs, Theorem 2.1.6 characterizing interval graphs, and the fact that a minimal dominating set of a threshold graph considered as a split graph G(K, S) is S itself or consists of one vertex from K and all its nonneighbors in S. 3) =~ 2): This follows from Theorem 5.2.1. 4) =~ 2): This follows from Theorem 2.1.6. 5) =~ 2): Since G is a domishold and domistable graph, it must contain a dominating vertex or an isolated vertex (otherwise, as in the proof of 3) =, 4) in Theorem 12.2.4, if x is any maximal vertex with respect to the dominal preorder, y C M ( x ) and z C N(x), then {y, z} is a minimal dominating set that is not stable). Delete any dominating or isolated vertex from G, and the
12.2
Notations and Main Results
319
resulting graph is clearly again a domistable and domishold graph. Hence this process can be repeated until all the vertices are removed. Since no vertex of an induced C4 can become isolated or dominating, it follows that G is C4-free. 2) => 1): This follows from Condition 6 of Theorem 12.2.4 and Condition 2 of Theorem 1.2.4. 9 A linear inequality n
E
WiXi ~__ 0
i=1
with Wl >_ " " >_ Wn > 0 is called domigraphic if W l , . . . , w ~ are weights and 0 is a threshold of a domishold graph. The condition n
y~'~wi >__O i=1
is obviously necessary for the inequality to be domigraphic since V(G) is a dominating set. In the case n - 1, it is sufficient too. For n - 2, this condition and w2 < 0 => wl < 0 are again sufficient. The following theorem gives a recursive characterization of domigraphic sequences for n >_ 3. Theorem
12.2.6 A linear inequality n
E wixi >_ 0 i=1
with W 1 ~ " ' " ~___W n > 0 and n >_ 3 is domigraphic if and only if one of the following conditions holds: 1.
W 1 ~
0 and ~i~2 wixi >_ 0 is domigraphic;
2. w~ < O, E ~ 2 wi < 0 and ~i~2 wixi >_ 0 - Wl i8 domigraphic; 3. Wl < O, W 1 -~- W2 ~__ 0 and ~i~3 wixi >_ 0 is domigraphic. P r o o f . Let the inequality ~-~iL1W i X i ~-- 0 be domigraphic and let G be the associated graph. Let x be a maximal vertex of the dominal preorder. Without loss of generality, we may assume that w(x) - w I (&s in the proof of 1) => 2) of Theorem 12.2.4). If x is dominating, then Condition 1 is satisfied. If x is isolated, then Condition 2 is satisfied. If G has no isolated or dominating vertex, then there exists another maximal vertex y such that {x, y} is a stable
Domishold Graphs
320
dominating pair (as in the proof of 2) ~ 3) of Theorem 12.2.4). Without loss of generality, we may assume that w(y) = w2. Then Condition 3 is satisfied. Conversely, suppose one of the Condition 1 - Condition 3 is satisfied. Let G' be the domishold graph associated with the domigraphic inequality in each case. Then, clearly, the linear inequality of the theorem corresponds to a domishold graph obtained from G' be adding an isolated vertex (Condition 2), or a dominating vertex (Condition 1), or a dominating stable pair (Condition 3). "
12.3
Equidominating Graphs
Payan [PayS0] defined equidominating graphs as follows: D e f i n i t i o n 12.3.1 A graph G = (V, E) is said to be e q u i d o m i n a t i n g when there exist a weight function w: V ~ Z + U {0} and a number t E Z + U {0} such that for every S C_ V, S is a minimal dominating set ~
~
w(x)-
t.
xES
The class of equidominating graphs neither includes nor is included in the class of domishold graphs. As evidence, observe that 2K2 is an equidominating graph but not a domishold graphs. To see that it is equidominating assign a weight of 1 to each vertex of one edge and a weight of 2 to each vertex of the other edge, and let t = 3. From Theorem 12.2.4, 2K2 is forbidden for domishold graphs. On the other hand, K2,z is a domishold graph (Theorem 12.2.4), but not an equidominating 'graph. To see this, let (A, B) be the bipartition of K2,3 into stable sets with ]A I = 2. Then A and any pair {x, y} with x C A and y C B are minimal dominating sets, and hence every vertex should receive the weight t/2. But then B is a minimal dominating set with total weight exceeding t. The following theorem characterizes the equidominating graphs that are also domishold graphs. T h e o r e m 12.3.2 ([Pay80]) The following conditions are equivalent for every graph G: 1. G is both a domishold graph and a equidominating graph;
12.3
Equidominating Graphs
321
2. G has weights and threshold that characterize simultaneously the dominating sets by inequality and the minimal dominating sets by equality; 3. G can be built from the empty graph or from a cocktail-party graph (the complement of a perfect matching) by repeatedly adding an isolated vertex or a dominating vertex; ~. G has no induced graph isomorphic to H i , . . . , H4 of Figure 12.2.
Figure 12.2: The forbidden induced subgraphs for domishold and equidominating graphs.
H1
H2
H3
H4
P r o o f . 2) =~ 1)" This is obvious. 1) =~ 3)" We proceed by induction on IV(G)]. Let G - (V, E) be a domishold graph with weight function w~ and threshold t~, as well as an equidominating graph with weight function w~ and threshold t~. By Theorem 12.2.4, G has an isolated vertex, a dominating vertex, or a stable dominating pair. C a s e 1" G has an isolated vertex x 9 For i - 1, 2, let WG_ ~ x be the restriction of w~ to G - x, and let t ia_~ - t ~ - wb(x). It is easy to check that G - x is domishold and equidominating with these modified weights and thresholds. Hence by induction G - x satisfies Condition 3, and so G does too. C a s e 2: G has a dominating vertex x. The argument is similar to Case 1 with t~_~ - t~. C a s e 3" G does not have an isolated or a dominating vertex, but has a stable dominating pair x, y. Observe that a set S C V(G) is a minimal dominating set of G if and only if S is a minimal dominating set of G - {x,y}, or S - {x,y}, or for some v e V ( G ) - {x,y}, S - {x,v} or 5' - {y,v}. Since we m a y assume that IV(G)I >_ 3, it follows that w~ - t2c/2. Hence S is a minimal dominating set of G if and only if ISI - 2. We assert that G is a cocktail-party graph. Since every two vertices of G form a dominating set, G does not contain any induced K1 tO K2 or an i n d u c e d / 3 . Hence G does not contain any isolated vertices or induced K1,2 or induced K3. Therefore every 9
322
D o m i s h o l d Graphs !
q
connected component of G has exactly two vertices, and hence G forms a perfect matching, and G satisfies Condition 3. 3) =~ 2)" The empty graph trivially satisfies Condition 2 with null w a and ta - O. The cocktail-party graph satisfies Condition 2 with wa - 1 and to - 2. Assume now that x is an isolated or dominating vertex of G such that G - x satisfies Condition 2 with weights wa-~ and threshold t a - ~ . Then we obtain wa and ta as follows: 9 If x is an isolated vertex, then S is a minimal dominating set of G if and only if x 6 S and S - {x} is minimal dominating set of G - x. We then set for all v 6 V -
wa(v)
--
wa_~(v)
wa(x)
=
y~ wa_~(v) + 1 - ta_~ v~V-{~}
ta
=
E w a - x ( v ) + 1. v~V-{~}
{x}
9 If x is a dominating vertex, then S is a minimal dominating set of G if and only if S - {x} or S is a minimal dominating set of G - x. Then we set for a l l v C V - { x }
-
a(x)
-
ta_
ta
--
tG-x.
In each case it can be verified that G is both a domishold graph and a equidominating graph with weights wa and threshold ta, and hence condition 2 is satisfied. 3) =~ 4)" This follows from the fact that the graphs H 1 , . . . , / / 4 neither are cocktail-party graphs nor do they contain isolated or dominating vertices. 4) =~ 3)" Assume that G contains no induced subgraph isomorphic to the graphs of Figure 12.2, and assume further that G contains no isolated or dominating vertices. We complete the proof by showing that G must be a cocktail-party graph. G is connected, for otherwise G contains HI. Hence by a well-known result about P4-free graphs (Lemma 14.3.6), G is not connected since G is H2-free. Each connected component of G has at least two vertices since G has no dominating vertices. If each component of G has exactly two
12.4
P s e u d o d o m i s h o l d Graphs
323
m
vertices, then G is a cocktail-party graph, as required. Otherwise, G has a component with at least two vertices Xl,X2 and another with at least three vertices yl, y2, y3 such that Y2 is adjacent to both yl and y3. But then, depending on whether or not 91 and y3 are adjacent, these five vertices induce Ha or//4 in G, a contradiction. .. The following corollary follows from Condition 3 or 4 of Theorem 12.3.2.
Corollary 12.3.3 The threshold graphs are equidominating (and domishold).
12.4
P s e u d o d o m i s h o l d Graphs
Chernyak and Chernyak [CC90] call a graph G - (V, E) a pseudodomishold graph if there is a function w" V , R + U {0}, w ~ 0, and a nonnegative real t such that for every S c_ V, SdomG
?, ~ w( x ) >_t xES
--(SdomG)
:, ~ w ( x ) _ O,
P4}.
Let tn denote the number of non-isomorphic n-vertex graphs in M without isolated vertices, dominating vertices, or stable dominating pairs (tn -- 0 for n < 4), and put tnX n n>l
For n > 6, it is possible to evaluate tn from the definition of M by partitioning the vertex set of the complementary graph into cycles and paths of length 3 or more. Let Un denote the number of non-isomorphic n-vertex HP graphs, and put u(x) - ~ unx n. n>l
The following result expresses u ( x ) i n terms of t(x). C o r o l l a r y 12.4.2 t ( ~ ) + x - ~3 u(x)
-
1-
2x-
x 2 + x 3-
x 4"
L2.4
325
Pseudodomishold Graphs
Figure 12.3" Forbidden configurations and induced subgraphs for HP graphs.
(a) Forbidden configurations. Dashed lines indicate nonedges and solid lines indicate edges. Other edges may or may not exist.
w
v
"i__ '. i I I I w
w
w
v
w
(b) Forbidden induced subgraphs
w
w
Domishold Graphs
326 An estimate of un is given by the following result.
Corollary 12.4.3
For
n > 6, (2.32) n-1 < un < (2.37) n-1.
We can compare the above estimates with an estimate of the number d~ of non-isomorphic n-vertex domishold graphs. Put
d ( x ) - E dnxn" n>l Then we have the following result.
Corollary 12.4.4 XmX 3 d(x) Further,
dn ~" c r n, w h e r e r ..~
-
1 - 2 x - x 2 + x 3"
2.25
is t h e m a x i m u m
root (in absolute
value)
of the equation x 3 - 2x 2 - x + 1 -
O.
We see that un is larger than dn, consistent with the fact that all domishold graphs are HP graphs, but not conversely.
Chapter 13 The Decomposition Method 13.1
Introduction
In Definition 11.4.2 we saw how to compose a split graph with an arbitrary graph, and in this chapter we investigate the opposite operation of decomposition. This subject was studied by Tyshkevich [Tys84, TysS0], Chernyak and Chernyak [CC90, COS1]and Tyshkevich and Chernyak [TC85b, TC85a]. It turns out that every graph has a unique decomposition into indecomposable components. Tyshkevich and Chernyak [TC85a] also introduced a generalization from the class of split graphs to the larger class of polar graphs that we saw in Section 7.6, and obtained a hierarchy of decompositions and a unique decomposition at each level of the hierarchy. The decomposition method proves to be a unifying approach to the study of many graph problems, since many properties of graphs hold for a graph if and only if they hold for each indecomposable component, perhaps with a small change which is easy to pinpoint. It simplifies the description of the structure of several graph families, leads to their enumeration, and serves as a tool to explore them. Section 13.2 presents the decomposition, the hierarchy, and the unique decomposition result. Section 13.3 applies this theory to threshold and domishold graphs, Section 13.4 applies it to box-threshold graphs and enumerates them, and Section 13.5 does the same for matroidal and matrogenic graphs. 327
The Decomposition Method
328
13.2
The Canonical D e c o m p o s i t i o n
In Section 7.6 we defined the class of polar graphs and its subclasses (c~,/3), and discussed the complexity of their recognition problems. In this section we present a theory of decomposition of polar graphs based on Tyshkevich and Chernyak [TC85b]. To review the definition, a graph G = (V, E) is said to be polar if V can be partitioned into subsets A and B, possibly empty, such that each of the graphs G[A] and G[B] is a union of cliques. This partition is called a polar partition. If in addition every connected component of G[A] (resp. G[B])is (a clique)of size at most c~ (resp./3), we say that G belongs to the class (c~,/3) of polar graphs. The polar partition may not be unique: for example, if G is the path P3 with consecutive vertices a, b, c, then we may take A = {b}, B = {a,c} or A = {a,b}, B = {c}, putting G in (1, 1)in both cases; if G is the cycle C4, we may take A to consist of three vertices, putting G in (2, 1), or we may take A to consist of two adjacent vertices, putting G in (1, 2), or we may take A to consist of two non-adjacent vertices, putting G in (2, 1) again. Other examples are G = C6 with A consisting of three mutually adjacent vertices and G E (1, 3), and G bipartite, with G C ( ~ , 1). The class (1, 1) is precisely the class of split graphs. We indicate a polar graph with its polar partition as G(A, B), and use the (c~,/3) notation for it as well. We say that polar graphs G(A,B) and G'(A', B') are polar isomorphic if there is an isomorphism of G to G' mapping A onto A' (and hence B onto B'). It can be checked that the graph in Figure 13.1 is not polar. Figure 13.1" A non-polar graph.
w
v
w
Clearly if G(A,B) r (o~,/3), then G(B, A) C (~, c~). Another easy result is the following one, establishing a hierarchy among the polar graphs. P r o p o s i t i o n 13.2.1 If a < (~' and/3 < /3', then (a,/3) C_ (a',/3'), and the
inclusion is proper if at least one of the inequalities is strict.
13.2
The Canonical Decomposition
329
P r o o f . The inclusion is trivial. We show that the graph
C = (/3' + I)K~, U (c~' + I )K~,, which clearly belongs to (c~',/3'), cannot belong to (c~',/~'- 1) if /3' > 1. Indeed, if G(A, B) E (c~',/~'- 1), then A must contain at least one vertex of each of the c~' + 1 copies of KZ,. Consequently, G[A] contains a stable set of size c~' + 1, which is not the case by G(A, B) E (c~',/3'- 1). Similarly if c~' > 1, G cannot belong to (c~'- 1, fl). 9 We now define the operation of graph composition, which specializes to the operation of Definition 11.4.2 for the class (1, 1). D e f i n i t i o n 13.2.2 The classes of polar and ordinary graphs are denoted by
7) and G, respectively. We define the c o m p o s i t i o n operator
O:Pxg~g as follows. If G(A,B) C 79 and H C ~ are vertex-disjoint, then G(A, B ) 0 H is obtained from G U H by adding every edge between A and H. The graphs G and H are called the c o m p o n e n t s of the composition. Note that if in the above definition H is polar and H(C, D) E T', then the composition Q = G O H is also polar Q(A U C, B U D). Thus O is a binary operation on the class of polar graphs. In general, composition is associative when all components except possibly the rightmost are polar. Therefore the algebra (T ~, O ) i s a semigroup, and each class (c~,/3) is a subsemigroup. D e f i n i t i o n 13.2.3 A graph G that can (cannot) be written in the form F(A, B ) 0 H is called d e c o m p o s a b l e ( i n d e c o m p o s a b l e ) . If in addition
we require that F(A, B) E (c~,/3), G is said to be decomposable 5ndecomposable) at t h e level (c~,/3). If in addition G is polar, G(C,D) E (c~,/3) and H E (c~,/3), then G(C,D) issaid to be decomposable (indecomposable) in t h e class (c~,/3). It can be verified that Cs is an indecomposable polar graph and the graph of Figure 13.1 is an indecomposable non-polar graph. To state the decomposition result below, we define an equivalence relation on the set of polar graphs as follows: 1. each polar graph G(A,B) is equivalent to itself (but when A and B are non-empty, not to any other polar graph even if they are polar isomorphic);
330
The Decomposition Method
2. every two graphs of the form G(O, B) are equivalent, and every two graphs of the form G(A, rg) are equivalent. It is easy to see that the composition Q is commutative on every class of equivalent graphs. T h e o r e m 13.2.4 1. every graph from the class (a, fl) can be represented as a composition of components (necessarily from (a, fl)) that are indecomposable in the class (a, fl). The decomposition is unique up to permutations of consecutive equivalent components.
2. every graph not from the class (a, fl) and decomposable at the level (a, fl) decomposes uniquely as H(C, D) @ F, where H(C, D) C (a, fl), F ~ (a, fl) and F is indecomposable at the level (a, fl). Before proving the theorem, we use it to make the following definition. D e f i n i t i o n 13.2.5 Let G(A,B) be decomposable in the class (a, fl). Consider its decomposition into indecomposables in this class. If in this composition we replace each maximal set of consecutive equivalent components by their composition, we obtain a representation of G in the form
G(A, B)
-
Xl
(~ . . .
(~ X n ,
X k --
Gk(Ak, Bk) e (a, fl),
(13.1)
where Xk and X k + l a r e not equivalent and each Xk satisfies one of the following three conditions: 1. Ak - O C Bk; 2. Ak r O - Bk; 3. Ak, Bk r 0 and Xk is indecomposable in the class (a, fl). The decomposition (13.1) is referred to as t h e c a n o n i c a l d e c o m p o s i t i o n of G ( A , B ) in t h e class (a, fl). Likewise, if F does not belong to the class (a, fl), its c a n o n i c a l d e c o m p o s i t i o n at t h e level (c~,/3) has the form F
= X 1 (~ . . .
(~ X n
(~
H,
where the Xk are as above, H ~ (a, fl) and H is indecomposable at the level
13.2
The Canonical Decomposition
331
13.2.4. The existence of the decompositions in question follows from the finiteness of the graphs, and what we need to prove is their uniqueness. P A R T 1" Let G(A, B) C (o~,/~). For each a C A, we let d(a) denote IN(a)N BI, and for each X C_ A, d(X) - ~-~aEXd(a). Similarly for each b E B, d(b) denotes IN(b)N AI, and for each Y C_ B, d(Y) - ~be~" d(b). Since the induced subgraph G[B] is a union of cliques of cardinalities at most /3, there exists a unique partition B - B1 tO -.. tO BZ, where G[Bj] ji(j and the nj are non-negative integers. Thus Bj uniquely partitions into cliques of cardinality j" Bj - Bj~ U ... U Byn,. Similarly A - A 1 U . . . U Ac~, where G[Ai] - m i K i and the m i are nonnegative integers, and Ai uniquely partitions into stable sets of cardinality i" Ai -- A l l I..J . . . I..J Aim,. Observe that since the Air are stable sets, for each canonical decomposition (13.1), each All is contained in one of the Bk, and similarly each Bjt is contained in one of the Bk. We represent G(A, B) in the form Proof of Theorem
/..
G(A,B) - X1 6) H(C,D),
Xa - GI(A1,B~),
where X1 is the first term of a canonical decomposition (13.1). We want to show that X1 is uniquely determined by G(A, B), from which it follows that H(C,D) is also determined, and the result follows by induction. We distinguish three exclusive cases. Case 1" A - / ~ and some Bjk satisfies d(Bjk) - O. Denote the union of all such Bjk by N. We must have A1 - 0, for otherwise (0, Bjk) is a further component of X1. Similarly B1 - N, and so G1 - G[N] and X1 is uniquely determined by G(A, B). Case 2" B r e and some Bik satisfies d(Aik) - i . IBI. Denote the union of all such Aik by N. In analogy to Case 1 we must have A1 - N, B 1 - O , G1 - G I N ] a n d X1 is uniquely determined. C a s e 3: The situations under the previous cases do not occur. We may also assume that A, B r O, for otherwise the theorem is trivial, and therefore we have that A1, B1 -r O and X1 is indecomposable. We note that /..
t..
/..
^ ^ d(Bjk) { < j" JAil, if Bjk C Be, j I./~ll, if Bjk ~ B1.
(13.2)
If, for example, the first line of (13.2) is false, then (;g, Bjk) is a right component of X1. The second line follows similarly because in that case Bjk C_ D. Equation (13.2) shows that B1 is uniquely determined by A1.
332
The Decomposition M e t h o d
Assume that G(A, B) can also be represented in the form ~..
G(A, B) - X~ fi) H'(C', D'),
A!
X~ - Gi (A'I, J~l),
where X~ is the first A term ofh another canonical decomposition. We assert that one of the setsA ! B1 and B~ contains the other one. For assume that, if A h possible, x C B a B 1 and y C B ~ B1. Then the assumptions on x imply that h A! ~. x C B~ND' and therefore Aa C_ ANN(x)C_ Aa. Similarly the assumptions on y imply that A1 C_ A 1. Thus A1 A-! A 1. But as noted above, A1 determines B1 h by (13.2), and therefore B1 -- BJ' contrary to assumption. This proves the assertion. Next we assert that .B 1 -- B 1. Assume that, if possible, B 1 C B1. A! A A Then by theA argument ofA ! the t .previous assertion, A 1 C_ A1, Aand since A1 ! determines B1, we ..have A 1 C A1. Moreover, every., vertex of A 1 of adjacent to every vertex of/31 - B[, and every vertex of B~ of non-adjacent to every v e r t e x of A 1 - A 1. This shows that (AI,Ba)is a component of (A1,_B1) , contradictinfi^ the indecomposability of X1. This proves our second assertion that B1 - B~, and therefore D - D'. We also note that
d(Ail)
IDI, if A~, c_ A1, _< i IDI, if Aa ~ m l ,
> i.
(13.3)
for otherwise X1 decomposes. Equation (13.3) shows that D uniquely deterA ~ A! ! mines A1, and thus A1 - A 1. Therefore X1 - X 1. P A R T 2" Assume that G ~ (c~,/3) and G has two decompositions
G - H(C,D) Q F - H'(C',D') (!) F', with H ( C , D ) , H ' ( C ' , D ' ) C (c~,~)and F , F ' ~ (c~,~). It follows that the set R - V ( F ) N V(F') is not empty, for otherwise F ' is an induced subgraph of H and so F ' E (c~,/3), or symmetrically. Assume that , if possible, R' = V ( F ' ) - R 7/=r We represent R' in the form R ' - C1 U D1 with C1 C_ C and D1 C_ D. But then we have the decomposition F ' - F(R')(CI,D1)fi)F(R), which contradicts the indecomposability of F ' at the level (c~,/3). This shows that R ' - e , i.e., V(F') C_ V(F). By symmetry we also have V(F) C_ V(F'), and so V(F') - V(F). Since F is an induced subgraph of G, V(F) uniquely determines F, and also C and D as the sets of vertices of G adjacent to all and to none of the vertices of V(F), respectively. The sets C and D uniquely determine H. 9
13.2
T h e Canonical Decomposition
333
C o r o l l a r y 13.2.6 The canonical decomposition of the graph G(A, B) in the class (a,/~) coincides with its canonical decomposition in every class (c~',/3') containing (a, fl). In other words, a polar graph has a unique canonical decomposition. Proof. This follows from Part 1 of Theorem 13.2.4. C o r o l l a r y 13.2.7 Every non-polar graph G has a unique representation of the form G = H(C,D) 0 F, where H(C,D) is polar and F is an indecomposable non-polar graph. Proof. This follows from Part 2 of Theorem 13.2.4. P r o p o s i t i o n 13.2.8 If (a',fl') D (c~,fl), then there exist graphs indecomposable at the level (a,/~) but decomposable at the level (a', fl'). Proof. We may assume without loss of generality that a' > a. Let F be a non-polar indecomposable graph, such as the graph of Figure 13.1. Define a graph G by its canonical decomposition at the level (a',/Y) as G = S~,(A,O)O F, where S~, is an edgeless graph on a' vertices. We assert that G is indecomposable at the level (a,~). Assume the contrary, and let G1(A1, B1) be the first component of the canonical decomposition of G at the level (c~, fl). Let G2(A2, B2) be the first component in the canonical decomposition of GI(A1,B1)in the class (a', fl'). Then G2(A2, B2)is the first component of the canonical decomposition of G at the level (a',/3'). Since A1 does not contain a stable set of size c~', the same is true of A2. But then G2(A2, B2) cannot be the same as S~,(A, rg), contradicting Theorem 13.2.4. Now we specialize to the class (1, 1), the class of split graphs. Two polar isomorphic split graphs are said to be split isomorphic. We would like to use the canonical decomposition at the level (1, 1) to give a criterion for split graphs to be isomorphic. We need the following lemma, which spells out the freedom in choosing a split partition of a split graph. L e m m a 13.2.9 Let G(A,B) and G'(A',B') be isomorphic split graphs on the same vertices. If they are not split isomorphic, then there exists a vetrex x such that A = A ' U { x } (and therefore B' = B U { x } ) , we have the decompositions
G(A,B) G'(A',B')
= H(A- {x},B)O/(l({X},~), = H ' ( A ' , B ' - {x})O SI( e , {x}),
334
The Decomposition Method
and H ( A - { x } , B ) and H ' ( A ' , B ' - { x } ) are split isomorphic, or the same holds with reversing the role of the primed and unprimed letters. P r o o f . The sets A - A ' = A N B' and A ' - A = A ' N B are both cliques and stable sets, hence their cardinality is at most 1. If they are both empty, then G(A,B) and G'(A',B') are split isomorphic, which is not the case by assumption. If both sets have cardinality 1, say A - A ' = {x} and A ' - A = {y}, then N(x) = N ( y ) = A N A', except that x and y are possibly adjacent; thus x and y are twins and by swapping them we turn (A, B ) i n t o (A', B'), so again G ( A , B ) a n d G'(A',B')are split isomorphic, contrary to assumption. Therefore one of the sets, say A - A', consists of a single vertex x and the other one, say A ' - A, is empty. Then the required properties are seen to hold. ,, We are now ready for characterizing split isomorphism, first proved by Tyshkevich.
Theorem 13.2.10 ([Tys80]) Let G(A,B) G'(A', B')
=
al(A1,B1)
@
-
G'1(A'I, B~) @
am(Am,Bm)
...
@
-..
@ G~n(A'n, B;)
be the decompositions of split graphs G and G' into indecomposable components in the class (1, 1). Then G and G' are isomorphic graphs if and only
if 1. m = n ; 2. Gi(Ai, Bi) and G'i(A~,B~) are split isomorphic for i -
1,...,m-
1;
3. either G,~(Am, Bin) and G~(A~, B~) are split isomorphic, or else one of them is K1 ({x }, ;g) and the other one is S1 (;g, {x }). P r o o f . The sufficiency is clear. For the necessity, we have two cases. If G and G' are split isomorphic, then Part 1 of Theorem 13.2.4 implies that our conditions hold with the first alternative in Condition 3. If G and G' are not split isomorphic, then Lemma 13.2.9 and Theorem 13.2.4 imply that our conditions hold with the second alternative in Condition 3. 9
Theorem 13.2.11 ([Tys80]) Let :
~'
-
~I(A1,B1)@ -
...
QGm(A,~,Bm)Q H
~'1 (A'I, B'I) @ "'" @ Gin (din, B'n) @ H'
13.3
Domishold Graphs and Decomposition
335
be the decompositions of non-split graphs G and G' with Gi(Ai, Bi) and G~(A~,B~) indecomposable in the class (1,1) and H and H' non-split indecomposable at the level (1, 1). Then G and G' are isomorphic graphs if and only if 1. m - n ; 2. G~(A,, B,) and G~(A~, B~) are split isomorphic for i = 1 , . . . , m; 3. H and H' are isomorphic. P r o o f . This is Part 2 of Theorem 13.2.4 specialized to the class (1, 1).
13.3
Domishold Graphs and Decomposition
We apply the theory of canonical decomposition presented in Section 13.2 to enumerate and classify the domishold graphs, which we saw in Chapter 12. First we do the same with the simpler threshold graphs. This section is mainly based on Tyshkevich and Chernyak [TC85b]. We use the notation I 1, graphs of the form (13.4) and (13.5) are not isomorphic. Graphs of the form (13.4) (or (13.5)) corresponding to different compositions n -- rtl - J r " .
Jr- n s
and
n -
ml
-Jr'"-Jr
are isomorphic if and only if s - t + 1, ns - 1, for i - 1 , . . . , t - 1, or vice versa.
mt
1 and ni
n t - - m t --
-
mi
T h e o r e m 13.3.3 Let cm denote the number of non-isomorphic threshold graphs without isolated vertices that have m edges, with Co - 1 by convention. Let D ( m ) denote the set of partitions of m into distinct (positive integer) parts. Then 1.
2.
E cm x m m>O
l-I(l+xJ) 9 j_>l
Proof. The two parts of the theorem are equivalent by the well-known generating function for the number of partitions of integers into distinct parts. We prove Part 1 by exhibiting a suitable bijection [TC85b]. Part 2 has been proved by different techniques by Peled [Pel80], see Chapter 17. The partition m -- ml
- - I - ' ' ' nt- rrtk
with
ml
>
'''
>mk
:>
0
is mapped to the graph K1 @
Sml--m2--1 (~ ''" (~
K1 @
Sink_l--ink--1 (~
K1 @ Sm k C
D(m),
13.3
Domishold Graphs and Decomposition
337
where empty graphs So are understood to be omitted. By Part 3 of Theorem 13.3.2, this mapping is injective. Moreover, for the same reason, every threshold graphs without isolated vertices can be represented in the form K~ Q Sl~ Q . . . Q K~, Q Sl, with positive ni and li, and it is easy to reconstruct the partition that maps to it. Thus the mapping is surjective. .. Turning now to domishold graphs, we denote by D the set of polar graphs in the class (2, 1) that are domishold. We easily deduce the following theorem from Theorem 13.2.4 and Condition 3 of Theorem 12.2.4: Theorem
13.3.4
1. All domishold graphs belong to the class (2, 1).
2. A graph in the class (2, 1) is domishold if and only if all its indecomposable components in this class have one or two vertices. .
o
The algebra (~, 5)) is a free semigroup generated by the polar graphs K l ( { X } , e ) , SI(•, { x } ) a n d S2(;g, {x,y})with x,y non-adjacent. The set of domishold graphs in the class (1, 1) coincides with the set of threshold graphs.
For the classification of domishold graphs, it is convenient to work with their complements, the codomishold graphs, which by Part 3 of Theorem 13.3.4 are the compositions of components of the form KI({X}, e ) , S l ( e , {x}) and M ( e , {x, Y}) with x, y adjacent. Using the notation Mtm - S~ Q M m - S[ Q M TM, we see that the codomishold graphs are the graphs that have one of the following forms of decomposition" M/lrn I @ I(n I @ M/2rn 2 @ tin 2 @ . . . @ MItrnt @ lin t
(13.6)
Mllrnl Q linl Q Ml2rn2 @ l(n2 @ "" Q l(nt_l @ Mltrrtt
(~a.7)
I'(nl @ Ml2m2 @ I'(n2 @ Ml3m3 @"" @ Mltmt @ I~nt
(13.8)
I(nl Q Mt~m~ Q K~2 Q Mt3.~3 Q " Q
K~_I Q Mt,.~,
(~a.9)
with t >_ 0. Here, as above, Kn denotes K~. We call a composition of the form (13.6) or (13.8) with nt >_ 3 or of the form (13.7) or (13.9) with It -t- rat > 2 &
standard decomposition.
338
The Decomposition Method
T h e o r e m 13.3.5 Every codomishold graph with at least three vertices has a unique standard decomposition. P r o o f . Observe that in each standard decomposition, every component with the possible exception of the last two is uniquely determined. For instance, if the graph has isolated vertices or edges, then [1 and ml are the numbers of isolated vertices and edges, respectively, n l is the number of dominating vertices remaining after the previous vertices are deleted, and so on. It follows that in any two standard decompositions, the numbers of components differ by at most 1. The following are the only cases. Case 1" One standard decomposition ends with Mtm q) K~. Then n >_ 3, and n is the number of vertices of degree at least 2 in the graph induced by the vertices of these last two components. Therefore n is uniquely determined, and with it the last two components. C a s e 2" One standard decomposition ends with K~ Q Mzm. Then l + m >_ 2, and n is the number of dominating vertices in the graph induced by the vertices of these last two components. So, again, n and the last two components are uniquely determined. C a s e 3" All standard decompositions have exactly one component, namely Mllml or K ~ . This case is trivial because the graph has at least three vertices. 9 Every threshold graph is a unigraph, i.e., determined up to isomorphism by its degree sequence (in fact, a threshold sequence has a unique labeled realization). In contrast, non-isomorphic domishold graphs can have the same degree sequence. For example, the complete bipartite graph K2,3 and the pentagon with one chord have the same degree sequence (3, 3, 2, 2, 2). The first one is domishold, since its complement has the standard decomposition M2,0 (~ K3; the second is not, since it contains an induced P4, which is forbidden by Theorem 12.2.4. However, a domishold graph without isolated vertices is an edge-unigraph, i.e., determined up to isomorphism by the list of degrees of its edges , where the degree of an edge is defined as the unordered pair of degrees of its endpoints. To show this, we make the following definition. D e f i n i t i o n 13.3.6 Let a graph G have an alternating ~-cycle t with edges ab, cd and nonedges ad, bc, and let deg(a) = deg(c). We then say that G admits a swap t, and the graph t G is obtained from G by swapping the edges and nonedges oft. See Figure 13.2.
13.3
Domishold Graphs and Decomposition
339
Figure 13.2: A swap.
a
d
b
a
c
d I
I
I I Ark
I I
b
d e g ( a ) - deg(c)
c
d e g ( a ) - deg(c)
Chernyak and Chernyak [CC81] have proved that if two graphs without isolated vertices have the same list of degrees of edges, then one can be obtained from the other by a sequence of swaps. Therefore it is enough for us to prove the following result. P r o p o s i t i o n 13.3.7 If G is a domishold graph admitting a swap t, then G
is isomorphic to tG. m
P r o o f . Clearly the complement G also admits t as a swap, and so it is enough to show that G and tG are isomorphic. By Theorem 13.3.5, G has a unique standard decomposition in one of the forms (13.6) - (13.9). Without loss of generality we may assume that the first component of this decomposition contains one of the vertices a, b, c, d of t, for we may delete all preceding components, if any. Considering the polar graph G(A, B), we examine the possible cases of which of the sets A, B contains each vertex of t, taking into account the fact that A is a clique and B induces a matching with isolated vertices. C a s e 1" a, b, c, d C B. Since a and c have the same degree, they must be in the same component, which must also contain their neighbors c and d. But then clearly tG is isomorphic to G. C a s e 2" a,b,c E B, d C A (the case a,c,d C B, b C A is similar). The neighbors a and b are in the same component, and c must be in a later component, since the vertex d of A is a neighbor of it but not of a. Therefore a, b C Miami, hence deg(a) - 1, and it follows that deg(c) - 1. This implies
The Decomposition Method
340
that d E K ~ , nl - 1, and c E Mt2m2. But now swapping a and cgives a graph isomorphic to G and also equal to tG. C a s e 3" b,c,d E B, a E A (the c a s e a , b,d E B, c E A i s similar). Since a E A is adjacent to b E B but not to d E B, d (and with it its neighbor c E B) must be in an earlier component than b. Hence c,d E Mt~n~, and therefore deg(c) - 1, from which deg(a) - 1. Hence b is the only vertex in its component, and this component is the last one. But this contradicts the definition of the standard decomposition, so this case is impossible. C a s e 4: a,b E A, c,d E B (the case c,d E A, a,b E /3 is similar). We must have c, d E Mt~nl, so deg(c) - 1, hence deg(a) - 1. This implies that A - {a, b}. Now if a and b are in the same component, then this component is K ~ and is the last component, contradicting the definition of the standard decomposition. If a and b are in different components, then since deg(a) - 1, the component containing a has no other vertices and is the last component, again contradicting the definition of the standard decomposition. Therefore this case is impossible. C a s e 5" a, c E A, b, d E /3 (the case b, d E A, a, c E B is similar). Since a is a neighbor of b and c is not, the component containing a is earlier than the one containing c. But this contradicts the fact that d is a neighbor of c but not of a, so this case is impossible. 9 We conclude this section by classifying the forcibly domishold degree sequences, i.e., those degree sequences all of whose realizations are domishold. We prove, in fact, that a forcibly domishold sequence must be unigraphic. As usual, we denote a star on n + 1 vertices by KI~. 13.3.8 A forcibly domishold degree sequence is unigraphic, and its realization is of one of the following types:
Theorem
1. a threshold graph; 2. rnK2 U Kin with rn, n > 1; 3. T(A, B ) Q inK2 U Kin with rn, n >_ 1, where T is a threshold graph. Conversely, every graph of the one of the above types is domishold and a unigraph. P r o o f . Let d be a forcibly domishold sequence with a realization G. We have to show that G i s o f t y p e 1 - 3 . I f G E (1,1), then G i s o f t y p e 1 b y Part 1 of Theorem 13.3.4, and we are done. Assume now that G ~ (1, 1).
13.4
Box-Threshold Graphs and Decomposition
341
We show below that if G is indecomposable at the level (1, 1), then it is of type 2, from which it follows that if G is decomposable at the level (1, 1), then it is of type 3. Since all realizations of d are domishold, and since every transfer (cf. Corollary 3.1.11) done on the complete bipartite graph K2,3 yields the nondomishold pentagon with a chord, G has no induced K2,3. Consider the standard decomposition of G. Since G is indecomposable at the level (1, 1), the first component of the decomposition is of the form M0ml, rrtl ~ 1. If there are no other components, then rn~ >_ 2 because G ~ (1, 1), and hence G is of type 2. Assume now that there are other components. Then G is triangle-free, since G is (K2 U K3)-free. Hence in the polar representation G(A,B), IAI _< 2. By the definition of a standard decomposition, the last component must then be of the form Mr,0 with It >_ 1. Further, t - 2 and nl 1 since G is triangle-free. This means that -
-
G
--
M0rrt
I
@ K1 @ M I 2 0
--
rnlK2 U Kll2,
so G is of type 2, as required. To prove the converse, it is enough to notice that if G is of type 1 3, then it is domishold, and every transfer done on it yields an isomorphic graph, hence it is a unigraph by Corollary 3.1.11. 9
13.4
B o x - T h r e s h o l d Graphs and D e c o m p o s i tion
This section is adapted from Tyshkevich and Chernyak [TC85a]. We characterize box-threshold (BT) graphs in terms of their decomposition into indecomposable components at the level (1, 1), and then use these results to enumerate them. All the polar graphs in this section are in the class (1, 1) (split) and the decompositions will always be at the level (1, 1). We use the notation Na(x) for the set of neighbors of a vertex z in a graph G, and ~ a for the vicinal preorder ~ on the graph G. If A and B are sets of vertices of a graph, we use NA(B) to denote A N ( U N(b)), and A ~- B to indicate bEB
that a ~- b for all a C A and b C B, and similarly for A ~ B. L e m m a 13.4.1 A composition is BT if and only if each component is BT.
342
The Decomposition Method
P r o o f . Let F = G(A, B) Q H. We have to show that F is BT if and only if G and H are BT. Note that A ~ V ( H ) ~ B. " O n l y if"" Assume that F is BT. To show that H is BT, consider vertices x,y of H with degH(x ) > degH(y ). Since Na(x) = Na(y), we must have degF(x ) > degp(y). Since F is BT, this implies that x L~ Y, and therefore x ~H Y, so H is BT. To show that G is BT, consider vertices x,y of G with dega(x ) > dega(y ). If x and y are both in A or both in B, then since NH(X) = NH(y), we must have degr(x ) > degp(y). Since F is BT, this implies that x ~ p y, and therefore x ~ a y, as required. If not, then necessarily x C A and y C B, which already implies that x ~ a y. " I f " : Assume that G and H are BT. To show that F is BT, assume that degF(x ) > degF(y ). If y E A, then x E A and dega(x ) > dega(y ). Since G is BT, we have x ~ a y and hence x ~ p y. If y E V(H), then x E V(H) or x C A. In the first case x ~ p y since H is BT, and in the second case x ~ p y since A ~ V(H). If y C B, we similarly reach the same conclusion x LF YTherefore F is BT. .. The next two lemmas identify the indecomposable BT graphs. Observe that if a split graph G(A, B) is biregular, then all the vertices of A have the same degree and all the vertices of B have the same degree. L e m m a 13.4.2 A split graph with more than one vertex is an indecompos-
able B T graph if and only if it is biregular and has no dominating or isolated vertices. P r o o f . " O n l y if": Let G(A,B) be an indecomposable graph in the class (1, 1) with more than one vertex and also a BT graph. By being indecomposable with more than one vertex, G has no dominating or isolated vertex. We show that it is biregular. Let A' be the set of vertices of A with largest degrees, and let B ' = N B ( A - A'). Since G is BT, we have A' ~ A - A' and therefore B' is completely joined to A'. Unless A ~ = A, we then have the decomposition
G(A, B) = G'(A', B - B') Q H ( A - A', B'), contradicting the indecomposability of G. So A' = A and all the vertices of A have the same degree. Similarly all the vertices of B have the same degree. " I f " : Let G(A, B) be a split biregular graph with no isolated or dominating vertices. To show that G is BT, notice that if deg(x) > deg(y), then x C B and y E A, which implies in turn that x ~ y. To show that G is indecomposable, assume the opposite and let G = G'(A', B ~) Q H be a decomposition
13.4
Box-Threshold Graphs and Decomposition
343
of G with G ~ and H non-empty. There must exist a vertex x C A' such that NB,(X) 7/= rg, for otherwise the vertices of B' are isolated in G, and if B' = rg then the vertices of A ~ are dominating in G. Similarly there must exist a vertex y C B' such that N ( x ) is a proper subset of A'. Now pick any vertex h of H, and we must have dega(x) > dega(h ) > dega(y ). But this contradicts the biregularity of G. L e m m a 13.4.3 A non-split graph with more than one vertex is an indecomposable B T graph if and only if it is regular and has no dominating or isolated vertices. P r o o f . " O n l y if"" Let G be an indecomposable non-split graph with more than one vertex and also a BT graph. By being indecomposable with more than one vertex, G has no dominating or isolated vertices. To show that G is regular, let A be the set of vertices with the largest degree in G, let H = (1 N(a), and let B = V(G) - (A U H). For each a E A and x E B U H aG_.A
we have deg(a) > deg(x), and so A ~ BU H. Each b e B has a non-neighbor in A, for otherwise b C H; therefore by A ~- B U H, b has no neighbors in B U H . I f H r 0 , then by A ~- H, A is a clique and G decomposes as a = G ' ( A , B ) O G[HI, where G[H 1 is the subgraph induced by H. This contradicts the hypothesis, and so H = O. If A is a clique, then G is a split graph G(A, B), contrary to hypothesis. Therefore there is a nonedge uv in A. S i n c e u ~- B, v has no neighbors in B. Let b b e a v e r t e x o f B , if any. Since it is not isolated, it has a neighbor z C A. Since A >- b, the closed neighborhood N[z] contains A U {b}. However, N[v] is a proper subset of A, and thus deg(z) > deg(v), contradicting the definition of A. This proves that B = 0 , and so G is induced by A and is therefore regular. " I f " : Let G be a regular graph without dominating or isolated vertices. By being regular, G is BT. To show that G is indecomposable, assume the contrary and let G = G'(A, B ) Q F, where G' and F are non-empty graphs. It follows that A ~- B, and hence deg(a) > deg(b) for all a E A and b E B. Now A r 0 , for otherwise the vertices of B are isolated, and B 7~ 0, for otherwise the vertices of A are dominating. This contradicts the regularity of G. " The next theorem characterizes BT graphs in terms of their indecomposable components at the level (1, 1).
The Decomposition M e t h o d
344
T h e o r e m 13.4.4 Let G = Xx Q ... Q X~ be the decomposition of a graph G into indecomposable components at the level (1, 1). Then G is B T if and
only if 1.
X I
, . . . ,
Xn-1 are split biregular graphs;
2. Xn is a split biregular or non-split regular graph. P r o o f . Note that if Xi has more than one vertex, then it cannot have dominating or isolated vertices by being indecomposable. The result then follows immediately from Lemmas 13.4.1, 13.4.2 and 13.4.3. [] We now enumerate the BT graphs using the above results. Let us denote by ap the number of non-isomorphic indecomposable split graphs on p vertices, by bp the number of non-isomorphic indecomposable non-split graphs on p vertices, and by cp the number of non-isomorphic BT graphs on p vertices. We also use the generating functions of these sequences, namely
a(x) - ~ apx p,
b(x) - ~ bpx p,
p>_l
p>_l
c(x) - ~ cpx p. p>_l
By the uniqueness of the canonical decomposition and Lemma 13.4.1, we have the following relation: 4x) =
1 -a(x)
Below we show how the ap and bp can be computed. C o m p u t i n g ap: Clearly a~ - 1 and for p > 1, by Lemma 13.4.2, ap is the number of non-isomorphic biregular split graphs on p vertices with no dominating or isolated vertices. Therefore ap --
Z Amn, m+n=p m,n>O
where A,~n is the number of non-isomorphic biregular split graphs with no dominating or isolated vertices, having the form G ( A , B ) with IAI = m, It l = S u c h ~ graph G ( A , B ) i s specified by the adjacency matrix M whose rows correspond to the vertices of A and whose columns correspond to the vertices of B. It is an m x n 0/1 matrix with constant row sums r, 0 < r < n, and constant column sums s, 0 < s < m. Let us denote by
13.4
Box-Threshold Graphs and D e c o m p o s i t i o n
345
Ad~n~ the set of all such matrices, and by Sm and S~ the symmetric groups of degrees m and n respectively. We define an action of the direct product Sm x S~ o n . / ~ m n by
(7r,a)M = PTrMP~ -1 ,
~ESm,
aES~,
where P~, Po are the permutation matrices of 7r, a, respectively. Then Amn is the number of orbits of this group action. By Burnside's Lemma, we have 1
A.~ =
~
F(rr, a),
TIt! TL! (Tr,o')e"~mXSn
where F(~r, a ) i s the number of matrices M C JPlmn that are fixed by (re, a), that is (Tr, a ) M = M. To compute F(r~, or), we write rr and cr as products of disjoint cycles 71- - -
where rri is a cycle of length
7f'l
999
7Ck~
0"
~
O" 1 9 9 9 0 " I ~
mi and aj is a cycle of length nj,
77"t1 :> " ' "
~
?Ytk~
Using the notation m = ( m l , . . . , m k ) a n d n = ( n l , . . . , n t ) , we observe that F(rr, a ) d e p e n d s only on m and n and not on the particular rc and a, and we may denote it by F ( m , n). There are m!/ml...mk permutations rr corresponding to m and similarly for n. Hence rtl >
"'"
~
rtl, m
-- ml
+'..
+ ink,
rt - - rtl + . . .
1
Am,~ - }-~. 77"tl . . .
?Tt k Tt 1 . . .
+ rtl.
r(m, n), TLl
where the summation extends over all the partitions m , n of the integers rn, n into positive parts. To compute F(m, n), consider the action of (rr~, aj) on the submatrix M~j of M consisting of the m i rows cycled by rri and the nj columns cycled by aj. The entries of Mij are partitioned into dij = g c d ( m / , nj) cycles, each one of length rninj/dij (i.e., the least common multiple of m i and nj). Each such cycle meets each row of Mij nj/dij times, and each column of Mij mi/dij times. The matrix M is fixed by (rr, a) if and only if in each submatrix Mij, each cycle is all-0 or all-1. If exactly t cycles of Mij are all-l, then Mij has constant row sums tnj/dij and constant column s u m s tmi/dij. We denote by e~...~k;,~..., , the number of m x n 0/1 matrices M fixed by (rr, a) for which the rows cycled by rri have constant row sums ri and the columns cycled by
The Decomposition Method
346
O'j have constant column sums sj. Then by the above discussion we obtain the generating function
E
"''Yk~kZ~ a ' ' ' z ~ ~ - -
Crl,...,rk;sl,...,sty~l
rl ,...,rk 81 ,...,3l
H
l t
~(~) + ~(~)
>_ t.
But these inequalities are clearly inconsistent with t > 0. The following theorem characterizes pseudothreshold graphs. T h e o r e m 14.2.7 For every graph G equivalent:
(V, E), the following conditions are
1. G is pseudothreshold; 2. P* n Q* - 0 and G is 3K2-free; 3. V can be partitioned into sets P, Q, R such that (a) every vertex in P is adjacent to every vertex in P U R, (b) no vertex in Q is adjacent to another vertex in Q u R, (c) there are no three mutually nonadjacent vertices in R. P r o o f . 3) =~ 1)" An appropriate assignment is
w(u)-
..
0, i f u c Q 1, i f u E R 2, if u C P
t-2.
14.2
Pseudothreshold Graphs
355
1) =~ 2): This follows from Lemma 14.2.4 and Lemma 14.2.6.
2) =~ 3): We prove this by means of a simple algorithm that terminates in O(rt 4) steps either by showing that Condition 2 does not hold or by constructing the partition described in Condition 3. The algorithm is as follows. First of all, find P* and Q*. (This can certainly be done in O(n 4) steps.) Then find out whether P * N Q* = ~. (If not, stop: Condition 2 does not hold.) Then set S = V - (P* U Q*); note that by the definition of P* and Q*, every vertex in S is adjacent to all the vertices in P* and to no vertex in Q*. Let So consist of all the vertices in S that are adjacent to no other vertex in S; define
P=P*,
Q=Q*USo,
R=S-So.
Find out whether there are three mutually nonadjacent vertices in R. If not, stop: by Lemma 14.2.5 P, Q and R have all the properties described in Condition 3. If, on the other hand, there are three mutually nonadjacent vertices Ul,U2,U3 C R, then each ui is adjacent to some vi E R. Using the fact that R N (P0 U Q0) = o , we may now easily verify that the set {Ul, u2, u3, Vl, v2, v3} induces a 3K2 in G. (To see this, note that the vj's are distinct, mutually nonadjacent and that each vj is adjacent to exactly one ui.) Hence Condition 2 does not hold. " The corollary below follows from the proof of the above theorem.
Corollary 14.2.8 If G is pseudothreshold, Equation (14.1) can be satisfied with t = 2 and w(u) C {0, 1,2} for all u.
"
We conclude this section with the following remarks.
Remarks 1. The degree sequences of pseudothreshold graphs on n vertices lie on the boundary of the convex hull of the degree sequences of all graphs on n vertices, because they satisfy the facet inequality
Z iEP
x ~ - ~ x~ _< IPl(n- 1 - I Q I )
(14.2)
iEQ
with equality, where P, Q, R is the partition of V as in Condition 3 of Theorem 14.2.7. This follows from Lemma 3.3.13.
356
Pseudothreshold and Equistable Graphs
2. The converse is not true. For instance, the graphs K1@K3,3 and K1@C6 have the same degree sequence (6, 4, 4, 4, 4, 4, 4) satisfying (14.2) with equality, where P is a singleton corresponding to the vertex of K1, Q is empty, and R consists of the other six vertices. But the first graph is not a pseudothreshold graph and the second one is. 3. The complement of a pseudothreshold graph need not be pseudothreshold. For instance, 3K2 is pseudothreshold. In view of the above remarks, the following questions are interesting.
Problem 1. Which degree sequences lie on the boundary? 2. Which degree sequences are potentially pseudothreshold?
14.3
Equistable Graphs
Payan [Pay80] introduced the equistable graphs, and later Mahadev, Peled and Sun IMPS94] obtained several interesting results on these graphs. We present the results of IMPS94] in this section. Definition 14.3.1 A graph G - (V,E) is said to be equistable if there exists a mapping ca" V --. N + such that for all S C_ V, S is a maximal stable set ~
c a ( S ) - ~ c a ( v ) - 1. vES
Payan [PayS0] showed that the threshold graphs and the domishold graphs are equistable, and announced that the P4-free graphs are equistable. In the next subsection we introduce a subclass of equistable graphs the strongly equistable graphs. We show that the strongly equistable graphs are closed under disjoint unions and joins, and therefore the P4-free graphs are strongly equistable. Later in this section, necessary conditions for equistability and sufficient conditions for strong equistability are obtained and used to characterize the equistability of some classes of perfect graphs, pseudothreshold, and outerplanar graphs. The structure of these equistable graphs is also given.
14.3
Equistable Graphs
357
Finally, we show that for a wide family of graphs, equistability implies strong equistability, and discuss the closure of various classes of strongly equistable graphs under graph substitution. An induced path on four vertices is denoted by P4(a, b, c, d) to indicate that its vertex set is {a, b, c, d} and its edges are ab, bc, cd.
14.3.1
Strongly Equistable Graphs
Let G = (1,1,E) be a graph. We denote by $ the set of maximal stable sets of vertices of G, and by 7- the set of all other nonempty sets of vertices of G. Put A - A(G) - {~v" V ~ R+'~v(S) - 1 for all S C $}. Thus A(G) is a set defined by a finite number of linear equations and nonnegativity inequalities, i.e., it is the intersection of an affine set with the first orthant. Since each vertex belongs to some maximal stable set, A ( G ) is bounded and therefore it is a polytope, possibly empty.
Definition 14.3.2 A graph G is said to be strongly equistable if for each T E 7- and each c 2, Vi
(14.11)
-
0
v j _> 2, vi.
( 4.12)
(14.9)
14.3
Equistable Graphs
365
To prove that G is strongly equistable, let T E T and c _< 1 be given, and assume that, if possible, w(T) - c for all w E ~4. Then the equation aJ(T) - c is a consequence of Equations (14.9)-(14.12), and therefore a linear combination of them. Therefore the equation w ( T ) - c takes the form t
E (~p[O,.)(p)-~-(.u(NQ(p))] Jr E ")/i[('u(~il ) + ("g(]r pEP t
-t- ~
+ (,M(Q)]
i=1 t
~_~ aij[w(kij) - co(kii)] + ~ ~ a'ij[w(k~j ) - w(k~l)]
i = 1 j_>2
(14.13)
i = 1 j_>2 t
pEP
i=1
' Since w(p) appears only once in the sysfor some multipliers 5p, ~/i, aij, aij. tern (14.9)-(14.12), and with a coefficient of 1, we have 5p = 1 if p E P N T, 5p=0ifpE P-T. Similarly for j _ > 2 , aij = 1 ifkij E K i n T , aij = 0 i f k~j E K~ - T, a~j - 1 if k~j C K[ N T, a}j - 0 if k~j E K~ - T. Therefore, in order for the coefficient of w(k~l) in the linear combination (14.13) to assume its correct value (1 if k~l E T, 0 if not), 3'~ must be equal to ]Is n T I. Similarly 7i must also be equal to IK[ N T I. Thus we can rewrite (14.13) in the following way: t
pEPnT
[~(p) + ~(xq(p))] + E ~ ( Q ) i=1 t
+ ~(/q n T) + ~(/(~
n T ) --
IP n TI + ' ~ ~,
(14.14)
i=1
where
~ - IIq n T I Note that
IK[ n TI.
(14.15)
t 1 ~/i - 71R n T I.
(14.16)
i=1
Therefore 1 IP N T] + ~]R n T] - c.
(14.17)
Further, in order for w(q), q E Q, to assume its correct coefficient in (14.14) (lifqEONT, 0ifqEO-T),wemusthave 1
INpnr(q)l + -~lR A T
]e 1 I - ~[ 0
for q E Q N T forqEQ-T.
(14.18)
Pseudothreshold and Equistable Graphs
366
From (14.15)-(14.17), c is an integer, and from c _< 1 we obtain c - 0 or C--1. C a s e 1" c - 0. Then by (14.17) P A T - R A T - 0 . Then the left-hand side of (14.18) is 0, which implies Q n T - O. Therefore T - O, contradicting TET. C a s e 2" c - 1. From (14.17), either P n T - {p} and R n T - O, or else P A T - O and R A T - {kij, k~l } (by (14.15)). In the first possibility, (14.18) can be rewritten as [I.N{p}(q).-
1 forqCQNT 0 for q E Q - T,
which means that Q N T - NQ(p). Thus T - { p } U N Q ( p ) , which is a m a x i m a l stable set, contradicting T C T. In the second possibility, the left-hand side of (14.18) is 1, which implies Q - T - o . Thus T - {kij, k~l} U Q, which is a m a x i m a l stable set, contradicting T E T. 9 D e f i n i t i o n 1 4 . 3 . 2 0 A planar graph is said to be o u t e r p l a n a r if it can be embedded in the plane so that every vertex lies on the boundary of the infinite region. Obviously, a graph is outerplanar if and only if each of its blocks is outerplanar. The following l e m m a is used to prove Theorem 14.3.23 below. L e m m a 1 4 . 3 . 2 1 Every 2-connected outerplanar graph with more than two vertices is a Hamiltonian graph. P r o o f . E m b e d the 2-connected outerplanar graph in the plane so t h a t all vertices lie on the b o u n d a r y B of the infinite region. Let C - V l , . . . , v k be ~ cycle all of whose edges lie on B. If C is not Hamiltonian, there is a vertex Ul ~ C adjacent to some vertex of C, say Vl. Since Ul is on B, Ul is in the exterior of C. Let v~ be any vertex of C, r :/: 1. By 2connectivity there is a cycle C' - V l t t l U 2 . . . containing the edge VlU 1 and the vertex v~. Let rn be the smallest index such that u,~ coincides with a vertex on C, say vj. Note that U l , . . . , u m - 1 are in the exterior of C and t h a t 2 _< j _< k. If j - 2, then V l , U l , . . . , u m , v 3 , . . . , v k is a cycle enclosing the interior of the edge VlV2, so that edge cannot lie on B, contradicting the choice of C. A similar contradiction occurs if j - k. If 2 < j < k, then either the cycle V l , U l , . . . , u m , v j + l , . . . , v k encloses v2 or else the cycle Vl, U l , . . . , Urn, V j - I , . . . , ?22 encloses vk, so either v2 or vk is not on B, a contradiction. 9
14.3
Equistable Graphs
367
D e f i n i t i o n 14.3.22 A flower Fn
-
Fn(?tl,...,?.tn;Yl,...,Vn)
is a graph consisting of an even cycle UlVlU2V2...UnVn whose chords are precisely the short chords UxU2, U2U3, . . . , U~Ul. Figure 1~,.3 illustrates the flowers F2 and F6.
Figure 14.3: Flowers.
V2
vx
U2
Y3 ..
U
V4 v
U
2
.Yl
6
,., V6
Ul
?,)2
~
~
U
2/ V5
F2
F6
T h e o r e m 1 4 . 3 . 2 3 Let G be an outerplanar graph. Then the following conditions are equivalent:
1. G is equistable; 2. G is strongly equistable; o
(a) each block of G is a flower, a square, or a complete graph on at most 3 vertices; (b) in a block of G that is a flower or a square, no vertex of degree 2 in the block is a cut vertex:
368
Pseudothreshold and Equistable Graphs
(c) in a block of G that is a complete graph, not all vertices are cut vertices.
P r o o f : 3) =~ 2)" By Theorem 14.3.4 we may assume that G is connected. If a block of G is a square, then by Condition 3b, this block is the entire G, and it is strongly equistable by being P4-free. If the block is a single vertex, then it is the entire G by connectivity and the same conclusion holds. We may therefore assume in addition that each block of G is a flower or a complete graph on two or three vertices. We construct a set S of vertices as follows" from each block that is a flower, put in 5' all the vertices of degree 2 in the block; from each other block put in S exactly one vertex that is not a cut vertex. Clearly the vertices of S in each block from a maximal stable set in the block, and further, by Condition 3b, S contains no cut vertex of G. Therefore S is a maximal stable set of G. It is easy to verify that S satisfies the condition of Theorem 14.3.10, and hence G is strongly equistable. 2) =~ 1)" This follows from Lemma 14.3.3. 1) =~ 3)" First we prove Condition 3a. Every block B of G is outerplanar. If B has at most four vertices, then B is clearly F2, a square, or a complete graph on at most three vertices. So assume that B has at least five vertices. By Lemma 14.3.21 B has a Hamiltonian cycle C - Zl . . . zk. We assume that G is embedded in the plane so that every vertex lies on the boundary of the infinite region. Therefore every chord of C lies in the interior of C. We now show the following two facts" F a c t 1 No f o u r consecutive vertices along C induce a P4. P r o o f . Assume without loss of generality that C has a P4(Xl, x2, z3, x4). By Theorem 14.3.8, G has a vertex v adjacent to both x2 and x3, but to neither Xl nor x4. Clearly v is in B, and hence v - xi for some i _> 6. Note that zi is the only vertex of G adjacent to both x2 and x3 by planarity. If x4xi-1 ~ E , extend { X l , X 4 ~ X i _ I } t o & maximal stable set S, and S will violate Theorem 14.3.8 with respect to P4(xl,x2, x3, x4). Therefore x 4 x i - , E E. See Figure 14.4. Now consider P4(x2, xi, xi-1, x4). The only vertex that can possibly be adjacent to both xi and xi-1 is x3, which is adjacent to x2, a contradiction to Theorem 14.3.8. This proves Fact 1. F a c t 2 Between every four consecutive vertices along C, there is exactly one chord, and this chord joins two vertices at distance 2 along C. P r o o f . Consider for example Xl,X2, X3, X4. By Fact 1, there is a chord between them, and it is enough to show that XlX4 is not a chord. If XlX 4 is & chord, then between x2, z3, x4, xs or between xk, x l , x 2 , x3 there is no chord,
14.3
Equistable Graphs
369
Figure 14.4: Illustrating the proof of Fact 1. Dashed lines indicate nonedges. X4 X3
9
X 2
Xl
Xi-1
contradicting Fact 1. This proves Fact 2. From Fact 2, it follows that k is even and B has a spanning flower fn(tll,
. . . , tln;
Yl,.
. . , Yn),
where n - k/2 >_ 3. It remains to show that the cycle U l . . . t t n is chordless in B. Assume for example that U l U i C E , where 2 < i < n. Then the P4(vl, Ul, u~, v~) and any maximal stable set S containing { V l , . . . , v~} contradict Theorem 14.3.8. This proves Condition 3a. To prove Condition 3b, consider first a block B that is a flower fn(tll,
. . . , tln;
Yl,
. . . ~ ~)n)
and assume for example that Vl is a cut vertex. Let x be any neighbor of t~1 outside B. Then the P4(x, vl, u2, v2) and any maximal stable set S containing {x, v2, v~} contradict Theorem 14.3.8. Similarly, consider a block B that is a square abcd, and assume for example that a is a cut vertex. Let x be any neighbor of x outside B. There is no vertex adjacent to both a and b, yet there is P4(x, a, b, c), again contradicting Theorem 14.3.8. The proof of Condition 3c is similar to the proof of Condition 3b. 9
370
14.3.4
Pseudothreshold and Equistable Graphs
Equistability, Strong Equistability, and Substitution
In Subsection 14.3.2 we saw among other things that for split graphs and block graphs, equistability is equivalent to strong equistability. Below we show that more generally, the same is true for all perfect graphs. Recall the definitions of .A, S, and T at the beginning of Subsection 14.3.1. Theorem
14.3.24
1. If G is a perfect graph and .A 7~ 0, then .A has an integer-valued weight function w. 2. A(G) has an integer-valued weight function if and only if G has a clique meeting all maximal stable sets. 3. If G is equistable, and if for every T C T there exists some co E A such that w(T) is not in the open interval (0,1), then G is strongly equistable. ~. In particular, if G is an equistable perfect graph, and more generally if G is an equistable graph with a clique meeting all maximal stable sets, then G is strongly equistable. P r o o f . 1" Consider the clique matrix M of the complementary graph G, whose rows are the incidence vectors of the maximal stable sets of G. By Ldvasz's Perfect Graph Theorem [Lov72], G is perfect. Hence by Chvs Theorem [Chv75], the extreme points of the polytope defined by Ma~ _< 1, aJ _> 0 are integer-valued. Note that A is a face of this polytope, and since by assumption A r 0 , .4 contains an extreme point of the polytope. 2: Let co be an integer-valued weight function in .4. Since every maximal stable set S satisfies aJ(S) - 1, S must contain exactly one vertex x with w(x) - 1. Let C be the set of all vertices x such that aJ(x) - 1. Clearly C meets all maximal stable sets. Moreover, C is a clique, otherwise some two vertices of C can be extended to a maximal stable set S with a~(g) >_ 2. Conversely, if C is a clique meeting all maximal stable sets, let aJ be the characteristic vector of C. Clearly w is an integer-valued weight function in
A. 3: Since G is equistable, there exists a non-negative weight function a / s u c h that J ( S ) - 1 if and only if S E $. To show that G is strongly equistable,
14.3
Equistable Graphs
371
let T E 7 " a n d c_< 1 be given. B o t h w a n d w ~ are in A. I f c < 0 there is nothing to prove. If c - 0, then co(T) r 0 since T r e (any vertex x C T can be extended to a set 5' C $; if w'(x) - O, then co'(S- x) - 1 even though S-x ~,5'). If0 < c < 1, then co(T) r c. If c - 1, t h e n a / ( T ) r csince
T s. If G is equistable, then ~4 -~ ~. If in addition G is perfect, then by Parts 1 and 2 above, G has a clique meeting all maximal stable sets. If G is equistable and has such a clique, then by Parts 2 and 3, G is strongly equistable. [] Note that the flower Fs is a non-perfect equistable graph having a clique meeting all maximal stable sets. The strongly perfect graphs [BC84] are the graphs in which every induced subgraph has a stable set meeting all maximal cliques. Their complements then have a clique meeting all maximal stable sets. Moreover, the strongly perfect graphs are perfect [BC84], and so are their complements. A clique of the form N[v] for some vertex v is said to be a simplicial clique. Clearly a simplicial clique meets all maximal stable sets. We now characterize the graphs satisfying the sufficient condition of Theorem 14.3.10 using simplicial cliques. 4:
T h e o r e m 14.3.25 For every graph G - (V, E), the following conditions are equivalent:
1. G has a maximal stable set S such that for every u, v C V - S, uv C E if and only if Ns(u) n N s ( v ) # ~; 2. every edge of G is in a simplicial clique. P r o o f : 1) ~ 2)" For every s C S, N[s] is a clique by the "if" part of Condition 1. These simplicial cliques cover E by the "only if" part of Condition 1. 2) =~ 1)" Let N[v~],... ,N[vk] be all the distinct simplicial cliques, and let S - {Vl,..., vk}. The set S' is stable, since if vivj C E, then N[vi] C_ N[vj] and N[vj] C_ N[vi], contradicting the distinctness of N[vi] and N[vj]. Further, S is maximal stable, because every vertex belongs to some simplicial clique by Condition 2. Again by Condition 2, if u, v C V - 5' and uv ~ E, then u and v cannot have a common neighbor vi, because N[vi] is a clique.
C o r o l l a r y 14.3.26 If G - (V, E) is a graph satisfying Condition 1 of Theorem 14.3.25, then for each x C V there exists some w C A with co(x) - 1.
~
~
~
C~
~
9 r bo
~
II ~
~ ~
Ch
~'~ ~ "
~"
~.~.~
~i,
9
-
mr
1, we have the same conclusion, otherwise wp~ could be decreased below 1, contradicting the optimality of the threshold assignment. Similarly wq + w~ _< t - 1. On the other hand, wp + wq + Wpq ~ t together with wpq < 1 imply that w p + w q > t - 1, and similarlyw~ + w ~ > t - 1. These four inequalities are inconsistent. ..
15.2
T h r e s h o l d Weights
379
The following result follows directly from the constraints (15.1). It is very useful in proving that certain graphs possessing symmetry properties are heavy. L e m m a 15.2.5 Let ab be an edge and cd a nonedge of a graph G. Every feasible threshold assignment (w; t) for G satisfying wc + Wd > Wa + Wb must satisfy Wab >_ 1. In particular, if e -- xy is an edge, zy is a none@e, and Wz > wx, then we > 1. The following corollary is an example of applying Lemma 15.2.5. C o r o l l a r y 15.2.6 All cycles of length 4 or more and their complements are heavy. Proof. Let G be the k-cycle Ck. Consider any optimal threshold assignment (w;t) of G. Each of the k cyclic permutations of the vertex weights of w around the cycle accompanied by a corresponding permutation of the edge weights is again an optimal threshold assignment. The average (w*; t) of these k optimal threshold assignments is yet another optimal threshold assignment (this follows from the convexity of the set of optimal threshold assignments). But (w*;t) enjoys the property that its vertex weights are constant. Let e be an arbitrary edge of G and let f be a nonedge adjacent to e, which exists since k > 4. By applying Lemma 15.2.5 to (w*; t), we see that w~ > 1. Thus G is heavy. The proof for the complements of cycles is similar. 9 The heavy graphs cannot be characterized by forbidden configurations, since they are not closed under taking induced subgraphs. In fact, every graph with at least one edge has an induced threshold subgraph with one edge, which is not heavy. The following theorem gives a convenient characterization of heavy graphs in terms of the optimum value of a linear programming problem having fewer variables and constraints than (15.1). T h e o r e m 15.2.7 A graph G [El
>
max
(V, E) is heavy if and only if
E dizi
iEv
s.t.
for each maximal stable set iES
Scv
z>_O,
(15.2)
T h r e s h o l d W e i g h t s and M e a s u r e s
380
where di is the degree of vertex i. Proof. The linear programming dual of (15.1) is max
~-~ ys S
s.t. S
e
Vi C V S ~i
(15.3)
e ~i
y~ _< 1
Ve C E
y>_O. The constraints of (15.3) imply that E s ys _ di
VicV
S ~i
y>_O. In other words, again using linear programming duality, if and only if IE]
_> min E Ys
=
max
s
s.t.
~ ys >_di Vi
E dizi iEv
s.t.
~ z i O.
A direct proof of Theorem 15.2.7 not using LP duality can be found in [MP91]. We apply Theorem 15.2.7 to show that several types of graphs are heavy. L e m m a 15.2.8 Ps (the path on 5 vertices) is heavy.
15.2
Threshold Weights
381
P r o o f . Denote the vertices as 1 , . . . , 5 along the path. Let z be feasible for (15.2). Then Zl + z3 + zs _< 1, z2 + z4 _< 1, and z3 _< 1. After multiplying these inequalities by 1, 2, and 1, respectively, and adding, we obtain Zl + 2 ( z 2 + z3 + z4) + zs _< 4, which is the condition for a heavy graph in Theorem 15.2.7.
T h e o r e m 1 5 . 2 . 9 If G - ( V , E) is a regular graph and x(G) < G is heavy.
IVI/2,
th~
P r o o f . Let r be the degree of every vertex, and let S 1 , . . . , S X be a partition of V into X - x(G) stable sets. If z is feasible for (15.2), then x k=l iESk
iEV
which is the condition of Theorem 15.2.7 for G to be heavy. T h e o r e m 1 5 . 2 . 1 0 Let G = (V,E) have a proper vertex-coloring in which each color has at least two vertices and all the vertices of the same color have the same degree. Then G is heavy. In particular, the complete k-partite graph K ~ .....~nk is heavy if mj >_ 2 for each j, the perfect matching inK2 is heavy if m >_ 2, and so on.
P r o o f . Let '--q'l,-.., Sk be a partition of V into stable sets, and let Dj be the common degree of the vertices of Sj for each j - 1 , . . . , k. If z is feasible for (15.2), then k
iEV
k
Dy j=l
E Dj iCSj
j=l
d /2 -IEI, iEV
which is the condition of Theorem 15.2.7 for G to be heavy. 9 Recall that N ( x ) denotes the open neighborhood of a vertex x, namely the set of all neighbors of x. The next theorem has a construction that preserves heaviness. T h e o r e m 15.2.11 Let G be a heavy graph, let x be a vertex of G, and let H be obtained from G by adding a new vertex y such that N(y) C N ( x ) in H (y is not adjacent to x). Then H is heavy. In particular, if v is a non-isolated vertex of a heavy graph and we add a new vertex of degree 1 adjacent to v, the graph remains heavy.
Thre s hold Weights and Measures
382
P r o o f . Let di denote the degree of vertex i C V ( H ) in H. The degree of vertex i E V(G) in G is then given by
d'i
if i ~ N(y) _ f di, d i - 1, if i c N(y).
In particular, d2 - d~ >_ dy. Let z be a feasible solution of (15.2) for H. Then the following constitutes a feasible solution of (15.2) for G:
, {
zi -
zi, if/=fix z~ + zu, i f i - x .
By applying the constraints of (15.2) to z on H and Theorem 15.2.7 to z / on G, we obtain
E
di Zi --
iEV(H)
iq~N(y)U{x,y}
_ 0, v C V and a threshold r < 1 such that ~ v e S Wv 1 for each edge u v C E . In other words, the minimum of r is smaller than 1 when subject to the maximal stable set and the minimal non-stable set constraints and the non-negativity of w. We therefore regard the minimum of r subject to the above constraints as a certain measure of the non-thresholdness of G. This motivates the following definition.
15.3
387
Threshold Measures
1 5 . 3 . 1 The t h r e s h o l d m e a s u r e v ( G ) of a graph G - (V, E) is the optimal value of the following linear programming problem.
Definition
min
s.t.
r
~ wv _ l for each edge uv C E w>O.
To emphasize the dependence of the optimum w on G, we sometimes denote it by w(G). Thus the optimum value of 15.5 is smaller than 1 if and only if G is a threshold graph. In that case, the minimum can be found by the very efficient and elegant algorithm of Section 1.3. The optimum value is equal to 1 if and only if G is a pseudothreshold graph. We have seen in Section 14.2 an algorithm to recognize these graphs and a characterization of their structure, from which it follows that when the optimum of (15.5) is 1, there is always an optimum solution for which the components of w have values of 0, 1/2 and 1. Figure 15.2 illustrates optimal solutions of (15.5) for a threshold graph and a non-threshold graph. Like (15.1), the linear programming problem (15.5) has exponentially many constraints, since in general there are as many maximal stable sets. This indicates that finding T(G) may be hard. We show below that this is indeed the case. We begin our general discussion of the threshold measure by determining the threshold measure of a disjoint union of two or more graphs. 15.3.2 w(G1 [..J G2) --~ (w(G1),w(G2)). Consequently T(G1 U G2) -- 7"(G1) -k- T(G2). This generalizes immediately to the union of any number of graphs.
Theorem
P r o o f . For k = 1,2, it is convenient to denote by Ak and Bk the edge-vertex incidence matrix and the maximal-stable-set-vertex incidence matrix of Gk. In other words, Ak has a column for each vertex and a row for each edge of Gk, and the corresponding entry is 1 or 0 according as the vertex is or is not an endpoint of the edge. Similarly Bk has a column for each vertex and a row for each maximal stable set, and the corresponding entry is 1 or 0 according
Threshold Weights and Measures
388
Figure 15.2" Threshold measures of (a) a threshold graph and (b) a non-threshold graph. Tz
8 9
5 4
T-1 4
1
1
9
l
9
~ ~1
12 / '
(b) as the vertex belongs or does not belong to the maximal stable set. We also let e k and fk denote all-1 column vectors of appropriate dimensions. Then the linear programming problem (15.5) for Gk can be written as follows, where tt k and u k denote row vectors of dual variables corresponding to the vector constraints of (15.6)" min #k
s.t.
Vk
A k w k >_ ek - - B k w k + rk.f k > 0 Wk>_O.
(15.6)
The linear programming dual of (15.6) is the following" T(Gk) lVk
-
max
13,k e k
s.t.
ttkAk -- v k B k _ O,
(15.7)
~'k >_ O.
To formulate the corresponding linear programming problems for G1 U G2, we use the fact that its maximal stable sets are precisely the unions of maximal
15.3
Threshold Measures
389
stable sets of G1 and G2. If i and j index the maximal stable sets of G1 and G;, respectively, then the maximal independent set constraint of (15.5) for G1 tO G2 reads ( B l W l ) i -~- (B2w2) j ~ T. We can therefore write (15.5) for G1 [-J G2 as follows, where jttl, jM2, Pl, P2 and uij are the dual variables to the corresponding constraints of (15.8), and where ~1 and rl2 are auxiliary column vectors introduced for convenience: r(GIUG2)
--
rain T s.t.
/-t2 Pl P2 Pij
A l W l ~ el A2w2 k e2 - B l W l + T~I -- 0 - B 2 w 2 + r12 - 0
(15.8)
--(?]1)i- (?]2)j -t'- T ~ 0 W 1 ~ 0,
W2 ~ 0.
The linear programming dual of (15.8) is as follows" 7-(G1 [,.J G2) Wl w2 (?]1)i
-
max s.t.
I-I,l e l -~ ~2e2 ~t 1A1 - PlB1 < 0 t t 2 A 2 - p2B2 < 0 ( P l ) i - E j 17ij = 0 (m)j
T
-
aj
= 0
~ i j vii = 1 tt 1 > 0 , tt 2 > 0 ,
aj >_ O.
The theorem asserts that if wl, r~ and w~, r~ are optima of (15.6) for k - 1,2, respectively, then (w~, w~), r~ + r~ defines an optimum of (15.8) when the auxiliaries rl~, rl~ are determined from the equations of (15.8). To show this, note that the constraints of (15.8) are satisfied, and then consider optima t t ; , v ; of (15.7). They satisfy the following complementary slackness conditions: tt*k(Akw*k -- ek) = 0 (15.10) v*k(--Bkw*k + r[~f k ) = O.
(15.11)
We define variables for (15.9) as follows:
It is easy to see that these variables define a feasible solution of (15.9). They also satisfy all the required complementary slackness conditions: (15.10),
390
Threshold Weights and Measures
(15.11), and the condition .;
+
-(,;),-
- 0
By theorem 15.3.2 we may assume without loss of generality that G is connected. We now consider how the threshold measure of a join of two graphs is related to their threshold measures. Unlike the case of the union, we only give estimates, for a reason that will be shown soon. T h e o r e m 1 5 . 3 . 3 Let G1 and G2 be disjoint graphs with largest stable set sizes OL1 and a2, respectively. Then C~lt~2 ~1 -t'- C~2
~ T(G1 (~ G 2 ) ~
~1 max(o~ 1 , c~2 ) 9
P r o o f . The upper bound for T(G10 G2) follows from the fact that assigning to every vertex a weight of g1 and taking T -- 71 max(a1, a2) defines a feasible solution of (15.5) for G1 | G2. To prove the lower bound, let Sk be a stable set of size ak in Gk for k = 1, 2. For every feasible solution w, T of (15.5) for G1 @G2, Sk possesses a vertex vk satisfying Wvk
max
wTb
s.t.
Ab < e
(15.14)
b binary, because A b
max
wTb
s.t.
A b w
b_>O
(15.15)
#>_0.
Since A T >_ O, the inequality in (15.15) is equivalent to the existence of tt >_ 0 satisfying AT# _> w and r - eTpt. Therefore (15.13) is equivalent to the following: T(G)
--
min
7"
s.t.
Aw>e
(15.16)
ATIt >_ w eT# - r
#>o,
w>O.
The constraints involving w in (15.16), namely A w >_ e, ATtt >_ w and w >_ 0, can be replaced by A A T # >_ e, thus eliminating w. Indeed, this condition is clearly necessary, and conversely, if it holds and tt >_ 0, then w defined by w - A T # will satisfy its constraints in (15.16). This proves (15.12) and the moreover statement of the theorem. ..
15.3
Threshold Measures
393
There are several variations of Theorem 15.3.6. By the symmetry of A A T, the linear programming dual of (15.12) is given by "r(G)
-
max s.t.
err A A T v 0.
(15.17)
Further, if G has edges, then w(G) > 0, which permits us to transform the variables/~ and r - e T # of (15.12) according to 1~-#
t T
1 T
and rewrite (15.12) as follows" =
max
t
=
s.t.
AAT)~ >_ te er,~ -- 1 ,~>0
max
min
( A A r,~)i
s.t.
eT~- 1 ,~>0.
(15.18)
Similarly, we can transform the variables v and r - e r u of (15.17) according to p=-- v S - - - -1 T
T
and rewrite (15.17) as follows" 1
=
min s.t.
s A A T p O
=
min
max
o
3
s.t.
"(AATp)j "---e Tp -1 p>_O.
(15.19)
The problem (15.19) is the linear programming dual of (15.18). The formulations (15.18) and (15.19) exhibit ~ 1 as the value of a two-person zero-sum game with payoff matrix A A T. The pure strategies of the players are the edges of G, and the payoff to the first player is the number of common endpoints of the two chosen edges. The optimal mixed strategies are the optimal ,~ and p.
:394
T h r e s h o l d W e i g h t s and M e a s u r e s
A vertex cover (or transversal) is a set of vertices meeting every edge. Using (15.17) and essentially reversing the steps in the proof of Theorem 15.3.6, we can obtain the following theorem, which together with Definition 15.3.1 gives a combinatorial rain-max relation for bipartite graphs:
T h e o r e m 15.3.7 If G -
( V , E ) is a bipartite graph with an edge-vertex incidence matrix A, then r ( G ) is the optimal value of the following linear programming problem:
max s.t.
r
~
zv >_ r
for each minimal vertex cover C c_ V
vec
(15.20)
zu+z~0.
Moreover, if v is an optimum of (15.17), then z of (15.20).
A T v is an o p t i m u m
P r o o f . By arguments similar to those of Theorem 15.3.6, we obtain the following from (15.17)" r(G) - m a x { r " A z < e, A T v < z, e T v -- r , z > O , v > 0}.
The existence of v > 0 satisfying A T v < z and e T v -- r is equivalent to T _ 1 red and b >_ 1 blue vertices. Assign a weight of ir to each red vertex and a weight of-g1
396
Threshold
Weights
and Measures
to each blue vertex. Then G is equitable if and only if the largest total weight of a stable set is 1.
P r o o f . Let i and j index the red and blue vertices of G - (V, E), respectively. We can write the linear programming problem (15.12) as follows, using the vertex variables w = ATtt, which by Theorem 15.3.6 constitute an o p t i m u m of (15.5): r(G)
-
min
~#~j i,j
s.t.
E
#ij -- Wi
J E
#ij -- Wj
(15.21)
i Wi nt- Wj ~ 1
ijcE
#~j >_ 0
ijEE
#~j - 0
ij~E.
If G is equitable, then (15.21) has an o p t i m u m satisfying wi + wj = 1 for all ij C E. Then by the connectivity of G all the wi are equal and all the wj b for all i are equal. Since ~ i wi - ~ i j pij - ~ j wj, it follows that wi - -;-4-g r wj = ~+b for all j, and r(G) - r+b" ~b Since w is a feasible solution of (15.5) every stable set S must satisfy ~ e s Wv < ;~b" This means that with the vertex weights given in the theorem, the total weight of every stable set is 1 or less. Since the stable set of all red vertices has a total weight of 1, the "only if" part of the theorem follows. Conversely, if G satisfies the condition of the theorem, then
1
~
max
s.t.
(
l~xi+
1 )
7" i
-b j
xi nt- xj ~ 1
~xj
ij E E
(15.22)
x binary. Since the constraint matrix of (15.22) is totally unimodular and there are no isolated vertices, the condition "x binary" can be relaxed to x >_ 0. Then
15.3
Threshold
397
Measures
by linear programming duality there exist
)~ij, ij
C E, satisfying
Aij
b
ijEE
1
~j >_ j:ijEE
for all i
r
(~5.23)
1
for all j
i:ijEE
A~j > 0
ij C E.
But every solution of (15.23) must satisfy all its inequalities except for the non-negativity constraints with equality, as can be seen by summing the /-inequalities or the j-inequalities. Therefore the vector A*=
rb A>O r+b -
satisfies AATA * - e, which proves the "if" part. 9 Next we characterize equitable bipartite graphs by a generalization of Hall's condition for the existence of a perfect matching in a bipartite graph. T h e o r e m 15.3.10 Let G be a connected bipartite graph having r >_ 1 red and b >_ 1 blue vertices. Then G is equitable if and only if every set X of red vertices satisfies 1
IN(X)l >_ -~ F ~h~
(15.24)
N ( X ) d ~ o t ~ th~ ~ t of ~ighbo~ of th~ w~tic~ of X .
P r o o f . Denote by R and B the sets of red and blue vertices, and assign weights of !~ and g1 to their vertices, respectively. Condition (15.24) states that the total weight of N ( X ) is at least the total weight of X, for each XcR. To prove the "if" part, let S be any stable set of vertices, and take X - S N R, so that S and N ( X ) are disjoint. Then B includes the set ( S N B ) U N ( X ) , whose total weight is at least the total weight of ( S N B ) U X = S by (15.24). Hence B is a stable set of maximum weight, and so G is equitable by Theorem 15.3.9. To prove the "only if" part, assume that G is equitable, and therefore the weight of every stable set is at most 1 by Theorem 15.3.9. For every set
Threshold Weights and Measures
398
-IXl+w(b-IN(X)l
1 X C R, the set X U ( B - N ( X ) ) i s stable, and hence 1 _> r giving (15.24). 9 From Theorem 15.3.10 and Hall's condition we immediately obtain the following.
Corollary 15.3.11 Let G be a bipartite graph with an equal number of red and black vertices. Then G is equitable if and only if G has a perfect matching. To characterize equitable trees, we need some notation. We continue the convention that i, i0 index red vertices and j, jo index blue vertices. For a tree T and any edge e = ij of T, deleting e from T produces two subtrees. We denote by Tij the subtree containing i and by Tji the subtree containing
j. P r o p o s i t i o n 15.3.12 Let T = (V, E) be a tree and let x be an assignment of vertex-weights, possibly negative, satisfying
E xi - E xj. i
(15.25)
j
Then there exist unique edge-weights Aij, ij E E satisfying ~ij
-
xi
for all i
-
xj
f o r aU j.
(15.26)
j:ijEE i:ijEE
An explicit formula for the ~iojo = E ieTio3o
xi-
,~ij i8 given by E
xj-
JET/0~ 0
E
xj-
jET30i 0
y~ xi.
(15.27)
iET~0,0
P r o o f . The existence and uniqueness of the Aij follow from an algorithm that solves (15.26) "from the leaves up". That is, choose a leaf (a vertex of degree 1), say i ~, and its unique neighbor j'. Set ~i,j, = xi, as dictated by (15.26), delete i' and the edge i'j' and replace x~ by x j , - x~,. The result is a smaller tree that still satisfies (15.25), to which we can apply induction. The basis of the induction is the one-edge tree, where (15.25) is utilized. To prove the explicit formula (15.27), we show below that the Aij defined by (15.27) satisfy
E j:iojEE
Xo.
(15.28)
15.3
399
Threshold Measures
Figure 15.5: Illustration of the proof of Proposition 15.3.12. i0
Tjxio
~/j2io
Tjkio
A similar proof holds for the other equation of (15.26). Let j l , . . . ,jk be all the blue neighbors of i0, as in Figure 15.5. Then
k
k(
E ~,oj- E ~,o~- E r=l
j:ioj6E
r=l
j
i
E x~j6Tj,.i o
Zx )
iETj,., o
xo)
proving (15.28). " We are now ready for a simple combinatorial characterization of equitable trees. T h e o r e m 1 5 . 3 . 1 3 Let T be a tree having r > 1 red and b >_ 1 blue vertices. For each edge e - iojo of T, define _ ~(~) r
r
b(~) b '
(15.29)
Tiojo containing io, respectively. Then T is equitable if and only if ~ >_ 0 for each edge e. In that case )~ is an optimum of the threshold measure problem (15.18).
Threshold Weights and Measures
400
P r o o f . Formula (15.29) is obtained from (15.27) by setting x~ xj __ ~. 1 Therefore by Proposition 15.3.12, )~ satisfies 1
A~j -
-
j:ijEE
all i
r
Aij
1 ~
-
1?- and
(15.30) all j,
i:ijEE
where T - (V, E). Thus if )~ _> 0, then )~ satisfies (15.23), and the argument of the "if" part of Theorem 15.3.9 shows that T is equitable. Conversely, if T is equitable, then as in the "only if" part of Theorem 15.3.9, the equations (15.30) admit a solution )~' _> 0. But by Proposition 15.3.12 these 1 equations have a unique solution given by (15.27), where xi - 1 and xj = -g, namely the ,k defined by (15.29). Therefore ,V - ,k and ,k _> 0 [] We conclude this section with two examples of equitable trees. T h e o r e m 15.3.14 Let T be a tree in which all red vertices have degree 2 (equivalently, T is obtained by subdividing each edge of a tree with a new vertex). Then T is equitable. Proof. To see vertex from j
T satisfies b - r + l . A l s o , each edge e - iojo of T satisfies r(e) - b(e). this, use the bijection mapping each red vertex i of Tiojo to the blue j of Tiojo such that i is the immediate successor of j along the path to i0. Therefore formula (15.29) gives A(e)-r(e)
(1 1) r
r + l
>0"
A complete k-ary tree is a rooted tree in which every internal vertex has exactly k children, and all the leaves have the same distance from the root. Theorem
15.3.15 Every complete k-ary tree is equitable.
P r o o f . Let T - (V, E) be a complete k-dry tree, and assume without loss of generality that its root is blue. Let h be the height of T (the distance from the root to each leaf) 9 The total number of vertices of T is IV] - k h]+~ l l l 9 The numbers r and b of red and blue vertices of T are functions of k and h. Specifically, if h is even, then b - P(k, h) "- 1 + k 2 +
k 2]Ch -
-
Q ( k , h) . -
kh + 2 - 1
k 4 --~-""" + ~h __
IVI - P ( k ,
1
h) ]~2
__1
~
1
15.4
401
Threshold and Majorization Gaps
and if h is odd, then k h+l -- 1 k:-i
b - R(k,h) "- g ( k , h - 1 ) -
k h+l - 1
- s(k, h) . - IVI - R(k, h) - k k~---=-V To compute the A~ of (15.29) for a given edge e, it is convenient to always use that subtree of T - e that contains the root, using either (15.29) or its analog with red and blue switched. Let 1 be the height of this subtree. If h is even, then
_
P(k,l) Q(k,1) V~(k~h) - Q(k, h)'
R(k,~)
S(k,1)
-d-(-i l h ) - P ( k, h ) '
for k even for k odd,
and if h is odd, then
~ -
P(k, 1) Q(k,l) 5 ( ; , ~ - R(k,h)' R(k,~) S(k,t)
for k even = 0,
for k odd.
In all cases, one obtains Ae _> 0.
15.4
Threshold
and
Majorization
Gaps
In this section we introduce two measures of non-thresholdness of a degree sequence: the threshold gap, based on the work of Hammer, Ibaraki and Simeone [HIS78, HISS1], and the majorizaion gap, based on the work of Arikati and Peled lAP94]. It turns out that these two measures coincide, and we investigate their properties and the relations between them. Recall from Section 3.1 that a vector d - ( d l , . . . , d~) with integer components is called a proper sequence if n - 1 >_ dn >_ " " >_ dl > 0. Recall also the definition of the corrected Ferrers diagram C(d) representing a proper sequence d (Definition 3.1.6), and of the corrected conjugate sequence d' and the corrected Durfee number m of d. Also recall (Theorem 3.2.2) that a proper sequence d is threshold if and only if C(d) is symmetric (i.e., d' - d), in which case C(d) is the adjacency matrix of the unique labeled realization of d. We define a distance between proper sequences as follows.
402
T h r e s h o l d Weights and M e a s u r e s
D e f i n i t i o n 15.4.1 The distance between proper sequences d and e of the same length n is defined as n
l i d - ell -
(15.31)
I & - e l. i=1
In other words, the distance is just 1/2 of the Ll-norm of the difference d - e . If ~'~'~in._=ldi and ~i~1 ei are even, as is the case when d and e are graphic sequences, then the distance lid- ~11 is an integer. The reason is that
9
di~ei
di>_ei
i
di<ei
Since 2in___l di and ~i~a ei are even, so are ~d,>~, di--~-~d,<e, di and ~d,~, ei, and therefore ~i~1 I d i - ei] is an even integer. We denote by 7~T,~ the set of proper threshold sequences of length n. The threshold gap of d is defined as follows. D e f i n i t i o n 15.4.2 If d is a proper sequence of length n, its t h r e s h o l d g a p is eE D"I'n
lid-
Thus the threshold gap of d is a measure of the non-thresholdness of d. In Subsection 15.4.1 we determine the threshold gap of a proper sequence and all the threshold sequences that realize it. Recall the definitions of the majorization ~ and strict majorization ~relations between sequences of length n (Definition 3.1.1). Recall also the definition of unit transformation (Definition 3.1.2). If a is obtained from b by a unit transformation, we say that b is obtained from a by a reverse unit transformation. By Condition 9 of Theorem 3.1.7, every degree sequence d is majorized by some threshold sequence, and by Condition 8 of Theorem 3.2.2 the majorization is strict if and only if d is not a threshold sequence. From this and from the Muirhead Lemma (Theorem 3.1.3) it follows that if d is any degree sequence, then some threshold sequence can be obtained from d by a finite number of successive reverse unit transformations. This motivates the following definition of the majorization gap of d as a measure of the non-thresholdness of d.
15.4
Threshold and Majorization Gaps
403
D e f i n i t i o n 15.4.3 The m a j o r i z a t i o n g a p R ( d ) of a graphical sequence d is the smallest number of reverse unit transformations required to transform d into a threshold sequence. Thus R(d) = 0 if and only if d is already a threshold sequence. In Subsection 15.4.2 we give a formula for R(d) and show that it is equal to the threshold gap of d. We also determine its m a x i m u m for a fixed number of edges or vertices, as well as exhibiting all the sequences that realize this maximum, which can be thought of as the most non-threshold in this sense. We also discuss the related questions of the difference gap and some others.
15.4.1
The Threshold Gap
Before we discuss the threshold gap, we introduce two binary operations on the proper sequences of length n. D e f i n i t i o n 15.4.4 If d and e are proper sequences of length n, their j o i n and m e e t are defined as the proper sequences d V e and d A e given by (dV e)i
-
max(di, ei)
i - 1,...,n
(dA e)i
-
min(di, ei)
i - 1,...,n.
It turns out that DT-~ forms a lattice under the operations of join and meet, having the largest element ( n - 1, n - 1 , . . . , n - 1) and the smallest element (0, 0 , . . . , 0). This follows from the following proposition. Proposition
15.4.5 DT~ is closed under joins and meets.
P r o o f . For integers k _> 1 and h >_ 0, we denote by S(k, h) the set of the smallest h positive integers excluding k, that is to say
S(k,h)-
{1,...,h}, {1, ,k-l,k+l,...,h+l},
ifhk.
If d is a proper threshold sequence whose unique labeled realization on the vertices 1 , . . . , n is G, then the neighborhood of each vertex i in G is Na(i) = S(i, di). Conversely, if a graph G on the vertices 1 , . . . , n has neighborhoods satisfying this condition, then G is a threshold graph with degree sequence d, so d is a threshold sequence.
404
Threshold
Weights and Measures
We prove the closure of 2)7",~ under joins; the results for meets can be proved similarly or deduced from the closure under complements. Let d, e, C 2)Tn, and let G and H be the threshold graphs on the vertices 1 , . . . , n in which vertex i has degree d~ and e~, respectively. Thus N o ( i ) = S(i, d~) and NH(i) = S(i, ei) for each i. Put f = dV e and F = G tO H. Then for each i we have
NF(i) -- S(i, d~) U S(i, e~) - S(i, max(d~, e~)) - S(i, f~). Hence f is a threshold sequence, i.e., f C ~)']"n. 9 We now define an easily-computed quantity t(d) associated with a proper sequence d, and show that it is equal to the threshold gap of d. D e f i n i t i o n 1 5 . 4 . 6 If d is a proper sequence of length n and corrected Durfee
number rn, we put 1
m
t(d) - -~ ~
!
1
d~[ -
Id~ -
7-51
n
E
!
Id - d l.
( 5.a2)
~=m+l
To justify the equality between the two expressions in the definition of t(d), consider the corrected Ferrers diagram C - C(d), and divide it into four submatrices M, A, B, 0 as illustrated in Figure 15.6. Figure 15.6: The corrected Ferrers diagram of a proper sequence. The submatrix M is full of l's except on its main diagonal; the submatrix O is full of O's. 1
m
m+l
n
M
B
A
O
m
m+l
Since each row and column sum of M is m - 1, d~ - di is equal to the difference between the i-th column sum of A and the i-th row sum of B. Since
15.4
Threshold and Majorization Gaps
405
the l's of A are at the top of the columns and the l's of B are at the left n ~" 1, if Cij r Cji of the rows, we have Id~- di I - ~j=~+~ u~j, where u~y - ~ O, otherwise Consequently m
m
i=1
n
i=1 j = m + l
A similar argument involving the rows of A and the columns of B shows that n
m
E Id:-d i=m+l
n
l-E E
Uij~
i=1 j = m + l
and therefore the equality in Definition 15.32 has been justified, and we also have an interpretation of t(d) as a measure of the deviation of C(d) from symmetry. Furthermore, we have ~i~=~ di - m ( m - 1) + a + b, where a and b are the numbers of l's in the submatrices A and B, respectively. Therefore if ~in=.l di is even, as is the case when d is graphic, a and b have the same parity, and it follows that t(d) is an integer. As a first step in proving that t(d) is the majorization gap of d, we associate with each proper sequence d of length n with corrected Durfee number m the sequences d and d defined by
d i - { d'i' f o r i - 1 , . . . , r n
di,
for i - m
+ 1,..., n
cli- { di, f o r / - 1 , . . . , r n d}, for i - m + l , . . . , n .
Referring to Figure 15.6, we see that C(d) is obtained from C(d) by replacing the submatrix B with A T, whereas C(d) is obtained by replacing A with B r. Therefore d and d are proper sequences and their Ferrers diagrams are symmetric, i.e., d, d C 7PTn. To find the corrected Durfee numbers r'n - re(d) and rh - re(d) of d and d, we notice that for C - C(d) we always have Cm+l,~n = 0 (for if Cm+l,,~ = 1 one would also have Cm,m+l = 1 by dm >_ din+l, and m could be increased by 1). In contrast, x - C~n,m+l can be either 0 or 1. From the definition of d it is clear that d - C(d) satisfies C',~+~,,~ = 0 and therefore rh - m. On the other hand, d' - C(d) s a t i s f i e s C m + l , m = x and Cm+2,m+l = 0, and therefore rh is m or m + 1 according as x is 0 or 1. We have therefore verified the following formula: ~-m
~_{ '
m, re+l,
if d i n - m - 1 if d i n > m - 1 .
(15.33)
Thre s hold Weights and Measures
406
It follows directly from (15.31) and (15.32) that for each proper sequence d, the proper threshold sequences d and d satisfy
lid- rill - l i d - rill -
t(d).
(15.34)
T h e o r e m 15.4.7 E v e r y proper sequence d of length n satisfies t(d) -
min l i d - cll.
cEDTn
P r o o f . Let g: C 77T~ be optimal, i.e., lid- ~11- mincev~rn lid- ell. Let d be the corrected Ferrers diagram of ~: and rh its corrected Durfee number. First we show that rn _< rh _< rh, where rh is the corrected Durfee number of d, given by (15.33). Assume that, if possible, rh < m. Put h = ~:,~+1 < rh, and define the sequence c by { ~i+1, fori-h+l,...,rh ci ~, for i - rh + 1 g:i, otherwise. Since C is symmetric and h < rh, c is a proper threshold sequence, and its symmetric corrected Ferrers diagram is obtained from ~' by enlarging its corrected Durfee,square from size rh to size rh + 1. We have C~+l - h < c~+1 - r h _< r n - 1 _< dm _< drh+l and therefore Ic~+l - d~+l I < 1~=~+1- d~+a I - ( ~ - h).
(15.35)
In addition, we have in any case that (i-
h+ 1,...,~).
(15.36)
By adding the inequalities (15.35) and (15.36), we obtain IIc-dll < II~-dll, contradicting the optimality of ~:. This proves that m < ~n, and a similar argument shows that rh < rh. We have shown that m < rh < rh, and therefore by (15.33) rh is either m or rh. We show below that l[~:- dll - l i d dll if r~ - m, and it can be similarly shown that I1~- dl[ - l i d - dll if ~ - ~ . Consequently, at least one of J and d is an optimal threshold sequence, and the required conclusion
that l i d - ~11-
t(d) will follow from (15.34).
As stated above, we consider the case ~ - m and have to show that I 1 ~ - d l l - lid-dll. Note that ~ ~nd d ~re proper threshold sequences with the
15.4
T h r e s h o l d a n d M a j o r i z a t i o n Gaps
407
same corrected Durfee n u m b e r m. If g: - d, we are done. Otherwise there must be some i - m + 1 , . . . , n such that g:i r a~i, because rows m + 1 , . . . , n of the s y m m e t r i c corrected Ferrers diagram determine it completely. Let k be the largest such i satisfying g:i > di, if any. Put h - g:k and define the sequence c' by
, ~ c.i-1, ci - ~ ci,
fori-k,h otherwise.
The sequence c' is proper, because by the maximality of k we have g:k > dk _> dk+l >_ ck+l, and by the s y m m e t r y of its corrected Ferrers diagram it is a threshold sequence. Clearly its corrected Durfee n u m b e r is again rn. Since c~ - g : k - 1 >_ dk - dk, we have I c ~ - d k [ - Ig:k--dkl-- 1. Hence I I c ' - d l l _< I]g;- dll because in any case I c ~ - dh] _< Ig:h- dhl + 1. It follows that c' is another optimal threshold sequence and we may replace g: with c'. R e p e a t e d application of this procedure eventually allows us to assume that ci _0 0, otherwise.
5~
1, i f d ' p - d p > O 0, otherwise,
Similarly, - 5~
-
-1, ~ ( e ) - 5~(d) -
ifd'~-d~>l O, otherwise.
I
1
l)+,~nd
412
Threshold
Weights and Measures
Figure 15.7: Illustrating a transfer from p to 7r. Solid lines represent l's and possibly . ' s .
T
~T
I
1 ....
I
C a s e 2: 7r 7~ T and p - or. Here, ~ r ( e ) - ~ ( d ) and 8 ; ( e ) - ~ ( d ) Case 1. Also 6 p ( e ) - ( e ' p - ep) + - ( d ' p - d p + 2) +. Hence
8 p ( e ) - 6p(d) -
2, if d'p - dp > 0 1, if d'p - - d p - - - - 1 O, otherwise.
C a s e 3" 7 r - T and p5r a. H e r e S p ( e ) - S p ( d ) )+ - ( d rI - d ~ - 2 ) Case 1. S i n c e 6 r ( e ) - ( G - e/r --2,
6r(e)-6r(d)
-
are as in
-1, O,
and 5o(e)-5~,(d) + we obtain
are as in
if d'~ - dr > 2 if d" - d~ - 1 otherwise.
C a s e 4" 7r - ~- and p - ~r. We now h a v e S r ( e ) - S r ( d ) as in Case 3, and 5 p ( e ) - 6p(d) as in Case 2. To complete the proof, note that 5i(e) - 5i(d) except for i E {~r, p, ~, ~-}. For future reference, note the following from the proof of L e m m a 15.4.12. Necessary and sufficient conditions for 6(e) - 5 ( d ) - 2 in Cases 1, 2, 3 are: 1. if 7r -~ T, p -~ or, then (a) d ' -
d~> 1,
15.4
Threshold and Majorization Gaps
(b)
r
413
0,
(c) f ~ - d ~ < O , (d) s
d~>_ 1.
2. if 7r 7~ r, p - (r, then (a) d ' -
(b)
d~>_ 1,
l
(c) d ' -
d~_> 1.
3. if 7c - r, p -r a, then
d'-
2,
(b) d ' p - d p < O ,
d'-
O.
L e m m a 1 5 . 4 . 1 3 Let d be a proper sequence. For 1 ( i , j (_ n, if di < j - 1, ! then dj < i. P r o o f . The proof is simple. L e m m a 1 5 . 4 . 1 4 Let d C D~ and let p be the largest index i such that di < d~. Then p < n. Further, 1. A s s u m e d'p (_ p -
!
1 and let q - d;. Then dq - p -
!
1 and dq ( p -
2.
2. A s s u m e dpt >_ p and let q - dpt + l. Then dq - p and dql _ d~, so p < n by definition of p. The fact that p < n justifies our mentioning of column p + 1 (of C) below. In both cases q is the position of the last 1 in I ! column p, so Cqp - 1. The assumption on p implies dp+ 1 _< dp+l _< dp < dp. ' 1 < dp, ' implying that Cq,p_t_ 1 r 1. We prove the two parts as Hence dp+ follows" !
1. See Figure 15.8. If d; _< p - 1 , then p > q and Cq,p+l r "k. It follows that Cq,p_t. 1 -- 0, and therefore (since Cqq - * and Cqp - 1) dq - p - 1. Also dp < d; - q gives Cpq - 0, and again C(q, q) - * implies d; _< p - 2.
T h r e s h o l d Weights and M e a s u r e s
414
Figure 15.8: Illustrating Part 1 of Lemma 15.4.14. q
P
~ * ~ 1 0 0
*
!
2. In this case, apply L e m m a 15.4.13 with i - p and j - 1 - dp. T h e n ! j - dip + 1 - q and dq < p, as required. In our case p < q. See Figure 15.9. We have seen t h a t Cq,;+l r 1, b u t Cqp - 1, so dq _> p. If q > p + 1, t h e n Cq,p+l = 0 and dq - p, as required. If q - p + 1, t h e n Cqp - Cp+l,p = 1 and Cqq - , . On the other h a n d , Cq,q+l m u s t be 0 if ! it exists, for otherwise Cp;;+l would be 1, c o n t r a d i c t i n g dq < p. T h u s again dq - p.
Figure 15.9: Illustrating Part 2 of Lemma 15.4.14.
Lemma 1,...,p-
p
q
,
0
1
*
Let d C Dn and let p be as in L e m m a 15.~.1~. FOr r 1, dr >_ p implies dr < dlr.
15.4.15
-
-
15.4
Threshold and Majorization Gaps
415
Figure 15.10" Illustrating the proof of Lemma 15.4.15. r
p
t
~ * ~ I 0
0
*
P r o o f . A s s u m e to the c o n t r a r y t h a t d~ > d'~. Let t - d~ + 1 > p (since d~ > r, t is the position of the last 1 in row r. See Figure 15.10). By c o n s t r u c t i o n C~t - 1, and by a s s u m p t i o n Ct~ - O, so d't ~_ r and dt < r - 1. B u t t h e n dlt > dt c o n t r a d i c t i n g the definition of p. 9 Lemma
15.4.16
Let d be a proper sequence. If there exists a k such that
d'k < d'k+l, then 1. d'k - k - l , d ' k +
~-k;
2. k is unique. P r o o f . 1" If d~ < k - 1 or d~ > k, then d~+ 1 < d~ since the l's of C(d) are left-justified. Hence d~ - k - 1. T h e n again by this reason and the a s s u m p t i o n t h a t d~ < d~+l, we obtain d~+ 1 - k. 2" T h e s a m e p r o p e r t y of C(d) implies t h a t the conditions d~ - i - 1 and d~+ 1 - i are possible for at most one i. ,, 1 5 . 4 . 1 7 (1) Let d C Dn have corrected Durfee number m, and let C(e) be obtained from C(d) by deleting the last 1 in row i, where 1 < i < rn and (a) if i < m, then di > di+l; (b) if i - r n , then dm >_ m. Then e C D , . Further, if d} > d,, then 5(e) > 5(d). (2) Let d G Dn, and let C(e) be obtained from C(d) by adding a 1 to the end of column i, where 1 < i < rn and (a) if i > 1, then d~ < d~_l;
Lemma
416
Threshold Weights and Measures
(b) if i = m, then dm >_ m. Then e e D , . Further, if d~ >_ d~, then 5(e) > 5(d). P r o o f . 1: The a s s u m p t i o n s on i g u a r a n t e e t h a t e is a proper sequence. The c o l u m n of the 1 t h a t is removed is j = 1 + di > m. This implies t h a t e E Dn. Further, if d~ > di, then Cji - 1, and thus dj ~_ i. But the a s s u m p t i o n s on i i m p l y d~ - i. Therefore by equation (15.38),
5(e) - 5(d) - (d'i - di + 1) + - (d'i - di) + + (d~ - dj - 1) + - (d} - dj) + - 1 + O. 2: This is similar to Part 1.
..
Let d c Dn with d~ > 0 and m - 2. Then S(d) 0 implies n _> 2. Using d2 < dl _< n - 1, d~ - 1, d2 _> 1, and di - 1, d~ - 2, for i - 3 , . . . , d2 + 1, we obtain n
~(d)
-
~ (d'~ - d~)+ i=1
=
(n-
_< ( n -
1-dl)+
-~-(d2- 1)(2-
1)
1 -d2) + + d2- 1
=
n-l-d2+d2-1
"~
rt ~2.
We need three more l e m m a s to prove T h e o r e m 15.4.10. Lemma
1 5 . 4 . 1 9 Let d C D~. A s s u m e that dn > O, d~ < d'a, and let p be the
largest i such that di < d~. Let q be such that dq > dqt and either (a) q < n or (b) q - n , dn > 2. Then Sk(d) < X k ( d ' ) - 1,
for k -
p,...,q-
1.
Since d C Dn, Sk(d) S~(d')d~ > d~,
( a s s u m p t i o n on p) ( a s s u m p t i o n on q) ( a s s u m p t i o n on p)
I dq >_ dq-+-i
(d~ < dl ~ d'n - 0 , d. > 0)
d. > d ' + l .
d~ > d~,
i-k+l,...,q-1 i-q+l,...,n-1
15.4
417
Threshold and Majorization Gaps
Adding all these inequalities, we obtain a contradiction, S~(d) > S ~ ( d ' ) + 1. W h e n q - n, the inequality for dq drops, but by assumption the inequality for dn can be strengthened by 1, and the same contradiction is obtained.
1 5 . 4 . 2 0 Let d C Dn, a s s u m e that dl < d'1, dn > O, and let p be as in Lemma 15.~. 19. Then
Lemma
Sk(d) p + 1. From L e m m a 15.4.14, dq - p. S i n c e p < q,
Threshold Weights
418
and
Measures
dp _> dq - p. Thus d l > . . . > dp. Using L e m m a 15.4.15, we obtain di _< d~ for i - 1 , . . . , p - 1. This and the assumption d l < d~ complete the proof. 9 P r o o f of T h e o r e m 15.4.10. From L e m m a 15.4.12 we have R(d) >_ [ ~ 1 , as a reverse unit transformation can decrease 5(d) by at most 2. We show that if 5(d) > 0, we can always construct a proper sequence e such that 1. e is obtained from d by a reverse unit transformation, 2. e is a degree sequence, and 3.
= e(d)-
2.
It then follows that 5(d)is even and R(d) - 6(d) 2 " To show that e is a degree sequence we use the Berge Condition of Theorem 3.1.7, i.e., & ( e ) is even and Sk(e)_< Sk(e')for k = 1 , . . . , n . To show that 5(e) = 5 ( d ) - 2 we use the observation made after the proof of L e m m a 15.4.12. Without loss of generality, we may assume that dn > 0, for if dn = 0, then we work with c - ( d l , . . . ,d~-l). We may also assume that d 1 < d~, for if d 1 - d i ( - 7 " t - 1), then we work with c - ( d 2 - 1 , . . . , d ~ - 1). We distinguish two cases. C a s e 1" There exists an i > 1 such that di < d~. Let p be the largest such i. The basic idea is to transfer the 1 at the end of column p to the end of row 1. Let s = dl + 2 be the destination column for the moving 1. We have two subcases now. ! ! C a s e 1.1" dp _< p - 1 . Then q - dp is the source row for the moving 1. Also dq = p - 1 by L e m m a 15.4.14 and hence p is the source column for the moving 1. Define e = d - uq + Ul. Then for i = 1 , . . . , n , (a) ei = di except for e l - - d l q- 1 and eq - - d q - 1, and (b) e Ii - - d Ii except for epI - d Ip - 1 and Cs,
1.
We first prove that e is a degree sequence. Since 5's(e') = S'n(e), it sumces to prove S'k(e) _< S'k(e'), for k = 1 , . . . , s - 1. This is done as follows: Sk(e)=Sk(d)+l_< Sk(d') = S k ( e ' ) for k = 1 , . . . , q - 1 (by L e m m a 15.4.21), Sk(e) = Sk(d) < Sk(d') = Sk(e') for k = q , . . . , p - 1,
15.4
Threshold and Majorization Gaps
Sk(e)=Sk(d) p. From L e m m a s 15.4.20 and 15.4.21 we have Sk(d) < Sk(d'),
k-
1,..., n-
1.
(15.39)
Let q - dp! + 1 be the source row for the moving 1, and define e as in Case 1.1. We first show that Sk(e) _< Sk(e') for k = 1 , . . . , s - 1. It follows from L e m m a 15.4.14 that dq - p and dq' 1, f ( n -
2. f ( n ) > n 3. g ( n - 1 )
First, some simple facts
1) < f ( n ) for n > 3;
2 for n _> 2, with strict inequality for n > 5;
< g(n) for n > 4, g(n) > n + l f o r n > 7 .
Next, three lemmas about corrected Ferrers diagrams. L e m m a 1 5 . 4 . 2 7 Let 0 r d C D n . Put s - dl + 1, p - d~s, q - d~p + 1, r - ds. Thus s > _ r e > p > 1 and q > m > r > 1. Assume that the following hold (See Figure 15.11):
1. s > m + 1, and if equality holds, then p < m; 2. q < s ; 3. r < p - r . Then: 1. there exists a sequence e C
Dn
such that
e I
--
dl-1 and 5(e) > 5(d)+2;
2. there exists a sequence f c D~ such that (a) (i) f l
-
dl -
1 or (ii) f l
-
dl
and f'~ - 1;
(b) 5 ( f ) > 5(d) + 1; (c) S n ( f ) and Sn(d) have the same parity. P r o o f . We begin by proving s t a t e m e n t 1. First note that p _< m _< q and r < p. Secondly, we m a y assume that the first r columns of C(d) are full, i.e., d~i - n - 1, i - 1,...,r, (15.41) for otherwise we can fill column 1, then 2, and so on up to r (by adding l's at the end of these columns) without going out of D~, and increase 5(d) at each step, by L e m m a 15.4.17.
424
Threshold Weights and Measures
Figure 15.11: Illustrating L e m m a 15.4.27. r
p
m
q
s
*
0
m
0
Counting in two ways the number of l's in rows 1 , . . . , s and columns r + 1 , . . . , p of C(d), we obtain p
~
d:-(p-r)(q-1)+ i=r+l
(15.42)
(di-r).
i=q+l
Since d e Dn, ~p(d) ~ ~p(d'). This implies, by (15.41) and (15.42), p(s - 1 )
p
dr+ 2 >_ 9.. > dp, i < p _< q, a contradiction. Also r < i < p, then q - 1 _< d~ < d~+1 -
15.4
425
Threshold and Majorization Gaps
di-s-1
I a d'~,
Hence
i - r + 1,...,p.
(15.44)
I I l l Again by Lemma 15.4.16, dlq+l ~ dq+ 2 ~ . . . ~ ds, f o r if d i < di+ 1 f o r some q < i < s, then d~+1 - i > q >_ m, a contradiction. Therefore for i - q + 1 , . . . , s , di _ d'~ - p . Hence
d'~ > d,,
i-
q + 1,...,s.
(15.45)
Using (15.41), (15.44) and (15.45), we have q
5(d) - r(n - s) + ~
(d'i - di) + + ~
i=p+l
(d'i - di).
(15.46)
i=q+l
Define a sequence e such that C(e) is obtained from C(d) by making the first p columns full and deleting the s-th column, i.e., , ei-
f n-l, / 0, d~
fori-l,...,p fori-s otherwise
It is easy to check (using s > m + 1) that e C Dn by Lemma 15.4.17. Further, s--1
~(~) -
~(~,,-
~)+
i=1 q
= p(n-s+l)+
s-1
y~ (d'i - d i ) + +
y~ (d'i - p )
i=p+l
=
i=q+l q
~(~ - ~) + ( p - ~)(~ - ~) + p + ~
(d'~ - d~) + +
i=p+l
(d'~ - p)
(~s d' - p)
i=q+l q
=
r(n-s)+(p-r)(n-s)+p+
Z
(d',- d,) + +
i----p+l
(d'i - di) i--q+l
=
~
( p - di)
i-q+l
~(d) + ( p - ~)(~ - ~) + p -
~ i=q+l
( p - d~)
(by (15.46))
Threshold Weights and Measures
426
=
5(d)+(p-r)(n-s)+p-
)_2 ( P - r ) + i=q+l
=
~(d) + (p - ~)(~ - ~) + p -
~_~ ( d i - r ) i=q+l
(~ - q ) ( p - ~) + ~
(d~-
~)
i=q+l
> 6(d)+ ( p - r)(n - s ) - r(n - 1 - s) = ~(d) + ( p - ~)(~ - ~ ) - r(~ - ~ ) + >
~ ( d ) + ~.
( ~ ~ < p-
(by (15.43))
~)
So 5(e) >_ 5(d)+ 2. This proves Statement 1. We now prove Statement 2. If S'~(e) has the same parity as S~(d) (before the achievement of (15.41)), then f = e has the required properties. If the parities differ, take f = e + u l . In this case C ( f ) can be obtained from C(d) by making the first p columns full and deleting all the l's except the first 1 in the s-th column. So f C Dn by Lemma 15.4.17. The other required properties of e follow from S~(I) = S~(e)+ 1, 5~(f) = 5 1 ( e ) - 1, and 5~(f) = 5~(e) for i = 2,...,n. ,, L e m m a 15.4.28 relaxes one of the assumptions of L e m m a 15.4.27. Lemma
15.4.28
Under the conditions of Lemma 15.4.27 except r 3, then there exists a sequence and 5(e) > 5(d) + 1;
e C Dn such that el - d l - 1
2. if m > 4, then there exists a sequence f c Dn such that (a) (i) f l = d l -
1 or (ii) f l - - d ,
and f~ = 1;
(b) 5 ( f ) > 5(d)-4- 1; (c) S ~ ( f ) and Sn(d) have the same parity. P r o o f . We begin by proving Statement 1. First, observe that m < n - 1, because m = n - 1 and p = m would imply Sin(d) > Sm(d'), contradicting d E Dn. Secondly, assume without loss of generality that (15.41) holds as in
428
Threshold Weights and Measures
t h e p r o o f of L e m m a 15.4.27. Define a s e q u e n c e e such t h a t C ( e ) is o b t a i n e d f r o m C(d) by m a k i n g t h e first m - 1 c o l u m n s full, if necessary, a n d d e l e t i n g t h e c o l u m n s - rn + 1, i.e.,
ei! m
rtml 07
d~, Then e E
Dn
~ for i - 1 , . . . , r n for i - m + 1 otherwise.
1
by L e m m a 15.4.17. Using t h e facts d} - n -
di
m, d~ - m - 1 for i - r + 1 , . . . , definition of m a n d r), we o b t a i n
5(d) - r ( n -
1 - m)+ m-
m,
d i 'n +
r < (m-
1 -
m,
1 for i - 1 , . . . , r, 1 (by _< m -
dm+l - r
1)(n - 1 - m ) +
m - r.
Also, 5(e) - ( r n -
1)(n - 1 - ( r n
- 1)) - (rn - 1)(n - 1 - r n ) + m - 1.
It is now clear t h a t if r > 1, t h e n 5(e) > 5(d). If r = 1, t h e s a m e c o n c l u s i o n holds, since m - 1 > 1 a n d n - 1 - m > 0. T h i s proves S t a t e m e n t 1. W e now prove S t a t e m e n t 2. If Sn(d) before t h e a c h i e v e m e n t of (15.41) a n d S~(e) h a v e t h e s a m e parity, t h e n f = e has t h e r e q u i r e d p r o p e r t i e s . O t h e r w i s e t a k e f = e + Um+l. O n c e again, C ( f ) can be o b t a i n e d f r o m C(d) by m a k i n g t h e first m - 1 c o l u m n s full a n d d e l e t i n g all t h e l ' s e x c e p t t h e first 1 in c o l u m n m + 1. T h e r e f o r e f C Dn by L e m m a 15.4.17. F u r t h e r , (~(f) = ( ~ ( e ) - 1, as (~l(f) = ( ~ l ( e ) - 1, a n d 5~(f) = 5~(e) for i = 2 , . . . , r n . Hence 5 ( f ) = (rn - 1)(n - 1 - rn) + m - 2. N o w it is clear t h a t if r > 2, t h e n 5 ( f ) > 5(d). If r - 2, t h e s a m e c o n c l u s i o n holds since m - 1 > r ( b e c a u s e m >_ 4) a n d n - 1 - m > 0. Finally, if r - 1, t h e n t h e conclusion holds again, since
5(f)
-(m-2)(n-l-rn)+n-l-m+m-2 >_
(m-2)(n-l-rn)+m-1
>
n-l-rn+m-1
(asn-2>rn)
(asm>_4andn-l-m>O)
=
W e are now r e a d y to prove T h e o r e m s 15.4.25 a n d 15.4.26.
15.4
Threshold and Majorization Gaps
429
P r o o f o f T h e o r e m 1 5 . 4 . 2 6 . T h e s t a t e m e n t is true for n - 1, so a s s u m e n > 2. If d n - 0, t h e n by induction on n, 8(d) < f ( n 1) < f ( n ) , and f ( n - 1) < f ( n ) for n > 3. We m a y therefore assume t h a t d~ > 0, and c o n s e q u e n t l y m > 2. If m 2, then 5(d) < n - 2 by L e m m a 15.4.18, and hence 8(d) _< f(n). E q u a l i t y holds if and only if d~ - d2 and n - 2 - f ( n ) , which implies n < 4. Hence we m a y assume m _> 3. P u t s - dl + 1, p - d~s, q - dp~ + 1, r - ds, and observe as before t h a t p _< m < q, and m < s Proposition. If s > m + 1, then there exists a sequence e C Dn such that el
-
-
dl
-
1 and 5(e) > 5(d). Consequently we may assume that s - m.
To prove the proposition, first assume t h a t q > s. Define a sequence e such t h a t C(e) is o b t a i n e d from C(d) by m a k i n g the first p columns full, and deleting the s-th column. T h e n e C D~ and 5(e) > 5(d) by L e m m a 15.4.17. Now a s s u m e q < s. T h e n r d~ < dq+l < p - l , and h e n c e r < p. I f s > m + l or p < m, t h e n the required sequence e exists by L e m m a 15.4.28. If s - m + 1 and p - m, t h e n the required sequence e exists by L e m m a 15.4.29. This proves the proposition, and we m a y assume t h a t s - m. F u r t h e r , we m a y a s s u m e t h a t the first m - 1 columns of d are full, for otherwise we m a k e t h e m full w i t h o u t leaving D~, t h e r e b y increasing 8(d) by L e m m a 15.4.17. Then d - ( m - 1) n ,
5(d)- (m-
1)(n-
1 -(m-
1)),
and 5(d) reaches a m a x i m u m when n--1
m-
2 ' n ~ 7-1,7,
1 -
nodd neven.
(15.47)
Therefore 5(d)_
7, and ~i(d) - g(n). If d n - - 0 , then ~(d) 0, and therefore m > 2. Also m - 2 implies, by L e m m a 15.4.18, that ~i(d) _< n - 2 < g(n), again contradicting our assumption, so we assume that m > 3 . P u t s - d l + l . Proposition. If s > m + 2, then there exists a sequence f E En such that
(~(f) > ~(d), contradicting ~(d) - g(n). Consequently we may assume that s-m ors-m+ l. We prove the proposition by showing that, when s > m + 2, 1. if d's - 1, then there exists an f E En such that 5(f) > 5(d); 2. if d'~ >_ 2, then there exists an f E E~ such that 6(f) >_ 5(d), and if equality holds, then fl - d~ (so that f and d have the same s) and f;-1. C a s e 1" d's - 1. The basic idea is to work with the sequence ( d l 1, d 2 , . . . , d~) and introduce a 1 at the end of the first row if the parity becomes odd. Put t - s - l , p - d ~ t , q d tp + l . T h e n p _ < m_< q. Assume that q >_ t. Define a sequence f such that C ( f ) is obtained from C(d) by deleting the last 1 in column s - 1 and the 1 in column s. Then f C E~ and 5(f) > 5(d) by L e m m a 15.4.17. Therefore we may assume q < t. Put r = dr. See Figure 15.13. C a s e 1.1: t > m + 1 or p < m. Define a sequence c such that C(c) is obtained from C(d) by deleting the last 1 in the first row. Note that c E D~ by L e m m a 15.4.17 and, further, c satisfies all the hypotheses of L e m m a 15.4.28. By the latter, there exists a sequence g E Dn such that 5(g) >_ 5(c) + 1. The required sequence f is defined by f
_ J" g, (g~ + 1 , 9 2 , . . . , g ~ ) ,
\
Sn(g) even S~(g) odd.
15.4
Threshold and Majorization Gaps
431
Figure 15.13" Illustrating Case 1 of the Proposition in the proof of Theorem 15.4.25. r
m
p
m
ts
.jrm
0
To see this, recall that in the proof of L e m m a 15.4.28, C(g) is obtained from C(c) by making the first p columns full and deleting the t-th column (if r _< p - r ) , or by making the first r columns full and deleting the t-th column (if r > p - r). Note that when S~(g) is odd, C(f) could be obtained from C(c) by making the appropriate columns full and then deleting all but the first 1 in column t. Therefore f C Dn in this case by L e m m a 15.4.17. Clearly f E Dn also holds when S~(g) is even. Further, Sn(f) is even in both cases. Thus f C E~. From the construction of c and f it is clear that 5(c) - (5(d)+ 1 and (5(f) _> ( 5 ( g ) - 1. Therefore 5(f) >_ 5 ( g ) - 1 _> (5(c) - 5 ( d ) + 1. 1.2" t - m + 1 and p - m. This is similar to Case 1.1, except that here we use L e m m a 15.4.29 instead of L e m m a 15.4.28. Case
C a s e 2" 2 _< d', _< rn. Put p - d'~ and q - d'p + 1. If q _> s, then we may remove the last two l's in the s-th column without leaving En and thereby increase (5(d) by L e m m a 15.4.17. We therefore assume that q < s. Then the required f exists by L e m m a 15.4.28.
This completes the proof of the proposition and hence we may assume that s - m or s - m + 1.
Threshold Weights
432
and
Measures
It is convenient here to define a new function. Let a _> 6 be a fixed integer such t h a t a - 2 m o d 4. For 0 _< k _< a, define k(a
h~(k) -
k(a-
-- k),
k)-
k
l,
even
k odd,
i.e.,
h~(k)- 2 L"~-"~ 2 J. Note t h a t the m a x i m u m of ha(k) occurs at k -
~2
1, 7, 7 + 1~, ~
and the
m a x i m u m is ~2-4 4 " Proposition. if and only if
If s - m or s - m + 1, then 5(d) _< h n - l ( m -
d
-
(m-l)
d
-
(m -
~
1), with equality
formodd
1) n - 1 ( m -
2)
or
d - m(m
-
1) n - 1
f o r m even.
To prove the proposition, consider first the case t h a t s - m. If columns 1 , . . . , m - 1 are m a d e full in this order, then d stays in Dn and 5(d) increases at each step by L e m m a 15.4.17. For odd m, the resulting d - ( m - 1) n also belongs to E~, and so it is the only sequence of E~ satisfying s - m t h a t m a x i m i z e s 5(d). T h e m a x i m u m in this case equals ( m - 1 ) ( n - 1 - ( m 1)) hn-1 (m1 ) . For even m, the above d does not belong to E~, and so the only sequence of E~ satisfying s - m t h a t maximizes 5(d) is the previous sequence in the filling-up process, n a m e l y d - ( m - 1 ) ~ - l ( m - 2). T h e m a x i m u m in this case equals ( m - 1)(n - 1 - ( m 1 ) ) - 1 - h~-i (m - 1). Now consider the case t h a t s - m + 1. We assert t h a t if d'~ _> 2, t h e n there exists f C E~ such t h a t 5( f ) > 5(d), and consequently we m a y a s s u m e t h a t d'~- 1. To prove the assertion we distinguish two cases. ! C a s e 1" 2 _< d~ < m. P u t p - d~, r - d~, q - dp + 1. We m a y a s s u m e t h a t q < s, for otherwise the last two l's in the s-th column of d m a y be r e m o v e d w i t h o u t leaving E~, t h e r e b y increasing 5(d) by L e m m a 15.4.17. T h u s q - m. If r _< p - r , then the required f exists by L e m m a 15.4.27, so we a s s u m e t h a t r > p - r. Using dli > rn and di - rn for i -- 1 , . . . , r, d~ - m - 1 and di - rn ' for i - r + 1 , . . . , p, d~ - di - r n - 1 for i - p + 1 , . . . , m , din+ 1 - - p, d m + l - r and d} - 0 for i - rn + 2 , . . . , n , we obtain 5(d) - ~ ( d ' i - di) + p i=1
r.
(15.48)
15.4
Threshold
and Majorization
Gaps
433
Let f be a sequence such that C(f) is obtained from C(d) by (a) deleting the s-th column and (b) adding a 1 at the end of the (r + 1)-st column if p is odd. Then f C En by L e m m a 15.4.17 and 7-
5(f)
>
~-~(d'~- ( d ~ - 1)) i=1 r
=
r
i=1
>
5(d).
(by (15.48) a n d r > p - r )
C a s e 2" d~, - m. If m > 4, then the required f exists by L e m m a 15.4.29. Now assume the special case m - 3. Then d - 32d'2-21n-d'2-1 with d~ > 2 (see Figure 15.14).
Figure 15.14: Illustrating a special case in the proof of Theorem 15.4.25. m
* 1
1 1
1 1 * 1 1 0
1
1
m s
:
d;-t-1
*
:
1 1 1
n
8
1 1
0
1
If d~ - 2, then Sn(d) - n + 6 is odd, since n - 3 mod 4, contradicting d C En. Therefore d~ >_ 3. Hence 5(d) - ( n - 4 ) + + ( d ; - 3 ) + + ( 2 3) + + ( 3 - 2 ) + - n+d~-6 _< 2 n - 7 , and writing n - 4 k + 3 , we have 5(d) S ~ ;-
~
~
~bm
~.~
a
T
~
~
~
"-" ,_,..,
~
~ ~
m,(:bm
~
~
~
~.~
~
~
=
~
~
9
~-"
~
..
~
~
~.
--
o
i,..., o
p'1
9
0
t~
t~
16.3
Bidimensions
453
-- { ( a l , d l ) , . . . , ( a k , dk)} is monochromatic in is an edge of Hb(R), to obtain a contradiction. either ai = mi or aiRmi, and similarly either ni therefore from mi+lQnni and the transitivity of taken cyclically. Therefore S is indeed an edge of
Hb(R). We show Observe that for = di or niRdi. It Qn that ai+lRdi, Hb(R).
that S each i, follows indices ..
L e m m a 1 6 . 3 . 1 6 If B is a biorder extending a relation R, then there exists a minimal biorder B m such that R C B m C t3. m
P r o o f . Consider the family/3 of biorders B' satisfying R C B' C_ B, partially ordered by inclusion. By Hausdorff Maximal Principle, equivalent to the Axiom of Choice, the poset /3 has a maximal chain C. The intersection C - N{B " B C C} of all the biorders in C is also a biorder. To see this, assume that a, b C A and d, e C D are such that aCd, bCe, aCe, bCd. We must have able and bB2d for some B1,/32 C C. Since C is a chain, we have either Ba C_ B2 or B2 C B1. Assuming the former, we have aBld, bBle, able, bBld in contradiction to the fact that B1 is a biorder. Thus C is a lower bound for C, and by the maximality of C, C must be a minimal element of B. ,, L e m m a 1 6 . 3 . 1 7 Let R C_ A • D and let B be a minimal biorder extending R. Then,
1. B A R 2. B -
B-
RBD;
B B - 1 B C R.
P r o o f . 1" BA being reflexive, one has R C BAR and BAR C BAB C_ B. Hence, by the minimality of B, the equality BAR - B will be established if we prove that BAR is a biorder. Let a, b C A and d, e C D be such that aBaRd and bBARe. Since BA is a weak order we may assume that we have aBAb. Since BA is transitive (i.e., BABA C_ BA), this implies aBARe and proves that BAR is a biorder. The equality B - RBD can be proved in a similar way. 2" It is sufficient to notice that for every biorder B, if (a, d) G B - B B - 1 B , then B - {(a, d)} is also a biorder (otherwise abe, bBd, and bBe for some b C A and e C D, yielding (a, d) C B B - 1 B ) . " The following lemma states a much deeper property of minimal biorders.
454
Threshold Graphs and Order Relations
L e m m a 16.3.18 Let B be a minimal biorder extending a relation R. If R is transitive, then B is transitive. Furthermore, B has the same loops as R. P r o o f . The transitivity of B is a result of the following computation based on Part 1 of Lemma 16.3.17:
B 2 -(BAR)(RBD)-
BA(RR)BD C_ BARBD - BBD C_ B.
As for the loops of B, let us remark that, B being transitive, a B B 1Bd implies a ~ d. According to Part 2 of Lemma 16.3.17, this implies that all the loops in B are loops in R. 9 L e m m a 16.3.19 For every quasi order Q, we have dB(Q) - d(Q). P r o o f . Since every weak order is a biorder, we have dB(Q) _ d(Q), let us consider a family of biorders {B~" i C I} such that Q - Ni B~ and Ill- dB(Q). According to Lemma 16.3.16, for each i E I there exists a minimal biorder B ~ such that Q c_ B ~ C_ Bi. We have thus Q -~B?. (16.7) i
Using Lemma 16.3.18 we see that each B ~ is transitive and reflexive. B ~ is thus a weak order (see Proposition 16.2.1). The desired inequality follows from the fact that Q is, according to (16.7), the intersection of Ill weak orders. [] Theorem
16.3.20 For every relation R C A x D with A N D - 0, we have d . ( R )
-
-
-
P r o o f . Applying Part 1 of Lemma 16.3.12 to R and QR, we obtain the first and third equalities. The second equality follows from Lemma 16.3.15. The last equality follows from Lemma 16.3.19 and Proposition 16.2.9. .. T h e o r e m 16.3.21 For every strict partial order P, d s ( P ) - di(P) (i.e., the interval dimension of P). P r o o f . This is similar to Lemma 16.3.19 (remember that the interval orders are the irreflexive biorders). []
16.4
Relations of Bidimension 2
16.4
455
R e l a t i o n s of B i d i m e n s i o n 2
In this section we study the relations of bidimension 2. We characterize them and the uniqueness of their representation as the intersection of two biorders in terms of the graphs defined as follows. 16.4.1 Let R be a relation. The g r a p h a s s o c i a t e d w i t h R is the graph G(R) whose vertices are the ordered pairs in R and whose edges are the sets {(a, e), (b, d)} such that aRd, are, bRe, bRd. The b a r e g r a p h a s s o c i a t e d w i t h R is the graph Gb(R) obtained from G(R) by deleting every vertex not contained in an edge. Definition
It is easy to see that every coloring of H(R) is a coloring of G(R), hence x(G(R)) < x(H(R)). For some relations R we may have strict inequality, as shown by the example of Cozzens and Leibowitz in Section 6.1. Remark.
F a c t 16.4.2
1. The edges of Gb(R) are the edges of cardinality 2 of Hb(R); 2. Gb(R) and Hb(R) have the same vertices;
P r o o f . Part 1 follows from Part 1 of Lemma 16.3.11. Part 2 follows from Part 2 of Lemma 16.3.11. Part 3 follows from Parts 1 and 2. Lemma
16.4.3 For every relation R C_ A • D with A N D - 0,
P r o o f . This is similar to Lemma 16.3.15. 9 Let C(P), I(P) denote the comparability and the incomparability graphs of a poset P. This means that P is the vertex set of both C(P) and I(P), and xy is an edge of C(P) (resp. I(P)) if and only if xPy V yPx (resp.
-~(xPyVyPx)). A poset P has dimension at most 2 if and only if I(P) is a comparability graph.
Lemma
16.4.4 ( D u s h n i k - M i l l e r T h e o r e m )
456
T h r e s h o l d Graphs and Order R e l a t i o n s
P r o o f . " O n l y if"" Let P - L1 VI L2 where L1 and L2 are linear orders. Then L11 VI L2 is a transitive orientation of I(P). "If"" Let P* be a transitive orientation of I(P). Define L1 - P U P* and L2 - P U ( p . ) - l . Then it can be checked that L1, L2 are linear orders and L1 N L2 - P. L e m m a 16.4.5 ([GH64]) A graph G - (V, E) is a comparability graph if and only if G has no sequence v0,...,v2t of vertices with vivi+l E E and v~v~+2 ~ E for i - 0 , . . . , 2 1 (indices mod 21 + 1). The following theorem gives a convenient characterization of relations of bidimension 2. Theorem
16.4.6 For every relation R C_ A x D with A N D - 0, dB(R) _ X(Gb(R)), by Theorem 16.3.20 and Part 3 of Fact 16.4.2. Thus if dB(R) _< 2, then X(Gb(R)) _< 2. Conversely, we show that if X(Gb(R)) 2, I(P) is not a comparability graph and hence, by Lemma 16.4.5, there exists a sequence m 0 , . . . , m21 of elements of P such that {mi, mi+l} is an incomparable pair and {m~, m~+2} is a comparable pair of P for each i, with indices taken mod 21. Since P is reflexive and miPmi+l and mi+lPmi, the pairs (mi, mi+l) and (mi+l, mi) are adjacent vertices of Gb(P). Consider the cyclic sequence of the 21 + 1 vertices (m~, m~+l) of Gb(P). By virtue of rni+lPmi+l, the consecutive vertices (rn~, rni+l) and (m~+~, mi+2) are adjacent if and only if m~Pm~+2. If they are not adjacent, then mi+2Pmi and hence (mi+l, mi) and (mi+2, mi+l) are adjacent, in which case we insert them between (mi, mi+l) and (m~+l,m~+2), obtaining a path of length 3 between (m~, m~+l) and (m~+l, m~+2). Having done all the appropriate insertions, we obtain an odd cycle in Gb(P). []
16.4
457
Relations of Bidimension 2
D e f i n i t i o n 16.4.7 A graph G - (V, E) is u n i q u e l y o r d e r a b l e when IVI 1 or there exist exactly two (opposite) partial orders on V whose comparability graph is G. D e f i n i t i o n 16.4.8 Given a graph G, we say that two of its edges are dir e c t l y e q u i v a l e n t if they are of the form ran, rap with np not an edge. The transitive closure of this reflexive symmetric relation is an equivalence relation on the set of edges of G, whose equivalence classes are called the i m p l i c a t i o n c l a s s e s of G. Fact 16.4.9 Let P be a poset of dimension 2. Then P is uniquely representable as an intersection of two linear orders if and only if I(P) is uniquely orderable. P r o o f . This follows from the proof of Lemma 16.4.4. Fact 16.4.10 (see [Gol80]) A graph is uniquely orderable if and only if it is a comparability graph with only one implication class. L e m m a 16.4.11 A partial order P of dimension 2 has a unique representation as an intersection of two linear orders if and only if the graph Gb(P) is connected. P r o o f . From Lemma 16.4.4 and Facts 16.4.9 and 16.4.10, it is enough to show that I(P) has only one implication class if and only if Gb(P) is connected. Associate with each edge mn of I(P) the two adjacent vertices (m, n), (n, rn) of Gb(P). If two edges rnn and rnp of I(P) are directly equivalent, then { (n, m), (m, p) } or {(p,m), (re, n)} are edges of Gb(P). It follows that if I(P) has only one implication class, then Gb(P) is connected. On the other hand, if {(re, n), (p,q)} is an edge Gb(P), then rnn and pq are in the same implication class of I(P). Indeed, rnq and np are nonedges of I(P) by definition of a 2-alternating cycle, whereas rnn, pq and nq are edges of I(P) by the transitivity of P, and hence mn is directly equivalent to nq, which is directly equivalent to pq. It follows that if Gb(P) is connected, then I(P) has only one implication class. .' L e m m a 16.4.12 For every relation R C_ A x D with A N D -
1. if Gb(QR) is connected then Gb(R) is connected;
r
we have
458
T h r e s h o l d Graphs and Order R e l a t i o n s
2. if (re, n) and (a,d) are as in Definition 16.3.13, then mn and ad are edges of I(QR) belonging to the same implication class. P r o o f . 1" Let (a, d) and (a', d') be any two vertices of Gb(R). Since they are also vertices of Gb(QR),there exists a path (a, d) -- (?Ttl, rt 1), (m2, rt2),... , (?T/k, ~tk) --
(a', d')
in Gb(Qn). Replacing each (mi, ni) by the associated (ai, di) as in Definition 16.3.13, we obtain a path from (a,d) to (a',d')in Gb(R) by Part 2 of Lemma 16.3.14. 2: Since ( m , n ) i s a vertex of Hb(QR), we have mQnn and Qn has a 2alternating cycle including the pairs (m, n) and (rn', n') of Qn. By the transitivity of Qn we have nQnm. Therefore mn is an edge of I(QR). By Part 3 of Lemma 16.3.14, (a, d) is a vertex of Hb(R), and hence aRd and aQnd. By Part 2 of Lemma 16.3.14, R has a 2-alternating cycle including the pairs (a, d) and (a', d') of R, and hence by Proposition 16.2.9, we have dQna. Therefore ad is an edge of I(Qn). To show that mn and ad are in the same implication class of I(Qn), we consider separately the four cases of Definition 16.3.13. In Case a), ad - mn and there is nothing to prove. In Case b), we have nRd, hence nQnd, and so nd is a nonedge of I(Qn), and a d - md and rnn are directly equivalent. Case c) is similar to Case b). In Case d), we have mQnd by the transitivity of Qn and dQnm by Proposition 16.2.9. Therefore md is an edge of I(Qn). Moreover, am and nd are nonedges of I(Qn) by virtue of a r m and nRd. Hence rnd is directly equivalent to both mn and ad. 9 T h e o r e m 16.4.13 Let R C A x D with A N D - O and dB(R) - 2. Then
R is uniquely representable as an intersection of two biorders contained in R U R R - 1 R if and only if the graph Gb(R) is connected. P r o o f . Notice that by Part 2 of Lemma 16.3.12, R is indeed representable as the intersection of two biorders included in R t_JRR -1R. The only issue is the uniqueness of the representation. " I f " : Let R = B1 N B2, where each Bi is a biorder contained in R U R R -1R. Every vertex (a,d) of Gb(R) belongs to an edge {(a,d), (a',d')} of Gb(R). The biorder B~ must contain one of (a,d) and (a',d'), say (a,d). Then (a, d) C B~ - B2 and (a', d') E B 2 - B~. Thus every vertex of Gb(R) belongs to one of B1 - R and B 2 - R, and no edge meets both of them. In other words,
16.4
459
R e l a t i o n s of B i d i m e n s i o n 2
(B1 - R, B2 - R) is a proper coloring of Gb(R). Since Gb(R) is connected, the coloring is uniquely determined and with it B1 and B2. " O n l y if"" Assume that Gb(R) is not connected. Then by Part 1 of Lemma 16.4.12, Gb(QR) is not connected. Hence Gb(P) is not connected, where P is the poset defined by P - Q n / E and xEy if and only if zQny A yQnx. Hence by Lemma 16.4.11, P has two representations P - L1 N L2 = L i N L~, where Li and L~i are linear orders and L1, L i agree on one implication class of I(P) and disagree on another. It follows that Qn - W1 N W2 = W; N W~, where W~, W[ are weak orders and W~, W~ agree on one implication class of I(QR) and disagree on another. By Part 2 of Lemma 16.4.12, each implication class of I(Qn) has an edge ad with (a, d) e A x D. Therefore by Lemma 16.2.10, Wi N (A x D) and W[N (A x D) define two representations of R, as required. ,, The following theorem exhibits many ways of representing a relation R of bidimension 2 as the intersection of two biorders, under a certain condition. Theorem
16.4.14
Let R C_ A x D, A N D - 0 ,
and assume that R admits
no 3-alternating cycles. Then 1. d . ( n )
-
2. if d , ( R ) O. P r o o f . Assume that (f; t l , . . . , t i n ) iS a constant threshold representation for T ~. Since there is only a finite number of strictly positive values f ( x ) - f ( y ) tj, we can find an e > 0 that is a lower bound for all these values. We then have
xPjy ---.. f(x) >_ f(y) + tj + and also
xPjy - .
f(x) >_ f(y) - tj
Threshold Graphs and Order Relations
462
It follows that f and the following weights for G(T') satisfy Condition 1 of Lemma 16.5.4: w(x~ y) -- I tj--tj,-4- e, ifif xPjy xP~y. Therefore every cycle of G has a non-positive weight. Thus m
m
E pj(tj + ~)+ E p}(-tj) < O, j-1
j=l
which implies m
E qjtj > O. j=l
Conversely, since there is only a finite number of simple cycles from 7', we can find an e > 0 such that each cycle satisfies rn
m
E qjtj - e E p j >_ O. j=l
j=l
Using the same weights on G and taking the argument in reverse order, one derives the existence of a mapping f such that (f; t ~ , . . . , tin) is a constant threshold representation for 7'. 9
binary relation P is irreflexive and for all a, b, c, d C X,
Corollary 16.5.6 A
a
semiorder if and only if P is
(aPb A cPd) ~
(aPd V cPb),
(aPb A bPc) ~
(aPd V dPc).
Figure 16.~ shows the configurations forbidden by these conditions. P r o o f . The "Only if" part follows from the definition. To prove the "if" part, we assume that P does not have the configurations of Figure 16.4, and show that the 1-tuple (1) is a constant threshold vector for P. By Lemma 16.5.5, what we need to show is that q > 0 for every simple cycle from P. We show this by induction on the length of the cycle. The statement is true for cycles of length 1, 2 or 3 since the configurations G1, G2 and G3 of Figure 16.4 are forbidden. Let C be a simple cycle of length >_ 4 with q _< 0, if that were possible. If all arcs of C are in P, then we must have G3 or G2, a
16.5
Multiple Semiorders
463
Figure 16.4" The forbidden configurations for a semiorder.
G1
G2
G3
G4
G5
contradiction. Hence C has arcs in P'. Assume that C has two consecutive arcs in P. Then it has consecutive vertices a, b,c, d with aPbPcP'd, and hence aPd holds since Gs is forbidden. We can use the arc (a, d) to shortcut C and obtain a shorter cycle with the same q _< 0, contradicting the induction hypothesis. The only remaining possibility is that C has no consecutive arcs in P, and since q _< 0, its arcs actually alternate between P and P' and q = 0. Consider consecutive vertices a, b, c, d with aPbP'cPd. Then aPd holds since G4 is forbidden, and we can use the arc (a, d) to shortcut C and obtain a shorter cycle with q = 0, again contradicting the induction hypothesis. 9 The following theorem characterizes multiple semiorders. T h e o r e m 16.5.7 Let T' - ( P 1 , . . . , P r o ) be an m-tuple of binary relations on X , with m > 1. There is a constant threshold representation for ~P ( P 1 , . . . , Pro) if and only if no m-cyclone from 7:) is balanced. P r o o f . " O n l y if"" Assume that P admits a constant threshold representation ( f ; t l , . . . , t m ) . Then
xpjy ~
f ( x ) - f(y) > tj,
xpjy ~
f ( x ) - f(y) >_ -tj.
Threshold Graphs and Order Relations
464
If we consider any k-cyclone, write these implications for all its arcs and sum the resulting inequalities, we obtain m
m
m
o > E pJJ + E p}(-tJ)- E(pJ- p})tj. j=l
j= l
j=l
Moreover, the inequality is strict since at least one arc from some Pj is used by the cyclone. This clearly implies that pj r p} for at least one j. Hence the cyclone is not balanced. "If"" Assume that no cyclone from P is balanced. Observe that each Pj is either reflexive or irreflexive (otherwise we construct a balanced 2-cyclone by taking one loop in Pj and another one in P~). Consider first the case m - 1. If P1 is irreflexive, we show that the 1-tuple (1) is a constant threshold vector for P1. By Lemma 16.5.5, it is enough to show that q > 0 for each cycle. If C were a cycle with q _< 0, we could add Iq[ loops from P~ to obtain a balanced 1-cyclone, a contradiction. Similarly, If P1 is reflexive, then ( - 1 ) is a constant threshold vector for P1. Now assume that m >_ 2. Since there is only a finite number of values f ( x ) - f(y), we may always look for a nonzero threshold tl, and by an appropriate change of scale, even assume that tl - +1. We treat the case that P1 is irreflexive, setting tl - 1 (the other case is similar). By Lemma 16.5.5, we have to show the existence of a common solution ( t 2 , . . . ,try) to all the inequalities
q2t2 + ' " + qmtm > --ql associated with simple cycles from P. By Helly's Theorem (see [Chv83, p. 266]) applied to the convex sets defined by these inequalities in the Euclidean space of all ( m - 1)-tuples (t2, t 3 , . . . , tin), it is sufficient to show that every m of these inequalities have a common solution, say
qi2t2 + " " + qimtm > -qil
(16.9)
with i - 1 , . . . , m. By the Farkas Lemma, this is equivalent to the following assertion" given real numbers 11,..., lm with li >_ 0 for each i, and li > 0 for at least one i, m
-
i=1
0
j -
2,...,
(16.10)
16.5
Multiple Semiorders
465
implies m
liqil
> 0.
(16.11)
i=1 !
Since qij -- Pij - - P i j is an integer, we only have to verify the assertion for rational numbers l~ (because the real tuples ( / 1 , . . . , lm) satisfying l~ >_ 0 and (16.10) form a polyhedron with rational extreme points and directions). It follows that we need only consider li's that are natural numbers. Now each of the equations in (16.9), for fixed i, comes from a cycle Ci from 7). Assuming that li is a natural number, we consider the cycle C[ obtained by traversing the cycle Ci li times, and then the union U of these C~. Then U is an m-cyclone that uses ~i~__1lip}j arcs from P~ and ~i~=1 lipij arcs from Pj. Equation (16.10) tells us that for j - 2 , . . . , m , these two quantities are equal. Since by assumption U is not balanced, we deduce m
liqil 7~ O. i=1
In order to establish (16.11), and thus complete the proof, it remains to show that (16.10) together with m
liqil < 0
(16.12)
i=1
lead to a contradiction. If (16.12) were true, we would of course have one of the qil strictly negative (because all li are nonnegative), and thus the mcyclone U would use at least one arc (x,y) from P1. Now we form a new m-cyclone U' by adding m
I ~ liqil I i=1
times the arc (x, x) to U. Since P~ is irreflexive, (x, x ) i s in P~, so the new arcs increase the left-hand side of (16.12) to zero. Thus U' is a balanced m-cyclone, a contradiction. 9 Let us make the following remark. The proof above used only a restricted version of Helly's Theorem, namely the convex sets were open half spaces in I~ m-1 (a similar result for closed half spaces can be found in [Chv83, p. 146]). This restricted version follows easily from the Farkas L e m m a as follows. Assume that A x > b is unsolvable, where A has m - 1 columns. Then Am > tb, t > 0 is also unsolvable. Hence by the Farkas L e m m a there exist y > O, Yo > O, (y, yo) r (0, 0), satisfying y T A -- O , - - y r b + Yo -- O, and,
466
moreover, we m a y find such (y, eliminating yo, we are left with c o m p o n e n t s and satisfying y r A most m inequalities among Ax
Threshold Graphs and Order Relations
Y0) with at most m positive components. By a non-zero y >_ 0 having at most rn positive -- 0, yTb >_ O. Therefore a subsystem of at > b is unsolvable.
Chapter 17 Enumeration 17.1
Introduction
In this chapter we look at enumeration results of some families of graphs. For convenience we assume that all our graphs have the vertex set N = {0,...,n1}. Recall that two graphs G = (N,E) and H = (N,F) are the same unlabeled graph when they are isomorphic, i.e., when there is a permutation 7r of N such that ij C E if and only if ~r(i)Tr(j) e F. We say that G and H are the same labeled graph when E = F. Many enumeration results are expressed in terms of ordinary and exponential generating functions. The ordinary generating function (ogf) of a sequence a0, al, a 2 , . . . is ~ > 0 a~ x~ and its exponential generating function (egf) is ~ > 0 anxn/n!" The elementary properties of these functions are discussed in most introductory combinatorics books. In this chapter we have occasion to use the Burnside Lemma and other basic enumeration results, which can be found in the same books. We have already seen in Section 13.4 and 13.5 that decomposition techniques can be used to find the egf of the number of unlabeled box-threshold, matroidal and matrogenic graphs. See also Section 12.4 for the egf of the number of unlabeled domishold and hereditary pseudodomishold graphs. In Section 17.2 we discuss enumeration of threshold graphs, both unlabeled and labeled. It turns out that the enumeration of unlabeled threshold graphs can be used to derive the q-binomial theorem, and the enumeration of labeled threshold graphs can be used to derive a theorem of Frobenius connecting the Eulerian polynomial with Stirling numbers of the second kind. In 467
468
Enumeration
Section 17.3 we enumerate difference graphs, both unlabeled and labeled. Recall that difference graphs are bipartite. We refer to a bipartite graph G = (N, E) together with a fixed partition of N into two stable sets as bipartitioned. Section 17.3 discusses labeled and unlabeled difference graphs, either bipartitioned or not.
17.2
E n u m e r a t i o n of Threshold Graphs
We begin with the enumeration of unlabeled threshold graphs, based on Peled [Pel80]. Denote by r,~mk the number of unlabeled threshold graphs having n vertices, m edges, and m a x i m u m clique size k, and consider the generating function
Tn(x, q) - ~ 7"nmkqmxk. m,k We evaluate theorem. Lemma
T~(x, q) in two ways, and compare them to obtain the q-binomial
17.2.1
Tn(x, q) = x(1 + qx)(1 + q2x)... (1
+
qn-lx).
(17.1)
P r o o f . For each binary vector u = ( u 0 , . . . , Itn-1) ~ {0, 1}" with u0 = 1, we construct a graph G(u) - ( N , E ) on the vertex set N - { 0 , . . . , n 1} where E is such that for all i < j we have ij E E if and only if uj = 1. By Condition 4 of Theorem 1.2.4, G(u) is a threshold graph for each u, and every threshold graph can be obtained in this way. Moreover, G(u) is isomorphic to G(v) if and only if u - v. Indeed, the "if" part is obvious. Conversely, assume that u r v yet G(u) is isomorphic to G(v), and let j be the largest subscript such that uj --fi vj. Then G(u') is isomorphic to G(v'), where u ' = ( u 0 , . . . , uj) and v ' = ( v 0 , . . . , vj), since removing an isolated or a dominating vertex does not affect isomorphism. But this is impossible, since one of G(u') and G(v') has an isolated vertex and the other one does not. From this we already see that the number of unlabeled threshold graphs on n vertices is 2 "-1. The number of edges of G(u) is ~2j=0 n-1 jUj. Further, the set K - {j 9 uj 1} is a m a x i m u m clique of G(u). Indeed, K is clearly a clique containing 0, and consider any other clique K ~ % K. Then K ~ contains some element i > 0
17.2
E n u m e r a t i o n of T h r e s h o l d Graphs
469
with ui = 0, and so i is adjacent only to elements j C K such that i < j, i.e., to fewer than IKI elements. Hence IK'I _< IKI. We conclude that the unlabeled graphs enumerated by Tnmk are in oneto-one correspondence with the binary sequences u such that n-1
u0-1,
n-1
m-Eju
j,
k-Eu
j =o
j,
j =o
and (17.1) follows. 9 We remark that by putting q = i in (17.1) and extracting the coefficient of x k, we find that the number of unlabeled threshold graphs having n vertices n--1 and maximum clique size k is ~m r~,~k - (k-l)" This result can also be obtained by counting symmetric corrected Ferrers diagrams having n rows and corrected Durfee number k (Definition 3.1.6). The diagram is determined by the sequence of its first k row sums, which are non-increasing, do not exceed n - 1, and the last of them is equal to k - 1. The number of such [ n-1 \ sequences is [k-l), as can be seen by standard techniques. The next lemma makes use of the Gaussian polynomials [n.]_
[3J
(q)~
,
where(q)n-(1-q)(1-q2).-.(1-qn).
(q)j(q)n-j
L e m m a 17.2.2 T~(x,q) -
.
J
1
"
P r o o f . If G is a split graph having a split partition (K, S) with K a clique and S stable, and if K is a maximal clique, then K is in fact a maximum clique, since no vertex of S can be adjacent to all the vertices of K. If in addition G is a threshold graph, then by Condition 3 of Theorem 1.2.4, G is determined up to isomorphism by n, IK[, and the degrees of the vertices of S. Hence the unlabeled threshold graphs enumerated by rnmk are determined by the partitions of the integer m - (k2) (the number of edges outside IKI)into at most n - k positive parts (the positive degrees in S) not larger than k - 1. It is well-known [And76, p. 33] that the ogf for the number of partitions of integers into at most A positive integers not larger than B is [A+B]. Therefore rn~nk is the coefficient of qm-(~) in In-l], k-1 which is equal to the coefficient of nl q m x k in t h e s u m ~ j =
[ j-1 n-l] q (~)xj 9
This proves 17 929
"
Enumeration
470
By comparing (17.1) with (17.2), changing the index of summation and increasing n by 1, we obtain the following. C o r o l l a r y 17.2.3 (The q - B i n o m i a l T h e o r e m )
(1 + x)(1 +
qz)...
(1 +
q~-ax) - j~o
q(~)xJ"
(17.3)
Identity (17.3) goes back to Cauchy [Cau93] and probably to Euler. Many important identities are easily derivable from (17.3) and vice versa. For example, by letting n --+ cx~, we can obtain the following formal-power-series identity of Euler" (1 - x)(1
- qx)(1
-
q2x) . . . .
q('~)(-x)J
j~0 (1 - q)(1 - q~ :- (1 -
qJ)"
(17.4)
From (17.4), again using (17.3), we obtain the companion identity of Euler:
1 (1 - x)(1
- qx)(1
oo -
q2x)...
xj
= j~0 (1 - q)(1 - q2)...
(1 -
qi)"
(17.5)
From (17.4) and (17.5) we can get, again using (17.3), the following identity of Heine: (1 - ax)(1 - qax)(1 - q 2 a x ) . . . = ~-,~ (1 - a)(i - qa~_2..!l_. - qJ-la)x j (1-x)(1-q/)(1-q2x)... ~o ( 1 - q ) ( 1 - u --). ( 1 - q J ) (17.6) Andrews [And76] derives many series-product identities from (17.6), including the q-binomial theorem itself. Knuth [Knu71] discusses an interesting connection between the fact that [~] is the number of k-dimensional subspaces of GF(q) ~ and the fact that it is the egf for the number of partitions of integers into at most k parts not exceeding n - k. The connection is via matrices in row-echelon form, whose shape can be interpreted as a threshold graph. We now present the work of Beissinger and Peled [BP87] on enumerating labeled threshold graphs and the identity of Frobenius. Let tnjk denote the number of labeled threshold graphs with n vertices, j isolated vertices, and exactly k distinct vertex-degrees. For convenience we set tojk = ~jo~ko,
(17.7)
17.2
E n u m e r a t i o n of T h r e s h o l d G r a p h s
471
where ~ is the Kronecker symbol. We denote by
tn - ~ tnjk
(17.8)
j,k
the total number of labeled threshold graphs on n vertices. We determine the generating function n
F(x,y,z)- ~ t~jk~.yJz k.
(17.9)
n,j,k
Theorem
17.2.4
The generatingfunction defined by (17.9) is given by 1/z+e~Y-1 F(x,y,z)- (1 - xz)1/z - e~ + 1" (17.10)
P r o o f . It is clear that for n >_ 2 one has t~jk - 0 unless 0 _< j _< n and 1 _< k _l
(17.11)
since any number of isolated vertices can be added to or removed from a threshold graph without affecting its thresholdness, as follows from Condition 4 of Theorem 1.2.4.
tnok- ~tnjk
for n >__2
(17.12)
j>__l
because, by Theorem 1.2.4, the complement of a threshold graph with n >_ 2 vertices, no isolated vertices, and k distinct vertex-degrees is a threshold graph with n vertices, one or more isolated vertices, and k distinct vertexdegrees. Equations (17.11) and (17.12) can be combined to obtain the recurrence
The boundary conditions (17.14) and (17.15) below apply to all classes of graphs, not just to threshold graphs. t~,~_,,k - 0
(17.14)
472
Enumeration
tnnk -- 5kl
(17.15)
for n > 1.
As an aid in determining F(x, y, z), consider the generating function
XnZk
yj(x, z) - E E t~j~ ~!
(17.16)
n>0 k>0
Then r 0 ( x , z) Xn
E E t~o~-fzk n)>0 k > 0
Xn
= 1 + o + E E t~o~., Z k
by (17.7),(17.14)
n>2 k>0
Xn
= 1+ E E Et~J~-~z k
by (17.12)
n>_2k>_Oj>_l
X~zk
= 1 + ~ ~ ~ ~njk n! j>_l n>_2 kkO
= 1 + ~-~ ( F j ( x ' z ) - O>_k~ t ~
k>_O ~-'~tljkxzk
-- l-Jr-j~>l ( F j ( x ' z ) - - E ~ j Ok>_Oe ~ k o Z k - -
by (17.16)
k>_O
by (17.7),(17.14),(17.15). Thus
Fo(x, z)
-
~
j>l
Fj(x, z)+ 1 - xz.
(17.17)
It follows from (17.11) and (17.16)that
xj Fj(x, z) - z~Fo(x, z) for j _> 1.
(17.18)
By summing (17.18) for j > 1 and using (17.17), we obtain
Fo(x,z)- 1 + xz - z(e ~ - 1)F0(x,z) and therefore F 0 ( x , ~) -
1-xz 1 - ( e ~ - 1)z
x-l/z e~ - (1 + l/z)"
(17.19)
17.2
E n u m e r a t i o n of T h r e s h o l d Graphs
473
By (17.18) we have
F(x,y,z)-
~ Fj(x,z)y j - Fo(x,z)(1 + z(e ~ y - 1)) j>o
and (17.10) follows. 9 To expand functions with denominators like that of (17.10), we use the Eulerian polynomials An(u), defined as follows (cf. Comtet [Com74] or Riordan [Rio58]): u-
1
U -- e v(u-1)
1
v~
U n>l
n'
= 1 + - ~ An(u)--r,
A o ( u ) - 1.
(17.20)
One combinatorial interpretation of the Eulerian polynomials is that for n _> 1, the coefficient of u m in An(u) is the number of permutations 7rl,..., 7r~ of 1 , . . . , n having m - 1 ascents, that is to say, positions j such that 7rj < 71"j+ 1 [GKP89, Equation (7.56)]. By substituting u - 2 in (17.20), we obtain 1
Xn
-
2
~
(17.21)
gn
n>O
where gn - A~(2)/2 for n >_ 1 and go - 1. The well-known numbers g~ were studied by Cayley in the 19th century and several references to them can be found in Knuth [Knu73] and Good [Goo75]. gn is the number of ordered partitions of the set { 0 , . . . , n - 1}, that is to say, the number of ways to partition the set into non-empty blocks and to arrange the blocks in a sequence. Equivalently, g~ is the number of those n-letter words over the alphabet { 1 , 2 , . . . } that for some k contain precisely the letters { 1 , . . . ,k}, repetitions allowed. To see this interpretation of gn, we only have to write 1
2-e ~
=
1
1 - ( e ~ - 1)
=
~-~(eX
-
1) k
k>0
and note that (e ~ - 1)k is the egf for the number of words over the alphabet { 1 , . . . , k} with each letter appearing at least once. Because of this interpretation of g~, we have
g n - ~ kk=0 !
nk '
(17.22)
474
Enumeration
where (~} denotes the Stirling number of the second kind, i.e., the number of partitions of { 0 , . . . , n - 1} into k unordered non-empty blocks. This agrees with the well-known expansion (see Comtet [Com74] or Riordan [Rio58])
1)
E k>0
"
k
"
Knuth [Knu73, Ex. 5.3.1.4] presents the following asymptotic for gn: g--5~-1 ( 1 ) nl9 -- 2(ln2)n+ 1 + k>, ~ Re (ln2 + 27rik) ~+1
n > - 1"
(17.23)
It is not hard to use (17.23) to obtain that
1
(1)
We can now enumerate the labeled threshold graphs. T h e o r e m 1'/'.2.5 The number of labeled threshold graphs on n vertices is given by 1, if n - 0,1 t~ 2(g~ - - ngn-1), if n _> 2. (17.24) P r o o f . According to (17.9) and (17.8), F(x, 1, 1) = E~>0 tnxn/n!" evaluate F(x, 1, 1)using (17.10), we obtain (1 - x)e ~ (2 2-e ~ -(l-x)2-e
) ~
1
If we
x~
-l-at-x-~-2E(gn-ngn_l)----(, n>2
~"
and the result follows. .. We now see a combinatorial argument that gives another proof of Theorem 17.2.5, as well as of the theorem of Frobenius. We denote by Tnk the set of labeled threshold graphs with the vertices { 0 , . . . , n - 1}, no isolated vertices, and exactly k distinct vertex-degrees, so that t~0k = IT~kl.
(17.25)
We also denote by Dnk the set of ordered partitions ( B 1 , . . . , Bk) of { 0 , . . . , n 1} into non-empty blocks B~ such that IB~] >_ 2. It is easy to see that
,D~k, - k ' { nk} - n(k - 1)'{k " - 1 1} "
(17.26)
17.2
475
E n u m e r a t i o n of Threshold Graphs
L e m m a 17.2.6 There exists a bijection between Dnk and Tnk. P r o o f . Both sets are empty if n - 1, so we may assume that n > 2. Given a partition d - ( B 1 , . . . , Bk) E Dnk, we construct a labeled graph G on the vertices 0 , . . . , n - 1 by joining each vertex of Bi to all the other vertices of B1 U - . . U Bi if i + k is odd, and to none of them if i -+- k is even. Then G is a threshold graph by Condition 4 of Theorem 1.2.4, and it has no isolated vertices, since the vertices of Bk are dominating. We show that G has exactly k distinct vertex-degrees by showing that two vertices v C Bi and w C Bj have the same degree if and only if i - j. The "if" part is obvious. Conversely, assume that i < j. If i + j is even, then i -t- 1 < j and every vertex of Bi+l is adjacent to exactly one of v and w, hence v and w have different degrees. If k + i is even and k + j is odd, then v has some neighbor in B1U---UB~ (because IB~I > 2), whereas w has no such neighbor; therefore v and w again have different degrees. A similar argument shows the same conclusion if k -4- 1 is odd and k -4-j is even. Therefore G C Tnk. Given G C Tnk, we construct an ordered partition d as follows. Let G1 - G. If Gi has been constructed as a threshold graph with two or more vertices, then it has either isolated vertices or dominating vertices, but not both, and we remove all these vertices to obtain Gi+l. By an argument similar to the one in the previous paragraph, we can show that two vertices of G have the same degree if and only if they are removed at the same time. Since G has exactly k distinct vertex-degrees, the last non-empty graph is Gk, and it has at least two vertices. We put the vertices of Gi - Gi+~ into the block Bk+~-i for i -- 1 , . . . , k and obtain an ordered partition d - ( B 1 , . . . , Bk) C Dnk. It is easy to see that the two mappings defined above between Dnk and Tnk a r e inverses of each other, and the proof is complete. 9 The second proof of Theorem 17.2.5 can be seen as follows. From (17.25), (17.26) and Lemma 17.2.6, we obtain " k
"
1} 1
(17.27)
for n >_ 1, and this equation also holds for n - 0 by our convention that took - ~k0. If we sum (17.27) over k, the left-hand side becomes t~/2 for n >_ 2 by (17.12), the right-hand side becomes g n - ng~_~ by (17.22), and (17.24) results. The reason we presented the first proof of Theorem 17.2.5 is that it enables us to prove the theorem of Frobenius [Frol0] (see also Comtet [Com74]),
IV
0
~.
I
....
I
._
~ ~,
~
I
I
~
~
IV
I
I
I
I
I--t,
~
~
I ~
oc--F-'
0
~
"~
i,~
~
~
9
~ ~
~
,~ o
o
~
I
CD
0
~
~.~
t'~
IV ~
I
9
~
.-
I
~
~
I
'ov
o
~
I I
;._,
o
~ ~
I
"--,.1
-~
0
~
O
c--c-
0
I
"
O0
-q
9
II
'--'
~1~
+
II
I
~ I
I
I
II
I
c~
I
~-~
I
o~
9
II
I
II
~
~
~
:~
~9
0
~ 9
~o~
IV
~
'ov
o ~ "~
i.-~~
~
0
~
~
o-'
~'~
~-
~
~
~
~
l:r'
0
eo
~
O~ ~
I
~
~
~. .
to
~
~
0 ~ o0
0~
0
""
~
0 , . ( k " k+l
}
-
"
k + l
'
we find
xs
1 + E ~2Ej! s>_l
" j_>0
{;} -
E k>0
([e~( e ~ - 1)]k) ~ - E k_>0
~.k!
{r + k -~
and therefore 2
j>0 J!
{ ~ } _ ~ > 0 ( : ) k ~ > ok?2 { r + l } { s - r + l } k+ 1 k+ 1 _ _
T h e o r e m 17.3.2
s>l.
The number of labeled difference graphs on n vertices is
(l+gn)/2. Proof. By Theorem 2.4.4 every difference graph consists of zero or more isolated vertices and a connected difference graph, and conversely. As in Theorem 17.3.1, the egf for the number of labeled non-empty bipartitioned connected difference graphs is ~k>l(e ~ - 1) 2k. Hence the egf for the number of labeled non-empty connected difference graphs is 71 E k ~ l ( ex - 1) 2 k where 1 the factor g comes from exchanging the two color classes. By adding 1 we also include the empty graph, and by then multiplying by e~ we allow for the
17.3
E n u m e r a t i o n of Difference G r a p h s
479
isolated vertices. Thus the egf for the number of labeled difference graphs is given by
(
e~ l + a . y -k>l ~(e~_l)2k 1
= z e~
(1+
) 1(
)
_ 2 e~ 1 + k>0 ~(eX-
1 )l(e~ 1 _ (e~_ 1)2 - ~
-+
1) 2k
1 ) 2_e~
and the number of labeled difference graphs on n vertices is the coefficient of W., namely (1 + gn)/2. " In the rest of this section we consider the enumeration of unlabeled difference graphs. We denote by | the set of unlabeled threshold graphs on n vertices, and by A, the set of unlabeled bipartitioned difference graphs on rz vertices. By (17.1) we have I O n l - 2n-l, as already noticed. We use the following connection between IZX~Iand IO~l. X n
L e m m a 17.3.3 [An[- IOnl + IAn_ll
7/
>_ 1.
Proof. Consider the mapping f 9 An -+ | given by f ( D ) - T, where D - (X,Y; E) C A~ is a bipartitioned difference graph with bipartition (X, Y), T - ( X U Y, E U E x ) , and E x is the edge-set of the complete graph on X. By Theorem 2.1.9 f is onto | but it is not one-to-one. To find f - x ( f ( D ) ) , we note that T and D are determined by their degree partitions according to Theorem 1.2.4 and Theorem 2.4.4. In T, the degree of every vertex of Y/ is [Xk U . . - U Xk-i+x[ and the degree of every vertex of Xi is IYk tO... U Yk-i+x [q-IX0 U . . . LI Xk[- 1. Since Xi and Y/are non-empty for i >_ 1, it follows that the degree partition of T is simply (Y0,..., Yk) unless one of the following conditions holds in D" 1. x 0 - o , IYkl - 1, in which case the degree partition of T is ( Yo , . . . , rk-1,
Yk
I,.,J X l
, X2 , . . . , Xk
);
2. I X 0 [ - 1, in which case the degree partition of T is ( ro , . . . , rk - l , Yk
[,-J X o
, X l , . . . , X k).
It follows that f - l ( f ( D ) ) - {D} unless Condition 1 or 2 holds for D, in which case f - l ( f ( D ) ) - {D,D*}, where D* is obtained from D by moving
480
Enumeration
the singleton Yk to X in Condition 1 and the singleton X0 to Y in Condition 2. Note that if Condition 1 holds for D, Condition 2 holds for D* and conversely. It follows that IOnl - Izx f- Izx'~l, where A" denotes the set of unlabeled bipartitioned difference graphs satisfying Condition 1 or 2. We complete the proof by indicating a bijection from A~ to An-l- It is given by deleting the singleton Yk in Condition 1 and the singleton X0 in Condition 2. The resulting bipartitioned difference graph D' has isolated vertices in X in Condition 1 and does not have them in Condition 2. Therefore we can uniquely reconstruct D from D' by adding a singleton Yk to Y if D 1 has isolated vertices in X and adding a singleton X0 to X if D' has none. This shows that indeed we have a bijection. 9 Theorem
17.3.4
The number of unlabeled bipartitioned difference graphs
on n vertices is 2 n. P r o o f . By Lemma 17.3.3 and the fact that IO~1- 2 n-l, we have [ A n [ - 2 n-1 q- 2 n-2 + " "
q- 20 + [ A o [ - 2n - 1 + 1 - 2 ~.
Another way to see this result is to modify the bijection between | and the set of binary words of length n - 1 into a bijection between A~ and the set of binary words of length n. We omit the details. 9 17.3.5 The number of unlabeled difference graphs on n >_ 1 vertices is 2~-2 + 2[~J -1.
Theorem
P r o o f . Again, according to Theorem 2.4.4, a difference graph consists of zero or more isolated vertices and a connected difference graph. Hence the required number is ~ - - 0 c~, where c~ is the number of unlabeled difference graphs with no isolated vertices on s vertices. According to Theorem 2.4.4, a bipartitioned difference graph with no isolated vertices is determined up to isomorphism by the cardinalities of the blocks X 1 , . . . , Xk and Y1,..., Yk, so the number of unlabeled bipartitioned difference graphs with no isolated vertices on s vertices is equal to the number of compositions of s into an even number of parts, i.e., sequences ( x l , . . . , x k , Y l , . . . , Y k ) of positive integers whose sum is s. For s _> 2, there a r e (;[21) compositions of s into 2k parts, and therefore a total of ~k (~k-~l) - 2s-2 compositions of s into an even number of parts. Since we are counting non-bipartitioned difference graphs
17.3
481
E n u m e r a t i o n of Difference Graphs
with no isolated vertices, Cs equals the number of inequivalent compositions of s into an even number of parts, where the composition ( x l , . . . , xk, yl,. 9 9 yk) is considered equivalent to itself and to ( y l , . . . ,xk, zl,... ,xk). We find c~ by the Burnside Lemma. All 2 ~-2 compositions into an even number of parts are invariant under the identity permutation. A composition into an even number of parts is invariant under the permutation 7r that switches color classes if and only if it has the form (if:l,...,3g.k, X l , . . . , X k ) , and ( x a , . . . , x k ) must then be a composition of s/2. Hence there are (~s even)compositions of s into 2k parts that are invariant under 7r, where the Iversonian (s even) stands for 1 if s is even and for 0 otherwise. By summing over all k >_ 1, we find that there are 2~/2-1(s even) compositions of s into an even number of parts that are invariant under 7r. Hence by the Burnside L e m m a c~ = 89(2 ~-2 + 2~/2-1(s even)) for s >_ 2. Clearly Co = 1 and cl - 0. therefore the number of unlabeled difference graphs on n >_ 2 vertices is given by n
~ - ~ Cs s~O n
1+ 0+ ~
(2 s-3 +
2s/2-2(s
even))
s~-2 n
= 2+ ~
(2 s-3 + 242-2(s even))
s----3
=2+2~-2-1+
(1+2+4+...2[~J
-2)
- 1 + 2 '~-2 + 2[~J -~ - 1 - 2 ~-2 + 2[~J -~. This value is also correct for n - 1.
9
Chapter 18 Extremal Problems 18.1
Introduction
Consider the following problems. How many connected threshold subgraphs can a graph have? What graphs with n vertices and rn edges have the largest number of connected threshold subgraphs? These problems still remain open. Paul ErdSs conjectured that for a fixed number of vertices, the complement of a matching contains the largest number of threshold subgraphs (private communication). Erd6s et al. [EGOZ89] consider the following problems. Given a graph G with n vertices and m edges, what is the largest number of edges that a chordal subgraph of G can have? What about an interval subgraph or a threshold subgraph? We report the results of these authors on interval subgraphs and threshold subgraphs in Section 18.2. Boesch et al. [BBB+90] considered the following problem. Which graphs on n vertices and m edges maximize the sum of squares of the degrees? It turns out that these graphs must be threshold graphs. However, not every threshold graph maximizes the sum of squares of the degrees. Only partial results are known in characterizing these threshold graphs. We report these results in Section 18.3. 483
484
Extremal Problems
18.2
Large Interval and Threshold Subgraphs of Dense Graphs
Given a graph with n vertices and m edges, what is the largest number of edges a chordal subgraph can have? What about an interval subgraph or a threshold subgraph? ErdSs et al. [EGOZ89] studied this problem for m >_ n2/4 + 1. They gave a complete answer for chordal subgraphs, and partial answers for interval and threshold subgraphs. We report their results for interval and threshold subgraphs. All results in this section are from [EGOZ89]. Let G be the class of graphs with n vertices and m edges, and 7 / b e a class of graphs closed under vertex deletion. Then there is a number f such that every member of ~ contains a member of 7-/with at least f edges, and there is a member of G containing no member of 7t with more than f edges. In practice, f may be hard to find, and it may be possible to bound it by finding a lower bound fl (there must be a member of 7-t with at least fl edges) and an upper bound f2 (there need not be a member of 7-t with more than f2 edges). The questions above are only interesting if the graphs contain sufficiently many edges. The complete bipartite graph Kn/2,n/2, for example, has no triangles and thus contains no interval graphs larger than a path ( n - 1 edges) and no threshold graph larger than a star (n/2 edges). We only consider graphs with n vertices and at least n2/4 + 1 edges (sufficient to guarantee the existence of a triangle), and refer to them as dense graphs. We define the size of a graph as the number of edges in the graph. The results in this section can be summarized as follows (all graphs are assumed to be dense graphs)" 9 Every graph has an interval subgraph of size > (1 + c)n. 9 There is a graph with no interval subgraph of size > 3n/2 - 1. 9 Every graph has a threshold subgraph of size > (1 + c)n/2. 9 There is a graph with no threshold subgraph of size > (1 + 0.5)n/2. The two constants c remain to be determined. By the star of a vertex a we mean the graph consisting of all vertices in the closed neighborhood of a, along with the edges joining a to all its
18.2
Large Interval and Threshold Subgraphs of Dense Graphs
485
neighbors. By the star of an edge ab we mean the union of the star of a and the star of b. We call a graph G an edge-star or an e-star if G has an edge ab such that G is the star in G on the edge ab. We need the following Lemmas. L e m m a 18.2.1 Let G be a threshold graph with no K4. Then G has an edge ab (called the r o o t edge) such that every edge of G is either incident to a (the r o o t v e r t e x ) , or is of the form bc for some c such that abc is a triangle of G. Proof. Let a, b be the first two vertices in the non-increasing order of degrees in the threshold graph G. " It is a known principle that if a graph has average degree A, it has a vertex of degree considerably larger than A or almost no vertices of degree much below A. The next lemma makes this more formal. L e m m a 18.2.2 For every 1 > e > 0 there is a d > 0 (depending only on e) such that every graph with n vertices and average degree A has a vertex of degree exceeding (1 + d)A or fewer than en vertices of degree at most (1 - e)A. Proof. Choose d < e2/(1 - e ) . Suppose the conclusion is false. Then the en smallest degrees do not exceed ( 1 - e)A each, for a total not exceeding e ( 1 - e)An, and the remaining ( 1 - e)n degrees do not exceed (1 + d)A each, for a total not exceeding (1 - e ) ( 1 + d)An. Thus the grand total degree for the whole graph is at most (1 + d - e 2 - ed)An < An, contradicting the fact that the total degree is An. 9 Another commonly-used fact is that in the neighborhood of the largestdegree vertex there must be a vertex of "not too small" degree. The next lemma states this more formally. L e m m a 18.2.3 Let G have average degree exceeding n/2. Let the largestdegree vertex of G have degree (1 + c)n/2. Then in its neighborhood there must be a vertex of degree exceeding
1-c+
2c 2 ) n i + c 2"
Proof. If not, then each of the (1 + c)n/2 vertices in the neighborhood has degree not exceeding that, while each of the (1 - c ) n / 2 vertices not in that
Extremal Problems
486
neighborhood has degree not exceeding (1 + c)n/2. Again, the total degree of all the vertices is no more than n2/2, a contradiction. .. Yet another general result is that if not too many vertices of small degree are deleted from a dense graph, the resulting graph is still dense. The next l e m m a formalizes this. L e m m a 18.2.4 Let 1 > e > O. If G is a dense graph on n vertices, and at most en vertices of degree at most ( 1 - e ) n / 2 are deleted from G, the resulting graph is still dense. P r o o f . The number of deleted vertices is k < en and the total number of edges deleted is at most k(1 - e ) n / 2 . The number of remaining edges is therefore at least
~ -~-+ 1 - k(1-
n
( ~ - k) ~
e)~ -
4
k(2~+ 1+
4
k) > ( ~ - k) ~ -
4
+ 1.
L e m m a 18.2.5 ([Edw]) If G has n vertices and at least n2/4 + 1 edges, then G has an edge that is on at least n/6 triangles. T h e o r e m 18.2.6 There is a constant c > 0 such that for sufficiently large n, every dense graph on n vertices has an interval subgraph with at least (1 + c)n edges. P r o o f . Let e - 0.1 and d - 0.01. If there is a vertex with at least (1 + d ) n / 2 neighbors, then the largest-degree vertex a has degree (1 + f ) n / 2 , where d _< f < 1. Then by Lemma 18.2.3, a has a neighbor b with degree exceeding (1 - f + 2_]_!_~ ~ The e-star on edge ab is an interval graph with more than l+fJ~" d~ < -l+f ~ this (1 + -l +~f ) n 1 edges. For sufficiently large n ~ and for c -~(1-r
n-en 6
2 = 5 ~(1 - r
n ->(l+c)g'
4e)/3.
F a c t 18.2.8 There is a dense graph on n vertices whose largest interval subgraph has exactly (3/2)n - 1 edges. P r o o f . Consider the dense graph G obtained by adding an edge ab to the complete bipartite graph Kn/2,n/2. Let H be any interval subgraph of G. Delete one vertex from H, making sure it is a or b if H contains one of them. The resulting induced subgraph of H is still an interval subgraph of a bipartite graph on n - 1 vertices, and hence is a forest and has at most n - 2 edges. Adding back the star on the deleted vertex adds at most (n/2) + 1 edges, showing that H has a total of at most (3/2)n - 1 edges. On the other hand, let abc be any triangle in G. It is easy to see that the union of the stars of a, b and c is an interval graph with exactly (3/2)n - 1 edges. .. F a c t 18.2.9 For n sufficiently large and any fixed e > O, there is a dense graph on n vertices having no threshold subgraph with at least (1.5 + e)n/2 edges. (Hence the c in Theorem 18.2.7 cannot exceed 1/2.) P r o o f . Let A and B be vertex sets of size (1 + c)n/2 and ( 1 - c)n/2 respectively and add all ( 1 - c 2 ) n 2 / 4 edges from A to B, where c = 1/3. Now divide A into two equal parts A1 and A2 and add (cn)2/4 + 1 edges between those two parts, distributed as regularly as possible. The degree of a vertex in B is simply (1 + c)n/2, while each vertex in A1 has (1 - c)n/2 edges leading to
488
Extremal Problems
B and at most just over ((cn)2/4)/((1 + c)n/4) = nc2/(1 + c) edges leading to A2; so no star can be a big enough threshold graph. Consider a threshold subgraph containing a triangle. The graph is tripartite so has no K4; hence by Lemma 18.2.1 it has a root vertex and a root edge. We distinguish three cases: (i) root edge from A1 to A2, root vertex in either set; (ii) root edge from B to A1, root vertex in B; (iii) root edge from B to A1, root vertex in A1. All other cases are equivalent to these. (i) The root edge is from A1 to A2 and lies on at most ( 1 - c)n/2 triangles (the other vertex of the triangle must be in B); the root vertex has in addition about nc2/(1 +c) edges from A1 to A2. The total size of this threshold graph is about (1 - c)n + nc2/(1 + c), which equals (1.5)n/2 (since c = 1/3). (ii) The root edge joins B to A1. It lies on about c2n/(1 + c) triangles, and the degree of the vertex at B is at most (1 + c)n/2, of which c2n/(1 + c) edges have already been counted. The total size of this graph is thus about (1 + c)n/2 + c2n/(1 + c), which again equals (1.5)(n/2). (iii) The root edge again joins A1 to B, but we consider also the edges from the vertex in A1 to B, of which there are (1 - c)n/2. The total size of the threshold graph is about 2nc2/(1 + c) + (1 - c)n/2 = n/2. ..
18.3
M a x i m i z i n g the S u m of Squares of Degrees
Recall that a graph having n vertices and e edges is called an (n, e)-graph. In this section we consider the following problem. Given n and e, which (n, e)graphs have the largest sum of squares of the degrees? We refer to these graphs as a-optimal. Boesch et al. [BBB+90] consider this problem and show that these graphs must be threshold. However, not every threshold graph is a-optimal. For every valid choice of n and e, the authors define two threshold (n, e)graphs, which always exist and are unique, and prove that at least one of them is a-optimal. This enables us to compute the optimal value for any valid choice of n and e. We give examples showing that not both need be a-optimal, as well as a-optimal graphs that are not of this form. This leaves open the problem of characterizing the threshold graphs that are ~r-optimal. In this section we report the results of [BBB+90]. If r is a function from graphs to integers, then a graph G is called r
18.3
M a x i m i z i n g t h e S u m of Squares of Degrees
489
optimal if r >_ r for all graphs G' with the same number of vertices and edges as G. One such function of interest here is a(G) - ~ n d~, where ( d l , . . . , dn) is the degree sequence of G. Throughout this section we assume that n and e denote, respectively, the number of vertices and edges of a graph. Thus 0 deg(a), consider the graph G' obtained from G by replacing edge ac with bc. It is easy to see that a(G') > a(G). Thus if a graph G is a-optimal, then for every two vertices a, b, deg(a) >_ deg(b) if and only if N[a] D_ N(b). Therefore G is a threshold graph by Condition 5 of Theorem 1.2.4. 9 Lemma
18.3.2 A graph G is a-optimal if and only if its complement G is
a-optimal. P r o o f . The result is immediate, since a(G) - ~"~in__l[(rt -- 1) - d~]2 or(G), where k(n, e)is a constant depending only on n and e. Lemma
18.3.3 Let G -
+
K1 0 H be an (n, e)-graph. Then
1. If G is a-optimal, then H is a-optimal. 2. Conversely, if H is a-optimal and if there is a connected a-optimal (n, e)-graph, then G is a-optimal. P r o o f . Part 1 follows from the fact that a(IQ | H) = k(n, e)+ a(H), where k is a constant depending only on n and e. To prove Part 2, let G t be a connected a-optimal graph. Then G ~ is a threshold graph of the form K1 | H', and both H and H' are ( n - 1, e - n + 1)-graphs. By part 1, H' is ~r-optimal, hence a(H) = a(H'), a(G) = cr(G'), and G is a-optimal. 9 D e f i n i t i o n 18.3.4 For fixed n and e, G1 and G2 are the following threshold
(n, e)-graphs (Figure 18.1)" (~l(p, q, r) --
I~p (~ ( ~q [,_JKl,r),
490
Extremal Problems
where n - p + q + r + l, e - ( ~ ) + p ( q + r + l ) + r ,
G2(p, q, r) - Sp I.J (I(q (]~ KI,~),
Figure 18.1: The (7, 6)-graphs of the forms G1 and G2.
GI(0, 0, 6) L e m m a 18.3.5
G2(2, 0,4)
1. G~ (p, q, r) is the complement of G2(p, q, r).
2. For every valid n and e, there exist (n, e)-graphs of the form G1 and G2.
~. If G1 (p, q, r) and G1 (pt qt rt) are (n, e)-graphs, they are isomorphic, and similarly for G2.
P r o o f . 1" This is obvious from the definition. 2" By part 1, it is enough to show that G2(p, q, r) exists whenever 0 < e < (2). Let k be the largest integer satisfying (k2) < e. Then k < n. If k - n~ then e -
(2) and we can t a k e q -
n-l,p-r-
0. I f k _< n - l ,
take
qe - ( k 2 ) _> 0, r - k - q > 0 (for otherwise e > ( ~ ) + k - (k+l), contradicting the maximality of k), and p - n - k - 1 > 0. 3: Let G1 (p, q, r) and GI(p', q', r') be (n, e)-graphs. If p - p', then clearly q _ ql and r - r' and we are done. Now assume that p' - p + 5 and 5 > 0. When p is increased by 1, (~) + p ( n - p)increases by n - p 1. Therefore
p,+,n
p 1,+,n
2,+
p
18.3
Maximizing the Sum of Squares of Degrees
491
Hence e - r' >_ (e - r) + (n - p - 1) - e + q, i.e., q + r' _< O, and we conclude that q - r' - 0 as well as 6 - 1. Therefore G1 (p, q, r) - G1 (p, 0, r) - Kp+a | S~_p_a,
G~ (p', q', r') - G~ (p', q', O) - Kp, | Sq, - Kp+, | Sn-p-1. We frequently use the following operation in the proof of Theorem 18.3.7 below (see Figure 18.2). Let G be a graph, a and b two vertices of G of degrees a and/3, respectively, T1 a set of m neighbors of a, all different from b and having the same degree tl, and T2 a set of m non-neighbors of b, all different from a and having the same degree t2. Then G ( a , b, Tx, T2) is obtained from G b y d e l e t i n g all edges from a to T1 and adding all edges from b to T2. A simple calculation shows that
a(G(a, b, T1, T2)) - ~r(G) - 2m(/3 - a + t2
-
tl
-~- m
-t- 1)
(18.1)
Figure 18.2: An operation used in the proof of Theorem 18.3.7.
G
T~
a
T~
T2
b
T2
G(a,b, T1,T2)
F a c t 1 8 . 3 . 6 If e < 3 and n is arbitrary, then both G1 and G2 are or-optimal
(n, e) -graphs. P r o o f . This follows from L e m m a 18.3.1, since if e _< 3, all the threshold (n, e)-graphs have the same ~r. 9 Theorem
1 8 . 3 . 7 For every fixed n and e, at least one of the (n, e)-graphs
G l a n d G2 is a-optimal.
492
Extremal Problems
P r o o f . We prove the result by induction on n. Obviously, the result holds for n - 1. Assume that for every e' q + r - 1), and
18.3
493
Maximizing t h e S u m of Squares of Degrees
Figure 18.3: The graph G used in the proof of Theorem 18.3.7.
s~
Y
G -
9 H'-
Is
@
G2(p, q, r)
H(a,b,T~,T2).
T h e n c r ( H ' ) - a ( H ) - 0 by (lS.1). Note t h a t H ' has the form ((/s I,.JS;)@ K~) U Sp_q+l, where K ' and S' are cliques and stable sets of the i n d i c a t e d cardinalities, all positive. P u t m - min(q + r - 2 , p - q + 1), and observe t h a t m > 0 by (18.3) and our a s s u m p t i o n . In H', let 9 b be the s a m e v e r t e x b as before, now of degree 2q + r - 1, 9 c be a v e r t e x in K~+r_ 1 of degree q + r - 1, 9 T; be a set of m vertices
*' in I(q+r_ 1 other
9 T~ be a set of m vertices
in Slp_q_t_l of
9 H"-
H'(c,b,T~,T~).
t h a n c, of degree q + r -
degree 0,
1
494
Extremal Problems
Then a( H") - (r( H') = 2 m ( m + 2 - r ) by (18.1). I f m = q + r - 2 , then ( r ( H " ) - a ( H ' ) = 2mq > 0 since we are in Case 1. I f m = p - q + l , then ~r(H")-a(H') = 2m(p+3-q-r) > 0 because p > q + r 1. This p roves (18.4). We now show t h a t r = 0. For otherwise, let m = min(p, r) > 0 by (18.2), and in G let 9 a be the vertex y of degree n - 1, 9 b be the vertex x of degree q + 1, 9 T1 be a set of m vertices from Sp of degree 1, 9 T2 be a set of m vertices from Kr of degree q + r, and
9 G ' - G(a,b, T1,T2). Then ~r(G')-a(G) = 2re(re+q-p) by (18.1). If m = p, then a(G')-cr(G) = 2mq > 0, and if m = r, then a ( G ' ) - a ( G ) = 2m(r + q - p ) > 0 by (18.4), so in both cases we have a contradiction to the a - o p t i m a l i t y of G. This proves t h a t r = 0. Thus H is of the form Kq+l Ig Sp. Now if p = 1, then G is in the form G2, as required. Otherwise p _> 2 by (18.2), and we let 9 a be the vertex y of degree n - 1, 9 b be a vertex of Sp of degree 1, 9 T1 - Sp - {b}, a set of vertices with degree 1, 9 T2 be a set of p - 1 vertices in Kq (which exist by (18.4) and r - 0) of degree q + 1, and
9 G ' - G(a,b, T1,T2). Then c r ( G ' ) - a ( G ) - 0 by (18.1). Thus G' is also a-optimal. Clearly, G' is an (n, e)-graph of the form G ~ ( p - 1, p, q - p + 1). C a s e 2" q - 0. Then H is of the form K~ t2 Sp+l, which is isomorphic to a2(p + 1, 1, 0). By (18.3) and q - 0 we certainly have r - 1 >_ 1, and we are again in Case 1. 9 R e m a r k . For (7, 6)-graphs, G2 is not a-optimal, since a ( G l ) > or(G2) (see Figure 18.1). By considering the complements of these two graphs as well,
18.3
Maximizing the Sum of Squares of Degrees
495
we conclude that neither G1 nor G2 need always be a-optimal. Moreover, the (6, 7)-graph G of Figure 18.4 is not of the form G1 or G2, yet is a-optimal along with G1 and G2, since all these graphs have the same cr value. Figure 18.4: Examples of it-optimal (6, 7)-graphs.
GI(1, 2,2)
G2(1, 1,3)
G
Chapter 19 Other Extensions 19.1
Introduction
In this chapter we discuss three unrelated concepts on threshold graphs. In Section 19.2 we look at a generalization of threshold graphs based on geometrical embeddings of graphs. The threshold graphs can be thought of as the graphs for which there exist a non-negative real weight w~ associated with each vertex u and a positive real threshold t such that u v is an edge if and only if w u w v > t . If we allow the weights and threshold to be any reals, we obtain the generalized threshold graphs characterized by Reiterman et al. in [RRS89]. If the weights are replaced by vectors and the product is replaced by the scalar product, we obtain a further generalization of the threshold dimension. In Section 19.3 we introduce a new class of graphs called C-tolerance intersection graphs, due to Jacobson et al. [JMM91]. Golumbic et al. [GM82, GMT84] defined the tolerance graphs by assigning to each vertex u an interval I~ and a non-negative tolerance t~, such that u v is an edge if and only if IIu n Iv I _> min(t~, t~). These graphs clearly generalize the interval graphs. Monma et al. [MRT88] defined the threshold tolerance graphs by assigning to each vertex u a non-negative weight w~ and a tolerance t~ such that u v is an edge if and only if w~ + w~ > min(t~, t~). Both these classes are special cases of C-tolerance intersection graphs. In Section 19.4 we study the universal graphs for the class T~ of threshold graphs on n vertices. A d-universal graph for a class C of graphs is a graph G such that every graph in d is an induced subgraph of G. Hammer and 497
498
Other Extensions
Kelmans [HK94] completely characterize those T~-universal graphs with the least number of vertices that are themselves threshold graphs.
19.2
Geometric Embeddings of Graphs
Given a simple graph G, a representation of G is a function that assigns to each vertex x of G a vector ~ C R d, such that two vertices x and y are adjacent in G if and only if x and y satisfy some specified geometrical condition. Representations of graphs of this type have been considered by numerous authors. Reiterman et al. [RRS89] consider the representation of a graph G = (V, E) such that for some real threshold t, xyCE
if and only if
~y_t.
The scalar product dimension dp(G) is the minimum number d such that G admits such a representation in R d. The authors show that dp(G) t , or equivalently, using w' - e ~, t' xy E E
-
e t,
if and only if
w~wy~ ~ >_ t',
wherew t:V--+N +,t ~EN +,andN +={xCN:x>0}. Relaxing f r o m N + to N, we have the following generalization of threshold graphs. D e f i n i t i o n 19.2.1 A graph G = (V, E) is a g e n e r a l i z e d t h r e s h o l d g r a p h when there are a threshold t C N and a label x C N 1 associated with each vertex x, such that for all distinct x, y C V, xy C E
if and only if
x y >_ t.
(19.1)
Thus the threshold graphs are precisely those generalized threshold graphs that can be represented by a positive labeling and a positive threshold. Generalized threshold graphs are not necessarily threshold. For instance, the
19.2
Geometric Embeddings of Graphs
499
graph 2K2 is a generalized threshold graph, as shown in Proposition 19.2.2. However, if in Definition 19.2.1 the labeling and the threshold are nonnegative, then G = (V, E) is in fact a threshold graph, as is easy to see. We can always ensure, by slightly decreasing t if necessary, that the inequality in (19.1) never holds with equality. P r o p o s i t i o n 19.2.2 A graph is a generalized threshold graph if and only if it is the disjoint union of two threshold graphs or the complement of a difference graph. P r o o f . Suppose G - (V, E) is the disjoint union of threshold graphs G1 = (V1, EI) and G2 - ( 89 E2), and let x ~ x C R (x e V), ti > 0 represent G~ for i - 1, 2, where x >_ 0 (x E V). By a simple rescaling we may assume that t 1 t 2 t . Then the labeling x ~ x ~ defined by x'-
~" x, ( -x,
for x E V~ for x C 1/2
together with the threshold t represent G. Suppose the complement G is a difference graph with color classes V1 and V2. By Theorem 2.4.3, if we add to G all the edges joining vertices of V1, it becomes a threshold graph T. Let x H x be a positive labeling and t > 0 a threshold representing T, such that x y =/= t for all distinct x, y. Then G is represented by x ~ x ~ and t ~, where x'-
~ x, [ -x,
for x C V1 for x E
t ! --
--t.
Conversely, let G be a generalized threshold graph represented by x H x and a threshold t such that the inequality in (19.1) never holds with equality. Put
{xc
0. Then there is no edge between V1 and 1/2. The induced subgraph G1 - (VI, El) is a threshold graph as the labeling is non-negative on V1. As for the induced subgraph G2 - (1/2, E2), we use the non-negative labeling x ~ - x and the same threshold t to see that G2 is also a threshold graph. Next, assume t _< 0. Then V1 and 1/2 are cliques and the non-negative labeling x ~ [x[ together with the threshold Itl represent a threshold graph
500
Other Extensions
(V, E") such that for all x C 1/1 and y C 1/2, we have xy E E if and only if xy ~ E". Thus by Theorem 2.4.3 G is the complement of a difference graph.
C o r o l l a r y 19.2.3 The generalized threshold graphs have Dilworth number at most 2. Proof. This follows from Propositions 19.2.2 and 2.4.2. 9 The following theorem gives a characterization of generalized threshold graphs by forbidden induced subgraphs. The proof can be found in [RRS89]. T h e o r e m 19.2.4 ([RRSS9]) A graph is generalized threshold if and only if it does not contain any of the graphs G1,..., G10 of Figure 19.1 as induced subgraphs.
Figure 19.1: The forbidden induced subgraphs for generalized threshold graphs.
": 9 i i i I t ! G1
G2
Ga
G4
G5
G6
G7
G8
G9
GlO
We now consider the dimension of generalized threshold graphs. Recall that the threshold dimension t(G) of a graph G = (1/, E) is the minimum number t of threshold spanning subgraphs G1,..., Gt of G whose union is E; in other words, two vertices form a nonedge in G if and only if they form a nonedge in each of G1,..., Gt. By taking exponents, we can say equivalently
19.2
Geometric Embeddings of Graphs
501
that the threshold dimension t(G) of G is the minimum number t with the property that there are a function f " V ~ IRt with f ( x ) > 0 (x E V) and a vector c > 0 in IRt, such that for distinct x, y C V,
xy~E
if and only if
f(x),J'(y) t.
The s c a l a r product d i m e n s i o n dp(G) of a graph G is the minimum number d such that G admits a representation in N d. The assignment x ~ x, where x is the row of the vertex-edge incidence matrix of G corresponding to x, together with the threshold t - 1, provide the simplest example of a representation of G. This shows that dp(G) is well-defined and gives the trivial upper bound dp(G) _< n, where n is the number of vertices (since the vectors a generate a subspace of dimension _< n over N). A better upper bound for dp(G) is established in the following theorem, which shows that the scalar product representation is more efficient than the threshold one, in the sense that it represents more graphs using small dimensions.
T h e o r e m 19.2.6 Every graph G satisfies dR(G) < t(G). P r o o f . Let G - (V,E), t(G) - s, and E - U~=IEk , where the (V, Ek) are threshold graphs. Let V - Dok U ... U D mkk be the degree partition of (V, Ek), k - 1 , . . . , s, as in Definition 1.2.3. Recall that for every distinct x , y C V with x C D~, y C D~,
xyCEk
if and only if
i+j>_mk+l.
502
Other Extensions
We construct a labeling x ~ x with values in IR~ by p u t t i n g 1
(x)k - M i- ~'~k f o r k - 1 , . . - , s ,
ifxcD/k,
where M > s is arbitrary. If xy C E, then xy C Ek for some k and x C D/k, y E D) for some i , j satisfying i + j _> mk + 1. Hence 1
1
x y > xkyk -- M i - ~ m k M j-~mk - M i+j-mk >_ M.
If xy ~ E for distinct x, y, then for every k - 1 , . . . , s we have xy ~ Ek. Let x E D~, y E D). Then i + j - i n k < 1. Thus xkyk -- M ~+j-mk _< 1, and hence xy-
~-~xkyk < S < M. k=l
This proves t h a t our labeling together with the threshold t - M represent G in N s, and hence dp(G) _ n/2 for n _> 4. W h e n G is the disjoint union of two or more graphs each of which has at least one edge, then again dp(G) is much smaller than t(G), as can be seen from the proposition below. Proposition
1 9 . 2 . 7 If G is the disjoint union of graphs G I , . . . , Gn, then
dp(G) m a x / t i is for the mom e n t unspecified. We show that if t is sufficiently large, then the assignment
19.2
503
Geometric Embeddings of Graphs
x ~ x', where ~ e ' - Ai(x + d i) for x C V/, i - 1 , . . . , n, represents G in IRm with the threshold t. Indeed, if x E !Vi and y C Vj, then
x'y' - A i ( x + d i ) A J ( y + d j) - ( x + d i ) T A iT A J ( y + d j) - ( x + d i ) T A J - i ( y + d j) m-2 (j -i)Tr ~(j = ~ xkyk + Xm-lY~n-1 COS ~ + ~ / t - tjxm-1 sin k=l
-
T/
v/t - t~ sin ~ (jy -m -i)Tr 1
i)Tr ?~
+ ~ / ( t - t i ) ( t - tj)cos (j - i)Tr
n
n
Now if i - j, then x'y' - x y + t - ti, and hence x y > t if and only if x y > ti, which happens for x :fi y if and only if xy is an edge of Gi. If i -r j, then obviously lira x y -- cos (j - i)Tr < 1. t ---*oo
t
r/
Hence x y < t for all x in Gi, y in Gj if t is sufficiently large. This verifies that we have defined a correct representation for G. 9 R e m a r k . In general, the upper bound of Proposition 19.2.7 cannot be improved. For instance, d e ( K 2 ) = de(2K2) = 1, but dp(3K2) > 1. On the other hand, it is not known whether de(V, UE~) 4. 2. d p ( P n ) - 2
>__ 5
3. dp(nK2) - 2 for every n >_ 3. ~. dp(Kn,~) - n for every n >_ 1. P r o o f . 1" Let Xl , . . . , X n be vertices of Cn along the cycle. For i - 1 , . . . , n, the vectors 27ri 27ri) R2 cos~,sin-C n
n
represent Cn in IR2 with the threshold t - cos 2_~. Hence dp(Cn) __min(t~,tv).
508
Other Extensions
In particular, if t~ _< IS~ l, then vertex 1 is dominating. If t 1 ) ISll and vertex 1 is not isolated, let u be adjacent to 1. Then tu
Observe that the Sv are nested sets. i _< j _< [~], then ISuNSvl-i+l
if j < [~J ifj >
Further, for u E Di and v C Dj, if
<m+l-min(tu,
tv),
so u and v are not adjacent. If i _> j > [~J, then
n Xvl-
+ 1 _> min(t~, t~),
so u and v are adjacent. Finally, if/_< [~J < j, then ISunSvl-i+l_>rn-j+2
if and only if
i+j>m.
Since by Condition 6 of Theorem 1.2.4, u and v are adjacent if and only if i + j > m, G is a min-tolerance chain graph with the assigned nested sets and tolerances. 2" Let G be a max-tolerance chain graph on the vertices V l , . . . , v ~ . We may assume without loss of generality that the associated nested sets are { 1 , . . . , r i } for i - 1 , . . . , n , where 1 _< r~ _< ... _< r~. Let the corresponding tolerance of vertex vi be ti. We may also assume that ti _< ri for all i, for if ti > ri, then vi is isolated and we can disregard vi. Consequently, we can associate the interval Ii - [ti, ri] with each vertex vi. Now vertices vi and vj are adjacent if and only if min(ri, rj) >_ m a x ( t i , t j ) . This holds if and only if Ii N Ij 5r 0 , and thus G is an interval graph. Conversely, if G is an interval graph with associated intervals [ti, ri], i - 1 , . . . ,n, we may assume without loss of generality that the ti and ri are positive integers. Then the sets { 1 , . . . ,ri} along with the tolerances ti represent G as a max-tolerance chain graph.
19.4
509
Universal Threshold Graphs
3: Recall that the complements of threshold tolerance graphs (coTT graphs) have the adjacency defined by
ij C E
if and only if
wi + wj < min(ti, tj).
Again we may assume that the ti are positive integers. Hence the coTT graphs are the same as the sum-tolerance chain graphs with nested sets { 1 , . . . , t i } and tolerances wi + 1/2. 9 R e m a r k . There is no known characterization of threshold tolerance graphs by forbidden subgraphs. However, a polynomial-time algorithm for recognizing this class of graphs is given in [MRT88].
19.4
Universal Threshold Graphs
In [HK94], Hammer and Kelmans studied the universal graphs for threshold graphs. Given a class C of graphs, a graph G is called a C-universal graph, or simply a C-UG, if every graph in C is isomorphic to an induced subgraph of G. If in addition G has the least number of vertices among all C-UGs, then G is called a minimum C-UG, or simply a C-MUG. Observe that if d is closed under complements, then G is a C-UG whenever its complement is, and similarly for C-MUGs. Likewise, if C* is the class of all induced subgraphs of the graphs in C, then every C-UG is a C*-UG, and similarly for MUGs. Let T~ denote the class of all threshold graphs on n vertices. We characterize all the :m-MUGs that are themselves threshold (i.e., the threshold T~-MUGs), a result originally proved by Hammer and Kelmans [HK94]. N o t a t i o n . We denote by (G, xy) the graph obtained from a graph G by adding two adjacent new vertices x, y, and joining y to all vertices of G. In other words, (O, xy) = (O U x) @ y. Similarly, we denote by (G,g-~y) the graph obtained by adding two nonadjacent new vertices x, y, and joining y to all vertices of G. In other words, (a,
= (G 9 y) u x.
Note that (G, xy) = (G, yx). Theorem
19.4.1 For every graph G, the following conditions are equivalent:
1. G is a (threshold)2/-~-UG;
510
Other E x t e n s i o n s
2. (G, xy) is a (threshold)~n+I-UG; 3. (G,-~y) is a (threshold)~n+I-UG. P r o o f . By Condition 4 of Theorem 1.2.4, if one of the three graphs is threshold, so are the other two. 1) =~ 2): Assume that G is a ~ - U G . Let T be any threshold graph on n + 1 vertices. If T contains an isolated vertex x, then T - {x} is isomorphic to an induced subgraph of G, and hence T is isomorphic to an induced subgraph of (G, xy). Otherwise T contains a dominating vertex y, and the proof is similar. 2) ~ 1): Assume that (G, xy) is a Tn+I-UG. Let T be any threshold graph on n vertices. The graph T' = T U {a} is isomorphic to an induced subgraph of (G, xy) on some vertex set X. Clearly, X does not contain y, since T' contains an isolated vertex. We may assume without loss of generality that x is the isolated vertex. It follows that T is isomorphic to the subgraph induced by X - {x} in G. The fact that 1) ,z---->, 3) follows from 1) ~ 2) by a simple complementarity argument. 9 Notation. Let a ~ = ( a l , . . . , a t ) b e a 0-1 sequence. Let H = G(a ~) be the graph constructed from G as described below: 1. s e t H : = G ; 2. for i = 1 , . . . , r , do H "-
(H, xiyi), (H, x-~),
if ai = 1, if ai = O.
1. Observe that G(a ~) is isomorphic to G(b ~) if and only if a ~ = b ~, and that, as with threshold graphs, there is a 1-to-1 correspondence between the set of 0-1 sequences a ~ and the set of graphs of the
F a c t 19.4.2
form 2. From Theorem 19.~. 1, it follows that G(a n) is a T~+k- UG if and only if G is a Tk- UG. 3. Since K1 is a threshold ~ - U G , K l ( a ~-a) is a threshold Tn-UG for each 0-1 sequence a n-1. Thus there are at least 2 n - 1 threshold T~-UGs, each with exactly 2 n - 1 vertices.
19.4
Universal Threshold Graphs
511
In fact, the 2 n-1 threshold Tn-UGs K l ( a n - l ) are all the threshold T~-MUGs, as we show below. T h e o r e m 1 9 . 4 . 3 If G is a Tn-MUG, then
1. the vertices of G can be partitioned into a stable set { S l , . . . , s n } and a clique { C l , . . . , Cn-1}. Hence G contains exactly 2n - 1 vertices. 2. If the si are indexed so that deg(sl) _ k - 1 for k - 1 , . . . , n - 1. Similarly, G contains as induced subgraphs all threshold graphs of the form I~UKn_~ for r - 1 , . . . , n - 1 . Now setting r - 1 we obtain deg(s~) _< 1. Then setting r - 2 we obtain deg(s2) _< 2. Continuing in this way, we obtain deg(sk)
