3-Colorability of plane hypergraphs

3-COLORABILITY OF PLANE HYPERGRAPHS S. G. Indzheyan

UDC 519. i

In this paper, we present sufficient conditions for 3-colorability of planar hypergraphs. Colorability in the sense of Erdos-Hajnal is considered. For the main definitions, see [I]. The problem 3-COLORABILITY OF A PLANAR GRAPH is NP-complete [2]. The problem of 3colorability of a planar hypergraph is obviously no less difficult, and it is natural to try and find at least some sufficient conditions. Let X = { ~ , xs.....xp} be a set and E a family of subsets from X. called a hypergraph with vertex set X and edge set E, w h e r e X = e ~ f.

The pair H = (X, E) is We assume that for any

e6E, le]~2. An ordinary graph is a particular case of the hypergraph H = (X, E) such that for each edge e 6 E we have lel = 2. Note that this definition of a graph coincides with Harary's definition of multigraph [3]. With each hypergraph H is associated its Koenig representation K(H) with the vertex set X U E and edge set {(x, e) E X x E/x6~. H is a planar hypergraph if its Koenig representation is planar graph. This means that a planar hypergraph may be fitted on a plane (or on a sphere) so that each vertex corresponds to a point and each edge e to a region which contains all the incident vertices (and only they) in the plane (the sphere). If ei~=es ~i, es6 E), then the regions corresponding to the edges e I and e 2 intersect only at the points corresponding to the common incident vertices. An edge of the hypergraph H is called an edge of degree i if lel = i. By H' we denote the graph made up of the edges of degree 2 of the hypergraph H and its incident vertices by m1(H) the number of edges of degree i of the hypergraph H, by m(H) the number of edges of H, and by n I = n1(G) the number of vertices of degree i of the graph G. Let s(G) be the maximum degree of the vertices of the graph G, 9(x) the degree of the vertex x. Definition. The planar graph G = (X, U) without 2-faces and with some labeled edges is called (n; t)-quasicolorable if its vertices may be colored in n colors so that the ends of the labeled edges have different colors and each face contains at least two differently colored vertices, with the possible exception of those faces r for which X(F) N X n ~ = ~ and

IX (r) l < t. Here X(F) is the set of vertices on the border of the face r, X~ = X(G~) is the set of vertices incident on the labeled edges, G~ = (XH, U~) is the graph generated by the labeled edges of G and their incident vertices, IX(F)I is the number of elements in the set X(F). Let W k be the class of planar graphs with the following property: any planar hypergraph H is 3-colorable if H'6Wk and for any edge e, lel~k. Clearly, W s _ W 4 ~ W ~ . . . ~ W h . Note that the class W defined in [4] coincides with W a. A conjecture advanced in [4] suggests that if G is a planar graph such that m (O)~.12.

(1)

THEOREM i. Each planar triangulation with at most 8 labeled edges which do not form a complete four-vertex K~ is (3; 3)-colorable. Translated from Kibernetika, No. 2, pp. 10-15, March-April, 1987. mitted March Ii, 1985.

158

0011-4235/87/2302-0158512.50

Original article sub-

9 1987 Plenum Publishing Corporation

Proof. Assume that this is not so and let G = (X, U) be a triangulation with the smallest number of vertices which satisfies the conditions of the theorem and is not (3; 3)colorable. Clearly, G contains no vertices of degree 4;

n s = 1.

Let the vertex x 6 G be of degree 3 (Pig. 3)~ If none of the vertices of degree ~ 5 coin~ cides with any of the vertices x~, x ~ or x s, we obtain that the graph G, in addition to the edges (x, x~), (x, x2), (x, xs), should have at least (3n~ - i + 2ns)/2 labeled edges, i.e., m ' i > 3 + (3.4-- |q- 2 ) / 2 > 8 , a contradiction. If only one of the vertices x~, x2, xs is of d e g r e e ~ 5 , then, in addition to these three labeled edges, there should be at least (3n~ - I + 2n s - 1)/2 labeled edges, i.e., m' i> 3 + (3.4 -- ] -? 2 -- I)/2 = 9, again a contradiction. Thus~ at ].east two vertices of degree ~ 5 should coincide with x ~ x=, xa, and so at least one of these three vertices should be of degree 4 (since n s = i). Let x~ be this vertex, and let y be the fourth vertex adjacent to x~ (Pig. 4). But then either p (x~)~ ~, or p (xs)~>6, since by Proposition 4 either (y, x s) or (y, x 2) belong to some subgraph K+ with all the other edges labeled. Thus, at most two vertices of degree ~ 5 may have a conm~on labeled edge with vertex x, which means that at least (3n~ - I + 2n s - 2)/2 additional edges should be labeled, i.e., m ' ~ > 3 + ( 3 . 4 - - ! nu 2 - 2 ) / 2 > 8 . The contradiction rules out Case 2. Case 3: n~ = 2. Then the inequality (1) becomes 2~, + ~ ~ 6, whence ~ ~ 3--~/2o At the same time, 16 I> 2m' ~> 3a~ + 3n, -- I + 2n~ i.e., 16 ~> 6 + 3n~ -- l + 2n~I> 6 + 3 (3 -- nJ2) + 29~ -i ! = 14 q- ha~2. Hence it follows that n 5 ~ . 4 and n 4 ~ |. Since n ~ 0 and | 6 ~ 3 n ~ / - 3a~ -- I + 2n~, where n~ = 2, we have n ~ 3 . Thus

1 ~ n,~< 3,

(2)

0 5~ i.e.~ only one vertex of degree ~ 5 may have common labeled edges with x and y. This means that, in addition to the five labeled edges incident on vertices of degree 3, there should be at least (2n 4 + 2n s - 2)/2 labeled edges, i . e . , m ' ~ 9 , a contradiction.

161

Fig. 4

Fig. 5

FiE. 6

Now assume that the vertices x and y do not have a common edge. situations are possible.

The following four

i. xi = Yi for i = i, 2, 3 (Fig. 5). Since the triangulation G should contain at least one other vertex in addition to x, x I, x 2, x 3, y, then x I say should be adjacent to one other vertex z, which is also adjacent to x 3. Clearly, there should exist a multiple edge (xl, x s) [since p(y) = 3], whence p (xi)>5, p ~ 8 ) ~ 5 and only for x 2 can we have p (x~)~5. This means that, in addition to the six labeled vertices incident on x and y, there should be at least (2n 4 + 2n s - 2)/2 labeled vertices, i . e . , m ' ~ 1 0 , a contradiction. 2. xl = Yl and x s = Ys (Fig. 6). If the vertices x I and x 3 are of degree~-~5, then p~)~6 and @ ~ ) ~ 6 . Indeed, since p ~,) = p(xs) = 5, the edge (x 2, Y2) exists. Since G is not (3; 3)-quasicolorable, we conclude that there exists at least one vertex z (in addition to x, y, x 2, Y2, xl = Yl, xs = Y3) which is adjacent to Y2, say. Since G is a triangulation, then z is also adjacent to x 2, and since p(x 3) = 5, there exists the multiple edge (x 2, Y2), i.e., p ~ , ) ~ 6 and p ~ , ) > 8 . This means that, in addition to the six labeled edges incident on the vertices of degree 3, there should be at least 2(n 4 + n s - 2)/2 labeled edges, i.e., m'>9, a contradiction. If three vertices of degree 5 or p(x s) > 5. Hence it follows that in addition to the six labeled edges, there should be at least (2n~ + 2n s - 4)/2 labeled edges, i.e., m'>9, again a contradiction. 3. xa = Y3 (Fig. 7). Since 0(x 3) > 5, then at most four vertices of degree ~ 5 may coincide with xl, x2, Yx, and Y2. Hence m ' / > 6 + ( 2 n 4 + 2ns--4)/2 = 9, a contradiction. 4. xl ~ Yl for i = I, 2, 3. Vertices of degree_s.5 may coincide with the vertices xl, Yl, i = I, 2, 3, i . e . , m ' > 8 + (2n4 + 2n~--5)/2>8 , also a contradiction. These contradictions rule out System I. System 2. Let n~ = 2. Then the inequalities (3) and (i), combined with n 3 = 2, imply that a 5 > 2 . Assume that x and y are of degree 3, whereas xl, x2, x 3 and Yl, Y2, Ys are the vertices adjacent to x and y, respectively. Let x and y have a common labeled edge (Fig. 4). As for System I, we can show that only one vertex of degree 5 may coincide with x I or x 3. This means that m ' ~ 5 ~ - ( 3 a 4 , 1 +2a5--2)/2>8, i . e . , m ' ~ 9 , a contradiction. Now assume that x and y have no common edge.

The following four situations are possible.

i. xi = Yi for i = i, 2, 3 (Fig. 5). Since the triangulation G is not (3; 3)-quasicolorable, there exists at least one vertex z, other than y, x, x I = Yl, x2 = Y2~ x3 = Ys. This means that there should exist a multiple edge (xl, xs). These considerations imply that only x 2 may satisfy the inequality p (x2)~5. Thus, m ' / > 6 + ( 3 n 4 i ~ 2 n S - - 2 ) / 2 > 8 , a contradiction. 2. xl = Yl and x 3 = Ys (Fig. 6). If the vertices x I and x s are of degree ~ 5 , then the edge (x2, Y2) exists, and since G is not (3; 3)-quasicolorable, there exists at least one vertex z, other than x, y, x 2, Y2, xl = Yl, x3 = Ys- This means that p(x 2) > 4 and P(Y2) > 4, i.e., m ' ~ 8 ~ (3n4-]-~2ab~4)/2>8. if three vertices of degree ~.5 coincide with the vertices Yl = xl, x2, Y2, Ys = xs, then clearly either p(x I) > 5 or p(x 3) > 5, w h e n c e ~ ' ~ 6 - ~ (3a4--] ~- 2n5 ~ 4 ) ~ > 8, which is also a contradiction. 3. xs = Ys (Fig. 7). diction. 162

Since p(x s) > 5, then m ' > 6

+ (3al -| ~ 2 n b - - 4 ) / 2 > S J

a contra-

2

@

Fig. 7

Fig. 8

Fig. 9

4. Xl ~ Yl for i = i~ 2, 3. Since the vertices of degree-~5+

3n4--I + 2

2n~

= 9,

a contradiction. Assume that the vertices x and y have no common edge.

Then four cases are possible.

I. xl = Yl for i = i, 2~ 3 (Fig. 5). Since the triangulation G is not (3~ 3)-quasicolorable, there should exist at least one vertex other than x, y, x I = Yl, i = It 2, 3, which means that there should exist a multiple edge (xl, x~), i.e., p(x l) > 4 and p(x~) > 4. Hence, m ' ~ 6 ~(3n4--I -~2n~--2)/2/>6-~ (3n~--3)/2 = 9 ~ a contradiction~ 2. xl = Y1 and x3 = Y3 (Fig. 6). Since p ~ 1 ) ~ 5 a n d at worst may coincide with x2, Y2, which means t h a t ~ ~ 6 3)/2 = 9, also a contradiction.

p @ 3 ) ~ 5 , the vertices of degree 4 ~- (3n~-- i -~ 2ns--~)/2 ~ 6-~ (3n4--

3. x~ = Y3 (Fig. 7). Since p(xs) > 4~ the vertex of degree 4 at worst may coincide with xl, x2~ Yl, and Y2~ each incident on one of the six labeled edges, which means that m' ~ 6 + ( 3 n 4 ~ 1 + 2n5 - 3)/2 ~ 6 + (3n4 - 4)/2 > 8 . 4.

xl r Yl for i = I, 2, 3.

As in 3 above, we can show that m' > 8, a contradiction~

The contradictions for all the three systems rule out Case 3. Case 4_i:n 3 = 3. Then the inequality (I) becomes 2n4-~-n~3. If no two vertices of degree 3 have a c o m m o n edge, the contradiction is obvious, since there are 9 labeled edges incident only on vertices of degree 3. Let the vertices x, y, z be of degree 3~ and let x and y have a common labeled edge. Assume that x is adjacent to xl, x2j x 3 and y to y~, y=, ys and that y = x=. As in System I of Case 2, we can show that either P(Xl) > 5 or 9(X~) > 5. At the same time, 2n~-~n~3, i.e.~ in addition to the three vertices of degree 3, there exist at least two vertices of degree ~ 5 , This means that, in addition to the eight labeled edges incident on vertices of degree 3, there should be at least one labeled edge, i.e~ m'~>9~ a contradiction. We have thus shown that for all possible values of n~, n~, ns satisfying the inequality (I), the number of labeled edges in the triangulation G is ~9, which contradicts the assump -, tion of the theorem. The contradiction proves Theorem I. Q.E.D. COROLLARY 1 (see [4

5]).

If G is a planar graph such that m ( G ) ~ 5 ,

then G6~Y~.

Note that the conditions used for our proof are unimprovable, which is clear from the exemple in Fig. 8. THEOREM 2. The planar h y p e ~ r a p h H = (X, E) is 3-colorable if ~ ( H ' ) ~ 2 and for any hypergraph edge e such that~ ~ X { ~ ' ) ~ , we have I ~ l ~ s ( H ) - - 5 . 163

As we have noted above, in order to prove this theorem it suffices to prove the following lemma. LEMMA.

Any planar graph G such that 8:(G~:~2 is (3; m(G E) - 5)-quasicolorable.

Proof. The proof is by induction on m(Gn~ For m(Gn)~-~8, the levm~a follows from Theorem i. Let the lemma be true for all m(G~) < t. We will prove it for m(G H) = t. Assume that the lemma is not true for m(G~) = t and G is the class of all planar graphs with minimum number of vertices and t labeled edges for which the lemma does not hold. Consider the graph G 6 G with minimum number of edges. Clearly, n2(G) = 0. As in Theorem I, we can show that all the edges incident on a vertex of degree 3 are labeled, i.e., n~(G) = 0. Consider a vertex Clearly, I ~ 4 . Without on x and all the faces that some faces r i and

x with at least one incident edge labeled. Let p(x) = s (Fig. 9). loss of generality, we may assume that two labeled edges are incident r1(x) , where i6[|,I--3], satisfy the condition IX(ri(x))l~t--5. Note rj may coincide. Consider the following two cases.

Case I. There are no multiple edges incident on x (Fig. 9). Remove the vertex x from G and add the edges(xs, xs), (X3,~)..... (xl-2,x~-0. The new graph G' contains t - 2 labeled edges, and therefore by the inductive hypothesis it is (3, t - 7)-quasicolorable. Having colored the graph G', restore G and color x in a color different from the colors of x I and xs It is easy to see that the result is a (3, t - 5)-quasicoloring for G. Case 2. There are multiple edges incident on x. Let x be incident on the edges (x, xl), ~, ~..,(x,x~) and let ~,Xl) , ~,xz) be labeled. Let r i be the face bordered by the edges (x, &+!), (x, Xi+=) i~ 6111,I-7 8]). If the vertices (x~§ &+2) are distinct for all the faces r i, then remove the vertex x from G and to each face r I add the edge Xl+l, X~+=. Note that some faces ri, rj with i # j may coincide. After this operation, each face F i is transformed into Fi such that IX(P~)l~t--& The resulting graph G' contains t - 2 labeled edges, and therefore, by the inductive hypothesis, it is (3; t - 7)-quasicolorable. Having colored G', restore G and color x in a color other than the colors of x I and xs leaving the colors of all the other vertices unchanged. It is easy to see that the result is a (3; t - 5)-quasicoloring for G, which contradicts the choice of G. Now assume that there is a face F i such that x~+l = x~+2. Let X~+l,y,,9~,;..,gm, x~+= = x~+1 be vertices other than x which lie on the border of the face r i, and all Yl are pairwise distinct. Remove the vertex x from G and add the edges (xi+1,~+@ to each face rj for which ~§ to each fact of type r i add the edge ~t+*,Y~ Since the resulting graph G' contains t - 2 labeled edges, it is (3; t - Z)-quasicolorable. Having colored G', we may obtain a (3; t - 5)-quasicoloring for G. The contradiction proves Lemma i. COROLLARY 2.

If G is a planar graph such that s(~.~