12.1 The Achromatic Number of a Graph

Transcription

1 Chapter 1 Complete Colorings The proper vertex colorings of a graph G in which we are most interested are those that use the smallest number of colors These are, of course, the χ(g)-colorings of G If χ(g) = k, then every k-coloring of G (using the colors 1,,, k as usual) has the property that for every two distinct colors i and j with 1 i, j k, there are adjacent vertices of G colored i and j If this were not the case, then the set of vertices colored i and the set of vertices colored j could be merged into a single color class, resulting in a (k 1)-coloring of G, which is impossible In fact, the chromatic number of a graph G can be defined as the smallest positive integer k for which there is a k-coloring of G having the property that for every two distinct colors, there are adjacent vertices in G assigned these colors In this chapter, we are primarily interested in vertex colorings of graphs having this property and in concepts related to this property 11 The Achromatic Number of a Graph By a complete coloring of a graph G, we mean a proper vertex coloring of G having the property that for every two distinct colors i and j used in the coloring, there exist adjacent vertices of G colored i and j A complete coloring in which k colors are used is a complete k-coloring If a graph G has a complete k-coloring, then G must contain at least ( k edges Consequently, if the size of a graph G is less than ( k for some positive integer k, then G cannot have a complete k-coloring If G is a k-chromatic graph, then every k-coloring of G is a complete coloring of G On the other hand, if there is a complete k-coloring of a graph G for some positive integer k, then it need not be the case that χ(g) = k For example, the -coloring of the path P 4 given in Figure 11 is a complete -coloring and yet χ(p 4 ) = Indeed, the 4-coloring of the path P 8 is a complete 4-coloring (see Exercise 1) Since every χ(g)-coloring of a graph G is a complete χ(g)-coloring and since every complete k-coloring of G is, by definition, a proper vertex k-coloring of G, it follows that the minimum positive integer k for which a graph G has a complete 9

2 0 CHAPTER 1 COMPLETE COLORINGS Figure 11: A complete -coloring of P 4 and a complete 4-coloring of P 8 k-coloring is χ(g) We saw in Figure 11 that a graph G can have a complete k-coloring for an integer k > χ(g) The largest positive integer k for which G has a complete k-coloring is the achromatic number of G, which is denoted by ψ(g) This concept was introduced by Frank Harary, Stephen Hedetniemi, and Geert Prins [96] It therefore follows that ψ(g) χ(g) for every graph G Certainly, if G is a graph of order n, then ψ(g) n and for the complete graph K n, ψ(k n ) = χ(k n ) = n Furthermore, while χ(p 4 ) = χ(p 8 ) = for the two paths P 4 and P 8 shown in Figure 11, ψ(p 4 ) = and ψ(p 8 ) = 4 Let s see why ψ(p 8 ) = 4 First, ψ(p 8 ) 4 since the 4-coloring shown in Figure 11 is a( complete 4-coloring, while ψ(p 8 ) < 5 since the size of P 8 is 7 which is less than 5 ) = 10 Thus ψ(p8 ) = 4 Hence there are graphs G for which ψ(g) = χ(g) and graphs G for which ψ(g) > χ(g) The paths P 4 and P 8 also illustrate the fact that a bipartite graph need not have achromatic number Every complete bipartite graph, however, does have achromatic number Theorem 11 Every complete bipartite graph has achromatic number Proof Suppose, to the contrary, that there is some complete bipartite graph G such that ψ(g) Since χ(g) =, it follows that ψ(g) = k for some integer k Let there be given a complete k-coloring of G Then two vertices in some partite set of G must be assigned distinct colors, say i and j Since this coloring is a proper coloring, no vertex in the other partite set of G is colored i or j Therefore, G does not contain adjacent vertices colored i and j, producing a contradiction From our earlier observations, it follows that if the size m of a graph G satisfies the inequality ( ) ( k m < k+1 ) for some positive integer k and there is a complete l-coloring of G, then l ψ(g) k The following theorem gives a rather simple bound for the achromatic number of a graph in terms of its size Proposition 1 If G is a graph of size m, then ψ(g) m Proof Suppose that ψ(g) = k Then ( ) k m = k(k 1) Solving this inequality for k, we obtain ψ(g) = k m

3 11 THE ACHROMATIC NUMBER OF A GRAPH 1 With the exception that χ(g) ψ(g) for every graph G, there are essentially no restrictions on the possible values of χ(g) and ψ(g) for a graph G Indeed, Vithal Bhave [17] observed that every two integers a and b with a b can be realized as the chromatic number and achromatic number, respectively, of some graph Theorem 1 For every two integers a and b with a b, there exists a graph G with χ(g) = a and ψ(g) = b Proof Let r = ( b ( a and let G = K a rk (Therefore, if a = b, then r = 0 and G = K a ) Since a, it follows that χ(g) = a It remains to show that ψ(g) = b Since the size of G is ( ( a + r = b, it follows that ψ(g) b Consider the b-coloring of G where the vertices of K a in G are assigned the colors 1,,,a such that distinct pairs {i, j} of colors, where 1 i < j b and j a + 1, are assigned to the r pairs of adjacent vertices in rk in G Since the number of such pairs {i, j} of colors is ( ) ( ) ( ) b a b a a(b a) + = = r, the b-coloring is a complete b-coloring and so ψ(g) b Thus ψ(g) = b While Theorem 1 shows that the number ψ(g) χ(g) can be arbitrarily large for a graph G, Shaoji Xu [190] established an upper bound for ψ(g) χ(g) in terms of the order of G Theorem 14 For every graph G of order n, ψ(g) χ(g) n 1 Proof Since ψ(g) = χ(g) = 1 if G is empty, we may assume that G is nonempty Suppose that χ(g) = k and ψ(g) = l Then k l So there exists a complete l-coloring of G but no larger complete coloring of G Hence V (G) can be partitioned into l color classes V 1, V,,V l such that for every two distinct integers i, j {1,,, l}, there is a vertex of V i adjacent to a vertex of V j Since χ(g) = k, at most k of the sets V 1, V,, V l can consist of a single vertex Therefore, at least l k of these sets consist of two or more vertices and so n (l k) + k = l k Therefore, l (n + k)/, which implies that ψ(g) χ(g) = l k n + k k = n k Since k, it follows that ψ(g) χ(g) n = n 1 The bound given in Theorem 14 is sharp We show that for every even integer n, there exists a bipartite graph G with achromatic number n + 1 and so

4 CHAPTER 1 COMPLETE COLORINGS ψ(g) χ(g) = ( n + 1) = n 1 For n =, the graph G = K has the desired properties; while for n = 4, the graph G = P 4 has the desired properties Hence we may assume that n 6 Let n = a +, where a, and suppose that G is a graph of order n with vertex set V (G) = {u, w} {v 11, v 1,, v 1a } {v 1, v,,v a } Two vertices v 1i and v 1j are adjacent in G if and only if i and j are of the opposite parity The set {v 1, v,, v a } is independent in G Two vertices v 1i and v j are adjacent in G if and only if i j and i and j are of the same parity The vertices u and v 1i are adjacent if and only if i is odd; while u and v i are adjacent if and only if i is even Also, w and v 1i are adjacent if and only if i is even; while w and v i are adjacent if and only if i is odd Finally, u and w are adjacent in G This completes the construction of G Thus the sets V r = {v r1, v r }, 1 r a, V a+1 = {u}, V a+ = {w} are independent in G Furthermore, G is a bipartite graph with partite sets U = {u} {v 1i : i is even} {v i : i is odd}, W = {w} {v 1i : i is odd} {v i : i is even} Since for every two distinct integers r and s with 1 r, s a +, a vertex in V r is adjacent to a vertex in V s, the coloring that assigns each vertex of V r (1 r, s a + the color r is a complete ( n + 1) -coloring The graph G is shown in Figure 1 for the case when n = 1 along with a complete ( n + 1) -coloring 6 u v 11 v 1 4 v 15 v 1 v 1 v 14 4 v 5 v v v w Figure 1: A bipartite graph G of order n = 1 with ψ(g) = n + 1 = 7 It is useful to know how the achromatic number of a graph can be affected by the removal of a single vertex or a single edge The following information was obtained by Dennis Paul Geller and Hudson V Kronk [79]

5 11 THE ACHROMATIC NUMBER OF A GRAPH Theorem 15 For each vertex v in a nontrivial graph G, ψ(g) 1 ψ(g v) ψ(g) Proof Suppose that ψ(g v) = k Let there be given a complete k-coloring of G v, using the colors 1,,,k If there exists a color i {1,,, k} such that v is not adjacent to a vertex colored i, then by assigning the color i to v, a complete k-coloring of G results Otherwise, there is a neighbor of v colored i for each i {1,,, k} By assigning the color k + 1 to v, a complete (k +1)-coloring of G results In either case, ψ(g) k = ψ(g v) Suppose next that ψ(g) = l Let there be given a complete l-coloring of G, using the colors 1,,,l, where the vertex v is colored l say If v is the only vertex of G colored l, then the (l 1)-coloring of G v is a complete (l 1)-coloring Otherwise, there are vertices of G other than v colored l If the l-coloring of G v is not a complete l-coloring, then for every vertex u that is colored l in G v, there exists some color i {1,,, l 1} such that no neighbor of u is colored i By recoloring each vertex of G v colored l with a color in {1,,, l 1} not assigned to some neighbor of the vertex, a complete (l 1)-coloring of G results In either case, ψ(g v) l 1 = ψ(g) 1 The following result is an immediate consequence of Theorem 15 Corollary 16 For every induced subgraph H of a graph G, ψ(h) ψ(g) As we saw in Theorem 15, the removal of a single vertex from a graph G can result in a graph whose achromatic number is either one less than or is the same as the achromatic number of G When a single edge is removed, there is a third possibility Theorem 17 For each edge e in a nonempty graph G, ψ(g) 1 ψ(g e) ψ(g) + 1 Proof Suppose that e = uv and that ψ(g e) = k Then there exists a complete k-coloring of G e, using the colors 1,,, k The vertices u and v are either assigned distinct colors or the same color If u and v are assigned distinct colors, then the complete k-coloring of G e is also a complete k-coloring of G Hence we may assume that u and v are assigned the same color, say k If both u and v are adjacent to a vertex colored i for each i {1,,, k 1}, then recoloring v with the color k + 1 results in a complete (k + 1)-coloring of G If exactly one of u and v, say v, has no neighbor colored i for some i {1,,, k 1}, then by recoloring v with the color i, a complete k-coloring of G is produced Suppose that no neighbor of u is colored i and no neighbor of v is colored j, where i, j {1,,, k 1} and i j, and for every vertex x colored k, there is no neighbor of x colored t for some t {1,,, k 1} Then by recoloring u by i, v

6 4 CHAPTER 1 COMPLETE COLORINGS by j, and any such vertex x by t, a complete (k 1)-coloring of G results In any case, ψ(g) k 1 = ψ(g e) 1 Next suppose that ψ(g) = l Then there exists a complete l-coloring of G, where the colors assigned to u and v are distinct, say u is colored l 1 and v is colored l If the resulting l-coloring of G e is not a complete l-coloring, then no two adjacent vertices in G e are colored l and l 1 Hence every vertex colored l may be recolored l 1, resulting in a complete (l 1)-coloring of G e In any case, ψ(g e) l 1 = ψ(g) 1 The bounds presented in Theorem 17 are sharp For example, for the graphs G and H of Figure 1 and the edges e in G and f in H, ψ(g) = and ψ(g e) =, while ψ(h) = and ψ(h f) = u e v u v G : G e : w x w x H : u w f v H f : u w v Figure 1: Showing the bounds in Theorem 17 are sharp A graph G is k-minimal (with respect to achromatic number) if ψ(g) = k and ψ(g e) < k for every edge e of G By Theorem 17, if G is a k-minimal graph, then ψ(g e) = ψ(g) 1 Since k-minimal graphs have achromatic number k, the size of every such graph is at least ( k Bhave [17] characterized graphs that are k-minimal in terms of their size Theorem 18 Let G be a graph with achromatic number k Then G is k-minimal if and only if its size is ( k Proof Assume first that the size of G is ( k Then for every edge e of G, the size of G e is ( ( k 1 Since the size of G e is less than k, it follows that ψ(g e) < k = ψ(g) and that G is k-minimal We now verify the converse Assume, to the contrary, that there is a k-minimal graph H whose size m is not ( ( k Since ψ(h) = k, it follows that m k However, because m ( ( k, we must have m k + 1 Let a complete k-coloring of H be given, using the colors 1,,, k Therefore, for every two distinct colors i, j {1,,,k}, there exist adjacent vertices of H colored i and j Since there are only ( k pairs of two distinct colors from the set {1,,, k}, there are two distinct edges f and f whose incident vertices are assigned the same pair of colors from {1,,, k} However then, the complete k-coloring of H is also a complete k-coloring of H f, contradicting the assumption that H is k-minimal

7 1 GRAPH HOMOMORPHISMS 5 The achromatic numbers of all paths and cycles were determined by Pavol Hell and Donald J Miller [104] Theorem 19 For each n, ψ(p n ) = max { k : ( k + 1 ) (k + n } According to Theorem 19, ψ(p 7 ) = and ψ(p 11 ) = 5 A complete -coloring of P 7 and a complete 5-coloring of P 11 are given in Figure 14 (see Exercise 9) P 7 : P 11 : Figure 14: The graphs P 8 and P 11 with ψ(p 7 ) = and ψ(p 11 ) = 5 The following is then a consequence of Theorem 19 (see Exercise 10) Corollary 110 For every positive integer k, there exists a positive integer n such that ψ(p n ) = k Theorem 111 For each n, ψ(c n ) = max { k : k k n } s(n), where s(n) is the number of positive integer solutions of n = x + x + 1 According to Theorem 111, ψ(c 10 ) = 5 and ψ(c 11 ) = 4 A complete 5-coloring of C 10 and a complete 4-coloring of C 11 are given in Figure 15 (see Exercise 11) 1 1 C 10 : C 11 : Figure 15: The graphs C 10 and C 11 with ψ(c 10 ) = 5 and ψ(c 11 ) = 4 1 Graph Homomorphisms One of the fundamental concepts in graph theory is isomorphism Recall that an isomorphism from a graph G to a graph H is a bijective function φ : V (G) V (H) that maps adjacent vertices in G to adjacent vertices in H and nonadjacent vertices in G to nonadjacent vertices in H Of course, if such a function should exist, then

8 6 CHAPTER 1 COMPLETE COLORINGS G and H are isomorphic graphs; while if no such function exists, then G and H are not isomorphic The related concept of homomorphism will be of special interest to us in our discussion of graph colorings A homomorphism from a graph G to a graph G is a function φ : V (G) V (G ) that maps adjacent vertices in G to adjacent vertices in G If φ is a homomorphism from G to G and u and v are nonadjacent vertices in G, then either φ(u) and φ(v) are nonadjacent, φ(u) and φ(v) are adjacent, or φ(u) = φ(v) The subgraph H of G whose vertex set is V (H) = φ(v (G)) and whose edge set consists of all those edges u v in G such that φ(u) = u and φ(v) = v for two adjacent vertices u and v of G is called the homomorphic image of G under φ and is denoted by φ(g) = H A graph H is a homomorphic image of a graph G if there exists a homomorphism φ of G such that φ(g) = H If H is a homomorphic image of a graph G under a homomorphism φ from G to a graph G, then φ is also a homomorphism from G to H Indeed, our primary interest is in the homomorphic images of a graph For the graphs G = P 6 and H of Figure 16, the function φ : V (G) V (H) defined by φ(u 1 ) = φ(u 6 ) = v 1, φ(u ) = φ(u 5 ) = v, φ(u ) = v, φ(u 4 ) = v 4 is a homomorphism In fact, H is the homomorphic image of G under φ In this case, the nonadjacent vertices u 1 and u 4 map into the nonadjacent vertices v 1 and v 4, the nonadjacent vertices u 1 and u 5 map into the adjacent vertices v 1 and v, and the nonadjacent vertices u 1 and u 6 both map into v 1 v 1 H : v P 6 : v v4 u 1 u u u 4 u 5 u 6 Figure 16: A graph G and a homomorphic image of G There are exactly four homomorphic images of the path P 4 These are shown in Figure 17 On the other hand, for each positive integer n, the only homomorphic image of K n is K n itself Figure 17: The homomorphic images of P 4 There is an alternative way to obtain the homomorphic images of a graph G As we noted, the only homomorphic image of the complete graph G = K n is K n Otherwise, G is not complete and thus contains one or more pairs of nonadjacent vertices An elementary homomorphism of a graph G is obtained by identifying two nonadjacent vertices u and v of G The vertex obtained by identifying u and

9 1 GRAPH HOMOMORPHISMS 7 v may be denoted by either u or v Thus the resulting homomorphic image G can be considered to have vertex set V (G) {u} and edge set E(G ) = {xy : xy E(G), x, y V (G) {u, v}} {vx : vx E(G) or ux E(G), x / V (G) {u, v}} Alternatively, the mapping ǫ : V (G) V (G ) defined by ǫ(x) = { x if x V (G) {u, v} v if x {u, v} is an elementary homomorphism from G to G The homomorphic image ǫ(g) of a graph G obtained from an elementary homomorphism ǫ is also referred to as an elementary homomorphic image Not only is G a homomorphic image of G, a graph H is a homomorphic image of a graph G if and only if H can be obtained by a sequence of elementary homomorphisms beginning with G For example, if we identify u 1 and u 6 in the graph G of Figure 16, which is redrawn in Figure 18, we obtain the graph G 1 shown in Figure 18, which is a homomorphic image of G Then identifying u and u 5 in G 1, we obtain G, which is a homomorphic image of G 1 The graph G is also a homomorphic image of G The graph G is isomorphic to the graph H of Figure 16, which is redrawn in Figure 18 u 1 u u 5 G : G 1 : u 1 u u u 4 u 5 u 6 u u 4 u 1 v 1 G : u H : v v v 4 u u 4 Figure 18: Some homomorphic images of a graph The fact that each homomorphic image of a graph G can be obtained from G by a sequence of elementary homomorphisms tells us that we can obtain each homomorphic image of G by an appropriate partition P = {V 1, V,, V k } of V (G) into independent sets such that V (H) = {v 1, v,, v k }, where v i is adjacent to v j if and only if some vertex in V i is adjacent to some vertex in V j The partition P of V (G) then corresponds to the coloring c of G in which each vertex in V i is assigned the color i (1 i k) In particular, if the coloring c is a complete k-coloring, then H = K k The 4-coloring of the graph G in Figure 18 shown in Figure 19 results in the color classes V 1 = {u 1, u 6 }, V = {u, u 5 }, V = {u }, and V 4 = {u 4 } and the homomorphic image H of Figure 18, which is also shown in Figure 19

10 8 CHAPTER 1 COMPLETE COLORINGS v 1 G : u 1 v u u u 4 u 5 u H : v v 4 Figure 19: A homomorphic image of a graph Therefore, if a graph H is a homomorphic image of a graph G, then there is a homomorphism φ from G to H and for each vertex v in H, the set φ 1 (v) of those vertices of G having v as their image is independent in G Consequently, each coloring of H gives rise to a coloring of G by assigning to each vertex of G in φ 1 (v) the color that is assigned to v in H For this reason, the graph G is said to be H-colorable This provides us with the following observation, which is a primary reason for our interest in graph homomorphisms Theorem 11 If H is a homomorphic image of a graph G, then χ(g) χ(h) We now show that the chromatic number of an elementary homomorphic image of a graph can exceed the chromatic number of the graph by at most 1 Theorem 11 If ǫ is an elementary homomorphism of a graph G, then χ(g) χ(ǫ(g)) χ(g) + 1 Proof Suppose that ǫ identifies the nonadjacent vertices u and v of G We have already noted the inequality χ(g) χ(ǫ(g)) Suppose that χ(g) = k and a k- coloring of G is given, using the colors 1,,,k Define a coloring c of ǫ(g) by c (x) = { c(x) if x V (G) {u, v} k + 1 if x {u, v} Since c is a (k + 1)-coloring of ǫ(g), it follows that giving the desired result χ(ǫ(g)) k + 1 = χ(g) + 1, Since the graph G of Figure 18 is an elementary homomorphism of G 1 and G 1 is an elementary homomorphism of G while χ(g) = and χ(g 1 ) = χ(g ) =, it follows that both bounds in Theorem 11 are attainable The following result indicates the conditions under which χ(ǫ(g)) = χ(g) for an elementary homomorphism ǫ of a graph G

11 1 GRAPH HOMOMORPHISMS 9 Theorem 114 Let u and v be nonadjacent vertices in a graph G and let ǫ be the elementary homomorphism of G that identifies the nonadjacent vertices u and v Then χ(ǫ(g)) = χ(g) if and only if there exists a χ(g)-coloring of G in which u and v are assigned the same color Proof Suppose first that there exists a χ(g)-coloring c of G in which u and v are assigned the same color Then the coloring c of ǫ(g) defined by { c(x) if x V (G) {u, v} c (x) = c(u) = c(v) if x {u, v} is a χ(g)-coloring of ǫ(g) Thus χ(ǫ(g)) χ(g) Since χ(g) χ(ǫ(g)) by Theorem 11, it follows that χ(ǫ(g)) = χ(g) Conversely, suppose that χ(ǫ(g)) = χ(g) = k Let there be given a k-coloring c of ǫ(g) Then the coloring which assigns the same color to a vertex x of G that c assigns to x in ǫ(g) is a k-coloring of G Since χ(g) = k, this χ(g)-coloring of G assigns the same color to u and v When computing the chromatic polynomial of a graph G, we saw in Section 8 that for nonadjacent vertices u and v of G, both the graph G + uv and the graph ǫ(g), where ǫ is the elementary homomorphism of G that identifies u and v, were of interest to us At least one of these graphs has the same chromatic number as G Theorem 115 Let ǫ be an elementary homomorphism of a graph G that identifies nonadjacent vertices u and v in G Then χ(g) = min{χ(ǫ(g)), χ(g + uv)} Proof If χ(g) = χ(ǫ(g)), then the statement is true since χ(g + uv) χ(g) On the other hand, if χ(g) χ(ǫ(g)), then χ(ǫ(g)) = χ(g)+1 by Theorem 11; while by Theorem 114, every χ(g)-coloring of G assigns distinct colors to u and v Thus every χ(g)-coloring of G is also a χ(g)-coloring of G + uv, completing the proof Beginning with a noncomplete graph G, we can always perform a sequence of elementary homomorphisms until arriving at some complete graph As we saw, a complete graph K k obtained in this manner corresponds to a complete k-coloring of G Consequently, we have the following Theorem 116 The largest order of a complete graph that is a homomorphic image of a graph G is the achromatic number of G We have seen that for every graph G of order n, χ(g) ψ(g) n With the aid of Theorem 11, we show that ψ(g) can never be closer to n than to χ(g)

12 40 CHAPTER 1 COMPLETE COLORINGS Theorem 117 For every graph G of order n, ψ(g) χ(g) + n Proof Suppose that ψ(g) = k Then there is a sequence G = G 0, G 1,, G t = K k of graphs where G i = ǫ i (G i 1 ) for 1 i t = n k and an elementary homomorphism ǫ i of G i 1 By Theorem 11, and so for 1 i t Therefore, χ(ǫ i (G i 1 )) χ(g i 1 ) + 1 χ(g i ) χ(g i 1 ) + 1 t χ(g i ) i=1 t [χ(g i 1 ) + 1] i=1 and so k χ(g) + t = χ(g) + (n k) Hence and so ψ(g) χ(g)+n k = ψ(g) χ(g) + n The following theorem is due to Harary, Hedetniemi, and Prins [96] and is an immediate consequence of Theorem 11 Theorem 118 (The Homomorphism Interpolation Theorem) Let G be a graph For every integer l with χ(g) l ψ(g), there is a homomorphic image H of G with χ(h) = l Proof The theorem is certainly true if l = χ(g) or l = ψ(g) Hence we may assume that χ(g) < l < ψ(g) Suppose that ψ(g) = k Then there is a sequence G = G 0, G 1,, G t = K k of graphs where G i = ǫ i (G i 1 ) for some elementary homomorphism ǫ i of G i 1 for 1 i t Since χ(g 0 ) < l < χ(g t ) = k, there exists a largest integer j with 0 j < t such that χ(g j ) < l Hence χ(g j+1 ) l By Theorem 11, and so χ(g j+1 ) = l χ(g j+1 ) χ(g j ) + 1 < l + 1, The Homomorphism Interpolation Theorem can be rephrased in terms of complete colorings, namely:

13 1 GRAPH HOMOMORPHISMS 41 For a graph G and an integer l, there exists a complete l-coloring of G if and only if χ(g) l ψ(g) The following result is the analogue of Theorem 11 for complementary graphs Theorem 119 If ǫ is an elementary homomorphism of a graph G, then χ(g) 1 χ(ǫ(g)) χ(g) Proof We first show that χ(ǫ(g)) χ(g) Suppose that χ(g) = k Then there exists a partition P = {V 1, V,,V k } of V (G) into k independent sets in G Next, suppose that ǫ is an elementary homomorphism of G for which ǫ(u) = ǫ(v) for nonadjacent vertices u and v of G Since u and v are not adjacent in G, it follows that u and v are adjacent in G and thus belong to different elements of P, say u V 1 and v V Define V 1 = V 1 {u}, and V i = V i for i k Then P = {V 1, V,,V k } is a partition of V (ǫ(g)) into k 1 or k subsets, depending on whether V 1 = or V 1 Since none of V 1, V,,V k contains v, these sets are independent in ǫ(g) We claim that V is independent in ǫ(g) as well If x and y are vertices in V distinct from v, then x and y are not adjacent in V since x and y are not adjacent in V Suppose that v and w are two adjacent vertices in V Since vw is an edge in ǫ(g), it follows that v and w are not adjacent in ǫ(g) Hence both u and w are not adjacent in G and v and w are not adjacent in G Therefore, v and w are adjacent in G, which contradicts the fact that v, w V Thus P is a partition of V (ǫ(g)) into k 1 or k independent sets and so χ(ǫ(g)) k = χ(g) Next, we show that χ(g) 1 χ(ǫ(g)) Suppose that χ(ǫ(g)) = l for some elementary homomorphism ǫ of G that identifies two nonadjacent vertices u and v in G, where the vertex in ǫ(g) obtained by identifying u and v is denoted by v Let there be given an l-coloring of ǫ(g) using the colors 1,,, l We may assume that the vertex v in ǫ(g) is assigned the color l Then no vertex in ǫ(g) adjacent to v is colored l Assign to each vertex in G distinct from u the same color assigned to that vertex in ǫ(g) and assign u the color l + 1 Since no vertex in G adjacent to v is assigned the color l, this produces an (l + 1)-coloring of G and so Therefore, χ(g) 1 χ(ǫ(g)) χ(g) l + 1 = χ(ǫ(g)) + 1 We now turn to bounds for the achromatic number of an elementary homomorphism of a graph G in terms of ψ(g) Theorem 10 If ǫ is an elementary homomorphism ǫ of a graph G, then ψ(g) ψ(ǫ(g)) ψ(g)

14 4 CHAPTER 1 COMPLETE COLORINGS Proof Let ǫ is an elementary homomorphism of a graph G that identifies the two nonadjacent vertices u and v and let the vertex in ǫ(g) obtained by identifying u and v be denoted by v Suppose that ψ(ǫ(g)) = k and let there be given a complete k-coloring c of ǫ(g) Assigning the vertices u and v in G the color c(v) in ǫ(g) produces a complete k-coloring of G, implying that ψ(g) k = ψ(ǫ(g)) We now show that ψ(g) ψ(ǫ(g)) Suppose that ψ(g) = l Then ψ(g u v) l by Theorem 15 Furthermore, G u v is an induced subgraph of ǫ(g) It follows by Corollary 16 that Hence ψ(ǫ(g)) ψ(g) ψ(ǫ(g)) ψ(g u v) l While the lower bound for ψ(ǫ(g)) in Theorem 10 may be unexpected, it is nevertheless sharp For example, ψ(g) = 5 for the graph G of Figure 110, while ψ(ǫ(g)) = for the elementary homomorphism of G that identifies the two nonadjacent vertices u and v in G (See Exercise G : u t w x y z v ǫ(g) : t x v w y z Figure 110: ψ(ǫ(g)) = ψ(g) For every homomorphic image H, with vertex set V (H) = {u 1, u,, u k } say, of a graph G, there exists a partition P = {V 1, V,, V k } of V (G) into k independent sets such that the mapping φ : V (G) V (H) defined by φ(v) = u i if v V i (1 i k) is a homomorphism Then two vertices u i and u j are adjacent in H if there exists a vertex in V i and a vertex in V j that are adjacent in G If H = K k, then the coloring that assigns the color i to every vertex in V i for each i (1 i k) is a complete coloring On the other hand, the existence of a complete k-coloring of G implies that K k is a homomorphic image of G Therefore, the greatest positive integer k for which K k is a homomorphic image of G is, in fact, the achromatic number of G Suppose that G is a noncomplete graph with ψ(g) = k Then there exists a complete k-coloring of G and, in turn, K k is a homomorphic image of G Hence there exists a sequence G = G 0, G 1,, G t = K k of graphs, where ǫ i (G i 1 ) = G i for an elementary homomorphism ǫ i of G i 1 (1 i t) By Theorem 10, k = ψ(g t ) ψ(g t 1 ) ψ(g 1 ) ψ(g) = k

15 1 GRAPH HOMOMORPHISMS 4 Therefore, ψ(g i ) = k for all i (1 i t) From this, we have the following corollary Corollary 11 For every noncomplete graph G, there is an elementary homomorphism ǫ of G such that ψ(ǫ(g)) = ψ(g) By the Nordhaus-Gaddum Theorem (Theorem 710), χ(g) + χ(g) n + 1 for every graph G of order n Since ψ(g) χ(g) for every graph G, a larger upper bound for ψ(g) + χ(g) may be expected, but such is not the case Theorem 1 For every graph G of order n, ψ(g) + χ(g) n + 1 Proof Suppose that ψ(g) = k There there exists a sequence G = G 0, G 1,, G t = K k of graphs, where G i = ǫ i (G i 1 ) for an elementary homomorphism ǫ i of G i 1 (1 i t) Then t = n k = n ψ(g) By Theorem 119, Thus χ(g i 1 ) χ(ǫ i (G i 1 )) + 1 χ(g i ) + 1 t χ(g i 1 ) i=1 t [ χ(gi )) + 1 ] i=1 and so χ(g) χ(g t ) + t = χ(k k ) + (n k) Hence completing the proof By Theorem 1, it follows that k + χ(g) = ψ(g) + χ(g) n + 1, χ(g) + χ(g) χ(g) + ψ(g) n + 1, thereby providing an alternative proof of the Nordhaus-Gaddum Theorem A natural problem now arises of establishing an upper bound for ψ(g) + ψ(g) in terms of the order of G Prior to providing a solution to this problem, we find an upper bound for ψ(g) + ψ(g) in terms of the order of G Theorem 1 For every graph G of order n, ψ(g) + ψ(g) n + 1

16 44 CHAPTER 1 COMPLETE COLORINGS Proof By Theorem 117, By Theorem 1, as desired ψ(g) χ(g) + n ψ(g) + ψ(g) (χ(g) + n) + ψ(g) = (χ(g) + ψ(g)) + n (n + 1) + n = n + 1, The following result is due to Ram Prakesh Gupta [87] and the proof is due to Landon Rabern [141] Theorem 14 For every graph G of order n, Proof By Theorem 1, ψ(g) + ψ(g) 4n ψ(g) + ψ(g) n + 1 and ψ(g) + ψ(g) n + 1 Adding these inequalities, we obtain Consequently, Therefore, giving the desired result (ψ(g) + ψ(g)) 4n + ψ(g) + ψ(g) 4n + 4n + ψ(g) + ψ(g) = 4n, We now show that the bound presented in Theorem 14 is sharp by giving an infinite class of graphs G of order n for which 4n ψ(g) + ψ(g) = For a positive integer k, let G be the graph of order n = k such that V (G) is the disjoint union of the three sets U, W, and X, where and U = {u 1, u,, u k }, W = {w 1, w,,w k }, and X = {x 1, x,, x k } E(G) = {u i u j : 1 i < j k} {u i w j : 1 i, j k} {w i x j : 1 i, j k, i j} A complete coloring c is defined on G by

17 1 GRAPH HOMOMORPHISMS 45 c(u i ) = i for 1 i k and c(w i ) = c(x i ) = k + i for 1 i k (See Figure 111(a) for the case k = 4) For the complement G, it therefore follows that E(G) = {w i w j : 1 i < j k} {x i x j : 1 i < j k} {u i x j : 1 i, j k} {w i x i : 1 i k} A complete coloring c is defined on G by c(x i ) = i for 1 i k and c(u i ) = c(w i ) = k + i for 1 i k (See Figure 111(b)) The colorings c and c of G and G, respectively, show that ψ(g) k and ψ(g) k and so ψ(g) + ψ(g) 4k Since ψ(g) + ψ(g) 4k by Theorem 14, it follows that ψ(g) = ψ(g) = k and so ψ(g) + ψ(g) = 4k = 4n w 1 x 1 K 4 w w x x K 4 w 4 x 4 u 1 u u u 4 K 4 (a) G w 1 x 1 w x K 4 K 4 w x w 4 x 4 u 1 u u u 4 K 4 (b) G Figure 111: A graph G of order n = 1 with ψ(g) + ψ(g) = 16 = 4n

18 46 CHAPTER 1 COMPLETE COLORINGS While the identification of two nonadjacent vertices in a graph G results in an elementary homomorphism of G, an elementary contraction of G is obtained by the identification of two adjacent vertices in G If a graph H is obtained from G by a sequence of elementary contractions, then H is a contraction of G (see Section 51) We saw that if a graph H is obtained from G by a sequence of elementary contractions, edge deletions, and vertex deletions, then H is a minor of G Also, a graph H is a contraction of a graph G if and only if there exists a surjective function φ : V (G) V (H) such that (1) the subgraph G[φ 1 (x)] of G induced by φ 1 (x) is connected for every vertex x of H and ( for every edge xy of H, there exist adjacent vertices u and v in G, where u φ 1 (x) and v φ 1 (y) Trivially, every graph is a contraction and a minor of itself We saw in Theorem 11 how an elementary homomorphism affects the chromatic number of a graph The corresponding result for elementary contractions is presented next Theorem 15 For every elementary contraction ǫ of a graph G, χ(g) 1 χ(ǫ(g)) χ(g) + 1 Proof Suppose that ǫ(g) is obtained from G by identifying the adjacent vertices u and v of G Let G = G uv We mentioned in Section 71 that χ(g ) = χ(g) or χ(g ) = χ(g) 1 Let ǫ be the elementary homomorphism that identifies the nonadjacent vertices u and v in G Then ǫ (G ) = ǫ(g) By Theorem 11, either χ(ǫ (G )) = χ(g ) or χ(ǫ (G )) = χ(g ) + 1 Therefore, χ(ǫ(g)) = χ(ǫ (G )) has one of the two values χ(g ) or χ(g ) + 1 Since either χ(g ) = χ(g) or χ(g ) = χ(g) 1, it follows that χ(ǫ(g)) has one of the three values χ(g) 1, χ(g), or χ(g) + 1 The graph G in Figure 11 has order 6 and chromatic number 4 The graph G 1 obtained by identifying the adjacent vertices u and v, the graph G obtained by identifying the adjacent vertices w and x, and the graph G obtained by identifying the adjacent vertices x and y are shown in Figure 11 as well Since χ(g 1 ) = 5, χ(g ) = 4, and χ(g ) =, the sharpness of the bounds in Theorem 15 is illustrated Corollary 16 If G is a k-chromatic graph, then for every positive integer l with l k, there exists a contraction H of G such that χ(h) = l Proof Suppose that the order of G is n Consider a sequence

19 1 GRAPH HOMOMORPHISMS 47 u G : x z v y w u u u x z y z v x z v y w w w G 1 G G Figure 11: Chromatic numbers and elementary contractions G = G 0, G 1,,G n l of graphs such that G i = ǫ i (G i 1 ) where ǫ i is an elementary contraction of G i 1 (1 i n l) Hence each graph G i (0 i n l) is a contraction of G Since the order of G n l is l, it follows that χ(g n l ) l Let j be the largest integer such that 0 j < n l for which χ(g j ) l + 1 and let H = G j+1 Hence χ(h) l Since H is obtained from G j by an elementary contraction, it follows by Theorem 15 that χ(h) χ(g j ) 1 (l + 1) 1 = l Thus χ(h) = l There are several other consequences of Theorem 15 concerning the chromatic number of contractions of a graph G and depending on whether G is critical Corollary 17 Every noncritical graph G has a nontrivial contraction H such that χ(h) = χ(g) Proof We may assume that G is a connected graph of order n such that χ(g) = k Since G is a noncritical graph, there exist one or more proper subgraphs of G having chromatic number k In particular, there is an edge e = uv of G such that χ(g e) = k Let ǫ be the elementary homomorphism of G e that identifies u and v in G e Then χ(ǫ(g e)) = k or χ(ǫ(g e)) = k + 1 by Theorem 11 Let ǫ 1 be the elementary contraction that identifies u and v in G Then χ(ǫ 1 (G)) = k 1, χ(ǫ 1 (G)) = k, or χ(ǫ 1 (G)) = k + 1 by Theorem 15 Since ǫ(g uv) = ǫ 1 (G), it follows that χ(ǫ 1 (G)) = χ(ǫ(g uv)) So χ(ǫ 1 (G)) = k or χ(ǫ 1 (G)) = k + 1 If χ(ǫ 1 (G)) = k, then H = ǫ 1 (G) has the desired properties Hence we may assume that χ(ǫ 1 (G)) = k + 1 Let G 0 = G and G 1 = ǫ 1 (G) Consider a sequence

20 48 CHAPTER 1 COMPLETE COLORINGS G = G 0, ǫ 1 (G) = G 1, G,,G n k of graphs such that G i = ǫ i (G i 1 ) for some elementary contraction ǫ i of G i 1 (1 i n k) Since the order of G n k is k, it follows that χ(g n k ) k Each graph G i (1 i n k) is a nontrivial contraction of G Let j be the largest integer such that 1 j < n k and χ(g j ) k + 1 and let H = G j+1 Thus χ(h) k By Theorem 15, χ(h) χ(g j ) 1 (k + 1) 1 = k Thus χ(h) = k The graph G = C 4 is -chromatic but not -critical Furthermore, ǫ(g) = K for every elementary contraction ǫ of G Therefore, the nontrivial contraction referred to in Corollary 17 may not be an elementary contraction Corollary 18 A graph G is critical if and only if for every elementary contraction ǫ of G, χ(ǫ(g)) = χ(g) 1 Proof Suppose that G is k-critical for some integer k Hence for every edge uv of G, χ(g uv) = k 1 Necessarily, every (k 1)-coloring of G uv assigns the same color to u and v, for otherwise this would imply that there is a (k 1)-coloring of (G uv) + uv = G, which is impossible Let ǫ be the elementary contraction of G that identifies u and v Then χ(ǫ(g)) = k 1 For the converse, let χ(g) = k and suppose that χ(ǫ(g)) = χ(g) 1 for every elementary contraction ǫ of G Let uv be an edge of G and let ǫ be the elementary contraction of G that identifies u and v Then χ(ǫ(g)) = k 1 and every (k 1)- coloring of ǫ(g) produces a (k 1)-coloring of G e, where each vertex in V (G) {u} is assigned the same color as in ǫ(g) and u is assigned the same color as v Hence χ(g uv) = k 1 and so G is k-critical A noncritical k-chromatic graph always has a k-critical contraction Corollary 19 For every noncritical graph G, there is a critical graph H that is a contraction of G such that χ(h) = χ(g) Proof Suppose that χ(g) = k Let G 0, G 1,,G t be a sequence of graphs of maximum length such that for each i with 1 i t, there exists an elementary contraction ǫ i of G i 1 such that ǫ i (G i 1 ) = G i and χ(g i ) k Let H = G t Therefore, H is a contraction of G and χ(ǫ (H)) = k 1 for every elementary contraction ǫ of H, which implies that χ(h) = k To show that H is k-critical, it suffices to show that χ(h e) = k 1 for every edge e of H Let uv be an edge of H and let ǫ be the elementary contraction of H that identifies u and v Thus χ(ǫ(g)) = k 1 Let a (k 1)-coloring of ǫ(h) be given We now assign the same color to each vertex of G v that was assigned to this vertex in ǫ(h) This produces a (k 1)-coloring of H uv and so H is k-critical Corollary 10 If ǫ is an elementary homomorphism of a k-critical graph G, then χ(ǫ(g)) = k Proof Suppose that ǫ identifies two nonadjacent vertices u and v of G By Theorem 119, either χ(ǫ(g)) = k or χ(ǫ(g)) = k + 1 Since ǫ(g) v = G u v and G is k-critical, χ(ǫ(g) v) < k Thus χ(ǫ(g)) k and so χ(ǫ(g)) = k