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The deficiency of a regular graph
Discrete Mathematics 306 (2006) 1947 – 1954 www.elsevier.com/locate/disc
The deficiency of a regular graph Alexander Schwartz Received 25 September 2002; received in revised form 11 November 2005; accepted 13 March 2006
Abstract A proper edge coloring c : E(G) → Z of a finite simple graph G is an interval coloring if the colors used at each vertex form a consecutive interval of integers. Many graphs do not have interval colorings, and the deficiency of a graph is an invariant that measures how close a graph comes to having an interval coloring. In this paper we search for tight upper bounds on the deficiencies of k-regular graphs in terms of the number of vertices. We find exact values for 1 k 4 and bounds for larger k. © 2006 Elsevier B.V. All rights reserved. Keywords: Deficiency; Interval coloring; Consecutive coloring
1. Introduction In this paper we study a relatively new invariant of finite, simple graphs, namely the deficiency. This is an extension of an older topic first introduced by Asratian and Kamalian in 1987 [1]. They studied interval colorings of graphs, proper edge colorings such that the colors incident at each vertex form a consecutive interval of integers. More recently, such colorings have also been called consecutive colorings. There are several papers on this topic, although almost all of them have been confined to bipartite graphs. This is likely because interval colorings have applications to scheduling problems which are mostly relevant in the bipartite case. Many graphs do not have interval colorings, and Giaro et al. [4] extended this concept to an invariant applicable to all graphs in 1999. The deficiency of a graph is a measure of how close it comes to having an interval coloring. Specifically, the deficiency of a graph is the minimum number of pendant edges one must attach so that the resulting graph has an interval coloring. Very little is known about the deficiencies of arbitrary graphs. Outside of the work on bipartite graphs, only very restricted subclasses have been considered, such as complete graphs and wheels [5]. The general problem is quite complicated; in particular, it is NP-hard [3]. In this paper, we examine the deficiencies of regular graphs. Our central aim is to find tight upper bounds on the deficiency of a k-regular graph in terms of the number of vertices. To this aim we define sk , which is the supremum of the ratio between the deficiency and the number of vertices over all k-regular graphs. We find the exact values of s1 , s2 , s3 , and s4 and bounds on sk for larger k. 2. Preliminaries Definition 2.1. Let S be a finite set of integers. The deficiency of S is the number of integers between min S and max S not appearing in S. E-mail address: [email protected]. 0012-365X/$ - see front matter © 2006 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2006.03.059
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Let G be a finite, simple graph, let c : E(G) → Z be a proper edge coloring, and let v ∈ V (G) be arbitrary. Let S ⊂ Z consist of the colors of edges incident on v. Then the number of deficiencies at v, denoted d(G, c, v), is the deficiency of S. Furthermore, let d(G, c) = v∈V (G) d(G, c, v), and let the deficiency of G be d(G) = minc d(G, c), where the minimum is taken over all proper edge colorings of G. Our primary goal is to bound the deficiency of a k-regular graph in terms of the number of vertices. First, we recall some classical graph theoretic results. Proposition 2.2 (Vizing). Let G be a graph where every vertex has degree at most . Then G admits a proper edge coloring with + 1 colors. Proposition 2.3 (Petersen). Any 3-regular graph with at most two bridges has a perfect matching. Corollary 2.4. If every vertex of a graph G has degree one or three and G has at most one bridge, then G has a perfect matching. Proposition 2.5 (Petersen). Any 4-regular graph contains a 2-factor. For any k-regular graph G, Proposition 2.2 gives an edge coloring c such that d(G, c, v)1 for all v ∈ V (G). In particular, if G has n vertices then d(G)n, and the following definition is justified. Definition 2.6. Let Rk be the set of all k-regular graphs, and set sk = supG∈Rk d(G)/|V (G)|. Clearly d(G)sk n for all k-regular graphs G on n vertices. Furthermore, by taking disjoint unions, we see that this bound is tight for large n. The goal of this paper is to bound and compute these sk ’s. It is quite easy to compute s1 and s2 , but the others are relatively difficult. Here and throughout we use Kn to denote the complete graph on n vertices. Proposition 2.7. s1 = 0 and s2 = 13 . Proof. Any 1-regular graph is simply a union of disjoint edges, and any edge coloring results in no deficiencies; thus, s1 = 0. Any 2-regular graph is a union of disjoint cycles. All even cycles can be properly colored by colors 1 and 2 with no deficiency. Any odd cycle can be properly colored by colors 1 and 2 except for a single edge colored 3, and the deficiency is equal to one. Then s2 13 because a triangle is the smallest odd cycle, and finally s2 = 13 because d(K3 ) = 1. 3. General bounds on the sk We now find upper and lower bounds for sk . In later sections, we use the lower bounds given here to find s3 and s4 . Proposition 3.1. sk (k − 1)/(k + 1). Proof. Let G be an arbitrary k-regular graph. Using Proposition 2.2, properly color the edges of G using the integers from 1 to k + 1. Swap the colors as necessary to give a coloring c where 1 and k + 1 occur the fewest times. Each vertex touches edges with k different colors, and exactly one color is missing at each vertex. By construction, there are at least 2|V (G)|/(k + 1) vertices where either 1 or k + 1 is missing, and these vertices have deficiency zero. All other vertices have deficiency one, so d(G, c)/|V (G)|1 − 2/(k + 1), completing the proof. Proposition 3.2. If k is even, then sk k/(2k + 2), and if k is odd, then sk (k − 1)/(k 2 + 1). Proof. It is proved in [5] that d(K2+1 ) = for all positive integers . This takes care of the even case, so suppose that k is odd. The case k = 1 is trivial, so assume that k 3. Let G be the graph obtained by subdividing an edge of Kk+1 with a vertex w. Suppose that c is a coloring of G with d(G, c, v) = 0 for all v = w. Reducing all the colors modulo k gives an edge coloring which is proper everywhere except possibly at w. Suppose that the two colors incident on w are also distinct. Then each of the k colors is used at most (k + 2)/2 = (k + 1)/2 times, but G has k(k + 1)/2 + 1
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edges, a contradiction. Thus, for all proper edge colorings c of G, either d(G, c, v) > 0 for some v = w or the two colors incident on w differ by at least k. We use this fact to construct a graph that gives the desired lower bound on sk . Take (k − 1)/2 copies of G and identify their vertices w to obtain a graph H on (k + 1)(k − 1)/2 + 1 vertices. Let c be any proper edge coloring of H, and let be the number of vertices v such that d(H, c, v) > 0 and v = w. Suppose < (k − 1)/2. Then, by the above argument, there exist at least (k − 1)/2 − > 0 disjoint pairs of edges incident with w whose colors differ by at least k. This implies that there are two edges incident on w whose colors differ by at least k + (k − 1)/2 − − 1, and d(H, c, w)(k + 1)/2 − because w has degree k − 1 in H. Hence d(H, c) (k + 1)/2. If (k − 1)/2, then clearly d(H, c) (k − 1)/2. Thus d(H, c) (k − 1)/2 for any coloring c, and equality can occur only when d(H, c, w) = 0. Let H1 and H2 be two copies of H with wi ∈ V (Hi ) corresponding to w for i = 1, 2. Let H be the disjoint union of H1 and H2 with the edge w1 w2 added. Then H is a k-regular graph on k 2 + 1 vertices. Let c be any proper edge coloring of H , and let the induced colorings on H1 and H2 be c1 and c2 , respectively. If d(H1 , c1 ) = (k − 1)/2, then d(H1 , c1 , w1 ) = 0, and the extra color on the edge w1 w2 cannot decrease the deficiency of H1 in H. Otherwise d(H1 , c1 )(k + 1)/2, and the extra color of the edge w1 w2 can decrease the deficiency of H1 by at most one. In either case v∈V (H1 ) d(H , c, v)(k −1)/2, and a similar argument for H2 gives d(H , c) k −1. Thus sk (k −1)/(k 2 +1), as desired. 4. The 3-regular case In this section, we prove that s3 = 1/5. Proposition 3.2 gives the lower bound s3 1/5, and s3 1/5 follows from Theorem 4.1. Theorem 4.1. Let G be a graph with n vertices of degree three and all other vertices of degree one. Then there exists an edge coloring c of G such that d(G, c)n/5 and such that d(G, c, v) 1 for all v ∈ V (G). Proof. We construct the coloring c by induction on the number of vertices of G. Clearly, we may assume that G is connected. Furthermore, suppose that G has a bridge e connecting two vertices of degree three. Let H1 and H2 be the components of G\{e}, and let Gi = Hi ∪ {e} for i = 1, 2. Both G1 and G2 have fewer vertices than G, so we can color G1 and G2 by induction. Adding a constant to each color in G2 in such a way that the colorings of G1 and G2 agree on e, we obtain a coloring of G with the required properties. So, assume that G is a connected graph whose every bridge is a pendant edge. Next we reduce to the case where G is triangle-free. Suppose that G contains a triangle. Each vertex in the triangle has degree three in G, so consider the third edges e1 , e2 , and e3 incident on the vertices in the triangle. If no two of the ei share a vertex, then collapse the triangle to a vertex v and color the resulting graph by induction. By assumption, this coloring must have deficiency zero or one at v. In either case, extend to a coloring of G as demonstrated in Figs. 1a and b, respectively. The deficiency is not increased, so the extended coloring has the desired properties. If e1 , e2 , and e3 all meet at a single vertex, then G, which is connected, must be the graph in Fig. 2 a. Color G as indicated. In the final case, exactly two of the ei meet at a vertex.
Fig. 1. Collapsing a triangle in G.
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Fig. 2. G contains a triangle.
Fig. 3. G has nearby vertices of degree one.
So suppose G contains two triangles which share a common edge. Then there exist two other edges e1 and e1 of G incident on the pair of triangles. If e1 and e1 meet at a vertex, then G must be the graph in Fig. 2b, because all bridges are pendant edges by assumption; color as indicated. Otherwise e1 and e1 lead to distinct vertices, which may or may not be adjacent in G. In the former case, consider the remaining edges of G incident on these two vertices, e2 and e2 . They either meet at a vertex, lead to adjacent vertices, or lead to nonadjacent vertices. The first case yields a triangle with the three incident edges leading to distinct vertices, which was previously considered. In the second case repeat the procedure, again finding edges leading to distinct vertices, and there must exist sufficiently large i such that ei and ei lead to distinct, nonadjacent vertices. Finally then, G contains a structure of the form illustrated in Fig. 2c, where there are zero or more squares at the bottom and the two half-drawn edges lead to nonadjacent vertices. So collapse the whole structure to a single edge joining these two vertices and color by induction. Extend this to a coloring of G as indicated; the colors at the bottom of the structure are determined by the parity of the number of squares. Thus, we may assume that G is triangle-free, as desired. Now suppose that G contains two vertices of degree one at distance at most three from each other. If these two vertices are at distance one, then G is a single edge; any coloring suffices in this case. If these two vertices are at distance two, then G must be the graph in Fig. 3a; color as indicated. If these two vertices are at distance three, then consider the remaining edges of G incident on the path between these two vertices. These two edges must lead to distinct vertices, because G is triangle-free. If these vertices are adjacent then consider the remaining edges of G incident on them, and proceeding in this fashion yields a structure of the form illustrated in Fig. 3b, where there are zero or more squares and the two half-drawn edges lead to distinct, nonadjacent vertices. Collapse this structure to a single edge joining these vertices and color by induction. Extend this to a coloring of G as indicated; as before, the colors at the bottom are determined by the parity of the number of squares. So we are reduced to the case where G is a connected, triangle-free graph, where every bridge is a pendant edge, and where any two vertices of degree one are at distance at least four in G. Suppose that G has more than one pendant
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Fig. 4. Coloring cycles in G\M.
Fig. 5. The c1 coloring of odd cycles in G2 .
edge. Let v1 w1 and v2 w2 be distinct pendant edges of G where v1 and v2 are the vertices of degree one. Then w1 and w2 must be distinct, nonadjacent vertices. Let H be the graph obtained by adding the edge w1 w2 to G and by removing vertices v1 and v2 , and color H by induction. Color each edge in G, except the two given pendant edges, with the same color as in H, and color the two pendant edges with the color of w1 w2 in H. It is clear that this coloring has all the required properties. Thus, we may assume that G has at most one pendant edge, and, by Corollary 2.4, the graph G has a perfect matching. Let M ⊆ G be such a matching, and color 1 each edge in M. Every vertex in G\M has degree two or zero, so it is a union of isolated vertices and disjoint cycles. Color even cycles with no deficiency as in Fig. 4a, and color odd cycles with deficiency equal to one as in Fig. 4b. Since G is triangle-free, each odd cycle contains at least five vertices, and therefore d(G)n/5, completing the proof. 5. The 4-regular case In this section we prove that s4 = 2/5. The lower bound s4 2/5 follows from Proposition 3.2, and Theorem 5.1 says s4 2/5. Theorem 5.1. If G is a 4-regular graph on n vertices, then there exists a proper edge coloring c such that d(G, c) 2n/5. Proof. By Proposition 2.5, there exist disjoint 2-factors G1 and G2 with G=G1 ∪G2 . Among all such decompositions, choose one where G1 has the minimal number of odd cycles. We now present two different colorings of G and show that at least one of them has the desired property. To obtain the first coloring c1 , properly color each cycle in G1 by the colors 1 and 3, and, in the case of an odd cycle, color one edge 5. In G2 , properly color the even cycles by the colors 2 and 4. For each odd cycle in G2 , at least one of the cases in Fig. 5 must occur; if several occur choose one at random, and color the odd cycle as shown. Now, let the number of odd cycles in Gi be ki for i = 1, 2. Besides the cases covered in Fig. 5, at most k1 vertices have their adjacent edges colored {1, 2, 4, 5}, which increases the deficiency by one for each vertex, and the remaining vertices have their adjacent edges colored either {1, 2, 3, 4} or {2, 3, 4, 5}. Furthermore, for the cases in Fig. 5, there is at most
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Fig. 6. H has a cycle.
Fig. 7. The c2 coloring of odd cycles in G1 .
Fig. 8. The c2 coloring of odd cycles in G2 .
one additional deficiency per odd cycle in G2 because we have already counted one deficiency for each pair of colors 1 and 5. Thus d(G, c1 ) k1 + k2 2k2 , because G1 has the fewest possible odd cycles. We perform a construction before describing c2 , the second coloring. Let H1 ⊂ G1 be a matching containing no edges of even cycles and intersecting with every odd cycle in G1 by exactly one edge. Construct the matching H2 ⊂ G2 as follows: for each even cycle in G2 add to H2 any matching covering this cycle. For each odd cycle C in G2 choose two consecutive vertices v1 and v2 having the same degree in H1 ; this is possible because every vertex of G has degree zero or one in H1 . Let v0 be the other neighbor of v1 in C, and add to H2 the matching covering all vertices of C except v0 . Clearly, the graph H = H1 ∪ H2 is bipartite and has degree at most two. Suppose that H contains a cycle, which must have 2 edges from H1 . Then, as illustrated in Fig. 6, we could replace odd cycles in G1 by a single cycle, a contradiction because G1 has the fewest possible odd cycles. Therefore, H is a union of disjoint paths. Using this fact, we describe c2 . Color 4 each edge in H1 , and properly color the other edges of G1 by the colors 2 and 3. Also, as illustrated in Fig. 7, do this in such a way that no edge in H2 would meet edges colored 2 and 4 at one end and colored 3 and 4 at the other end. This is always possible because H has no cycles. We complete our description of c2 by coloring the edges in G2 . First, for each odd cycle in G2 , color 5 the edge v0 v1 and color the edge v1 v2 according to the cases in Fig. 8. Since v1 and v2 have the same degree in H1 , one of these cases must occur. Note that d(G, c2 , v0 ) = 1 and d(G, c2 , v1 ) = d(G, c2 , v2 ) = 0. Color 1 all the remaining edges of G2 \H2 . Finally, color the remaining edges in H2 as in Fig. 9. By construction, one of the five illustrated cases must occur, and we always increase the deficiency by at most one per pair of vertices. Hence, each cycle C in G2 increases the deficiency by at most |C|/2.
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Fig. 9. Coloring H2 .
Table 1 Values of and bounds on the sk k
sk
1 2 3 4
0 1/3 1/5 2/5
Parity
Bounds
Odd
k−1 k 2 +1
sk k−1 k+1
Even
k 2k+2
sk k−1 k+1
For each 3 let a be the number of -cycles in G2 . Then the number of vertices in G is n = ∞ d(G, c1 ) + 4d(G, c2 ) 2 4 a2m+1 + 5 5 5 m=1
=
∞
ma 2m+1 +
m=1
∞
=3 a ,
and
ma 2m
m=2
∞ ∞ 4m + 2 4m a2m+1 + a2m 5 5
m=1
=
∞
m=2
∞ 2 2n a = . 5 5 =3
Thus, d(G, ci )2n/5 for at least one of i = 1, 2, and the proof is complete. 6. Conclusion We have found the values of the first four sk , and we have general bounds as well. Our full results are shown in Table 1. Besides computing larger sk , another interesting question concerns finding k-regular graphs on n vertices with deficiency exactly sk n. It is easy to construct interval colorings for all 3-regular graphs on eight or fewer vertices, so any 3-regular graph with nonzero deficiency must have at least 10 vertices. Both the graph constructed in the proof of Proposition 3.2 and the Petersen graph are 3-regular graphs on 10 vertices with deficiency 2 = 10s3 . Similarly, any 4-regular graph must have at least five vertices, and K5 is a 4-regular graph on five vertices with deficiency 2 = 5s4 . This suggests the following question. Question 6.1. Let n be the fewest number of vertices on which a k-regular graph with nonzero deficiency exists. Is it always true that there exists a k-regular graph on n vertices with deficiency sk n? Note that an affirmative answer to Question 6.1 would give an algorithm to compute each sk in finite time. Although the answer is true for k 4, there is no a priori reason to expect the answer is true in general. Of course, an answer in either direction would be interesting. Our original goal was to find tight bounds on the deficiencies of k-regular graphs in terms of the number of vertices n. The bound sk n is tight for large n if disconnected graphs are permitted. It is interesting to bound asymptotically the deficiency of connected k-regular graphs. Finally, a problem first posed in [5] was to bound the deficiencies of arbitrary graphs in terms of the number of vertices.
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Acknowledgments This research was performed under the supervision of Prof. Joseph Gallian, Geir Helleloid, and Philip Matchett at the Research Program for Undergraduates held at the University of Minnesota Duluth. This work was supported by NSA Grant MDA904-02-1-0060 and NSF Grant DMS-0137611. References [1] [3] [4] [5]
A.S. Asratian, R.R. Kamalian, Interval colorings of the edges of a multigraph, Appl. Math. 5 (1987) 25–34 (in Russian). K. Giaro, The complexity of consecutive -coloring of bipartite graphs: 4 is easy, 5 is hard, Ars Combin. 47 (1997) 287–298. K. Giaro, M. Kubale, M. Małafiejski, On the deficiency of bipartite graphs, Discrete Appl. Math. 94 (1999) 193–203. K. Giaro, M. Kubale, M. Małafiejski, Consecutive colorings of the edges of general graphs, Discrete Math. 236 (2001) 131–143.
The deficiency of a regular graph Alexander Schwartz Received 25 September 2002; received in revised form 11 November 2005; accepted 13 March 2006
Abstract A proper edge coloring c : E(G) → Z of a finite simple graph G is an interval coloring if the colors used at each vertex form a consecutive interval of integers. Many graphs do not have interval colorings, and the deficiency of a graph is an invariant that measures how close a graph comes to having an interval coloring. In this paper we search for tight upper bounds on the deficiencies of k-regular graphs in terms of the number of vertices. We find exact values for 1 k 4 and bounds for larger k. © 2006 Elsevier B.V. All rights reserved. Keywords: Deficiency; Interval coloring; Consecutive coloring
1. Introduction In this paper we study a relatively new invariant of finite, simple graphs, namely the deficiency. This is an extension of an older topic first introduced by Asratian and Kamalian in 1987 [1]. They studied interval colorings of graphs, proper edge colorings such that the colors incident at each vertex form a consecutive interval of integers. More recently, such colorings have also been called consecutive colorings. There are several papers on this topic, although almost all of them have been confined to bipartite graphs. This is likely because interval colorings have applications to scheduling problems which are mostly relevant in the bipartite case. Many graphs do not have interval colorings, and Giaro et al. [4] extended this concept to an invariant applicable to all graphs in 1999. The deficiency of a graph is a measure of how close it comes to having an interval coloring. Specifically, the deficiency of a graph is the minimum number of pendant edges one must attach so that the resulting graph has an interval coloring. Very little is known about the deficiencies of arbitrary graphs. Outside of the work on bipartite graphs, only very restricted subclasses have been considered, such as complete graphs and wheels [5]. The general problem is quite complicated; in particular, it is NP-hard [3]. In this paper, we examine the deficiencies of regular graphs. Our central aim is to find tight upper bounds on the deficiency of a k-regular graph in terms of the number of vertices. To this aim we define sk , which is the supremum of the ratio between the deficiency and the number of vertices over all k-regular graphs. We find the exact values of s1 , s2 , s3 , and s4 and bounds on sk for larger k. 2. Preliminaries Definition 2.1. Let S be a finite set of integers. The deficiency of S is the number of integers between min S and max S not appearing in S. E-mail address: [email protected]. 0012-365X/$ - see front matter © 2006 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2006.03.059
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Let G be a finite, simple graph, let c : E(G) → Z be a proper edge coloring, and let v ∈ V (G) be arbitrary. Let S ⊂ Z consist of the colors of edges incident on v. Then the number of deficiencies at v, denoted d(G, c, v), is the deficiency of S. Furthermore, let d(G, c) = v∈V (G) d(G, c, v), and let the deficiency of G be d(G) = minc d(G, c), where the minimum is taken over all proper edge colorings of G. Our primary goal is to bound the deficiency of a k-regular graph in terms of the number of vertices. First, we recall some classical graph theoretic results. Proposition 2.2 (Vizing). Let G be a graph where every vertex has degree at most . Then G admits a proper edge coloring with + 1 colors. Proposition 2.3 (Petersen). Any 3-regular graph with at most two bridges has a perfect matching. Corollary 2.4. If every vertex of a graph G has degree one or three and G has at most one bridge, then G has a perfect matching. Proposition 2.5 (Petersen). Any 4-regular graph contains a 2-factor. For any k-regular graph G, Proposition 2.2 gives an edge coloring c such that d(G, c, v)1 for all v ∈ V (G). In particular, if G has n vertices then d(G)n, and the following definition is justified. Definition 2.6. Let Rk be the set of all k-regular graphs, and set sk = supG∈Rk d(G)/|V (G)|. Clearly d(G)sk n for all k-regular graphs G on n vertices. Furthermore, by taking disjoint unions, we see that this bound is tight for large n. The goal of this paper is to bound and compute these sk ’s. It is quite easy to compute s1 and s2 , but the others are relatively difficult. Here and throughout we use Kn to denote the complete graph on n vertices. Proposition 2.7. s1 = 0 and s2 = 13 . Proof. Any 1-regular graph is simply a union of disjoint edges, and any edge coloring results in no deficiencies; thus, s1 = 0. Any 2-regular graph is a union of disjoint cycles. All even cycles can be properly colored by colors 1 and 2 with no deficiency. Any odd cycle can be properly colored by colors 1 and 2 except for a single edge colored 3, and the deficiency is equal to one. Then s2 13 because a triangle is the smallest odd cycle, and finally s2 = 13 because d(K3 ) = 1. 3. General bounds on the sk We now find upper and lower bounds for sk . In later sections, we use the lower bounds given here to find s3 and s4 . Proposition 3.1. sk (k − 1)/(k + 1). Proof. Let G be an arbitrary k-regular graph. Using Proposition 2.2, properly color the edges of G using the integers from 1 to k + 1. Swap the colors as necessary to give a coloring c where 1 and k + 1 occur the fewest times. Each vertex touches edges with k different colors, and exactly one color is missing at each vertex. By construction, there are at least 2|V (G)|/(k + 1) vertices where either 1 or k + 1 is missing, and these vertices have deficiency zero. All other vertices have deficiency one, so d(G, c)/|V (G)|1 − 2/(k + 1), completing the proof. Proposition 3.2. If k is even, then sk k/(2k + 2), and if k is odd, then sk (k − 1)/(k 2 + 1). Proof. It is proved in [5] that d(K2+1 ) = for all positive integers . This takes care of the even case, so suppose that k is odd. The case k = 1 is trivial, so assume that k 3. Let G be the graph obtained by subdividing an edge of Kk+1 with a vertex w. Suppose that c is a coloring of G with d(G, c, v) = 0 for all v = w. Reducing all the colors modulo k gives an edge coloring which is proper everywhere except possibly at w. Suppose that the two colors incident on w are also distinct. Then each of the k colors is used at most (k + 2)/2 = (k + 1)/2 times, but G has k(k + 1)/2 + 1
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edges, a contradiction. Thus, for all proper edge colorings c of G, either d(G, c, v) > 0 for some v = w or the two colors incident on w differ by at least k. We use this fact to construct a graph that gives the desired lower bound on sk . Take (k − 1)/2 copies of G and identify their vertices w to obtain a graph H on (k + 1)(k − 1)/2 + 1 vertices. Let c be any proper edge coloring of H, and let be the number of vertices v such that d(H, c, v) > 0 and v = w. Suppose < (k − 1)/2. Then, by the above argument, there exist at least (k − 1)/2 − > 0 disjoint pairs of edges incident with w whose colors differ by at least k. This implies that there are two edges incident on w whose colors differ by at least k + (k − 1)/2 − − 1, and d(H, c, w)(k + 1)/2 − because w has degree k − 1 in H. Hence d(H, c) (k + 1)/2. If (k − 1)/2, then clearly d(H, c) (k − 1)/2. Thus d(H, c) (k − 1)/2 for any coloring c, and equality can occur only when d(H, c, w) = 0. Let H1 and H2 be two copies of H with wi ∈ V (Hi ) corresponding to w for i = 1, 2. Let H be the disjoint union of H1 and H2 with the edge w1 w2 added. Then H is a k-regular graph on k 2 + 1 vertices. Let c be any proper edge coloring of H , and let the induced colorings on H1 and H2 be c1 and c2 , respectively. If d(H1 , c1 ) = (k − 1)/2, then d(H1 , c1 , w1 ) = 0, and the extra color on the edge w1 w2 cannot decrease the deficiency of H1 in H. Otherwise d(H1 , c1 )(k + 1)/2, and the extra color of the edge w1 w2 can decrease the deficiency of H1 by at most one. In either case v∈V (H1 ) d(H , c, v)(k −1)/2, and a similar argument for H2 gives d(H , c) k −1. Thus sk (k −1)/(k 2 +1), as desired. 4. The 3-regular case In this section, we prove that s3 = 1/5. Proposition 3.2 gives the lower bound s3 1/5, and s3 1/5 follows from Theorem 4.1. Theorem 4.1. Let G be a graph with n vertices of degree three and all other vertices of degree one. Then there exists an edge coloring c of G such that d(G, c)n/5 and such that d(G, c, v) 1 for all v ∈ V (G). Proof. We construct the coloring c by induction on the number of vertices of G. Clearly, we may assume that G is connected. Furthermore, suppose that G has a bridge e connecting two vertices of degree three. Let H1 and H2 be the components of G\{e}, and let Gi = Hi ∪ {e} for i = 1, 2. Both G1 and G2 have fewer vertices than G, so we can color G1 and G2 by induction. Adding a constant to each color in G2 in such a way that the colorings of G1 and G2 agree on e, we obtain a coloring of G with the required properties. So, assume that G is a connected graph whose every bridge is a pendant edge. Next we reduce to the case where G is triangle-free. Suppose that G contains a triangle. Each vertex in the triangle has degree three in G, so consider the third edges e1 , e2 , and e3 incident on the vertices in the triangle. If no two of the ei share a vertex, then collapse the triangle to a vertex v and color the resulting graph by induction. By assumption, this coloring must have deficiency zero or one at v. In either case, extend to a coloring of G as demonstrated in Figs. 1a and b, respectively. The deficiency is not increased, so the extended coloring has the desired properties. If e1 , e2 , and e3 all meet at a single vertex, then G, which is connected, must be the graph in Fig. 2 a. Color G as indicated. In the final case, exactly two of the ei meet at a vertex.
Fig. 1. Collapsing a triangle in G.
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Fig. 2. G contains a triangle.
Fig. 3. G has nearby vertices of degree one.
So suppose G contains two triangles which share a common edge. Then there exist two other edges e1 and e1 of G incident on the pair of triangles. If e1 and e1 meet at a vertex, then G must be the graph in Fig. 2b, because all bridges are pendant edges by assumption; color as indicated. Otherwise e1 and e1 lead to distinct vertices, which may or may not be adjacent in G. In the former case, consider the remaining edges of G incident on these two vertices, e2 and e2 . They either meet at a vertex, lead to adjacent vertices, or lead to nonadjacent vertices. The first case yields a triangle with the three incident edges leading to distinct vertices, which was previously considered. In the second case repeat the procedure, again finding edges leading to distinct vertices, and there must exist sufficiently large i such that ei and ei lead to distinct, nonadjacent vertices. Finally then, G contains a structure of the form illustrated in Fig. 2c, where there are zero or more squares at the bottom and the two half-drawn edges lead to nonadjacent vertices. So collapse the whole structure to a single edge joining these two vertices and color by induction. Extend this to a coloring of G as indicated; the colors at the bottom of the structure are determined by the parity of the number of squares. Thus, we may assume that G is triangle-free, as desired. Now suppose that G contains two vertices of degree one at distance at most three from each other. If these two vertices are at distance one, then G is a single edge; any coloring suffices in this case. If these two vertices are at distance two, then G must be the graph in Fig. 3a; color as indicated. If these two vertices are at distance three, then consider the remaining edges of G incident on the path between these two vertices. These two edges must lead to distinct vertices, because G is triangle-free. If these vertices are adjacent then consider the remaining edges of G incident on them, and proceeding in this fashion yields a structure of the form illustrated in Fig. 3b, where there are zero or more squares and the two half-drawn edges lead to distinct, nonadjacent vertices. Collapse this structure to a single edge joining these vertices and color by induction. Extend this to a coloring of G as indicated; as before, the colors at the bottom are determined by the parity of the number of squares. So we are reduced to the case where G is a connected, triangle-free graph, where every bridge is a pendant edge, and where any two vertices of degree one are at distance at least four in G. Suppose that G has more than one pendant
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Fig. 4. Coloring cycles in G\M.
Fig. 5. The c1 coloring of odd cycles in G2 .
edge. Let v1 w1 and v2 w2 be distinct pendant edges of G where v1 and v2 are the vertices of degree one. Then w1 and w2 must be distinct, nonadjacent vertices. Let H be the graph obtained by adding the edge w1 w2 to G and by removing vertices v1 and v2 , and color H by induction. Color each edge in G, except the two given pendant edges, with the same color as in H, and color the two pendant edges with the color of w1 w2 in H. It is clear that this coloring has all the required properties. Thus, we may assume that G has at most one pendant edge, and, by Corollary 2.4, the graph G has a perfect matching. Let M ⊆ G be such a matching, and color 1 each edge in M. Every vertex in G\M has degree two or zero, so it is a union of isolated vertices and disjoint cycles. Color even cycles with no deficiency as in Fig. 4a, and color odd cycles with deficiency equal to one as in Fig. 4b. Since G is triangle-free, each odd cycle contains at least five vertices, and therefore d(G)n/5, completing the proof. 5. The 4-regular case In this section we prove that s4 = 2/5. The lower bound s4 2/5 follows from Proposition 3.2, and Theorem 5.1 says s4 2/5. Theorem 5.1. If G is a 4-regular graph on n vertices, then there exists a proper edge coloring c such that d(G, c) 2n/5. Proof. By Proposition 2.5, there exist disjoint 2-factors G1 and G2 with G=G1 ∪G2 . Among all such decompositions, choose one where G1 has the minimal number of odd cycles. We now present two different colorings of G and show that at least one of them has the desired property. To obtain the first coloring c1 , properly color each cycle in G1 by the colors 1 and 3, and, in the case of an odd cycle, color one edge 5. In G2 , properly color the even cycles by the colors 2 and 4. For each odd cycle in G2 , at least one of the cases in Fig. 5 must occur; if several occur choose one at random, and color the odd cycle as shown. Now, let the number of odd cycles in Gi be ki for i = 1, 2. Besides the cases covered in Fig. 5, at most k1 vertices have their adjacent edges colored {1, 2, 4, 5}, which increases the deficiency by one for each vertex, and the remaining vertices have their adjacent edges colored either {1, 2, 3, 4} or {2, 3, 4, 5}. Furthermore, for the cases in Fig. 5, there is at most
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Fig. 6. H has a cycle.
Fig. 7. The c2 coloring of odd cycles in G1 .
Fig. 8. The c2 coloring of odd cycles in G2 .
one additional deficiency per odd cycle in G2 because we have already counted one deficiency for each pair of colors 1 and 5. Thus d(G, c1 ) k1 + k2 2k2 , because G1 has the fewest possible odd cycles. We perform a construction before describing c2 , the second coloring. Let H1 ⊂ G1 be a matching containing no edges of even cycles and intersecting with every odd cycle in G1 by exactly one edge. Construct the matching H2 ⊂ G2 as follows: for each even cycle in G2 add to H2 any matching covering this cycle. For each odd cycle C in G2 choose two consecutive vertices v1 and v2 having the same degree in H1 ; this is possible because every vertex of G has degree zero or one in H1 . Let v0 be the other neighbor of v1 in C, and add to H2 the matching covering all vertices of C except v0 . Clearly, the graph H = H1 ∪ H2 is bipartite and has degree at most two. Suppose that H contains a cycle, which must have 2 edges from H1 . Then, as illustrated in Fig. 6, we could replace odd cycles in G1 by a single cycle, a contradiction because G1 has the fewest possible odd cycles. Therefore, H is a union of disjoint paths. Using this fact, we describe c2 . Color 4 each edge in H1 , and properly color the other edges of G1 by the colors 2 and 3. Also, as illustrated in Fig. 7, do this in such a way that no edge in H2 would meet edges colored 2 and 4 at one end and colored 3 and 4 at the other end. This is always possible because H has no cycles. We complete our description of c2 by coloring the edges in G2 . First, for each odd cycle in G2 , color 5 the edge v0 v1 and color the edge v1 v2 according to the cases in Fig. 8. Since v1 and v2 have the same degree in H1 , one of these cases must occur. Note that d(G, c2 , v0 ) = 1 and d(G, c2 , v1 ) = d(G, c2 , v2 ) = 0. Color 1 all the remaining edges of G2 \H2 . Finally, color the remaining edges in H2 as in Fig. 9. By construction, one of the five illustrated cases must occur, and we always increase the deficiency by at most one per pair of vertices. Hence, each cycle C in G2 increases the deficiency by at most |C|/2.
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Fig. 9. Coloring H2 .
Table 1 Values of and bounds on the sk k
sk
1 2 3 4
0 1/3 1/5 2/5
Parity
Bounds
Odd
k−1 k 2 +1
sk k−1 k+1
Even
k 2k+2
sk k−1 k+1
For each 3 let a be the number of -cycles in G2 . Then the number of vertices in G is n = ∞ d(G, c1 ) + 4d(G, c2 ) 2 4 a2m+1 + 5 5 5 m=1
=
∞
ma 2m+1 +
m=1
∞
=3 a ,
and
ma 2m
m=2
∞ ∞ 4m + 2 4m a2m+1 + a2m 5 5
m=1
=
∞
m=2
∞ 2 2n a = . 5 5 =3
Thus, d(G, ci )2n/5 for at least one of i = 1, 2, and the proof is complete. 6. Conclusion We have found the values of the first four sk , and we have general bounds as well. Our full results are shown in Table 1. Besides computing larger sk , another interesting question concerns finding k-regular graphs on n vertices with deficiency exactly sk n. It is easy to construct interval colorings for all 3-regular graphs on eight or fewer vertices, so any 3-regular graph with nonzero deficiency must have at least 10 vertices. Both the graph constructed in the proof of Proposition 3.2 and the Petersen graph are 3-regular graphs on 10 vertices with deficiency 2 = 10s3 . Similarly, any 4-regular graph must have at least five vertices, and K5 is a 4-regular graph on five vertices with deficiency 2 = 5s4 . This suggests the following question. Question 6.1. Let n be the fewest number of vertices on which a k-regular graph with nonzero deficiency exists. Is it always true that there exists a k-regular graph on n vertices with deficiency sk n? Note that an affirmative answer to Question 6.1 would give an algorithm to compute each sk in finite time. Although the answer is true for k 4, there is no a priori reason to expect the answer is true in general. Of course, an answer in either direction would be interesting. Our original goal was to find tight bounds on the deficiencies of k-regular graphs in terms of the number of vertices n. The bound sk n is tight for large n if disconnected graphs are permitted. It is interesting to bound asymptotically the deficiency of connected k-regular graphs. Finally, a problem first posed in [5] was to bound the deficiencies of arbitrary graphs in terms of the number of vertices.
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Acknowledgments This research was performed under the supervision of Prof. Joseph Gallian, Geir Helleloid, and Philip Matchett at the Research Program for Undergraduates held at the University of Minnesota Duluth. This work was supported by NSA Grant MDA904-02-1-0060 and NSF Grant DMS-0137611. References [1] [3] [4] [5]
A.S. Asratian, R.R. Kamalian, Interval colorings of the edges of a multigraph, Appl. Math. 5 (1987) 25–34 (in Russian). K. Giaro, The complexity of consecutive -coloring of bipartite graphs: 4 is easy, 5 is hard, Ars Combin. 47 (1997) 287–298. K. Giaro, M. Kubale, M. Małafiejski, On the deficiency of bipartite graphs, Discrete Appl. Math. 94 (1999) 193–203. K. Giaro, M. Kubale, M. Małafiejski, Consecutive colorings of the edges of general graphs, Discrete Math. 236 (2001) 131–143.