The smallest k-regular h-edge-connected graphs without 1-factors

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1 The smallest k-regular h-edge-connected graphs without 1-factors John Ganci, Douglas B. West March 30, 2003 Dedicated to the memory of Bill Tutte Abstract The minimum order of a k-regular h-edge-connected graph of even order without 1-factors is determined. For h = k 2, the value is k 2 + 3k 2 when k is odd and k 2 + 2k 2 when k is even. The general formula is (α + 1) (t + 2) 2, where α is the least odd integer greater than k, and t is 2h 2h+2 k h when k h is even and is k h 1 when k h is odd. Constructions exist for all even orders at least this large. 1 Introduction Petersen [3] proved that every 3-regular graph with no cut-edge has a 1-factor. This has many generalizations. Showing that every k-regular (k 1)-edge-connected graph of even order has a 1-factor is a standard elementary exercise (# in [7]). In addition, every edge in such a graph belongs to some 1-factor. (Berge [1, page 162], # in [7]). The Petersen graph shows that Petersen s result is sharp in that edge-disjoint 1-factors need not exist. For k-regular graphs, the edge-connectivity condition also is sharp: k-regular graphs that are not (k 1)-edge-connected need not have 1-factors. Constructing such graphs is another standard exercise (#3.3.7 in [7]). In this note, we explore this sharpness by finding the minimum (even) order of k-regular h-edge-connected graphs without 1-factors, for all pairs (k, h) with 0 h k 2. When h = k 2, the minimum order is k 2 +3k 2 if k is odd and is k 2 + 2k 2 if k is even. Examples exist for all even orders above the minimum. Texas Instruments, 6500 Chase Oaks Blvd MS 8440, Plano, TX 75023, jganci@ti.com Department of Mathematics, University of Illinois, Urbana, IL 61801, west@math.uiuc.edu 1

2 The extremal graphs are not unique but have some structural properties that emerge from the proof of the bound. Regularity is crucial. Weakening the regularity requirement to allow vertices of degrees l 2 and l permits the example K l 2,l, which is l 2-edge-connected and has no 1-factor but has only 2l 2 vertices. After obtaining these results, we learned of prior work on a related problem. Pila [4] found the smallest even order of k-regular h-connected graphs without 1-factors. When 1 < h < k 2, our proof shows that extremal graphs in the h-edge-connected case cannot be h-connected. Hence Pila s result differs from ours for 1 < h < k 2. It has many special cases for small h and k. For h-edge-connected graphs, we find a uniform formula for the threshold. When h {1, k 2}, the two results agree. Wallis [6] settled the question earlier for the case h = 1. 2 The Lower Bound Tutte s 1-Factor Theorem [5] states that a graph G has a 1-factor if and only if, for all S V (G), the number of components of G S with odd order is at most S. A set S such that G S has more than S odd components is called a Tutte set in G. Let n(g) denote the order of a graph G. Theorem 1 Let G be a k-regular h-edge-connected graph of even order, and let α be the least odd integer greater than k. If h k 2 and G has no 1-factor, then n(g) (α+1)(t+2) 2, where t = 2h 2h+2 when k h is even and t = when k h is odd. k h k h 1 Proof. Since G has no 1-factor, G has a Tutte set S. Let s = S. Since G has even order, parity in counting vertices ensures that G S has at least s + 2 odd components. Let m be the number of edges of G with endpoints in both S and V (G) S. Since vertices have degree at most k, we have m ks, and equality requires S to be an independent set. Now we obtain a lower bound on m. From S to an odd component of G S there are at least h edges, since G is h-edge-connected. Since the degree-sum in an odd component of G S must be even, there must be at least h + 1 edges to S if h and k have opposite parity. Among the odd components of G S, let a be the number having at most k vertices, and let b be the number having at least α vertices (note that α = k + 2 when k is odd, and α = k + 1 when k is even). Altogether G S has a + b odd components. Since vertices of G have degree at least k, the cut from S to a component of G S with r vertices has at least r(k + 1 r) edges. For 1 r k, this is always at least k (with 2

3 equality only when the component is K 1 or K k ). We conclude that m ka + hb. Since b s + 2 a, we obtain ks ka + h(s + 2 a), which simplifies to s 2h + a, since s k h is an integer. When k h is odd, we have the stronger bound ks ka + (h + 1)(s + 2 a), which simplifies to s 2h+2 k h 1 + a. Counting S and the odd components of G S yields n(g) s+a+bα. Since b s+2 a, we have n(g) (s + 2 a)(α + 1) + 2a 2. Now s t + a and a 0 complete the proof. 3 Constructions We consider k-regular graphs. When h = k 1 there is always a 1-factor. When h = k 2, the lower bound in Theorem 1 is (α + 1)k 2. As h decreases, the lower bound decreases to 3α + 1 when h = 1 (except that when k = 4 the value is 4α + 2 at h = 1). When h = 0, the lower bound is 2α when k h is even and 3α + 1 when k h is odd. Our aim now is to show that for each pair (k, h) with 0 h k 2, the lower bound in Theorem 1 is sharp. We first describe a graph achieving equality when h = k 2. This graph arises in a different way when h = k 2 in the general construction, so we leave the verification to there. When k is odd, let H = C k 2 + 2K 2, where + denotes disjoint union, and mf is the graph consisting of m disjoint copies of F. When k is even, let H = k 2 2 K 2 + 3K 1. In each case, H is a (k 2)-edge-connected graph with α vertices, of which k 2 have degree k 1 and the rest have degree k. Take k copies of H. Add an independent set S of size k 2 and matchings from S to the set of vertices of degree k 1 in each copy of H. Our general construction begins with a well-known family studied by Harary. For m > l, the Harary graph H l,m is formed as follows. Let V (H l,m ) be the set of congruence classes of integers modulo m. Place these vertices around a circle in the natural way. Make each vertex adjacent to the l/2 nearest vertices in each direction along the circle. If l is odd, then add the edges {(i m/2, i): 0 i (m 1)/2 }. This last step adds one incident edge to each vertex, except that 0 receives two more incident edges when lm is odd. By construction, H l,m is l-regular, except that when l and m are odd it has one vertex of degree l + 1. Harary [2] proved that H l,m also is l-connected. We add edges to Harary graphs to obtain the components of G S in our construction. Construction 2 Let h, k, m be nonnegative integers such that m > k > h and that m is odd and k h is even. The graph F h,k,m formed by adding the edges {(i, i + m/2 ): 1 i (m h)/2 } to the Harary graph H k 1,m is an h-connected graph with m vertices, of which h have degree k 1 and all the rest have degree k. 3

4 Proof. We start with a (k 1)-connected graph and add edges. Since h < k, the resulting graph is h-connected. The added edges do not already belong to H k 1,m. They form a matching and increase the degrees of 2 (m h)/2 vertices to k. Since m is odd and k h is even and H k 1,m already has one vertex of degree k if k and h are even, F h,k,m has m h vertices of degree k and h vertices of degree k 1. Theorem 3 For h k 2, there is a k-regular h-edge-connected graph of even order n with no 1-factor if and only if n (α + 1)(t + 2) 2, where α and t are as defined in Theorem 1. Proof. Theorem 1 shows that the condition is necessary. For sufficiency, we may assume that k h is even, since otherwise we use the construction of a k-regular (h + 1)-edgeconnected graph of order n (when k h is odd, the value of t that results is the same as for k and h + 1). Let r = n (α + 1)(t + 2) 2; note that r is even. We build a graph G with an independent t-set S such that G S = (t+1)f h,k,α +F q,k,α+r, where q = (k h)t h. Since k h is even, we have t = 2h = 2h+k h 2 and k h k h h = (k h) 2h + k h 2 h q (k h)2h h = k 2, k h k h and hence F q,k,α+r exists. Let C denote this component of G S. We add edges from S to each component of G S. Altogether we add kt edges, with k at each vertex of S. From the other end, h of these edges go to each copy of F h,k,α, leaving kt h(t + 1) edges to arrive at C. This is exactly q, so each vertex with degree k 1 in G S receives one edge to raise its degree to k. Since k h 2, we have t h, and hence we can distribute the endpoints in S so that each vertex of S has at least one neighbor in each component of G S. By construction, G is k-regular with order n, and S is a Tutte set in G. It remains to show that G is h-edge-connected. Since q h, each component of G S is h-connected, so deleting fewer than h edges leaves these induced subgraphs connected. Also each vertex of S has k neighbors in G S and hence remains connected to some component of G S. Hence it suffices to produce h pairwise edge-disjoint paths from each component of G S to any other. Choose any two such components X and Y. From X and Y we have h edges to S; these are the end-edges of the desired h paths, paired up arbitrarily. For each such pair, the path is complete if their endpoints x and y in S are equal. Otherwise, we complete an x, y-path through one of the h components of G S other than X and Y, using distinct components for distinct pairs. 4

5 One would like to characterize the extremal graphs (k-regular, h-edge-connected, order (α +1)(t+2) 2, no 1-factors). The proof of Theorem 1 provides some necessary conditions. Equality in the bound requires equality in all the inequalities it uses. Consider the case k h = 2. Every extremal graph G has an independent set S of size t such that G S has t + 2 components, each with order α. There are exactly h edges from S to each component of G S, and each component of G S has exactly (αk h)/2 edges. As in the proof of Theorem 3, it then suffices that (1) each component of G S be h-edge-connected, and (2) each vertex of S have a neighbor in each component of G S. Condition (1) is not necessary. When (k, h) = (7, 5), there is a 7-regular 5-edge-connected graph with 68 vertices in which each component of G S has degree list (7, 7, 7, 7, 7, 7, 6, 6, 4) and hence is not 5-edge-connected. A weaker condition that is clearly necessary is that each component of G S have no edge cut of size less than k that separates a vertex from the set of vertices of degree less than k. Condition (2) also is not necessary. For (k, h) = (5, 3), with each of the five components of G S having degree list (5, 5, 5, 5, 4, 4, 4), trade neighbors for two vertices x, y S and two components of G S, so x and y each has no neighbor in one component of G S. The resulting graph is 5-regular and 3-edge-connected. We do not have a characterization of the extremal graphs, even when h = k 2. In the smallest case, which is (k, h) = (3, 1), the extremal graph is unique. It is the standard example of a 3-regular connected graph with no 1-factor. With 16 vertices, it must have a cut-vertex with a cut-edge to each of three components of order 5. These components must have degree list (3, 3, 3, 3, 2), which can only be obtained by subdividing one edge of K 4. References [1] Berge, C., Graphs and Hypergraphs. (North Holland, 1973), translation and revision of Graphes et Hypergraphes (Dunod, 1970). [2] Harary, F., The maximum connectivity of a graph. Proc. Nat. Acad. Sci. U.S.A. 48 (1962), [3] Petersen, J., Die Theorie der regulären Graphen. Acta. Math. 15 (1891), [4] Pila, J., Connected regular graphs without one-factors. Ars Combinatoria 18 (1983), [5] Tutte, W. T., The factorization of linear graphs. J. Lond. Math. Soc. 22 (1947),

6 [6] Wallis, W. D., The smallest regular graphs without one-factors. Ars Combinatoria 11 (1981), [7] West, D. B., Introduction to Graph Theory, 2nd edition. (Prentice Hall, 2001). 6