Toughness, minimum degree, and spanning cubic subgraphs

Erratum Toughness, Minimum Degree, and Spanning Cubic Subgraphs D. Bauer,1 T. Niessen,2 and E. Schmeichel3 1

Department of Mathematical Sciences, Stevens Institute of Technology, Hoboken, New Jersey 07030 2 Institut fu¨r Statistik, RWTH Aachen, 52056 Aachen, Federal Republic of Germany 3 Department of Mathematics and computer science, San Jose State University, San Jose, California 95192

DOI 10.1002/jgt.20033

The above article was published in the Journal of Graph Theory, Vol. 45, No. 2, pp. 119–141, February 2004 with incorrect figures. The electronic version has been replaced with the correct figures; the corrected print version follows. The article originally posted in Wiley InterScience is available from the publisher should anyone require access to it. We regret any inconvenience this may have caused.

ß 2004 Wiley Periodicals, Inc. 144

Toughness, Minimum Degree, and Spanning Cubic Subgraphs D. Bauer,1 T. Niessen,2 and E. Schmeichel3 1

DEPARTMENT OF MATHEMATICAL SCIENCES STEVENS INSTITUTE OF TECHNOLOGY HOBOKEN, NEW JERSEY 07030 E-mail: [email protected] 2

INSTITUT FU¨R STATISTIK, RWTH AACHEN 52056 AACHEN, FEDERAL REPUBLIC OF GERMANY E-mail: [email protected] 3

DEPARTMENT OF MATHEMATICS AND COMPUTER SCIENCE SAN JOSE STATE UNIVERSITY SAN JOSE, CALIFORNIA 95192 E-mail: [email protected]

Received January 19, 2000; Revised July 8, 2003

DOI 10.1002/jgt.10149

Abstract: Degree conditions on the vertices of a t-tough graph G (1 t < 3) are presented which ensure the existence of a spanning cubic subgraph in G. These conditions are best possible to within a small additive constant for every fixed rational t 2 [1,4=3) [ [2,8=3). ß 2003 Wiley Periodicals, Inc. J Graph Theory 45: 119–141, 2004

Keywords: toughness; minimum vertex degree; 3-factors

ß 2003 Wiley Periodicals, Inc. 119

120 JOURNAL OF GRAPH THEORY

1. INTRODUCTION In this paper, we present minimum degree conditions on the vertices of a t-tough graph G, 1 t < 3, which ensure the existence of a spanning cubic subgraph in G. Our conditions are best possible to within a small additive constant for every fixed rational t 2 ½1; 4=3Þ [ ½2; 8=3Þ. We begin with a few definitions and some notation. Additional definitions will be given later as needed. A good reference for any undefined terms is [5]. We will consider only undirected graphs with no loops or multiple edges. Let !ðGÞ denote the number of components of a graph G. A graph G is t-tough if jSj t!ðG SÞ for every subset S of the vertex set VðGÞ of G with !ðG SÞ > 1. The toughness of G, denoted ðGÞ, is the maximum value of t for which G is t-tough (taking ðKn Þ ¼ 1, for all n 1). We let ðGÞ denote the minimum vertex degree in G. A k-factor in G is a k-regular spanning subgraph of G. If A; B VðGÞ and v 2 VðGÞ, we use eðv; BÞ toP denote the number of edges joining v to a vertex of B, and eðA; BÞ to denote v2A eðv; BÞ. For A VðGÞ, we will use hAi to represent the subgraph of G induced by A. Toughness in graphs was introduced by Chva´tal in [6]. Clearly, being 1-tough is a necessary condition for a graph to be hamiltonian. Chva´tal conjectured that there exists a constant t0 such that every t0 -tough graph is hamiltonian, and this conjecture remains open. He also conjectured that every k-tough graph on n vertices with n k þ 1 and kn even has a k-factor. It was later established that this conjecture is both true and best possible, as the following result shows. Theorem 1.1 ([7]).

Let k 1 be an integer.

(i) Every k-tough graph on n k þ 1 vertices with kn even contains a k-factor. (ii) For any " > 0, there exists a ðk "Þ-tough graph G on n vertices with n k þ 1 and kn even which does not contain a k-factor. Let G be a t-tough graph with 1 t < k. It seems interesting to consider what conditions on G in addition to toughness (e.g., a minimum degree condition) would guarantee that G contains a k-factor. In [3] (see also [2]) Bauer and Schmeichel considered this problem for k ¼ 2. Their effort to determine a minimum degree condition that ensures a 2-factor in a t-tough graph (1 t < 2) stemmed from an intense interest in the (now disproved [1]) conjecture that all 2tough graphs are hamiltonian. In fact, examples discovered in [3] were forerunners of the examples that eventually led to the discovery, for any " > 0, of the ð94 "Þ-tough nontraceable graphs found in [1]. Their investigation led to the following results. Theorem 1.2 ([3]).

Let 1 t < 3=2.

t (i) If G is a t-tough graph on n 3 vertices with ðGÞ ð21 þ tÞn, then G contains a 2-factor. (ii) There exists an infinite family of t-tough graphs G on n vertices with t 5 ðGÞ ð21 þ tÞn 2 which do not contain a 2-factor.

SPANNING CUBIC-SUBGRAPHS

121

Note that the degree bound in Theorem 1.2 is best possible to within a small additive constant for every t 2 ½1; 3=2Þ. By contrast, the degree bound in the following theorem is merely asymptotically optimal for infinitely many t 2 ½3=2; 2Þ. Theorem 1.3 [3]. (i) Let 3=2 t < 2. If G is a t-tough graph on n 3 vertices with ðGÞ t t2 1 ð21 þ tÞð7t 7 t2 Þn, then G contains a 2-factor. (ii) Let t ¼ ð2r 1Þ=r, where r 1 is any integer. For any " > 0, there exists t2 1 an infinite family of t-tough graphs G with ðGÞ ðð2t 1þtÞð7t7t2 Þ "Þn, such that G does not contain a 2-factor. Our goal in the present paper is to give tight minimum degree conditions which will guarantee that a t-tough graph, 1 t < 3, contains a 3-factor (i.e., a spanning cubic subgraph). Define the function f : ½1; 3Þ ! R by 8 for 1 t < 4=3; < t=ð4t 2Þ f ðtÞ ¼ ð4 tÞ=ð2t þ 4Þ for 4=3 t < 8=3; : 2ð3 tÞ=ðt þ 2Þ for 8=3 t < 3: Note that f is a continuous and monotone decreasing function. Our first result is the following. Theorem 1.4. Let 1 t < 3. If G is a t-tough graph on n 4 vertices with n even and ðGÞ > f ðtÞn þ 2, then G contains a 3-factor. We will also explore the quality of the degree bound in Theorem 1.4. For t 2 ½1; 4=3Þ [ ½2; 8=3Þ, the degree bound is best possible to within a small additive constant as the next result shows. Theorem 1.5. For any rational t 2 ½1; 4=3Þ [ ½2; 8=3Þ, there exists an infinte family of t-tough graphs G on n 4 vertices, with n even and ðGÞ > f ðtÞn 8, which do not contain a 3-factor. For any rational t such that 4=3 t < 2, we have constructed examples which prove the following theorem. Theorem 1.6. For any rational t, 4=3 t < 2, there exist infinitely many t-tough graphs G of even order n with no 3-factor satisfying ðGÞ ð7t t 6Þn 3. f ðtÞ Since t=ð7t 6Þ < 1:09 for t 2 ½4=3; 2Þ, we see that the bound in Theorem 1.4 is tight to within a factor of 1:09 for any rational t 2 ½4=3; 2Þ. When 8=3 t < 3, the quality of the degree bound in Theorem 1.4 remains 2 unsettled. Letting gðtÞ ¼ 2ð16tt344 t2 Þ, we have the following result.

Theorem 1.7. Let t ¼ ð3r 1Þ=r, where r 5 is any odd integer relatively prime to 3. For any " > 0, there exists an infinite family of t-tough graphs G on n vertices, n even, with ðGÞ > ð f ðtÞgðtÞ "Þn and such that G does not contain a 3-factor.


Since 1=2 gðtÞ 1, for t 2 ½8=3; 3Þ, Theorem 1.7 implies that the degree bound in Theorem 1.4 is tight to within a factor of 2 for infinitely many t 2 ½8=3; 3Þ. In the concluding section of the paper, we will put forth the conjecture that ðGÞ f ðtÞgðtÞn is the asymptotically best possible condition for a t-tough graph, 8=3 t < 3, to contain a 3-factor. The remainder of the paper is structured as follows. In Section 2 we establish some preliminary results we require. Sections 3–5 are devoted to proofs of Theorems 1.4–1.6. In the final section, Section 6, we discuss possible extensions of some of our results to k-factors for k > 3, but only for very small and very large values of t. We also indicate that for the remaining values of t, any such extensions are likely to be difficult. 2. PRELIMINARY RESULTS In this section we prove a number of lemmas which are needed in the sequel. Though our main interest is in 3-factors, we will state and prove these results for k-factors, since this does not substantially lenghthen or complicate the proofs. Belck [4] and Tutte [8] have characterized graphs G which contain a k-factor. For disjoint subsets A; B of VðGÞ let C ¼ VðGÞ A B and let oddk ðA; BÞ denote the number of components H of hCi which satisfy kjHj þ eðH; BÞ 1ðmod 2Þ. We will call any such component H an odd component of hCi. Define X dGA ðyÞ kjBj oddk ðA; BÞ: k ðA; BÞ ¼ G;k ðA; BÞ ¼ kjAj þ y2B

Theorem 2.1. Let G be any graph on n vertices, and k 1. (i) For any disjoint sets A; B VðGÞ, k ðA; BÞ knðmod 2Þ [8]; (ii) G does not contain a k-factor if and only if k ðA; BÞ < 0 for some disjoint pair of sets A; B VðGÞ ½4; 8. We call any disjoint pair A; B VðGÞ, a k-Tutte-pair for G if k ðA; BÞ < 0. Following [7], we call a k-Tutte-pair A; B minimal if either B ¼ ; or k ðA; B0 Þ 0 for all proper subsets B0 of B. Lemma 2.1 ([7]). Let k 2, and let A; B be a minimal k-Tutte-pair for a graph G with no k-factor. If B 6¼ ;, then ðhBiÞ k 2, where ðhBiÞ is the maximum degree of hBi. Lemma 2.2. Let G be a t-tough graph, t 1, on n k þ 1 vertices with kn even and which contains no k-factor. Let A; B be any minimal k-Tutte-pair in G. Then t k jAjjBjþ 2. Proof. If B ¼ ;, then by Theorem 2.1 we have !ðG AÞ oddk ðA; ;Þ kjAj þ 2, contradicting t 1. So B 6¼ ; and ðhBiÞ k 2 by Lemma 2.1.


123

Therefore hBi has chromatic number at most k 1. Partition B into k 1 independent sets B1 ; B2 ; . . . ; Bk1 so that B1 is as large as possible. We may also assume that each vertex in A is complete (i.e., has degree n 1), and each component of hCi is complete. We now establish two claims. Claim 2.1. For 1 i k 1, jAj þ eðBi ; B Bi Þ þ eðBi ; CÞ tjBi j:

ð1Þ

If Bi ¼ ; we are done, so assume Bi 6¼ ;. Let X ¼ NG ðBi Þ. Case 1. !ðG XÞ 2. Here we have jAj þ eðBi ; B Bi Þ þ eðBi ; CÞ jNG ðBi Þj ¼ jXj t!ðG XÞ tjBi j: Case 2. !ðG XÞ ¼ 1. Here we have jBi j ¼ 1 and jNG ðBi Þj ¼ n 1. If G is complete, then G has a k-factor. Otherwise, ðGÞ < n 1, and the left side of (1) is greater than or equal to jNG ðBi Þj > t!ðG XÞ ¼ t ¼ tjBi j. This proves Claim 2.1. Our second claim focuses on B1 . Claim 2.2. jAj þ eðB1 ; B B1 Þ þ eðB1 ; CÞ !ðhCiÞ tjB1 j:

ð2Þ

Let C1 ; C2 ; . . . ; Cs be the components of hCi which contain a vertex v with eðv; B1 Þ ¼ 1, and let v 1 ; v 2 ; . . . ; v s denote one such vertex in C1 ; C2 ; . . . ; Cs , respectively. Let Z denote the set of vertices v 2 C with eðv; B1 Þ ¼ 0, and let denote the number of components H in hCi with eðH; B1 Þ ¼ 0. Define X ¼ A [ ðB B1 Þ [ ðC fv 1 ; v 2 ; . . . ; v s g ZÞ so that !ðG XÞ jB1 j þ . Case 1. !ðG XÞ 2. Then, since G is t-tough, jXj t!ðG XÞ tðjB1 j þ Þ. From the maximality property of B1 , we conclude jBj jB1 j eðB1 ; B B1 Þ:


We also have jCj jZj þ ð!ðhCiÞ Þ s eðB1 ; CÞ: This follows since the first two terms count the number of vertices in C adjacent to at least one vertex in B1 , and the remaining terms lower bound the number of components of C containing a vertex adjacent to two or more vertices in B1 . Hence jAj þ eðB1 ; B B1 Þ þ eðB1 ; CÞ !ðhCiÞ þ jXj t!ðG XÞ tðjB1 j þ Þ tjB1 j þ :

ð3Þ

This establishes Claim 2.2 in this case. Case 2. !ðG XÞ ¼ 1. Then clearly jB1 j ¼ 1 and ¼ 0. Also, the maximality property of B1 implies that hBi is complete. It follows that s ¼ !ðG ðA [ BÞÞ ¼ !ðhCiÞ and eðB1 ; CÞ !ðhCiÞ. Thus if jAj þ eðB1 ; B B1 Þ tjB1 j, we would be done. So we assume jAj þ jB B1 j ¼ jAj þ eðB1 ; B B1 Þ < tjB1 j ¼ t: In the remainder of this proof, let !ðhCiÞ be denoted by !. If ! ¼ 0, then G is complete and has a k-factor. Suppose ! 2. Then the fact that G is t-tough requires jA [ Bj t! 2t. However jA [ Bj ¼ jAj þ jB B1 j þ jB1 j < t þ 1 2t, a contradiction. Hence we suppose ! ¼ 1. If all vertices of B were joined to all vertices of C, then G would be complete and we would be done. So assume that the vertex in B1 is not joined to every vertex in C. Now if we let X 0 ¼ A [ ðB B1 Þ [ NC ðB1 Þ, we have !ðG X 0 Þ 2 ¼ jB1 j þ 1, and so jX 0 j t!ðG X 0 Þ tðjB1 j þ 1Þ. However jX 0 j ¼ jXj þ 1. Now from (3) (with ¼ 0), we have jAj þ eðB1 ; B B1 Þ þ eðB1 ; CÞ ! jXj ¼ jX 0 j 1 tðjB1 j þ 1Þ 1 tjB1 j: This establishes Claim 2.2. We now sum (1) for 2 i k 1, and (2) to get ðk 1ÞjAj þ 2eðhBiÞ þ eðB; CÞ ! tjBj: Since kjAj þ 2eðhBiÞ þ eðB; CÞ kjBj oddk ðA; BÞ 2


125

by Theorem 2.1, we obtain tjBj kjBj jAj ð! oddk ðA; BÞÞ 2 kjBj jAj 2: Dividing by jBj > 0, we obtain the desired result.

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Lemma 2.3. Let k 2, and let G be a t-tough graph, t 1, on n k þ 1 vertices with kn even and no k-factor. If A; B is any k-Tutte-pair for G, then B 6¼ ;. Moreover, if l ¼ minfdGA ðvÞjv 2 Bg, then l k and ( n 2 if l ¼ k; þ 1Þ þ k k ðGÞ kðt ðk lÞn 2 þl if l < k: 2k l Proof. We again let ! ¼ !ðG ðA [ BÞÞ. Then by Theorem 2.1, ! kjAj þ

X

dGA ðyÞ kjBj þ 2:

ð4Þ

y2B

We first show B 6¼ ;. If B ¼ ;, we have by (4) that ! kjAj þ 2. But since G is 1-tough, this implies jAj !ðG AÞ ¼ ! kjAj þ 2, a contradiction. Next we show that l k. If l k þ 1, we find by (4) that ! kjAj þ jBj þ 2, again contradicting the 1-toughness of G. We now consider two cases. Case 1. l ¼ k. Here we have ! kjAj þ 2 2 by (4). Using the fact that G is t-tough yields t! jAj þ jBj. Hence we have with (4) and ðGÞ jAj þ l ¼ jAj þ k n jAj þ jBj þ ! ðt þ 1Þ! ðt þ 1ÞðkjAj þ 2Þ ðt þ 1ÞðkððGÞ kÞ þ 2Þ: Now a simple rearrangement yields the result in this case. Case 2. l k 1. By (4), we have ! kjAj þ

X

dGA ðyÞ kjBj þ 2 kjAj þ ljBj kjBj þ 2

y2B

kjAj ðk lÞðn jAj !Þ þ 2 ¼ ð2k lÞjAj ðk lÞðn !Þ þ 2 ð2k lÞjAj ðk lÞn þ ! þ 2: 2 Thus ðGÞ jAj þ l ðk 2klÞn þ l. l

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Corollary 2.1. Let k 2, and let G be a 1-tough graph on n k þ 1 vertices with kn even and no k-factor. Let ðA; BÞ be any k-Tutte-pair in G. Then ðGÞ jAj þ k.

3. PROOF OF THEOREM 1.4 Let G be a graph without a 3-factor, and let ðA; BÞ be a minimal 3-Tuttepair for G. Let C ¼ VðGÞ ðA [ BÞ. Since B 6¼ ; by Lemma 2.3, we have ðhBiÞ 1 by Lemma 2.1. Thus we may assume that hBi ¼ rK2 [ sK1 , for some r; s 0. We will prove three theorems in this section, which together will establish Theorem 1.4. Theorem 3.1. Let 1 t < 4=3. If G is a t-tough graph on n 4 vertices with n even, and if ðGÞ > 4t t 2 n þ 2, then G contains a 3-factor. Proof. Let G be a graph satisfying the hypotheses of the theorem but without a 3-factor. Let ðA; BÞ be a minimal 3-Tutte-pair for G. Construct the completion G of G by adding edges to G so that each vertex in A becomes complete (i.e., has degree n 1), and so that each component of hCi becomes a complete graph. Since G ;3 ðA; B0 Þ ¼ G;3 ðA; B0 Þ for all B0 B, we have that G contains no 3factor, and that ðA; BÞ is a minimal 3-Tutte-pair for G. Set t ¼ ðG Þ t. Then we have ðG Þ ðGÞ > tn=ð4t 2Þ þ 2 t n=ð4t 2Þ þ 2, since t=ð4t 2Þ is monotone decreasing for 1 t < 3. Thus G also satisfies the hypotheses of the theorem and does not contain a 3-factor. Hence we may assume G is G itself. Since t < 4=3, we have ðGÞ > 25 n þ 2. If n ¼ 4, the theorem is trivially true, and thus we assume n 6. Let B0 ¼ fx 2 B j dGA ðxÞ ¼ 0g. We wish to show that B0 6¼ ;. By Lemma 2.3, B 6¼ ; and l ¼ minfdGA ðxÞ j x 2 Bg 3. If l ¼ 0, there is nothing to show. But if 1 l 3, Lemma 2.3, together with n 6, easily yields the contradiction ðGÞ 25 n þ 2. Thus B0 6¼ ;, and so ðGÞ ¼ jAj. Let Z ¼ fv 2 C j eðv; BÞ ¼ 0g. Let denote the number of odd components H in hCi such that H \ Z ¼ ;. Since each such component H satisfies eðH; BÞ jHj þ 1, we get eðB; CÞ jCj jZj þ : Thus 2r 2r þ eðB; CÞ ðjCj jZj þ Þ ð3jBj þ odd3 ðA; BÞ 2 3jAjÞ ðjCj jZjÞ :

ð5Þ


127

Set X ¼ A [ ðC ZÞ. Then !ðG XÞ 2 (otherwise jB0 j ¼ 1, B ¼ B0 ; C ¼ Z ¼ ;, and so G ¼ Kn has a 3-factor). Thus we have t jXj=!ðG XÞ ¼ ðjAj þ jCj jZjÞ=!ðG XÞ. But then we find by (5) !ðG XÞ jBj r 1 jBj ð3jBj 3jAj 2 þ odd 3 ðA; BÞ ðjCj jZj þ ÞÞ 2 1 ¼ ð3jAj ðn jAj jCjÞ þ 2 odd3 ðA; BÞ þ jCj jZj þ Þ 2 n 2jAj þ 1 þ jCj jZj; since jZj odd3 ðA; BÞ : 2 This implies (since B0 6¼ ; and jCj jZj 0) jAj þ jCj jZj jAj þ ðjCj jZjÞ !ðG XÞ 2jAj n2 þ 1 þ ðjCj jZjÞ jAj ðGÞ ¼ ; n 2jAj 2 þ 1 2ðGÞ n2 þ 1

1t

and thus ðGÞ 4t t 2 ðn 2Þ, a contradiction. This proves Theorem 3.1.

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Before proving our next theorem, we require a somewhat technical lemma. Let denote the number of odd components of G ðA [ BÞ which contain a vertex x with eðx; BÞ ¼ 0. Lemma 3.1. Let 1 < t < 3, and let G be a t-tough graph on n vertices with n even and without a 3-factor. Then for any minimal 3-Tutte-pair ðA; BÞ in G, we have 3 4 3 3 t r ðt 1Þs ðt 1Þ: odd3 ðA; BÞ eðB; CÞ 1 2 3 2 2 Proof. We first show that r þ s 2. Suppose otherwise. Then jBj 2. But now ! ¼ !ðG ðA [ BÞÞ jAj þ 1 (if ! 1, this is immediate; if ! 2, then ðGÞ > 1 implies ! < jA [ Bj jAj þ 2). Combined with Theorem 2.1(ii), we get X dGA ðyÞ 3jBj þ 2; jAj þ 1 ! odd3 ðA; BÞ 3jAj þ y2B

and hence X X X dG ðyÞ ¼ dGA ðyÞ þ eðB; AÞ dGA ðyÞ þ 2jAj 3jBj 1: y2B

y2B

y2B

But then B contains a point of degree at most 2, contradicting ðGÞ > 1.


There are ¼ odd3 ðA; BÞ odd components H of hCi such that every vertex of H is adjacent to a vertex in B. For each such odd component H, we have eðH; BÞ jHj þ 1. Let X ¼ A [ fv 2 C j eðv; BÞ > 0g. Then jXj jAj þ eðB; CÞ , while !ðG XÞ r þ s þ 2. Since G is t-tough, we have t ðjAj þ eðB; CÞ Þ=ðr þ s þ Þ, and so by Theorem 2.1, 3tðr þ s þ Þ 3jAj þ 3eðB; CÞ 3 ð3jBj þ odd3 ðA; BÞ 2 2r eðB; CÞÞ þ 3eðB; CÞ 3 ¼ 6r þ 3s þ odd3 ðA; BÞ 2 2r þ 2eðB; CÞ 3: Using the fact that þ ¼ odd3 ðA; BÞ, we obtain 2odd3 ðA; BÞ 2eðB; CÞ 2 þ r 3ðt 1Þðr þ s þ Þ; or finally 3 4 3 3 odd3 ðA; BÞ eðB; CÞ 1 t r ðt 1Þs ðt 1Þ: 2 3 2 2

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Theorem 3.2. Let 4=3 t < 8=3. If G is a t-tough graph on n 4 vertices with t n even and ðGÞ > ð2t4 þ 4Þn þ 2, then G contains a 3-factor. Proof. Suppose that G does not contain a 3-factor, and let ðA; BÞ be a minimal 3-Tutte-pair for G. Setting X ¼ A [ C, and noting (as in the proof of Lemma 3.1) that r þ s 2, we see that t jXj=!ðG XÞ ¼ðn 2r sÞ=ðr þ sÞ, or n ðt þ 2Þr þ ðt þ 1Þs. Set ¼ n=ðjAj þ 1Þ. By Theorem 2.1, we have 3jAj 4r 3sþ eðB; CÞ odd3 ðA; BÞ ¼ 3jAj 3jBjþ 2rþ eðB; CÞ odd3 ðA; BÞ 2, or n 4r þ 3s þ odd3 ðA; BÞ 3jAj þ eðB; CÞ þ 2 ¼ 3 1 þ eðB; CÞ þ 2 3 ððt þ 2Þr þ ðt þ 1ÞsÞ þ eðB; CÞ 1: By Lemma 3.1, we then obtain 3 3 4 3 t r þ ðt 1Þs: 4r þ 3s ððt þ 2Þr þ ðt þ 1ÞsÞ þ 2 3 2 Comparing coefficients of r and s on both sides of the above inequality, we see that either 3 3 4 3 3 t or 3 ðt þ 1Þ þ ðt 1Þ; 4 ðt þ 2Þ þ 2 3 2


129

or equivalently ð2t þ 4Þ=ð4 tÞ or ð2t þ 2Þ=ð3 tÞ. However, ð2t þ 2Þ=ð3 tÞ ð2t þ 4Þ=ð4 tÞ for t 1, and thus we may suppose ð2t þ 4Þ=ð4 tÞ. But then by Corollary 2.1 ðGÞ 3 jAj ¼

4t n n 1; 1 2t þ 4

or ðGÞ ð4 tÞn=ð2t þ 4Þ þ 2, a contradiction. This proves Theorem 3.2.

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Theorem 3.3. Let 8=3 t < 3. If G is a t-tough graph on n 4 vertices with n tÞ even and ðGÞ > ð2ð3 t þ 2 Þn þ 2, then G contains a 3-factor. Proof. Suppose G does not have a 3-factor, and let ðA; BÞ be a minimal 3Tutte-pair in G. Taking X ¼ A [ C, and again noting (as in the proof of Lemma 3.1) that r þ s 2, we find that t ðn 2r sÞ=ðr þ sÞ, or n ðt þ 2Þr þ ðt þ 1Þs. The above degree condition can be written as t >3

ðGÞ 2 t þ 2 ðGÞ 1 t þ 2 >3 : 2 n 2 n

ð6Þ

But by Lemma 2.2, we also have t 3

jAj þ 2 : jBj

ð7Þ

Using (6), (7), and Corollary 2.1 we get ðGÞ 1 t þ 2 jAj þ 2 ðGÞ 1 > ; 2 n jBj jBj or jBj >

2n : tþ2

ð8Þ

Using (8) together with n ðt þ 2Þr þ ðt þ 1Þs yields 2r þ s ¼ jBj >

2n 2ððt þ 2Þr þ ðt þ 1ÞsÞ 2ðt þ 1Þ ¼ 2r þ s; tþ2 tþ2 tþ2

or s > ð2tt þþ22Þs, a contradiction. This proves Theorem 3.3.

&

4. PROOF OF THEOREM 1.5 We will prove two theorems in this section, which together will establish Theorem 1.5.


Theorem 4.1. For any rational number t, 1 t < 4=3, there exist infinitely many t-tough graphs G of even order n with no 3-factor satisfying ðGÞ ð4t t 2Þn 1. Proof. Consider a large even integer a which is a multiple of t. Set r ¼ ð3a þ 2Þ 3a=t and s ¼ 4a=t ð3a þ 2Þ, and form the graph G ¼ Ka þ ðrK2 [ sK1 Þ. It is immediate that ðGÞ ¼ a=ðr þ sÞ ¼ t. Also, taking A ¼ Ka and B ¼ rK2 [ sK1 , we have 3 ðA; BÞ ¼ 2, and so G has no 3-factor. Finally, noting that ðGÞ ¼ a and n ¼ jVðGÞj ¼ a þ 2r þ s ¼ 4a þ 2 2a=t, we have ðGÞ a t ¼ ; and so ¼ 2a n 2 4a t 4t 2 t t ðGÞ ¼ ðn 2Þ n 1; 4t 2 4t 2 since t 1.

&

Theorem 4.2. For any rational number t, 2 t < 8=3, there exist infinitely many t-tough graphs G of even order n with no 3-factor satisfying ðGÞ t 15 ð2t4 þ 4Þ n 2 : þ 4ðt 1Þ Proof. Let a 2ð8 2tÞ be any even integer, and consider the graph 8 3t G ¼ Ka þ ðHðr1 Þ [ r2 K2 Þ, where Hðr1 Þ is the graph shown in Figure 1, and r1 að8 3tÞ4ðt 1Þ and r2 are integers defined as follows. Let x ¼ að3t84Þþ6t 2 2t , y ¼ 8 2t and define r1 ¼ dxe þ 6 and r2 ¼ byc 2. It is easy to see that ðGÞ is either r2 aþ 1 or r1aþþr22rþ1 1 : But we find

a a a 8 2t > > 2; ¼ r2 þ 1 byc 1 y 8 3t since 2 t < 83, and thus we have ðGÞ ¼ r1aþþr22rþ1 1 : We want ðGÞ ¼ r1aþþr22rþ1 1 t, or equivalently ðt 2Þr1 þ tr2 a t:

FIGURE 1.

The graph H(r1).

ð9Þ


131

For G not to have a 3-factor, it suffices (taking A ¼ Ka and B ¼ ðr1 þ r2 ÞK2 ) by Theorem 2.1 to have 3a þ 2ðr1 þ r2 Þ þ 2r1 3ð2r1 þ 2r2 Þ 2, or equivalently 2r1 þ 4r2 3a þ 2:

ð10Þ

We wish to show that our choices of r1 and r2 above satisfy both (9) and (10). 1 x ¼ ðdxexÞþ6 We have ryr ðybycÞþ2. Since 0 dxe x; y byc 1, it follows that 2 r1 x 7 t and thus 2 yr2 2. But since 2 t < 8=3, we have 72 < t2 2

r1 x t : < y r2 t 2

ð11Þ

Note that ðt 2Þx þ ty ¼ a t and 2x þ 4y ¼ 3a þ 2. Thus we obtain ðt 2Þr1 þ tr2 ¼ ða tÞ þ ½ðt 2Þðr1 xÞ tðy r2 Þ a t by (11), which is (9). We also have 2r1 þ 4r2 ¼ ð3a þ 2Þ þ ½2ðr1 xÞ 4ðy r2 Þ 3a þ 2 by (11), which is (10). In summary, we know G is t-tough and does not contain a 3-factor. Finally, since ðGÞ ¼ a þ 1 and n ¼ a þ 4r1 þ 2r2 , we have ðGÞ 1 a 16tþ8 ¼ n 82t 24 a þ 4r1 þ 2r2 16tþ8 82t 24

a aþ

4ðað3t4Þþ6t 82t ¼

þ 7Þ þ

2ðað83tÞ4ðt1Þ 82t 1

1þ

4ð3t4Þþ2ð83tÞ 82t

¼

2Þ ð16tþ8 82t Þ 24

4t : 2t þ 4

Thus we get 4t 16t þ 8 4t þ 24 ðGÞ n þ1 2t þ 4 8 2t 2t þ 4 4t 50 8t 4t 15 n þ1 n ; ¼ 2t þ 4 tþ2 2t þ 4 2

since t 2.

&

5. PROOF OF THEOREM 1.6 Theorem 5.1. For any rational number t; 4=3 t < 2, there exist infinitely many t-tough graphs G of even order n with no 3-factor satisfying ðGÞ t Þn 3. ð7t6


Proof. Let t 2 ½43 ; 2Þ be given, and let a be any large even integer such that a=t is an integer. Consider the graph defined by Ka þ ðHðr1 Þ [ r2 K2 Þ, where r1 ¼ 3a=2 2a=t þ 3, r2 ¼ a=t 1, and Hðr1 Þ is the graph shown in Figure 1. It is easy to see that ðGÞ ¼ r2 aþ1 ¼ t < 2 and that G does not contain a 3-factor, since 2r1 þ 4r2 ¼ 3a þ 2. Finally, since ðGÞ ¼ a þ 1 and n ¼ a þ 4r1 þ 2r2 , we have ðGÞ 1 a ¼ n 10 a þ 4r1 þ 2r2 10 a t 3a 2a a ¼ ; ¼ 7t 6 a þ 4 2 t þ 3 þ 2 t 1 10 and so ðGÞ

t t 3t þ 6 t ðn 10Þ þ 1 ¼ n n 3; 7t 6 7t 6 7t 6 7t 6

since t 4=3.

&

t Since f ðtÞ=ð7t6 Þ < 1:09 for t 2 ½4=3; 2Þ, the examples in Theorem 5.1 already show the degree bound in Theorem 1.4 when t 2 ½4=3; 2Þ is tight to within 9%. However, even better examples can be constructed when t < 18=11. The key is to generalize the graphs Hðr1 Þ in Figure 1. Given a path with 3k edges for some k 1, call the vertices in positions 1; 4; 7; . . . ; 3k þ 1 the ‘‘base vertices’’ of the path. To form the graph Hðk; rÞ, take r disjoint copies of such a k-path, and add edges to form a clique on the ðk þ 1Þr base vertices in all the copies. Note that Hðr1 Þ is just the special case Hð1; r1 Þ. The best examples so far when t < 18=11 have the form Ka þ ðHðk; r1 Þ [ r2 K2 Þ, for appropriate k > 1 (i.e., k ¼ 3 when t 2 ½4=3; 40=29Þ, and k ¼ 2 when t 2 ½40=29; 18=11ÞÞ. Given t and a, it is easy though tedious to determine r1 and r2 in terms of t and a by applying the fact that the graph is t-tough and has no 3-factor. We leave the details to the reader. Clearly more work needs to be done before an optimal degree condition when t 2 ½4=3; 2Þ can be determined, and we are not in a position to even conjecture one at this time.

6. PROOF OF THEOREM 1.7 Let r 5 be an odd integer relatively prime to 3. We first describe the con-tough graphs with no 3-factor. We will struction of an infinite family of 3r1 r then establish that this family provides the examples necessary to prove Theorem 1.7. Since gcdð3; rÞ ¼ 1, it follows by the Chinese Remainder Theorem that there exist infinitely many integers j such that j 0ðmod 3Þ and j r 1ðmod rÞ. For


133

ð j þ 1Þ 8j3 is also an integer. For each any such integer j, we have that m ¼ 3r1 r such j, consider the graph G ¼ Gðr; jÞ ¼ Kmþ2 j=3 þ Hðr; jÞ, where Hðr; jÞ is shown in Figure 2. Taking A ¼ Kmþ2j=3 ; C ¼ K2rð3mþ2Þþ2j [ ð3m þ 2ÞKr , and B ¼ VðGÞ A C, it is easy to verify using Theorem 2.1 that Gðr; jÞ has no 3-factor. 3r1 we get In a moment we will show that ðGðr; jÞÞ ¼ 3r1 r . Setting t ¼ r 8j r ¼ 1=ð3 tÞ and m ¼ tðj þ 1Þ 3 . We find that m þ 23 j þ 2 ðGðr; jÞÞ ¼ jVðGðr; jÞÞj m þ 23 j þ 5rð3m þ 2Þ þ 4j j½ð3 tÞðt 2Þ þ ð3 tÞðt þ 2Þ ð3 tÞðt 2Þ ¼ ; < 2 2 j½16t 34 t þ ð10 þ 18t t Þ 16t 34 t2 since

ð3tÞðt2Þ 16t34t2

> ð3tÞðtþ2Þ 10þ18tt2 for all t such that 14=5 t < 3.

ðGðr; jÞ ð3tÞðt2Þ Since limj!1 jVðGðr; jÞÞj ¼ 16t34t2 ¼ f ðtÞgðtÞ, it follows that for any " > 0, > f ðtÞgðtÞ " for j sufficiently large.

ðGðr; jÞÞ jVðGðr; jÞÞj

It only remains to show that ðGðr; jÞÞ ¼ 3r1 r . Taking X VðGðr; jÞÞ to be the vertices in Kmþ2 j=3 together with the vertices v 1 ; . . . ; v 2 j in Hðr; jÞ shown in jXj ¼ 3r1 Figure 2, we have ðGðr; jÞÞ !ðGXÞ r . For the reverse inequality, we require the following two lemmas, whose proofs are given in the Appendix. Lemma 6.1. Let H1 ðjÞ denote the graph shown in Figure 3. Then for every Y VðH1 ð jÞÞ , jYj þ 1: !ðH1 ð jÞ YÞ 2 Moreover, the inequality is strict if every vertex of K2j belongs to Y.

FIGURE 2.

The graph H(r, j ).


FIGURE 3.

The graph H1( j ).

Lemma 6.2. Let H2 ðr; mÞ denote the graph shown in Figure 4. Then for every Z VðH2 ðr; mÞÞ, we have !ðH2 ðr; mÞ ZÞ

j

r k jZj þ 1: 3r 1

Moreover, the inequality is strict if every vertex of the K2rð3mþ2Þ belongs to Z. Assuming these two lemmas, we now prove ðGðr; jÞÞ ð3r 1Þ=r. Let X ¼ A [ Y [ Z be a toughness determining set for Gðr; jÞ, where Y VðH1 ð jÞÞ and Z VðH2 ðr; mÞÞ, identifying H1 ð jÞ (resp, H2 ðr; mÞ) with the right (resp, left) portion of Hðr; jÞ in Figure 2. Then ðGðr; jÞÞ ¼

jXj jAj þ jYj þ jZj ¼ ; !ðG XÞ !ðH1 ð jÞ YÞ þ !ðH2 ðr; mÞ ZÞ

where ¼

0 if either K2 j Yor K2rð3mþ2Þ Z; 1 otherwise:

FIGURE 4. The graph H2(r,m ).


135

So by Lemmas 6.1 and 6.2 (in particular, noting when the inequalities are strict) and the definition of we obtain jYj þ jZj þ 3r1 ð j þ 1Þ 2 j r j k ðGðr; jÞÞ jYj r 3r1 jZj þ 1 2 þ ðj þ 1Þ jYj 2j þ jZj þ 3r1 r : jYj r jþ jZj þ ð j þ 1Þ 2 3r 1 Clearly, we may assume jYj jH12ð jÞj ¼ 2 j. If jYj ¼ 2 j, then ð j þ 1Þ 3r 1 jZj þ 3r1 r : ¼ ðGðr; jÞÞ r r 3r1 jZj þ ð j þ 1Þ If jYj < 2 j, then jYj 2 j; jYj 2 j < 0 and

jYj2 j jYj 2 j

¼ 2 < 3r1 r and thus

ð j þ 1Þ 3r 1 jYj 2j þ jZj þ 3r1 r > ; ðGðr; jÞÞ jYj r r jZj þ ð j þ 1Þ j þ 2 3r1 as desired.

7. CONCLUDING REMARKS For t 2 ½8=3; 3Þ, we conjecture that the degree bound in Theorem 1.4 can be t2 4 significantly improved. Recall that gðtÞ ¼ 2ð16t34t 2 Þ, and that 0:5 gðtÞ 1 for t 2 ½8=3; 3Þ. Conjecture 7.1. Let 8=3 t < 3. If G is a t-tough graph on n vertices, n even, with ðGÞ f ðtÞgðtÞn, then G contains a 3-factor. Of course, Theorem 1.7 implies that the coefficient of n in Conjecture 7.1 would be best possible. All attempts to prove Conjecture 7.1 have failed so far, and it now appears that it may be quite formidable. One reason for the apparent difficulty is that we do not have a strong enough structural result for graphs without 3-factors. Such a structural result does exist, however, for graphs without 2-factors (e.g., see Lemma 8 in [3]). It is also interesting to consider generalizations of our results for the existence of k-factors when k 4. One result which generalizes is Theorem 1.4 when 1 t < 4=3. The proof of the following result is very similar to that of Theorem 3.1.


Theorem 7.1. Let 1 t < 2 2k. Let G be a t-tough graph on n k þ 1 vertices with kn even and ðGÞ >

ðk 2Þt n þ 2: 2ðk 1Þt 2

Then G contains a k-factor if n is sufficiently large. The degree bound in Theorem 7.1 is tight. This can be shown using graphs analogous to those used in the proof of Theorem 4.1, namely the graphs Ka þ ðxK2 [ yK1 Þ for appropriately chosen integers x and y. There is also a very natural extension of Conjecture 7.1 to larger values of k. It is based on an obvious extension of the examples Gðr; jÞ constructed in the proof of Theorem 1.7. Let r > k be any odd integer relatively prime to k. By the Chinese Remainder Theorem, we can find infinitely many j such that j 0ðmod kÞ and j r 1ðmod rÞ. For any such j, let m be the integer k1 ðkr1 r Þð j þ 1Þ ð k Þ j. Consider the graph Gðk; r; jÞ ¼ Kpðk;r;jÞ þ Hðk; r; jÞ, where pðk; r; jÞ ¼ m þ ðk 1Þj=k and Hðk; r; jÞ is shown in Figure 5. It is readily verified that Gðk; r; jÞ does not contain a k-factor, and it can be shown that ðGðk; r; jÞÞ ¼ kr1 r . Setting hðt; kÞ ¼

ðk tÞðt k þ 1Þ ðk tÞðt þ k 1Þ ð2k 1Þðkðk tÞ 1Þ

(note that hðt; 3Þ ¼ f ðtÞgðtÞ), it can be shown that for any " > 0, ðGðk; r; jÞÞ > ðhðt; kÞ "Þn holds for sufficiently large j. We now state the following conjecture (analogous to Conjecture 7.1), which by the above example would be best possible in some sense.

FIGURE 5.

The graph H(k,r,j ).


137

Conjecture 7.2. Let k 1=k t < k. If G is a t-tough graph on n vertices with kn even and ðGÞ hðt; kÞn, then G has a k-factor. Though the range ½k 1=k; kÞ of t in Conjecture 7.2 is very small, the problem appears to be very interesting in this final part of the interval ½1; kÞ. But given the apparent difficulty of Conjecture 7.1 (which is Conjecture 7.2 in the special case k ¼ 3), it appears that Conjecture 7.2 is completely out of reach at the present time. Finally, we admit that we do not yet see how to extend Theorem 3.2 to obtain tight degree conditions for the existence of k-factors in t-tough graphs when 2 2k t < k 1k. Unfortunately, for large k this entails almost the entire range of applicable values of t.

ACKNOWLEDGMENT The authors thank an anonymous referee for a careful reading of the paper and for suggesting the examples at the end of Section 5.

APPENDIX: PROOFS OF LEMMAS 6.1 AND 6.2 Proof of Lemma 6.1 Label the vertices of H1 ð jÞ as shown in Figure 6. Let S ¼ fs11 ; s12 ; . . . ; sj 2 g. If S Y, then jYj jSj ¼ 2 j and so we find wðH1 ð jÞ YÞ j jYj 2 as asserted. Suppose, therefore, S Y 6¼ ;. Let S denote the component of H1 ð jÞ Y containing the vertices of S Y. Consider the components C1 ; . . . ; C‘ of H1 ð jÞ Y besides S. Without loss of generality, we may suppose Ci consists of the vertices in fqi1 ; qi2 g Y, for 1 i ‘. For Ci to be distinct from S, Y must contain at least two vertices from fqi1 ; qi2 ; si1 ; si2 g: Thus, jYj jYj 2‘, and so wðH1 ð jÞ YÞ 1 ¼ ‘ jYj 2 , or wðH1 ð jÞ YÞ 2 þ 1, as asserted.

FIGURE 6.


Proof of Lemma 6.2 For simplicity, we will refer to H2 ðr; mÞ as just H in the sequel. Label the vertices of H as shown in Figure 7. Let S ¼ fs111 ; s211 ; . . .g; Q ¼ fq111 ; q211 ; . . .g, and P ¼ fp11 ; p12 ; . . .g. Recall that r > 3. For 1 i 3m þ 2 and 1 j r, the vertices with lower indices i; j will form the ði; jÞth column of H. The ith group of columns are the r columns with indices ði; 1Þ; ði; 2Þ; . . . ; ði; rÞ. Suppose Lemma 6.2 fails for some Z VðHÞ. Consider the components of H Z consisting of K2 ’s in Q, and let I denote the set of indices of the columns in which there K2 ’s occur. Then Z contains the 3jIj vertices of the form fpij ; s1ij ; s2ij g for ði; jÞ 2 I. We may assume that for all i, at least one of the indices in the ith group is not in I (else replace Z by Z fpi1 g to get another Z for which Lemma 6.2 fails). In the sequel, we will assume ði; 1Þ 62 I, for 1 i 3m þ 2. Let HI denote the subgraph of H, which results when we delete all the vertices in the columns with indices in I. By the above assumption, HI contains at least one column from each of the 3m þ 2 groups of columns. We may also assume that S 6 Z (In proof, if S Z, then clearly we can assume Q \ Z ¼ ;. Since p11 ; q111 ; q211 2 VðHI Þ it follows that if S Z, then Z fs111 ; s211 g [ fq111 ; q211 g would be a subset of VðHÞ for whichLemma 6.2 fails, since we have increased r r jYj þ 1 by 3r1 2 < 1. But now S 6 Z). wðH ZÞ by 1 and 3r1 Denote !ðH ZÞ by c, so that c ¼ jIj þ M for some M 0. Upon removing ZI ¼ Z \ VðHI Þ from HI , exactly M components must result, none a K2 in Q. Let S denote the component of HI ZI containing vertices of S Z. The remaining M 1 components of HI ZI consist of the vertices in some Pi ¼ fpi1 ; . . . ; pir g, and possibly some vertices in NQ ðPiÞ. Let Pi denote such a component, and without loss of generality, suppose these M 1 components are P1 ; . . . ; Pm1 .

FIGURE 7.


139

Denote the total number of columns in HI in the first M 1 groups of columns by rðM 1Þ where jIj ¼ c M. Since there is at least one column from each of the first M 1 groups of columns in HI , we have rðM 1Þ M 1:

ð12Þ

Claim. ZI contains at least two vertices from each of these rðM 1Þ columns. Proof of the Claim. Suppose the ijth column is among the rðM 1Þ columns in question. If pij 62 Z and Z contains at most one vertex in the ijth column, then there exists 2 f1; 2g such that sij ; qij ; pij 62 Z and so Pi ¼ S, a contradiction. Hence, we will assume pij 2 Z. Suppose now that none of the other vertices in the ijth column besides pij belongs to Z. If we replace Z by Z [ fs1ij g [ fq2ij g, it is easy to see !ðH ZÞ has r r ÞjZj þ 1 has increased by ð3r1 Þ 2 < 1. So increased by 1, while ð3r1 1 2 Z [ fsij g [ fqij g is another subset of VðHÞ which violates Lemma 6.2, but for which the Claim is true for the ijth column. Simply iterate this construction until we have a Z which satisfies the Claim for all rðM 1Þ columns. This proves the Claim. By the Claim, we have jZI j 2ðrðM 1Þ Þ:

ð13Þ

Since jZj ¼ jZI j þ 3ðc MÞ, we have by (13) that jZj 2ðrðM 1Þ Þ þ 3ðc MÞ. The fact that Lemma 6.2 fails for Z implies r r

jZj þ 1 2½rðM 1Þ þ 3ðc MÞ þ1; c> 3r 1 3r 1 or

c r > 2ðrðM 1Þ Þ 3M þ1; 3r 1 3r 1 or

c > ð2r 1Þ rðM 1Þ þ 1 2rðM þ Þ; or

ð2r 1Þc > ð2r 1Þ rðM 1Þ þ 1 þ 2rðc M Þ:

Dividing through by 2r 1 > 0, we find that 2r ðc M Þ c > rðM 1Þ þ 1 þ 2r 1 |fflfflfflfflfflffl{zfflfflfflfflfflffl} >1

> rðM 1Þ þ 1 þ c M ;


or ðM 1Þ > rðM 1Þ ; which violates (12). This proves !ðH ZÞ r r ÞjZj þ 1, and thus !ðH ZÞ bð3r1 ÞjZjc þ 1, since wðH ZÞ is an integer. ð3r1 Finally, suppose we have S Z with !ðH ZÞ ¼

j

r k jZj þ 1: 3r 1

Then, we must have Q \ Z ¼ ;, since if qij 2 Z, then Z 0 ¼ Z fsij g would be

r jZ 0 j þ 1, which contradicts the above a subset of VðHÞ with wðH Z 0 Þ > 3r1 result. Suppose that jPi \ Zj ¼ gi ; for i ¼ 1; 2; . . . ; 3m þ 2. Then we leave ci components among the vertices of Pi [ Q when we remove Z, where ci ¼

gi þ 1 if gi < r; gi if gi ¼ r:

Thus, we may assume gi r 1, for all i. But then we find

jZj ¼ wðH ZÞ

jSj þ

3mþ2 X

i¼1 3mþ2 X

2rð3m þ 2Þ þ

gi ¼

3m þ 2 þ

ci

i¼1

3mþ2 X

gi

i¼1 3mþ2 X

>

gi

i¼1

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} >1

0

zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{ 3mþ2 3mþ2 X X 2rð3m þ 2Þ þ gi þ ð3m þ 2Þðr 1Þ gi ð3m þ 2Þ þ

i¼1 3mþ2 X

gi þ ð3m þ 2Þðr 1Þ

i¼1

which is equivalent

r to wðH ZÞ < jZj þ 1. wðH ZÞ ¼ 3r1

i¼1 3mþ2 X

¼

3r 1 ; r

gi

i¼1

r 3r1

jZj, and contradicts the assumption that &

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141

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