Isolated Toughness and Existence of [a, b]-factors in Graphs

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1 Isolated Toughness and Existence of [a, ]-factors in Graphs Yinghong Ma 1 and Qinglin Yu 23 1 Department of Computing Science Shandong Normal University, Jinan, Shandong, China 2 Center for Cominatorics, LPMC Nankai University, Tianjing, China 3 Department of Mathematics and Statistics Thompson Rivers University, Kamloops, BC, Canada Astract For a graph G with vertex set V (G) and edge set E(G), let i(g) e the numer of isolated vertices in G, the isolated toughness of G is defined as I(G) = min{ S /i(g S) S V (G), i(g S) 2}, if G is not complete; and I(K n ) = n 1. In this paper, we investigate the existence of [a, ]-factor in terms of this graph invariant. We proved that if G is a graph with δ(g) a and I(G) a, then G has a fractional a-factor. Moreover, if δ(g) a, I(G) > (a 1) + and G S has no (a 1)-regular component for any suset S of V (G), then G has an [a, ]-factor. The later result is a generalization of Katerinis well-known theorem aout [a, ]-factors (P. Katerinis, Toughness of graphs and the existence of factors, Discrete Math. 80(1990), 81-92). 1 Introduction All graphs considered in this paper are simple undirected graphs. Let G = (V (G), E(G)) e a graph, where V (G) and E(G) denote vertex set and This work is sponsored y Taishan Scholar Project of Shandong, Young Scientist Award Program of Shandong, China (2006BS01015) and Natural Sciences & Engineering Research Council of Canada (Grant numer OGP ). Q. Yu s yu@tru.ca or yu@nankai.edu.cn 1

2 edge set of G, respectively. We use d G (x) and δ(g) to denote the degree of x in G and the minimum degree of G. For any S V (G), the sugraph of G induced y S is denoted y G[S]. i(g) and c(g) are used for the numer of isolated vertices and the numer of components in G, respectively. For S, T V (G) let E(S, T ) = {uv E(G) u S, v T }, E(S) = {uv E(G) u, v S}, and e(s, T ) = E(S, T ). Other notation and terminology not defined in this paper can e found in [8]. Let g(x) f(x) e two nonnegative integer-valued functions defined on V (G) and let H e a spanning sugraph of G. We call H a (g, f)-factor of G if g(x) d H (x) f(x) holds for each x V (G). Similarly, for any two nonnegative integers a and, H is an [a, ]-factor of G if a d H (x) for each x V (G). Let a = = k > 0 for each x V (G), it is called a k-factor. Fractional factors can e considered as the rationalization of the traditional factors y replacing integer-valued function y a more generous fuzzy function (i.e., a [0, 1]-valued indicator function). It defined as follows: let h : E(G) [0, 1] e a real-value function and d h G (x) = e E x h(e), where E x = {xy xy E(G)}. Then d h G (x) is called the fractional degree of x under h in G and h is called an indictor function if g(x) d h G (x) f(x) holds for each x V (G). Let Eh = {e e E(G), h(e) 0} and G h e a spanning sugraph of G such that E(G h ) = E h. The G h is referred as a fractional (g, f)-factor. The fractional k-factors and fractional [a, ]-factors can e defined similarly. For any function f(x), we denote f(s) = x S f(x). Since fractional factors are relaxations of usual factors, many results aout traditional factors also are valid for fractional factors. We quote some of them used later elow. Lemma 1.1. (Heinrich et al. [4]) Let g(x) and f(x) e nonnegative integral-valued functions defined on V (G). If either one of the following conditions holds (i) g(x) < f(x) for every vertex x V (G); (ii) G is ipartite; then G has a (g, f)-factor if and only if for any suset S of V (G) g(t ) d G S (T ) f(s) where T = {x x V (G) S, d G S (x) g(x)}. Lemma 1.2. (Anstee [1]) Let G e a graph and let g(x) f(x) e two nonnegative integer-valued functions on V (G). Then G has a fractional (g, f)-factor if and only if for any S V (G), g(t ) d G S (T ) f(s) holds, where T = {x x V (G) S, d G S (x) g(x)}. 2

3 Lemmas 1.1 and 1.2 imply that when G is ipartite or g(x) < f(x) for every vertex x V (G), then G has a fractional (g, f)-factor if and only if G has a (g, f)-factor. This is an interesting phenomenon, which indicates that a real (g, f)-factor polytope is integral under one of the given conditions. Well-known Tutte s Theorem states that G has a 1-factor if and only if o(g S) S for all S V (G), where o(g S) is the numer of odd components in G S. The following theorem provides a characterization of fractional 1-factors y using the numer of isolated vertices. This result has een one of most widely quoted results in fractional factor theory. Lemma 1.3. (see [8]) A graph G has a fractional 1-factor if and only if i(g S) S holds for any S V (G). Chvátal [2] introduced the notion of toughness for studying hamiltonian cycles and regular factors in graphs. The toughness of a graph G, t(g), is defined as t(g) = { S min{ c(g S) S V (G), c(g S) 2} if G is not complete; V (G) 1 otherwise. The toughness numer has ecome an important graph invariant for studying various fundamental properties of graphs. In particular, Chvátal conjectured that k-toughness implies a k-factor in graphs and this conjecture was confirmed positively y Enomoto et al. [3]. In this paper, we investigate the existence of fractional factors in relation to the isolated toughness of G. The parameter of isolated toughness is motivated from Chvátal s toughness y replacing c(g S) with i(g S) in the definition. The isolated toughness, I(G), was first introduced y Ma and Liu [7] and is defined as I(G) = { S min{ i(g S) S V (G), i(g S) 2} if G is not complete; n 1 if G = K n. Since c(g S) i(g S) for any S V (G), the isolated toughness is no less than toughness for any graph. Therefore, to weaken a condition in a theorem we could consider to replace the condition in terms of toughness y isolated toughness. In this paper, we provide several sufficient conditions in terms of isolated toughness for graphs to have fractional factors and [a, ]- factors. We pursue Chvátal s idea along the same line for fractional factors and prove the following Theorem 1.1. Let G e a graph and a e a positive integer. If δ(g) a and I(G) a, then G has a fractional a-factor. 3

4 In [5], Katerinis otained a sufficient condition for the existence of [a, ]- factor y using toughness and proved that if the toughness of G is at least (a 1) + a and a V (G) is even when a =, then G has an [a, ]-factor. We replace the condition in Katerinis result from toughness to isolated toughness and prove the same conclusion. Thus, we generalize this wellknown result. Theorem 1.2. Let G e a graph with δ(g) a and the isolated toughness I(G) > (a 1) + (2 a < ). If, for any suset S of V (G), G S has no (a 1)-regular sugraph as a component, then G has an [a, ]-factor. In view of Lemmas 1.1 and 1.2, to show Theorem 1.2 we need only to prove the existence of fractional [a, ]-factors. So in Section 3, we only provide a proof for the following weaker version of Theorem 1.2. Theorem 1.3. Let G e a graph with δ(g) a and the isolated toughness I(G) > (a 1) + (2 a < ). If, for any suset S of V (G), G S has no (a 1)-regular sugraph as a component, then G has a fractional [a, ]-factor. 2 Proof of Theorem 1.1 This section is devoted to the proof of Theorem 1.1. To do it, we require the following lemma which is a necessary and sufficient condition for existence of fractional [a, ]-factors given y Liu and Zhang [6]. Lemma 2.1. Let G e a graph and a e positive integers. Then G has fractional [a, ]-factors if and only if for any S V (G) a T d G S (T ) S where T = {x x V (G) S and d G S (x) a}. Proof of Theorem 1.1: When a = 1 the result is secured y Lemma 1.3. Next we assume a 2. If G is a complete graph with at least a + 1 a vertices, then G has a fractional a-factor y setting h(e) = V (G) 1 for each e E(G). Now suppose that G is not a complete and satisfies the hypothesis of the theorem, ut has no fractional a-factors. By Lemma 2.1, there exists a vertex-set S 0 V (G) such that a T 0 d G S0 (T 0 ) > a S 0 (2.1) where T 0 = {x x V (G) S 0 and d G S0 (x) a}. 4

5 If S 0 =, then a T 0 d G (T 0 ) = a T 0 a T 0 = 0 > 0 = a S 0 as δ(g) a, a contradiction. Thus S 0. Let U 0 = V (G) (S 0 T 0 ), n = e G (T 0, U 0 ) and m = e G (T 0 ) = e G (T 0, T 0 ). Then (2.1) ecomes a S 0 a T 0 + d G S0 (T 0 ) = a S 0 a T 0 + 2m + n < 0. (2.2) Let A 1 e a maximal independent suset of T 0 and let T 1 = T 0 A 1. Moreover, let A i e a maximal independent suset of T i 1 (2 i a), where T i 1 = T i 2 A i 1. Since (G[T 0 ]) a, we have A a =. So we assume 2 i a 1. At first we prove the following two claims. Claim 1. S 0 + T 1 + e G (U 0, A 1 ) a A 1. Let U 1 = {u u U 0, e G (u, A 1 ) > 0}, U 2 = {u u U 0, e G (u, A 1 ) = 1} and U 3 = U 1 U 2. Then U 1 e G (U 1, A 1 ) = e G (U 0, A 1 ) and U 3 U 1. Let G 1 = G (S 0 T 1 U 1 ). Then i(g 1 ) A 1 as A 1 is a maximal independent set of T 0. Case 1. i(g 1 ) 2. We have S 0 T 1 U 1 a i(g 1 ) a A 1 y the assumption I(G) a. On the other hand, S 0 T 1 U 1 = S 0 + T 1 + U 1 S 0 + T 1 + e G (U 0, A 1 ). Therefore, S 0 + T 1 + e G (U 0, A 1 ) a A 1 and Claim 1 follows. Case 2. i(g 1 ) = 1. In this case A 1 = 0 or 1 as i(g 1 ) A 1. When A 1 = 0, Claim 1 clearly follows from A 1 = T 0 = = T 1. When A 1 = 1, let A 1 = {x}. Since S 0 + T 1 + e G (U 0, A 1 ) d G (x) a, we have S 0 + T 1 + e G (U 0, A 1 ) a A 1 = a. Claim 1 holds. Case 3. i(g 1 ) = 0. That is, i(g (S 0 T 1 U 1 )) = 0. Thus A 1 = and T 0 = as i(g 1 ) A 1. Claim 1 holds. Claim 2. For all i, 2 i a 1, S 0 + e G (T 0 T i 1, A i ) + T i + e G (U 0, A i ) a A i. If A i =, then T i = and Claim 2 holds. Assume that A i and let X i = N G (A i ) (T 0 T i 1 ), Y i = N G (A i ) U 0 and G i = G N G (A i ). Clearly, A i is a suset of isolated vertices in G i and i(g i ) = i(g N G (A i )) A i. Consider the following cases. Case 1. i(g i ) 2. Since I(G) a, we have S 0 X i Y i T i N G (A i ) a i(g i ) a A i. On the other hand, S 0 X i Y i T i S 0 + e G (T 0 T i 1, A i ) + e G (U 0, A i ) + T i. Therefore, S 0 + e G (T 0 T i 1, A i ) + e G (U 0, A i ) + T i a A i and Claim 2 holds. Case 2. i(g i ) = 1. Since A i, we have A i = 1. Let A i = {x i }. Then S 0 + e G (T 0 T i 1, A i ) + e G (U 0, A i ) + T i d G (x i ) a = a A i and Claim 2 follows. 5

6 Comining Claim 1 and Claim 2, we have a T 0 = i=1 a A i i=1 S 0 + i=1 T i + i=2 e G(T 0 T i 1, A i ) + i=1 e G(U 0, A i ) = (a 1) S 0 + i=1 T i + m + n. (2.3) (2.2) and (2.3) imply that a S 0 + 2m + n < a T 0 (a 1) S 0 + T i + m + n. i=1 Thus On the other hand, T i > S 0 + m m. (2.4) i=1 m = e G (T 0 ) = 1 i<j e G (A i, A j ) = e G (A i, j=i+1 A j) = e G (A j, T j ). Since A j is a maximal independent set of T j 1, e G (A j, x) 1 for any x T j. Therefore e G (A j, T j ) T j and m = e G(A j, T j ) T j, a contradiction to (2.4). The proof is complete. 3 Proof of Theorem 1.3 A vertex covering (or riefly, a covering) of a graph G is a suset C of V (G) such that every edge of G has at least one end in C. To prove Theorem 1.3, we start with the following Lemma. Lemma 3.1. Let H e a graph with 1 δ(h) (H) and a 3 e a positive integer. Suppose that each component of H has at least one vertex of degree no more than a 2. Let S 1, S 2,, S e the partition of V (H), where x S j iff d H (x) = j. Then there exists a maximal independent set I and thus a covering C = V (H) I such that i=1 (a j)c j (a 2) (a j)i j where I S j = i j, C S j = c j, j = 1, 2,, a 1. 6

7 Proof. We proceed y induction on V (H). If V (H) = 2, then H = K 2. Without loss of generality, let H = xy, I = {x} and C = V (H) {x} = {y}. Then (a j)c j = a 1 (a 2)(a 1) = (a 2) (a j)i j, so the lemma clearly holds. Suppose the lemma holds for V (H) n 1. Now we consider V (H) = n 3. Let m = δ(g) = min{j S j }, so 1 m a 2. Choose y S m and let T = {x N H (x) N H (y)} {y}. Then T 1 and T is an independent set. If V (H) = T N H (y), setting I = T and C = V (H) T = N H (y), then for any x T we have d H (x) = m and thus (a j)c j T m(a m) T (a 2)(a m) = (a 2) (a j)i j, the lemma holds. If V (H) T N H (y), let H = H (T N H (y)). Clearly, δ(h ) 1, (H ) a 1 and for each component R of H, there is at least one vertex v V (R) such that d R (v) a 2. Define S j = S j V (H ). From the induction hypothesis, there exist a maximal independent set I and a covering set C = V (H ) I such that (a j)c j (a 2) (a j)i j, where i j = S j I, c j = S j C for all j = 1, 2,, a 1. Let I = I T and C = V (H) I = C N H (y). Clearly, I is a maximal independent set of H. Then (a 2) (a j)i j = (a 2) (a j)i j+ T (a 2)(a m) (a j)c j+ T (a 2)(a m). Because d H (y) = m and m = min{j S j }, we have (a j)c j (a j)c j + T m(a m) (a j)c j + T (a 2)(a m). The induction is complete. 7

8 Now we proceed to the proof of Theorem 1.3. Proof of Theorem 1.3: If I(G) a and δ(g) a, then, y Theorem 1.1, G has a fractional a-factors. From Lemma 1.2, for any S V (G), a T d G S (T ) a S S since a <, where T = {x x V (G) S and d G S (x) a 1}. Therefore, G has a fractional [a, ]-factor. Now assume a > I(G) (a 1) +. If there exists no fractional [a, ]-factor in G. Then, y Lemma 1.2, there exists a vertex set S V (G) such that a T d G S (T ) > S (3.1) where T = {x x V (G) S and d G S (x) a}. Note that the set T = {x x V (G) S and d G S (x) = a} T and a T d G S (T ) = 0, so a T d G S (T ) = a T T + a T d G S (T T ) d G S (T ) = a T T d G S (T T ). Therefore, in the rest of the proof, we replace T y T T = {x x V (G) S, d G S (x) a 1} instead. If S =, then clearly T = since δ(g) a. Thus y (3.1), we have a T d G S (T ) = 0 > S = 0, a contradiction. So assume S from now on. For each i (0 i a 1), let T i = {x x T and d G S (x) = i} and T i = t i. So T 0 is the set of isolated vertices. Define a sugraph H = G[T 1 T 2 T ]. By the assumption, H has no (a 1)-regular sugraph as a component. Since d H (x) = i for any x T i, {T i i = 1, 2, a 1} forms a vertex partition of H. Therefore, y Lemma 3.1, there exists a maximal independent set I and a covering C = V (H) I such that (a j)c j (a 2) (a j)i j (3.2) where I T j = i j and C T j = c j (j = 1, 2,, a 1). Set W = G (S T ), U = S C (N G S (I) V (W )). Then and U S + ji j (3.3) i(g U) t 0 + i j. (3.4) Case 1. If i(g U) 2, y the definition of I(G), then U i(g U)I(G). (3.5) 8

9 Comining (3.3), (3.4) and (3.5), we otain S I(G)(t 0 + i j ) ji j. (3.6) Since a T d G S (T ) = at 0 + (a j)t j and t j = i j + c j. From (3.1), we have at 0 + (a j)i j + Thus (3.6) and (3.7) imply (a j)c j > S. (3.7) at 0 + (a j)i j + (a j)c j > (I(G)t 0 + I(G) i j ji j) = I(G)t 0 + (I(G) j)i j. (3.8) Since I(G) > ()+, we have I(G) a > a 1 > 0 as 2 a < and thus (a j)c j > (I(G) j a + j)i j > (3.9) (a 1 j + j)i j. From (3.2) and (3.9), there exist at least one index j 0 (1 j 0 a 1) such that (a 2)(a j) > a 1 j + j. However, this is impossile ecause a 1 j + j (a 2)(a j) = ( a 1)(a j 1) 0 as 2 a < and j a 1. Case 2. If i(g U) = 0, then y (3.4), we have 0 = i(g U) t 0 + i j, and so t 0 = i j = 0 for all j = 1, 2, a 1. Thus T = and a T d G S (T ) = 0 > S, a contradiction. Case 3. If i(g U) = 1, y (3.4), we have 1 = i(g U) t 0 + i j. If t 0 = i j = 0 for all j = 1, 2,, a 1, we follow the same arguments as in the case of i(g U) = 0. If t 0 = 1, then for all j = 1, 2,, a 1, i j = 0 and T is an isolated vertex. Thus, y (3.1), a T d G S (T ) = a > S, a contradiction. If there is some j 0 {1, 2,, a 1} such that i j0 = 1. Then H is a complete graph and d H (v) a 2 for each v V (H) ecause G S has no (a 1) regular sugraph as a component. Without loss of generality, we assume that I = {v}. Clearly, I is a maximal independent set of H. So U = S C (N G (v) V (W )) S +d G S (v) δ(g) a > I(G) (3.10) 9

10 where C = V (H) I. On the other hand, y (3.3), we have U S + j 0. By (3.2) and (3.7), we otain (a j)c j (a 2)(a j 0 ) and S < (a j)c j + (a j 0 ), and thus y comining these inequalities and (3.10), we have (a j)c j + (a j 0 ) > ( U j 0 ) > (I(G) j 0 ), which implies that (a 2)(a j 0 ) > (I(G) j 0 ) (a j 0 ), a contradiction due to I(G) a > a 1 and (a j 0 1)( a + 1) > 0. Hence, G has a fractional [a, ]-factor. 4 Remarks In this section, we construct two families of graphs to show the hypotheses in Theorem 1.3 can not e weakened. So the result is the est possile. Remark 4.1. The condition of I(G) > (a 1) + can not e replaced y I(G) (a 1) +. Consider a graph G constructed from K, K 1 and K as follows: let V (K 1 ) = {v 1, v 2,, v } and {u 1, u 2,, u } V (K ) such that each u i is from a different copy of K. Set V (G) = V (K ) V (K 1 ) V (K ) and E(G) = {u i v i i = 1, 2,, } E(K ) E(K ) E, where E = {uv i u V (K ) {v} and i = 2, 3,, 1} {vv 1 } for a fixed vertex v in K. It is not hard to see that I(G) = (a 1) + and for each suset S V (G), G S has no (a 1)- regular sugraph as a component. Choosing S = V (K ) {v}, then T = {v 2, v 3,, v, v} if a = 2 and T = {v 1, v 2, v 3,, v, v} if a > 2. In either case, we have a T d G S (T ) = a(+1) 2 a > (a 2) = S, this implies that G has no fractional [a, ]-factor. Remark 4.2. If we delete the condition for any suset S of V (G), G S has no (a 1)-regular sugraph as component in Theorem 1.3, then a graph G might not have fractional [a, ]-factor, even if the conditions I(G) > (a 1) + and δ(g) a hold. Consider a graph G constructed from K (n 1)(), (n + 1)K 1 and K (n+3)() as follows: let V ((n + 1)K 1 ) = {v 1, v 2,, v n+1 } and {u 1, u 2,, u n+1 } V (K (n+3)() ). Set V (G) = V (K (n 1)() ) V ((n+1)k 1 ) V (K (n+3)() ) and E(G) = E(K (n 1)() ) E(K (n+3)() ) {u i v i i = 1, 2,, n + 1} {vv i v V (K (n 1)() ) and i = 1, 2,, n + 1}. It is not hard to check that 10

11 I(G) = V (K (n 1)()) V (K (n+3)() ) V ((n+1)k 1) = a 1 + (n+1)() n+1 > (a 1) + and I(G) (a 1) + when n. Let S = V (K (n 1)() ), then T = V ((n + 1)K 1 ) and thus a T d G S (T ) = (a 1)(n + 1) > (n 1)(a 1) = S, this implies that G has no fractional [a, ]-factor. Acknowledgement: The authors are indeted to the anonymous referees for their constructive suggestions. References [1] R. P. Anstee, Simplified existence theorems for (g, f)-factors, Discrete Applied Math. 27(1990), [2] V. Chvátal, Tough graphs and hamiltonian circuits, Discrete Math. 5(1973), [3] H. Enomoto, B. Jackson, P. Katerinis and A. Saito, Toughness and the existence of k-factors, J. Graph Theory 9(1985), [4] K. Heinrich, P. Hell, D. Kirkpatrick and G. Z. Liu, A simple existence criterion for (g, f)-factors, Discrete Math. 85(1990), [5] P. Katerinis, Toughness of graphs and the existence of factors, Discrete Math. 80(1990), [6] G. Z. Liu and L. Zhang, Fractional (g, f)-factors of graphs, Acta Math. Scientia 21B(4)(2001), [7] Y. H. Ma and G. Z. Liu, Isolated toughness and existence of fractional factors in graphs, Acta Appl. Math. Sinica 26(2003), (in Chinese). [8] E. R. Scheinerman and D. H. Ullman, Fractional Graph Theory, John Wiley and Son Inc., New York,