Binding Number and Connected (g, f + 1)-Factors in Graphs

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1 Binding Number nd Connected (g, f + 1)-Fctors in Grphs Jinsheng Ci, Guizhen Liu, nd Jinfeng Hou School of Mthemtics nd system science, Shndong University, Jinn 50100, P.R.Chin helthci@163.com Abstrct. Let G be connected grph of order n nd let, b be two integers such tht b. Letg nd f be two integer-vlued functions defined on V (G) such tht g(x) f(x) b for every x V (G). A spnning subgrph F of G is clled (g,f +1)-fctor if g(x) d F (x) f(x) + 1 for every x V (F ). For subset X of V (G), let N G(X) = N G(x). The binding number of G is defined by bind(g) = x X min{ N G(X) X X V (G),N G(X) V (G)}. Inthispper,itis proved tht if bind(g) > (+b)(n 1), f(v (G)) is even nd n (+b), n then G hs connected (g,f +1)-fctor. 1 Introduction In this pper, we consider only finite undirected grphs without loops nd multiple edges. The set of vertices of grph G is denoted by V (G); the set of edges by E(G). We use ν(g) to denote the crdinlity of V (G). If H is subgrph of G nd S is subset of V (G), we denote by N H (S) the set of vertices in H which re djcent to some vertex in S,ndset N H (S) = d H (S). In prticulr, when H = G nd S = u, thenletn G (u) =N(u) ndletd G (u) =d(u). The binding number of X V (G),N G (X) V (G)}. We write E G (X, Y )={xy E(G) x X, y Y }, nde G (X, Y )= E G (X, Y ). As usul, we use δ for the minimum degree of G. For other definitions nd nottions not defined here, we refer the reders to [1]. Let G be connected grph of order n nd let, b be two integers such tht b. Letg nd f be two integer-vlued functions defined on V (G) such tht g(x) f(x) b. A spnning subgrph F of G is clled (g, f +1)-fctor if g(x) d F (x) f(x)+1 for every x V (F ). If F is connected, we cll it connected (g, f+1)-fctor. The purpose of this pper is to present sufficient condition which gurntees the existence of connected (g, f+1)-fctors in grph G relted to the binding number G is defined by bind(g) =min{ NG(X) X of grph G. For convenience, we write d G S (T )= d G S (x),f(s) = f(x) x T x S nd f(t )= f(x). x T Corresponding uthor. Y. Shi et l. (Eds.): ICCS 007, Prt III, LNCS 4489, pp , 007. c Springer-Verlg Berlin Heidelberg 007

2 314 J. Ci, G. Liu, nd J. Hou Mny uthors hve investigted fctors, connected fctors nd fctoriztions [3,5,7]. In this pper, we study conditions on the binding number nd the order of grph G which gurntee the existence of connected (g, f + 1)-fctor in G. We begin with some known results. There is well-known necessry nd sufficient condition for grph G to hve (g, f)-fctor which ws given by Lovász. Theorem 1 ([6]). A grph G hs (g, f)-fctor if nd only if δ(s, T )=f(s)+d G S (T ) g(t ) h(s, T ) 0 for ny disjoint subsets S nd T of V(G), where h(s, T ) denotes the number of components C of G (S T ) such tht g(x) =f(x) for ll x V (C) nd e G (T,V (C)) + f(x) is odd. Furthermore, if g(x) =f(x) for every x x V (C) V (G), then δ(s, T )=f(v (G))(mod ). M. Kno gve result which shows the reltion of binding numbers nd existence of f-fctors. Theorem []. Let G be connected grph of order n nd let nd b be integers such tht 1 b nd b. Let f be n integer-vlued function defined on V (G) such tht f(x) b for every x V (G). Suppose tht n (+b) nd f(x) 0(mod ). If x V (G) bind(g) > ( + b 1)(n 1) n ( + b)+3, nd for ny nonempty independent subset X of V (G) then G hs n f-fctor. N G (X) (b 1)n+ X 1, + b 1 The following two results re essentil to the proof of our min theorem. Theorem 3 [4]. Let G be grph nd let g nd f be two positive integer-vlued functions defined on V(G) such tht g(x) f(x) d G (x) for ll x V (G). If G hs both (g,f)-fctor nd Hmiltonin pth, then G contins connected (g,f+1)-fctor. Theorem 4 [8]. If bind(g) 3, then G hs Hmiltonin cycle. Liu nd Zhng proposed the following problem. Problem [5]. Find sufficient conditions for grphs to hve connected [,b]- fctors relted to other prmeters in grphs such s binding number, neighbor union nd connectivity.

3 Binding Number nd Connected (g,f + 1)-Fctors in Grphs 315 We now give our theorem which prtilly solve the bove problem. Theorem 5. Let G be connected grph of order n, nd let nd b be two integers such tht b. Letg nd f be two integer-vlued functions defined on V (G) such tht g(x) f(x) b for every x V (G) nd f(v (G)) is even. If n (+b), bind(g) > (+b)(n 1) n, then G hs connected (g, f +1)-fctor. Proof of the Theorem 5 In this section, we ssume tht G stisfies the conditions of Theorem 5. Since bind(g) > (+b)(n 1) n > 3, ccording to Theorem 4, grph G hs Hmiltonin cycle. Hence by Theorem 3, to prove Theorem 5, we need only to prove tht grph G hs (g, f)-fctor. At first, we need some lmms. Lemm 1. Suppose tht G stisfies the conditions of Theorem 5. Then N G (X) > + X 1 +b for ll nonempty subset X of V(G) with N G (X) V (G) nd δ(g) > +b. Proof. Let bind(g) >l,ndy = V (G) \ N G (X). Since N G (Y ) V (G) \ X, we obtin (n X ) N G (Y ) > l Y = l(n N G (X). Therefore we hve N G (X) > ((l 1)n+ X )/l, ndδ(g) > ((l 1)n +1)/l. Then replce l by (+b)(n 1) n b,wehve δ(g) > + b. (1) Since n (+b), similrly we obtin (n + b n)n + n X N G (X) > ( + b)(n 1) (n 1) n + n X = ( + b)(n 1) + X 1. + b So N G (X) > + X 1. () + b We begin to prove tht grph G hs (g, f)-fctor. By contrdiction, we suppose tht G hs no (g, f)-fctors. Then ccording to Theorem 1, there exist disjoint subsets S nd T of V (G) stisfying f(s)+d G S (T ) g(t ) h(s, T ) < 0. (3)

4 316 J. Ci, G. Liu, nd J. Hou Let s = S nd t = T. Then by Theorem 1 nd (3) we hve s + d G S (T ) bt ω 1, (4) where ω denotes the number of components of G (S T ). Obviously, ω n s t. (5) Let m denote the minimum order of components of G (S T ). Then we know tht m n s t. (6) ω By the definition of m, itisobvioustht The following Clims hold. Clim 1. T. Proof. Otherwise T =. δ(g) m 1+s + t. (7) Cse 1. S =. Since G is connected nd δ(s, T ) < 0, h(s, T ) = 1. So obviously f(x) =g(x) for every x V (G) by the definition of h(s, T ). Hence δ(s, T )= 1. On the other hnd, since f(v (G)) is even, ccording to Theorem 1, δ(s, T )iseven.this is contrdiction. Cse. S. Then by (4) nd (5), we get s +1 ω n s. (8) So from (1), (6), (7) nd (8) we know tht n s <δ(g) m 1+s + b ω 1+s n s s +1 1+s n 1 Since n 1 s s 0 by (8), it follows tht n n 1. This is contrdiction since. So T. (n 1 s s)(s 1). s +1

5 Binding Number nd Connected (g,f + 1)-Fctors in Grphs 317 Since T by Clim 1, let h = min{d H S (x) x T }. Then obviously δ(g) h + s. (9) Clim. h 0. Proof. If h =0,letX = {x T d G S (x) =0}. Then X 0.From()we get + X 1 s N G (X) >. (10) + b On the other hnd, from (4),(5) we get s b X +(1 b)(t X ) (b 1)(n s t) 1. Thus s (b 1)n+ X 1. + b 1 From (10) nd the bove inequlity, we get + X 1 + b < (b 1)n+ X 1, + b 1 which mens X > n+ 1. This is contrdiction. So h 0. Clim 3. h<b. Proof. Otherwise, h b. We consider two cses. Cse 1. h>b. Then by (4), s +(h b)t ω 1. (11) And so ω s + t +1 s + t +1. (1) Since T by Clim 1, ω. First suppose tht m. Then from (7) nd (1) we get <δ(g) m 1+s + t m + ω + b m + ω + = mω n. (m )(ω )

6 318 J. Ci, G. Liu, nd J. Hou This is contrdiction. Hence we my ssume m = 1. Then from (1) nd (5) we hve s + t n 1. Let C 1 be the lest component of G (S T ). Then C 1 contins only one vertex x. Hence + b This lso is contrdiction. n 1 <δ(g) d(x) s + t. Cse. h = b. From (4) we hve ω s +1. (13) So by (6) nd (13) m n s t ω From (1) nd (9) it is obvious tht n s 1 s +1. δ(g) h + s = b + s. + b So s> + b b(n b) b =. + b Let f(s) = n s 1 s+1.thenf (s) = 1 n+ (s+1).nmely,f (s) < 0. Hence, m< n b(n b) +b 1 b(n b) +b +1. Since n (+b), it follows tht m<1. This is contrdiction. Therefore, from Cse 1 nd Cse we hve known tht h<b. Hence, by Clim nd Clim 3 we hve 1 h b 1. By (4),(5) nd b h 1, we obtin s +(h b)t (b h)(n s t) 1, which implies s (b h)n 1 + b h. (14)

7 Binding Number nd Connected (g,f + 1)-Fctors in Grphs 319 Then from (1),(9) nd (14) we obtin (b h)n 1 δ(g) s + h + h. (15) + b + b h Let f(h) = (b h)n 1 +b h (+b) + h. Then by n, f n 1+( + b h) (h) = ( + b h) 1 < ( + b h) < 0. Hence the mximum vlue of f(h) is (b 1)n 1 +b So by (1) nd (15) we hve (b 1)n 1 < +1, + b + b 1 which implies n < ( + b) ( + b). Thus n< (+b). This contrdicts the ssumption tht n (+b). Hence we get the desired contrdiction nd conclude tht G hs (g, f)-fctor. Since G hs both Hmiltonin pth nd (g, f)-fctor, ccording to Theorem 3, grph G hs connected (g, f +1)-fctor. The possible further problem in this topic is the following. Question. Find the proper binding number nd order of G which gurntees the existence of connected k-fctors in G. References 1. Bondy, J. A., Murty, U. S. R.: Grph Theory with Applictions. Mcmilln, London (1976).. Kno, M., Tokushige, N.: Binding number nd f-fctors of grphs. J. Combin. Theory Ser. B 54(199) Kouider, M., Vestergrd, P. D.: Connected fctors in grphs - survey. Grphs nd Combintorics 1 (005) Li, G., Xu, Y., Chen, C., Liu, Z.: On connected (g,f + 1)-fctors in grphs. Combintoric 5 (4) (005) Liu, G., Zhng, L.: Fctors nd fctoriztions of grphs(in Chinese). Advnces in Mth. 9(000) Lovász, L.: Subgrphs with prescribed vlencies. J. Combin. Theory 8 (1970) Plummer, M. D.: Grph fctors nd fctoriztion. Chpter 5.4 in Hndbook on Grph Theory Eds.: J. Gross nd R. Yellen CRC Press, New York (003) Woodll, D. R.: The binding number of grph nd its Anderson number. J. Combin. Theory Ser. B 15 (1973) 5-55