AbstractLet G be a connected graph of order n, a and b be integers such that 1 ≤ a ≤ b and 2 ≤ b, and f: V(G) → {a, a + 1, …, b} be a function such that Σ(f(x); x ∈ V(G)) ≡ 0 (mod 2). We prove the following two results: (i) If the binding number of G is greater than (a + b −1)(n−1)(an−(a + b) + 3) and n ≥(a + b)2a, then G has an f-factor; (ii) If the minimum degree of G is greater than (bn − 2)(a + b), and n ≥(a + b)2a, then G has an f-factor

Kano, Mikio

Tokushige, Norihide

Elsevier - Publisher Connector 

JOURNAL OF COMBINATORIAL THEORY, Series B 54, 213-221 (1992) 
Binding Numbers and f-Factors of Graphs 
MIKIO KANO 
Akashi College of Technology, 
Akashi 674, Japan 
AND 
NORIHIDE TOKUSHIGE 
Department of Computer Science, Meiji University, 
Higashimita, Tama-ku, Kawasaki 214, Japan 
Communicated by the Editors 
Received May 25, 1988 
Let G be a connected graph of order n, a and b be integers such that 
1 <a< b and 2 <b, and f: V(G) + {a, a+ 1, . . . . b} be a function such that 
x(f(x); XE V(C)) ~0 (mod 2). We prove the following two results: (i) I f  the 
binding number of G is greater than (a+ b - I)(n - l)/(an - (a+ b)+ 3) and 
n 2 (a + b)2/a, then G has an f-factor; (ii) I f  the minimum degree of G is greater 
than (bn - 2)/(a + b), and n > (a + 6)*/a, then G has an f-factor. 0 1992 Academic 
Press, Inc. 
1. INTRODUCTION 
We consider a finite graph G with vertex set V(G) and edge set E(G), 
which has neither loops nor multiple edges. For a vertex x of G, the 
neighborhood N&X) of x in G is the set of vertices of G adjacent to x, and 
the degree deg,(x) of x is ING(x)l. We denote by 6(G) the minimum 
degree of G. For a subset X of V(G), let 
N,(X):= (J N&X). 
x E x 
We say that X is independent if NG(X) n X= 12(. The binding number 
bind( G ) of G is defined by 
bind(G) : = min ’ y+)’ I E3 #Xc V(G), N&k’) f v(G)} 
213 
0095-8956192 $3.00 
582b/54/2-f 
Copyright 0 1992 by Academic Press, Inc. 
All rights of reproduction in any form reserved. 
214 KANO AND TOKUSHIGE 
(cf. [12]). It is trivial by the definition that bind(G) > c implies that for 
every subset X of I’(G), we have N&X) = V(G) or ) N&X)( > c I XJ. It is 
also obvious that if bind(G) > 1, then G is connected. Let k be a positive 
integer and f be an integer-valued function defined on I’(G) (i.e., 
f: V(G) + (..., 0, 1, 2, . ..}). Th en a spanning k-regular subgraph of G is 
called a k-factor of G, and a spanning subgraph F of G is called an f-factor 
if deg,(x) =f(x) for all XE V(G). 
In this paper, we study conditions on the binding number and on the 
minimum degree of a graph G which guarantee the existence of an f-factor 
in G. We begin with some known results. 
THEOREM A (Anderson [ 11). If a graph G has euen order and 
bind(G) > 413, then G has a l-factor. 
THEOREM B (Woodall [ 121). If bind(G) B 3/2, then G has a Hamilton 
cycle, in particular, G has a 2-factor. 
Recently, Katerinis and Woodall [B] and Katerinis [6] found the 
following sufficient conditions for a graph to have a k-factor. These condi- 
tions were also obtained by Egawa and Enomoto [3] independently. 
THEOREM C. Let k 2 2 be an integer and G be a graph of order n. 
Assume n 2 4k - 6 and kn is even. Then the following two statements hold: 
(i) If bind(G) > (2k - l)(n - l)/(kn - 2k + 3), then G has a k-factor 
L-81. 
(ii) If 6(G) > n/2, then G has a k-factor [6]. 
It is shown that the conditions in (i) and (ii) are best possible. Let us 
note that if k 2 3 and n > 4k - 5, then 
2-~<W-W-l)<2 
k‘ kn-2k+3 ’ 
We now give our theorem, which is an extension of the above Theorem C. 
Moreover, the theorem gives a result concerning the following question: If 
bind(G) > c 2 2, what factor does a graph G have? 
THEOREM 1. Let G be a connected graph of order n, a and b be integers 
such that 1 < a < b and 2 d b, and f: V(G) -+ {a, a + 1, . . . . b}. Suppose that 
n > (a + b)*/u and C,, VCcj f (x) = 0 (mod 2). Zf one of the following three 
conditions is satisfied, then G has an f-factor. 
(i) bind(G)>(u+b-l)(n-l)/(un-.(u+b)+3); (1) 
BINDIN~~NUMBER~AND~-FA~T~R~OFGRAPHS 215 
(ii) S(G) > (bn - 2)/(a + b); (2) 
(iii) S(G)>((b-l)n+a+b-2)/(a+b-1) (3) 
and for every non-empty independent subset X of V(G), 
I No( 2 
(b-l)n+IXI-1 
a+b-1 
We now show that the conditions (1) and (2) are best possible. If a 
graph G consists of n (n 2 2) disjoint copies of a graph H, then we write 
G=nH. The join G=A+B has V(G)=V(A)uV(B) and E(G)= 
W)uE(B) u x ( YI XE V(A) and YE V(B)}. Let c= [b/a), m be a positive 
integer, and G = Kzmb ~ 2m _ 2r + (ma - 1) K,, where K, denotes the complete 
graph of order 1. Define a function f: V(G) -+ {a, a + 1, . . . . b > by 
ifxE V(K2mb-zm-2c) 
otherwise. 
Then G has no f-factor since for S = V( KZmb _ 2m _ *<) and T = V(G)\ S, we 
have 
y,(S, T) = 26 - 2ac - 2 < 0 
Moreover, we have 
(see Lemma 1). 
bind(G) = 
(a+b-l)(n-1) 
na-((a+b)+3+2(ac-b)’ 
Note that for X= V(G)\(V(K,,,_,,-,,)u {u}), where V(K,)= {u, u}, we 
obtain 
I N&3l n-l (a+b- l)(n- 1) 
1x1 =2(ma-1)-l= na-(a+b)+3+2(ac-b) 
= bind(G). 
Therefore, if b is divisible by a, then condition (1) is best possible. 
Next, suppose that a + b is even and there exist positive integers s and 
t such that bs=at +2 and s+ t is even. Let G= (am+s) K1 + Kbm+,, 
where m is a positive integer, and let f be a function on V(G) deiined by 
m={f: 
if XE V((am+s) K,), 
if XE V(Kbm+r). 
Then G has no f-factor and 
bn-2 
G(G)=bm+t=a+b. 
Hence condition (2) is also best possible in this sense. 
216 KANOAND TOKUSHIGE 
Note that (iii) of Theorem 1 is an extension of results in [9, 131, which 
are obtained from (iii) by setting a = b. Similar results on l-factor can be 
found in [2]. Moreover, a similar sufficient condition for a graph to have 
an [a, b]-factor, which is a spanning subgraph F such that a 6 deg,(x) < b 
for all vertices x, can be found in [S], and similar sufficient conditions for 
a bipartite graph to have k-factors are given in [7,4]. 
2. PROOFS 
Let G be a graph and 5’ and T be disjoint subsets of V(G). Then G - S 
denotes the subgraph of G induced by V(G)\S, and e,(S, T) denotes the 
number of edges of G joining a vertex in S to a vertex in T. Our proof of 
Theorem 1 is analogous to those of [3, 8, 9, 131 and depends on the 
following lemma, which is called thef-factor theorem. 
LEMMA 1 (Tutte [lo, 111). Let G be a graph andf: V(G) + (0, 1,2, . ..> 
such that %x Y(G) f(x) = 0 (mod 2). Then G has an f-factor if and only if 
yG(s, T) : = 1 f(x) + 1 (deg,-,(x) -f(x)) - h,(X T) B 0 
XES XE T 
for all S, Tc V(G), Sn T= @, where h,(S, T) denotes the number of 
components C of G - (S u T) such that C,, v(c, f (x) + eG( V(C), T) = 
1 (mod 2). 
Moreover, the following useful congruence expression holds: 
YG(‘% T)= c f(x)=0 (mod 2). 
XE Y(G) 
(5) 
LEMMA 2 [12]. Let G be a graph of order n. If bind(G) > c, then 
6(G)> ((c- l)n+ 1)/c, and [No(X)1 > ((c- l)n+ lXj)/cfor ailnon-empty 
subsets X of V(G) with No(X) # V(G). 
Proof. Let Y := V(G)\N,(X). Since NG( Y) c V(G)\X, we have n- 1x1 
21NG(Y)l>c I YI =c(n-INo(X)(). Hence IZVG(X)I >((c-l)n+lXl)/c, 
and so d(G)>((c-l)n+l)/c. 
Suppose that (1) in Theorem 1 holds. Then, by Lemma 2, we have 
6(G),(b-1)n+u+b-3,(b-l)n 
u+b-1 ‘a+b-1’ (6) 
and 
1 NC@?1 > 
(b-l)n+ulXI+(b-3)(n-IXJ)/(n-1) 
u+b-1 
>(b-l)n+IXI-2 
u+b-1 
BINDINGNUMBERSAND$FACTORSOFGRAPI~ 217 
for every independent subset X of V(G). Hence G satisfies (3) and (4). 
Therefore (i) of Theorem 1 is an immediate consequence of (iii) of the 
theorem, and so we shall prove (ii) and (iii) of the theorem. 
Proof of (iii) of Theorem 1. Suppose that G satisfies the conditions (3) 
and (4), but has no f-factor. By Lemma 1 and (5), there exist disjoint 
subsets S and T of V(G) such that 
where w  denotes the number of components of G - (Su T). Note that 
SuT#@ since ~(a,@)=-h(@,@)=O, which follows from the 
assumption that G is connected and C,, V(GJ-(~) ~0 (mod 2). In 
particular, we have 
a ISI + c (deg,_,(x)-6)-w< -2. 
iE f 
(7) 
We choose S and T so that 1 SI + 1 TI is as large as possible subject to 
y(S, T) < 0. Let s : = 1 S ( and t : = I TI . It is clear that 
w<n-s-t. (8) 
If w  >O then let m denote the minimum order of components of 
G - (S u T). Then 
n-s-t 
rndp (9) 
W 
and 
G(G)<m-l+s+t. 
Moreover, it follows from the choice of S and T that 
(10) 
ifa=bthenm33 (11) 
(cf. [8]). If T#@, let 
h := min{deg,-,(x)lxE T}. 
Then obviously 
d(G)dh+s. (12) 
We consider five cases and derive a contradiction in each case. 
Case 1. T=@. By (7) and (8), we have 
as+2<w<n-s. (13) 
218 KANO AND TOKUSHIGE 
Hence we have by (6), (lo), (9), and (12) that 
(b-1)n 
a+b-1 
<d(G)<m-1 +s< Y-l+, 
n-s 
<-- 
as + 2 
l+s 
n-2 (n-2-as-s)(as-a+ 1) =-- 
a+1 (a+ l)(as+2) 
Since n-2-sa-sa0 by (13), it follows that 
(b-l)n n-2 
a+b-l<a+l’ 
which implies a(b - 2) n < -2(a + b - 1). This is clearly impossible since 
b > 2. 
Case 2. T#@ and h=O. Let Z:= {xETIdeg,-,(x)=O}#@ and 
z = ) Z I. Since Z is independent, we have by (4) 
‘“,‘);“;- l d IN,(Z)1 <s. (14) 
On the other hand, we have by (7), (8), and the fact that b - 12 1 
as-bz+(l -b)(t-z)-(b- l)(n-s--t)< -2. 
Hence 
s< 
(b-l)n+z-2 
a+b-1 ’ 
This contradicts (14). 
Case 3. T#(a and 1 <h<b-1, By (7), (8), and the fact that 
b-hal, we have 
as+(h-b)t-(b-h)(n-s-t)< -2. 
Thus 
s<(b-h)n-2 
’ a+b-h ’ (15) 
219 
On the other hand, we obtain by (3) and (12) that 
(b-l)n-1 
a+&-1 
+ l<@G)<s+h. 
This inequality together with (15) gives us 
(b-l)n-1 
a+b-1 
+l-h<@-h)n-2 
’ a+b-h ’ 
Hence 
(h-l)an<(h-l)(a+b-l)(a+b-h)-(a+b+h-2). 
This implies h > 2 and 
an<(a+b-l)(a+b-h)- 
(a+b+h-2) 
h-l . 
This contradicts our assumption that n 2 (a + b)2/a. 
Case 4. Tf @ and h = b. We have w  > as + 2 by (7), and so we obtain 
by (9) that 
n-s-t n-s-l 
m< -< 
W as+2 ’ (16) 
If b> 3 then we get the following inequality from an> (a+b)*> 
(a+b+l)(a+b-1): 
an(b-2)>(a+b-l)(ab+b2-2a-b-2). (17) 
By (3) and (12), we have 
(b-l)n-l+ 
a+b-1 
and so 
s>(b-l)n-l-(b-l) 
’ a+b-1 
l<&G)<h+s=b+s 
(18) 
-n-3 I an(b-2)+(a+b-1)(3-(a+l)(b-l))-(a+l) 
a+1 (a+b- l)(a+ 1) 
n-3+(a+b-l)(b’-a-2b)+2b+a-3 
>- 
a+1 (a+b-l)(a+ 1) ’ (by (17)) 
220 KANO AND TOKUSHIGE 
Hence, if b > 3 then s > (n - 3)/(a + l), and so m < 1 by (16), a contra- 
diction, If u=b=2 then s>(n-4)/3 by (18), and so m-c3 by (16). This 
contradicts (11). Therefore we may assume that a = 1 and b = 2. By (16) 
and (18), we have m = 1. Thus it follows from (10) and (12) that 
d(G)<s+t and 6(G)<b+s=s+2. 
Hence, by (7) and (8), we obtain 
6(G)<s+2=us+2<w<n-s-t6n-6(G). 
Hence 6(G) < n/2. This contradicts (3). 
Case 5. T#Qr and h>b. By (7), we have as+(h-b)t-wd -2, and 
so 
w>as+t+2bs+t+2. 
Suppose that m 2 3. Then, by (10) and (9), we have 
(19) 
G(G)<m-l+s+t<m+w-3 
This contradicts (4). Thus we may assume that m < 2. It follows from (8) 
and (19) that s+ t + 1 <n/2. Then by (3) and (lo), we have 
(b-l)n 
a+b-1 
<S(G)<s+l+l$. 
Thus n(2b -a - 1) < 0. This is impossible. Consequently, (iii) is proven. 
Proof of (ii) of Theorem 1. This is almost identical to the proof of (iii). 
Since n > (a + b)‘/u, we have 
bn-2 (b-l)n+u+b-2 
-----a 
a+b a+b-I ’ 
and so (4) still holds by (3). Thus Cases 1, 3, 4, and 5 carry over without 
modification from (iii) to (ii) because we don’t use (4) in these cases. The 
only case that needs to change is the following: 
Case 2. T# fa and h = 0. By (7) and (8), we have 
-2>u.s--bt-(n-s-t)+us-bt-b(n-s-t), 
BImmci NUMBERS ANDJFA~TOFCSOF GRAPHS 221 
and so s6 (bn -2)/(a+b). Then (3) and (11) give 
bn-2 bn-2 
-<6(G)6h+s=sd---- 
a+b a+b’ 
a contradiction. Consequently the proof is complete. 
ACKNOWLEDGMENT 
The authors thank the referees for their helpful suggestions by which proofs became much 
shorter and clearer. 
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Binding numbers and f-factors of graphs 

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Binding numbers and f-factors of graphs

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