A Degree Condition for Graphs Having All (a,b) -Parity Factors
aa r X i v : . [ m a t h . C O ] S e p A Degree Condition for Graphs Having All ( a, b )-ParityFactors
Haodong Liu and Hongliang Lu ∗† School of Mathematics and StatisticsXi’an Jiaotong UniversityXi’an, Shaanxi 710049, China
Abstract
Let a and b be positive integers such that a ≤ b and a ≡ b (mod 2). We saythat G has all ( a, b )-parity factors if G has an h -factor for every function h : V ( G ) →{ a, a + 2 , . . . , b − , b } with b | V ( G ) | even and h ( v ) ≡ b (mod 2) for all v ∈ V ( G ). Inthis paper, we prove that every graph G with n ≥ b + 1)( a + b ) vertices has all ( a, b )-parity factors if δ ( G ) ≥ ( b − b ) /a , and for any two nonadjacent vertices u, v ∈ V ( G ),max { d G ( u ) , d G ( v ) } ≥ bna + b . Moreover, we show that this result is best possible in somesense. Keywords: degree condition, all ( a, b ) -factors We consider simple graphs in this paper. Let G be a graph with vertex set V ( G ) and edgeset E ( G ). Given x ∈ V ( G ), the set of vertices adjacent to x is said to be the neighborhood of x , denoted by N G ( x ), and d G ( x ) = | N G ( x ) | is called the degree of x . We write N G [ x ] for N G ( x ) ∪ { x } . For a vertex set D ⊆ V ( G ), let N D ( v ) := N G ( v ) ∩ D denote the set of verticeswhich are adjacent to v in D . For any vertex set X ⊆ V ( G ), let G [ X ] denote the vertexinduced subgraph of G induced by X and the subgraph induced by vertex set V ( G ) − X is also denoted by G − X . A graph F is a spanning subgraph of G if V ( F ) = V ( G ) and E ( F ) ⊆ E ( G ). Let h : V ( G ) → N and let S ⊆ V ( G ). We write h ( S ) := P v ∈ S h ( v ). ∗ [email protected] † Supported by the National Natural Science Foundation of China under grant No.11871391 and Funda-mental Research Funds for the Central Universities G , let g, f be two non-negative integer-valued functions such that g ( v ) ≤ f ( v ) and g ( v ) ≡ f ( v ) (mod 2) for all v ∈ V ( G ). We say that a spanning subgraph F of G is a ( g, f ) -parity factor if d F ( v ) ≡ f ( v ) (mod 2) and g ( v ) ≤ d F ( v ) ≤ f ( v ) for all v ∈ V ( G ). A ( g, f )-parity factor is an f -factor if f ( v ) = g ( v ) for all v ∈ V ( G ). If f ( v ) = k for all v ∈ V ( G ), then an f -factor is a k -factor . Let a, b be two integers such that a ≤ b and a ≡ b (mod 2). If f ( v ) = b and g ( v ) = a for all v ∈ V ( G ), then a ( g, f )-parity factor is an( a, b ) -parity factor . We call a graph G having all ( a, b ) -parity factors if G has an h -factorfor any functions h with a ≤ h ( v ) ≤ b and a ≡ h ( v ) (mod 2) for all v ∈ V ( G ). If G has an h -factor for any functions h with a ≤ h ( v ) ≤ b and h ( V ( G )) ≡ G having ( a, b )-factors.Tutte [9] give a characterization for a graph to have an f -factor. Lov´asz [5, 6] gavea criterion for a graph to have an ( g, f )-parity factor. Kano and Tokushige [3] gave aminimum degree condition for a graph to have all ( a, b )-factors. In 1998, Niessen gave acharacterization for a graph to have all ( a, b ) -parity factors . Theorem 1.1 (Tutte, [9])
Let G be a graph and let f : V ( G ) → N . G contains an f -parity factor if and only if for all disjoint sets S, T of V ( G ) , η ( S, T ) = f ( S ) − f ( T ) + d G − S ( T ) − q G ( S, T ; f ) ≥ , where q G ( S, T ; f ) denotes the number of f -odd components C of G − S − T such that f ( V ( C )) + e G ( V ( C ) , T ) ≡ . Moreover, η ( S, T ) ≡ f ( V ( G )) (mod 2) . Theorem 1.2 (Niessen, [7])
Let G be a graph and let g, f : V ( G ) → N such that g ( v ) ≤ f ( v ) and g ( v ) ≡ f ( v ) (mod 2) for all v ∈ V ( G ) . G has all ( g, f ) -parity factors if and onlyif η ( S, T ) = g ( S ) − f ( T ) + d G − S ( T ) − q G ( S, T ; f ) ≥ , for all disjoint sets S, T of V ( G ) , where q G ( S, T ; f ) denotes the number of f -odd components C of G − S − T such that f ( V ( C )) + e G ( V ( C ) , T ) ≡ . Nishimura [8] gave a degree condition for a graph to have a k -factor. Theorem 1.3 (Nishimura, [8])
Let k be an integer such that k ≥ , and let G be aconnected graph of order n with n ≥ k − , kn even, and minimum degree at least k . If G satisfies max { d G ( u ) , d G ( v ) } ≥ n/ for each pair of nonadjacent vertices u, v in G , then G has a k -factor. In this paper, we give a sufficient condition for a graph to have all ( a, b )-parity factors.2 heorem 1.4
Let a, b, n be three integers such that a ≡ b (mod 2) , a < b , na ≡ and n ≥ b + 1)( a + b ) . Let G be a connected graph of order n . If δ ( G ) ≥ b − ba and max { d G ( u ) , d G ( v ) } ≥ bna + b (1) for each pair of nonadjacent vertices u and v in V ( G ) , then G has all ( a, b ) -parity factors. By contradiction, suppose that the result does not hold. By Theorem 1.2, there exist twodisjoint subsets S and T of V ( G ) such that η ( S, T ) = a | S | − b | T | + X x ∈ T d G − S ( x ) − q ( S, T ) ≤ − , (2)where q ( S, T ) = q ( S, T ; a ) is the number of connected components C of G − S − T suchthat a | V ( C ) | + e G ( V ( C ) , T ) ≡ C is also called an a -odd components). Forsimplicity, we write s := | S | , t := | T | and w := q ( S, T ; a ). By Theorem 1.1, we have η ( S, T ) ≡ a | V ( G ) | ≡ η ( S, T ) = as − bt + X x ∈ T d G − S ( x ) − w ≤ − . (3)We claim that S ∪ T = ∅ . (4)Otherwise, suppose that s = t = 0. According to (3), we have η ( S, T ) = − w ≤ −
2, i.e., w ≥
2, a contradiction since G is connected.When w ≥
1, let C , C , . . . C w denote the a -odd components of G − S − T , and let m i = | V ( C i ) | for 1 ≤ i ≤ w . Without loss of generality, suppose that m ≤ m ≤ . . . ≤ m w .Let U = S ≤ i ≤ w V ( C i ).Without loss of generality, among all such subsets, we choose S and T such that U isminimal and V ( G ) − S − T − U is maximal. Claim 1. d G − S ( u ) ≥ b + 1 and e G ( u, T ) ≤ a − u ∈ U .By contradiction. Firstly, suppose that there exists a vertex u ∈ U such that d G − S ( u ) ≤ . Let T ′ = T ∪ { u } . By inequality (3), we have η ( S, T ′ ) = as − b | T ′ | + X x ∈ T ′ d G − S ( x ) − q ( S, T ′ )= ( as − bt + X x ∈ T d G − S ( x ) − q ( S, T ′ )) + d G − S ( u ) − b ≤ as − bt + X x ∈ T d G − S ( x ) − q ( S, T ′ ) ≤ as − bt + X x ∈ T d G − S ( x ) − ( q ( S, T ) − ≤ η ( S, T ) + 1 . Recall that η ( S, T ′ ) ≡ η ( S, T ) ≡ η ( S, T ′ ) ≤ η ( S, T ) ≤ −
2, whichcontradicts the choice of U .Secondly, suppose that there exists a vertex u ∈ U such that e G ( u, T ) ≥ a . Let S ′ = S ∪ { u } . Then we have η ( S ′ , T ) = a | S ′ | − bt + X x ∈ T d G − S ′ ( x ) − q ( S ′ , T )= as + a − bt + X x ∈ T d G − S ( x ) − e G ( u, T ) − q ( S ′ , T ) ≤ as − bt + X x ∈ T d G − S ( x ) − q ( S ′ , T ) ≤ as − bt + X x ∈ T d G − S ( x ) − ( q ( S, T ) − η ( S, T ) + 1 . With similar discussion, we have η ( S ′ , T ) ≤ η ( S, T ) ≤ −
2, which contradicts the choice of U . This completes the proof of Claim 1.Note that when we choose appropriate S and T , each components of G − S − T containsat least b − a + 1 vertices by Claim 1. Claim 2.
Let C i , . . . , C i τ be any τ components of G [ U ] and let U ′ = S τj =1 V ( C i j ). d G [ T ∪ U ′ ] ( u ) ≤ b − τ for every vertex u ∈ T .Suppose that there exists u ∈ T such that d G [ T ∪ U ′ ] ( u ) ≥ b + τ . Let T ′ = T − { u } . By43), one can see that η ( S, T ′ ) = as − b | T ′ | + X x ∈ T ′ d G − S ( x ) − q ( S, T ′ )= as − bt + b + X x ∈ T d G − S ( x ) − d G − S ( u ) − q ( S, T ′ ) ≤ as − bt + b + X x ∈ T d G − S ( x ) − ( b + τ ) − ( q ( S, T ) − τ )= as − bt + X x ∈ T d G − S ( x ) − q ( S, T ) ≤ − , contradicting to the maximality of V ( G ) − S − T − U . This completes the proof of Claim2. ✷ By the definitions of U and C i , one can see that | U | ≥ m + m ( w −
1) and so we have m ≤ ( | U | − m ) / ( w − | U | + s + t ≤ n . We have m ≤ n − s − t − m w − . (5)By Claim 1, e G ( u, T ) ≤ a − u ∈ U . Thus for u ∈ U , we have d G ( u ) ≤ ( m j −
1) + s + r, (6)where r = min { a − , t } . Let u ∈ V ( C ) and u ∈ V ( C ). It follow from (6) thatmax { d G ( u ) , d G ( u ) } ≤ ( m −
1) + s + r (7)since m ≤ m . Claim 3. S = ∅ .By contradiction. Suppose that S = ∅ . Recall that S ∪ T = ∅ by (4). So we have t ≥ b ≥ a + 2 and δ ( G ) ≥ ( b − b ) /a , it follows by (3) that w ≥ X v ∈ T d G ( v ) − bt + 2 ≥ t ( b − ab − ba ) + 2 ≥ bta + 2 . (8)Combining (5) and (7), since s = 0 and r = min { a − , t } , we havemax { d G ( u ) , d G ( u ) } ≤ n − t − m w − a − . (9)By (8), we have n − t − m w − − a − ≤ a ( n − t − m ) bt + a + a − . h ( t ) = n − t − m bt + a is a monotone decreasing function on t . Since t ≥ n > ( a − b + a ) b − a , we have n − t − m w − a − < an − aa + b + a − < bna + b , which implies that max { d G ( u ) , d G ( u ) } < bna + b , contradicting (1). This completes the proof of Claim 3. Claim 4. T = ∅ .Suppose that T = ∅ , i.e., t = 0. By Claim 1, m i ≥ b + 2 for 1 ≤ i ≤ w . By (3), we have w ≥ as + 2 . Thus n ≥ w ( b + 2) + s ≥ ( b + 1)( as + 2) + s, i.e., s ≤ n − b +1)( b +1) a +1 . Recall that r = min { a − , t } = 0. We havemax { d G ( u ) , d G ( u ) } ≤ m − s ≤ n − sw − s ≤ nas + 1 + s. Let h ( s ) = nas +1 + s . Notice that h ( s ) is a convex function on variable s . We know thatthe maximum value of the function h ( s ) = nas +1 + s be obtained only at the boundary of s .Recall that 1 ≤ s < na ( b +1)+1 . We have h ( s ) ≤ max { h (1) , h ( na ( b + 1) + 1 ) }≤ max { na + 1 + 1 , n ( b + 1) a + 1 + b + 2 } . Notice that na +1 + 1 < bna + b and n ( b +1) a + b + 2 < bna + b since n > a + b ). Hence we havemax { d G ( u ) , d G ( u ) } < bna + b , which contradicts (1). ✷ Let h := min { d G − S ( v ) | v ∈ T } , and let x ∈ T be a vertex satisfying d G − S ( x ) = h .We write p = | N T [ x ] | . Furthermore, if T − N T [ x ] = ∅ , we put h := min { d G − S ( v ) | v ∈ T − N T [ x ] } and let x ∈ T − N T [ x ] such that d G − S ( x ) = h . Observe that x and x arenot adjacent when x exists andmax { d G ( x ) , d G ( x ) } ≤ max { h + s, h + s } ≤ h + s. (10)For completing the proof, we discuss four cases. Case 1. h ≥ b . 6y (3), we have w ≥ as − bt + X v ∈ T d G − S ( v ) + 2 ≥ as + ( h − b ) t + 2 ≥ as + 2 , i.e., w ≥ as + 2 . (11)Note that the number of vertices in graph G satisfies n ≥ w + s + t . Then from (11), wehave n ≥ as + 2 + s + t and it infers that s < n − a + 1 . Recall that h ( s ) = nas +1 + s is a convex function. Hence we havemax { d G ( x ) , d G ( x ) } ≤ m − s + r< nas + 1 + s + a − ≤ max { h (1) + a − , h ( n − a + 1 ) + a − }≤ max { na + 1 + a − , n − a + 1 + a } < bna + b (since n ≥ a + b ) , which contradicts to (1).Now we may assume that h < b in the following discussion. Case 2. T = N T [ x ].Then we have t = | N T [ x ] | and t ≤ h + 1 ≤ b . Note that δ ( G ) ≥ b − ba . Since7 + h ≥ δ ( G ), we have s ≥ ( b − ba − h ). By (3), one can see that w ≥ as + ( h − b ) t + 2 ≥ as + ( h − b )( h + 1) + 2 ≥ a ( b − ba − h ) + ( h − b )( h + 1) + 2= ( b − − ( a + b − h − h + 1 ≥ ( b − − ( a + b − a + b a ≡ b (mod 2)) ≥ a + b a < b ) , i.e., w ≥ a + b . Now we have max { d G ( u ) , d G ( u ) } ≤ m + s + a − ≤ n − s − tw − s + a − ≤ n − s − t ) a + b + s + a − . Let h ( s ) = n − s ) a + b + s + a −
2. We discuss two subcases.
Case 2.1. s ≤ n a .Notice that h ( s ) is a linear function on s and 1 ≤ s ≤ n/ a . Then we have h ( s ) ≤ h ( n a )= 2( n − n/ a ) a + b + n a + a − < bna + b (since n ≥ a + b )( a − b − ) . Thus we have max { d G ( u ) , d G ( u ) } < bna + b , a contradiction. Case 2.2. s > n a . 8y Claim 1, we have m ≥ b − a + 1 ≥
3. Hence n ≥ s + t + ( b − a + 1) w ≥ s + t + 3 w ≥ s + t + 3 as + 3( h − b ) t + 6 (by (3)) > n + n a − (3 b − h − t + 6 ≥ n + n a − (3 b − h − h + 1) + 6 (since t ≤ h + 1 ≤ b ) ≥ n + n a − b ≥ n (since n ≥ a ( b + 1) ) , a contradiction.Now we may assume that t > p . Case 3. h ≥ b .Recall that p = | N T [ x ] | . Notice that except p vertices, each other vertex v in T satisfies d G − S ( v ) ≥ h . Thus by (3), we have w ≥ as + X v ∈ T d G − S ( v ) − bt + 2 ≥ as + ( h − b ) p + ( h − b )( t − p ) + 2 (12) ≥ as + ( h − b ) p + 2 , i.e., w ≥ as + ( h − b ) p + 2 . (13)Now we discuss two subcases. Subcase 3.1. h ≤ ( b +1) .From the assumption of theorem, we know that max { d G ( x ) , d G ( x ) } ≥ bna + b , whichimplies that h + s ≥ bna + b . Then s ≥ bna + b − h ≥ bna + b − ( b + 1) . (14)9ow we have n ≥ w + s + t ≥ ( a + 1) s + ( h − b ) p + 2 + t (by (13)) > ( a + 1) s + ( h − b + 1) p + 2 (since p < t ) ≥ ( a + 1) s + ( h − b + 1)( h + 1) + 2 (since p ≤ h + 1 ≤ b ) ≥ ( a + 1)( bna + b − ( b + 1) h − b + 1)( h + 1) + 2 (by (14)) ≥ ( a + 1)( bna + b − ( b + 1) − b ≥ n + 2 (since n ≥ b + 1)( a + b )) , a contradiction. Subcase 3.2. h > ( b +1) .Recall that p ≤ h + 1 ≤ b , p < t . Then we have w ≥ as + ( h − b ) p + ( h − b )( t − p ) + 2 (by (12)) ≥ as + ( h − b )( h + 1) + h − b + 2 ≥ as − ( b + 1) b + 1) − b + 2 ≥ as + 2 , i.e., w ≥ as + 2 . (15)By Claim 1 and (15), we know that n ≥ s + t + ( b − a + 1) w ≥ (3 a + 1) s + 6 , which implies that s ≤ n − a + 1 . (16)Consider u ∈ V ( C ) and u ∈ V ( C ). By (15) and (16), we havemax { d G ( u ) , d G ( u ) } < m − r + s ≤ n − s − tas + 1 + s + a − < nas + 1 + s + a − ≤ max { na + 1 + a − , n − a + 1 + a + 2 } , ≤ s ≤ n − a +1 and h ( s ) = nas +1 + s is a convex function.Observe that when n ≥ a + b ), max { na +1 + a − , n − a +1 + a + 2 } < bna + b . Thus we havemax { d G ( u ) , d G ( u ) } < bna + b when n ≥ a + b ), which contradicts (1). Case 4. ≤ h ≤ h ≤ b − x and x are not adjacent, we have s ≥ bna + b − h . By (3), wehave w ≥ as + X v ∈ T d G − S ( v ) − bt + 2 ≥ as + ( h − b ) p + ( h − b )( t − p ) + 2= as + ( h − b ) t + ( h − h ) p + 2 . (17)One can see that n ≥ s + t + w ≥ ( a + 1) s + ( h − b ) t + ( h − h ) p + 2 + t (by (17))= ( a + 1) s + ( h − h ) p + ( h + 1 − b ) t + 2 . (18)Note that p ≤ h +1, h +1 − b ≤ s ≥ bna + b − h , t ≤ n − s ≤ ana + b + h and ( h − h )( h +1) ≤ h . By taking these inequalities into (18), we have n ≥ ( a + 1)( bna + b − h ) + ( h − h )( h + 1) − ( b − h − ana + b + h ) + 2= ( a + 1) bna + b − ( b − ana + b + 2 + h ( ana + b + h − a − b ) + ( h − h )( h + 1)= n + 2 + h ( ana + b − a − b ) + h − ( h − h )( h + 1) (since n ≥ ( a + b ) a ) ≥ n + 2 , a contradiction.This completes the proof. ✷ Remark.
In Theorem 1.4, the bound in the assumption max { d G ( u ) , d G ( v ) } ≥ bna + b is tight.For showing this, we construct following graph: Let s, t, n be integers where s = ⌊ bn − a + b ⌋ and t = n − s . Let K n denote a complete graph of order s and let G = K n − E ( K t ). Obviously,every pair of nonadjacent vertices u, v in G satisfies max { d G ( u ) , d G ( v ) } ≥ bna + b −
1. Let S = V ( K s ) and T = V ( K n ) − S . Then we have η ( S, T ) = as − bt + X v ∈ T d G − S ( v ) − q ( S, T ) = as − bt < , G doesn’t satisfy the condition of Theorem 1.2. Thus it is not true for graph G to haveall ( a, b )-parity factors. References [1] A. Amahashi, On factors with all degree odd,
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