The Ramsey numbers of paths versus wheels: a complete solution
aa r X i v : . [ m a t h . C O ] D ec The Ramsey numbers of paths versus wheels: a completesolution ∗ Binlong Li a,b and Bo Ning a a Department of Applied Mathematics, Northwestern Polytechnical University,Xi’an, Shaanxi 710072, P.R. China b Department of Mathematics, University of West Bohemia,Univerzitn´ı 8, 306 14 Plzeˇn, Czech Republic
Abstract
Let G and G be two given graphs. The Ramsey number R ( G , G ) is the leastinteger r such that for every graph G on r vertices, either G contains a G or G containsa G . We denote by P n the path on n vertices and W m the wheel on m + 1 vertices.Chen et al. and Zhang determined the values of R ( P n , W m ) when m ≤ n + 1 and when n + 2 ≤ m ≤ n , respectively. In this paper we determine all the values of R ( P n , W m )for the left case m ≥ n + 1. Together with Chen et al’s and Zhang’s results, we give acomplete solution to the problem of determining the Ramsey numbers of paths versuswheels. Keywords:
Ramsey number; Path; Wheel
AMS Subject Classification:
We use Bondy and Murty [2] for terminology and notation not defined here, and considerfinite simple graphs only.Let G be a graph. We denote by ν ( G ) the order of G , by δ ( G ) the minimum degreeof G , and by ω ( G ) the component number of G . We denote by P n and C n the path andcycle on n vertices, respectively. The wheel on n + 1 vertices, denoted by W n , is the graphobtained by joining a vertex to each vertex of a C n . ∗ Supported by NSFC (No. 11271300) and the Doctorate Foundation of Northwestern PolytechnicalUniversity (No. cx201202 and No. cx201326). E-mail addresses: [email protected] (B. Li),ningbo [email protected] (B. Ning). G and G be two graphs. The Ramsey number R ( G , G ), is defined as the leastinteger r such that for every graph G on r vertices, either G contains a G or G containsa G , where G is the complement of G . If G and G are both complete, then R ( G , G )is the classical Ramsey number r ( ν ( G ) , ν ( G )). Otherwise, R ( G , G ) is usually calledthe generalized Ramsey number .In 1967, Gerencs´er and Gy´arf´as [9] computed the Ramsey numbers of all path-pathpairs, and gave the first generalized Ramsey number formula. (In fact, this question ofdetermining Ramsey numbers of paths versus paths appeared in a paper of Erd¨os [5] in1947, and the right upper bound was also determined there.) After that, Faudree et al.[8] determined the Ramsey numbers of paths versus cycles. We list these results as bellow,both of them will be used in this paper. Theorem 1 (Gerencs´er and Gy´arf´as [9]) . If m ≥ n ≥ , then R ( P n , P m ) = m + ⌊ n/ ⌋ − . Theorem 2 (Faudree et al. [8]) . If n ≥ and m ≥ , then R ( P n , C m ) = n − , for n ≥ m and m is odd ; n + m/ − , for n ≥ m and m is even ;max { m + ⌊ n/ ⌋ − , n − } , for m > n and m is odd ; m + ⌊ n/ ⌋ − , for m > n and m is even . Recently, graph theorists have begun to investigate the Ramsey numbers of pathsversus wheels. Baskoro and Surahmat [1] conjectured the values of R ( P n , W m ) when n ≥ m −
1, and got some partial results. Chen et al. [3] completely determined the valuesof R ( P n , W m ) when n ≥ m −
1. Salman and Broersma [11] further generalized Chen etal.’s result. Zhang [12] firstly obtained all the values of R ( P n , W m ) when n + 2 ≤ m ≤ n .We list the results of Chen et al.’s and Zhang’s in the following. Theorem 3 (Chen et al. [3]) . If ≤ m ≤ n + 1 , then R ( P n , W m ) = n − , m is odd ;2 n − , m is even . Theorem 4 (Zhang [12]) . If n + 2 ≤ m ≤ n , then R ( P n , W m ) = n − , m is odd ; m + n − , m is even . m ≥ n + 1, some upper bounds and lower bounds of R ( P n , W m ) weregiven [11, 12]. Furthermore, for some n, m , the exact values of R ( P n , W m ) were alsodetermined in [11, 12].In this paper we will prove the following formula, which can be used to determine allthe values of R ( P n , W m ) for the left case m ≥ n + 1. Theorem 5. If n ≥ and m ≥ n + 1 , then R ( P n , W m ) = ( n − · β + 1 , α ≤ γ ; ⌊ ( m − /β ⌋ + m, α > γ, where α = m − n − , β = ⌈ α ⌉ and γ = β β + 1 . Together with Theorems 4 and 5, we give a complete solution to the problem of deter-mining the Ramsey numbers of paths versus wheels.
Before our proof we will first list one result due to Zhang [12] and give some additionalterminology and notation. Second, we will prove a series of lemmas which support ourproof of the main theorem.The following result is a rewriting of two corollaries in [12]. It helps us to deal withthe cases n = 3 , Theorem 6 (Zhang [12]) . If n ≥ and m ≥ n + 1 , then R ( P n , W m ) = m + n − , if m = 1 mod ( n − m + n − , if m = 0 , n − . For integers s, t , the interval [ s, t ] is the set of integers i with s ≤ i ≤ t . Note that if s > t , then [ s, t ] = ∅ . Let X be a subset of N . We set L ( X ) = { P ki =1 x i : x i ∈ X, k ∈ N } ,and suppose 0 ∈ L ( X ) for any set X . Note that if 1 ∈ X , then L ( X ) = N . For an interval[ s, t ], we use L [ s, t ] instead of L ([ s, t ]).In the following of the paper, n always denotes an integer at least 2 and m an integerat least 3. We denote by par( n ) the parity of n , i.e., par( n ) = ⌈ n/ ⌉ − ⌊ n/ ⌋ .For integers n, m , let t ( n, m ) be the values of R ( P n , W m ) defined in Theorem 5, thatis, t ( n, m ) = ( n − · β + 1 , α ≤ γ ; ⌊ ( m − /β ⌋ + m, α > γ, α = m − n − , β = ⌈ α ⌉ and γ = β β + 1 . Lemma 1. If m ≥ n + 1 , then t ( n, m ) = min { t : t / ∈ L [ t − m + 1 , n − } .Proof. Set T = { t : t ∈ L [ t − m + 1 , n − } . Note that if t ∈ T , then t − ∈ T . So it issufficient to prove that t ( n, m ) = max( T ) + 1.Note that t ∈ T ⇔ t ∈ L [ t − m + 1 , n − ⇔ t ∈ [ k ( t − m + 1) , k ( n − , for some integer k ⇔ t ≤ kk − m −
1) and t ≤ k ( n − , for some integer k ⇔ t ≤ k ( n −
1) for some integer k < α + 1 , or t ≤ (cid:22) m − k − (cid:23) + m − , for some integer k ≥ α + 1 . This implies that T = { t : t ≤ k ( n − , k ≤ β } ∪ (cid:26) t : t ≤ (cid:22) m − k − (cid:23) + m − , k ≥ β + 1 (cid:27) . Thus max( T ) = max (cid:26) ( n − β, (cid:22) m − β (cid:23) + m − (cid:27) = ( n − · β, α ≤ γ ; ⌊ ( m − /β ⌋ + m − , α > γ. We conclude that t ( n, m ) = max( T ) + 1. Lemma 2.
Let G be a graph on at least three vertices.(1) If G is 2-connected and δ ( G ) ≥ ⌈ n/ ⌉ , then G contains a cycle of order at least min { ν ( G ) , n } .(2) If x ∈ V ( G ) , G is connected and d ( v ) ≥ n − for every vertex v ∈ V ( G ) \{ x } , then G contains a path from x of order at least n .(3) If x, y ∈ V ( G ) , G + xy is 2-connected and d ( v ) ≥ n − for every vertex v ∈ V ( G ) \{ x, y } , then G contains a path from x to y of order at least n .(4) If x, y ∈ V ( G ) , G + xy is 2-connected and d ( v ) ≥ ⌈ n/ ⌉ for every vertex v ∈ V ( G ) \{ x, y } , then G contains a path from x of order at least min { ν ( G ) , n } .(5) If G is connected and δ ( G ) ≥ ⌊ n/ ⌋ , then G contains a path of order at least min { ν ( G ) , n } .
6) If x ∈ V ( G ) , G is connected, and d G − x ( v ) ≥ n − for every vertex v ∈ V ( G ) \{ x } ,then G contains a path from x of order at least n .(7) If x ∈ V ( G ) , G is 2-connected and d G − x ( v ) ≥ ⌊ n/ ⌋ for every vertex v ∈ V ( G ) \{ x } ,then G contains a path from x of order at least min { ν ( G ) , n } .Proof. The assertions (1), (2) and (3) are results of Dirac [4], Erd¨os and Gallai [6], respec-tively. Now we prove the other assertions.(4) Let G ′ = G + xy . Since every two nonadjacent vertices of G ′ contain one withdegree at least ⌈ n/ ⌉ , by Fan’s Theorem [7], G ′ contains a cycle C with order at leastmin { ν ( G ) , n } . If C does not contain the added edge xy , then C is a cycle of G and G contains a path from x of order at least min { ν ( G ) , n } ; if C contains the added edge xy ,then P = C − xy is a path of G from x of order at least min { ν ( G ) , n } .(5) We add a new vertex x and join x to every vertex of G . We denote the resultinggraph as G ′ . Thus every vertex in V ( G ′ ) has degree at least ⌊ n/ ⌋ + 1 = ⌈ ( n + 1) / ⌉ . By(1), G ′ contains a cycle of order at least min { ν ( G ′ ) , n + 1 } , and G contains a path of orderat least min { ν ( G ) , n } .(6) Let H be a component of G − x , and let x ′ be a neighbor of x in H . Note thatevery vertex in H has degree at least n − H . By (2), H contains a path P from x ′ of order at least n −
1. Thus P = xx ′ P is a path from x of order at least n in G .(7) Let G ′ = G − x . If G ′ contains a vertex with degree 1, then n ≤ δ ( G ′ ) ≥ G ′ is 2-connected. By (1), G ′ contains a cycle C of order at leastmin { ν ( G ′ ) , n − par( n ) } . Let P be a path from x to C , let x ′ be the end-vertex of P on C , and let x ′′ be a neighbor of x ′ on C . Then P = P ∪ C − x ′ x ′′ is a path from x of orderat least min { ν ( G ) , n } .Now we assume that G ′ is separable. Then every end-block of G ′ is 2-connected. Let B be an end-block of G ′ , and b be the cut-vertex of G ′ contained in B . Since G is 2-connected, x is adjacent to some vertex, say x ′ , in B − b . By (3), B contains a path P from x ′ to b of order at least ⌊ n/ ⌋ + 1, and by (2), H − ( B − b ) contains a path P from b of order at least ⌊ n/ ⌋ + 1. Thus P = xx ′ P bP is a path from x of order at least n . Lemma 3. If G is a disconnected graph such that(1) m ≤ ν ( G ) ; and(2) every component of G has order at most ⌊ m/ ⌋ ,then G contains a C m . roof. Let G ′ be an induced subgraph of G with order m . Clearly every component of G ′ has order at most ⌊ m/ ⌋ . Thus every vertex of G ′ has degree at least ⌈ m/ ⌉ in G ′ . ByLemma 2, G ′ contains a C m . Lemma 4.
Let G be a graph.(1) If n ≤ ν ( G ) ≤ ⌊ n/ ⌋ − and G contains no P n , then G contains a path of order ν ( G ) + 3 − n .(2) If ν ( G ) ≥ ⌊ n/ ⌋ − and G contains no P n , then G contains a path of order ν ( G ) + 1 − ⌊ n/ ⌋ .(3) If n ≥ is even, ν ( G ) ≥ n/ − , and G contains no C n then G contains a pathof order ν ( G ) + 1 − n/ .Proof. The lemma can be deduced by Theorems 1 and 2.
Lemma 5.
Let G and G be two disjoint graphs. If(1) G contains a path of order p ≥ ; and(2) m ≤ min { ν ( G ) , ν ( G ) + ν ( G ) , p + 2 ν ( G ) − } ,then G ∪ G contains a C m .Proof. We first assume that ν ( G ) ≥ ⌊ m/ ⌋ . If m is even, then ν ( G ) ≥ m/ ν ( G ) ≥ m/
2. Let x , x , . . . , x k be k = m/ G , and let y , y , . . . , y k be k vertices in G . Then C = x y x y · · · x k y k x is a C m in G ∪ G . If m is odd, thenthen ν ( G ) ≥ ( m + 1) / ν ( G ) ≥ ( m − /
2. Note that G has two nonadjacentvertices. Let x , x , . . . , x k be k = ( m + 1) / G such that x x k / ∈ E ( G ), andlet y , y , . . . , y k − be k − G . Then C = x y x y · · · x k − y k − x k x is a C m in G ∪ G .Now we assume that ν ( G ) ≤ ⌊ m/ ⌋ −
1. Let V ( G ) = { y , y , · · · , y k } , where k = ν ( G ). Since 2 ≤ m + 1 − k ≤ p , G contains a path P of order m + 1 − k . Let s, t bethe two end-vertices of P . Note that ν ( G ) − ν ( P ) ≥ m − k − m − k = k −
1. Let x , x , . . . , x k − be k − V ( G − P ). Then C = sy x y x · · · x k − y k tP is a C m in G ∪ G . Lemma 6.
Suppose m ≥ n + 1 . Let G be a disconnected graph containing no P n . If(1) m ≤ ν ( G ) ; and(2) the order sum of every ω ( G ) − components in G is at least m + ⌊ n/ ⌋ − ν ( G ) ,then G contains a C m . roof. If every component of G has order at most ⌊ m/ ⌋ , then we are done by Lemma 3.Now we assume that there is a component H with order at least ⌊ m/ ⌋ + 1.Let G = H , and G = G − H . Note that m ≤ ν ( G ), m ≤ ν ( G ) = ν ( G ) + ν ( G )and ν ( G ) ≥ m + ⌊ n/ ⌋ − ν ( G ).Note that ν ( G ) ≥ ⌊ m/ ⌋ + 1 ≥ n . If ν ( G ) ≤ ⌊ n/ ⌋ −
2, then by Lemma 4, G contains a path of order p = 2 ν ( G ) + 3 − n . Since p + 2 ν ( G ) − ν ( G ) + 3 − n + 2 ν ( G ) − ν ( G ) + 2 − n ≥ m + 2 − n ≥ m, by Lemma 5, G contains a C m . If ν ( G ) ≥ ⌊ n/ ⌋ −
1, then by Lemma 4, G contains apath of order p = ν ( G ) + 1 − ⌊ n/ ⌋ . Since p + 2 ν ( G ) − ν ( G ) + 1 − j n k + 2 ν ( G ) − ν ( G ) + ν ( G ) − j n k ≥ ν ( G ) + m + j n k − ν ( G ′ ) − j n k = m, by Lemma 5, G contains a C m . Lemma 7.
Let G be a graph, X an independent set of G , R = G − X . If(1) | X | ≥ ;(2) every component of R is joined to at most one vertex in X ;(3) R contains a path of order p ≥ ; and(4) m ≤ min { ν ( G ) , p + 2 | X | − } ,then G contains a C m .Proof. Let P be a path in R with the largest order. Clearly ν ( P ) ≥ p .If ν ( P ) ≥ m −
1, then let P ′ be a subpath of P of order m −
1. Let s, t be the twoend-vertices of P ′ . Since each of s and t is adjacent to at most one vertex in X and | X | ≥
3, there is a vertex x in X nonadjacent to both s and t . Thus C = sxtP ′ is a C m in G . Now we assume that ν ( P ) ≤ m − s, t be the two end-vertices of P . If P contains all vertices in R , then ν ( P ) = ν ( R ).Let x be a vertex in X nonadjacent to s , and x ′ be a vertex in X \{ x } nonadjacent to t .Note that | X | = ν ( G ) − ν ( R ) ≥ m − ν ( P ). Let x , x , . . . , x k be k = m − ν ( P ) vertices in X such that x = x and x k = x ′ , then C = sx x · · · x k tP is a C m in G . Now we assumethat V ( R ) \ V ( P ) = ∅ .Let U = V ( R − P ). Note that each of s, t is adjacent to every vertex in U , and thisimplies that U ∪ { s, t } is contained in a component of R . Thus U ∪ { s, t } is joined to atmost one vertex in X . Let y be the vertex in X that is joined to U ∪ { s, t } . If such avertex does not exist, then let y be any one vertex in X .7ote that m − ν ( P ) ≤ m − p ≤ | X |−
3. If m − p is odd, then | X | ≥ ( m − ν ( P )+1) / x , x , . . . , x k be k = ( m − ν ( P ) + 1) / X \{ y } , and let u , . . . , u k − be k − U ∪ X \{ x , x , . . . , x k } . Then C = sx u x u · · · x k − u k − x k tP is a C m in G . If m − ν ( P ) is even, then m − ν ( P ) ≤ | X | − | X | ≥ ( m − ν ( P )) / x , x , . . . , x k be k = ( m − ν ( P )) / X \{ y } , and let u , . . . , u k − be k − U ∪ X \{ x , x , . . . , x k } . Then C = sx u x u · · · x k − u k − x k − x k tP is a C m in G . Lemma 8.
Let G be a graph, X , X two independent sets of G (possibly joint), X = X ∪ X , R = G − X . If(1) | X | = | X | ≥ , | X \ X | = | X \ X | ≥ ;(2) every component of R is joined to at most one vertex in X i , i = 1 , ;(3) R contains a path of order p ≥ ; and(4) m ≤ min { ν ( G ) , p + 2 | X | − } ,then G contains a C m .Proof. We first define an adjustable segment of a cycle C . If X ∩ X = ∅ , then letting x , x ′ , x ′′ ∈ X , x , x ′ , x ′′ ∈ X and u ∈ V ( R ), we call a subpath A an adjustable segmentof C with the center u if one of the following is true:(1) A = x x ′ ux ′ x with x ′′ , x ′′ / ∈ V ( C );(2) A = x x ′ x ′′ ux ′ x with x ′′ / ∈ V ( C );(3) A = x x ′ ux ′′ x ′ x with x ′′ / ∈ V ( C ); or(4) A = x x ′ x ′′ ux ′′ x ′ x .If X ∩ X = ∅ , then letting x , x ′ ∈ X \ X , x , x ′ ∈ X \ X and x ∈ X ∩ X , wecall a subpath A an adjustable segment of C with the center x if one of the following istrue:(1) A = x xx with x ′ , x ′ / ∈ V ( C );(2) A = x x ′ xx with x ′ / ∈ V ( C );(3) A = x xx ′ x with x ′ / ∈ V ( C ); or(4) A = x x ′ xx ′ x .If X ∩ X = ∅ , then let P be a path in R with the largest order; if X ∩ X = ∅ , thenlet P be a non-Hamilton path in R with the largest order.If ν ( P ) ≥ m −
5, then let P ′ be a subpath of P of order m − s, t be the twoend-vertices of P ′ . If X ∩ X = ∅ , then let x be a vertex in X ∩ X , x a vertex in X \ X nonadjacent to s , x ′ a vertex in X \ X \{ x } , x a vertex in X \ X nonadjacent to t and x ′ a vertex in X \ X \{ x } . Then C = sx x ′ xx ′ x tP ′ is a C m in G . If X ∩ X = ∅ ,then let u be a vertex in V ( R − P ′ ), x a vertex in X nonadjacent to s , x ′ a vertex in8 \{ x } nonadjacent to u , x a vertex in X nonadjacent to t and x ′ a vertex in X \{ x } nonadjacent to u . Then C = sx x ′ ux ′ x tP ′ is a C m in G .Now we assume that ν ( P ) ≤ m −
6. By a similar argument in the analysis above, wecan get a cycle C in G of order at least ν ( P ) + 5 such that(a) C contains P as a subpath;(b) C contains an adjustable segment A (with end-vertices x , x );(c) every edge of C has a vertex in R , unless it is an edge in A .Now we choose a cycle C in G satisfying (a)(b)(c) with order as large as possible butat most m . If ν ( C ) = m , then we are done. So we assume that ν ( C ) ≤ m −
1. We claimthat V ( R ) ⊂ V ( C ). Assume the contrary. Let v be a vertex in U = V ( R ) \ V ( C ).If ( X ∩ X = ∅ and) A = x x ′ ux ′ x with x ′′ ∈ X \ V ( C ), x ′′ ∈ X \ V ( C ), then C ′ = C − x x ′ ∪ x x ′′ x ′ is a required cycle with order ν ( C ) + 1, a contradiction. Using the sameanalysis, we can conclude that A = x x ′ x ′′ ux ′′ x ′ x (if X ∩ X = ∅ ) or A = x x ′ xx ′ x (if X ∩ X = ∅ ).If X ∩ X = ∅ , then P is a longest path of R ; if X ∩ X = ∅ , then noting that u, v ∈ V ( R − P ), P is a longest path of R as well. Thus ν ( P ) ≥ p and U ∪ { s, t } iscontained in a component of R . If there is a vertex y in X that is joined to U ∪ { s, t } ,then we use y instead of the vertex x ′ , x ′ or x in C , for the case y ∈ X \ X , y ∈ X \ X ,or y ∈ X ∩ X , respectively. Thus we assume that every vertex in X \{ x ′ , x ′ , x } is notjoined to U ∪ { s, t } .If every vertex in X is in V ( C ), then noting that there are at most 5 vertices in X each of which has a successor on C such that it is not in R − P , we have ν ( C ) ≥ ν ( P ) + | X | + ( | X | − ≥ p + 2 | X | − ≥ m, a contradiction. So we assume that there is a vertex x ′ in X which is not in C . Let v ′ be the predecessor of x in C . Clearly v ′ ∈ U ∪ { s, t } . Then C ′ = v ′ x ′ vx x ′′ −→ C [ x ′′ , v ′ ](if X ∩ X = ∅ ) or C ′ = v ′ x ′ vx x −→ C [ x, v ′ ] (if X ∩ X = ∅ ) is a required cycle of order ν ( C ) + 1, a contradiction. Thus as we claimed, every vertex in R is in C . This impliesthat C is a cycle in G satisfying(d) there is an edge x x ′ ∈ E ( C ) such that x , x ′ ∈ X ;(e) there is an edge x x ′ ∈ E ( C ) such that x , x ′ ∈ X ;(f) V ( R ) ⊂ V ( C ).Now we choose a cycle C in G satisfying (d)(e)(f) with order as large as possible butat most m . If ν ( C ) = m , then we are done. So we assume that ν ( C ) ≤ m −
1. If everyvertex in X is in C , then ν ( C ) = ν ( R ) + | X | ≥ m, x ′ in X which is not in C . If x ′ ∈ X ,then C ′ = C − x x ′ ∪ x x ′ x ′ is a required cycle of order ν ( C ) + 1; if x ′ ∈ X , then C ′ = C − x x ′ ∪ x x ′ x ′ is a required cycle of order ν ( C ) + 1, a contradiction.Thus the lemma holds.The proof of the next lemma is similar as the proof of Lemma 8, but more involved. Lemma 9.
Let G be a graph, R be an induced subgraph of G , X , X two independentsets of G − R (possibly joint), X = X ∪ X . If(1) | X | = | X | ≥ , | X \ X | = | X \ X | ≥ ;(2) every component of R has order at least 2;(3) every component of R is joined to at most one vertex in X i , i = 1 , ;(4) for any component H of R , there are at least q vertices in G − R each of which iseither in X or not joined to H ;(5) R contains a path of order p ≥ ; and(6) m ≤ min {⌈ ν ( R ) / ⌉ + 4 , ν ( R ) + q − , p + 2 q − } ,then G contains a C m .Proof. We use the concept of an adjustable segment defined in Lemma 8. If X ∩ X = ∅ ,then let P be a path in R with the largest order; if X ∩ X = ∅ , then let P be anon-Hamilton path in R with the largest order.If ν ( P ) ≥ m −
5, then similar as in Lemma 8, we can find a C m in G . Thus we assumethat ν ( P ) ≤ m −
6. By a similar argument as in Lemma 8, we can get a cycle C in G oforder at least ν ( P ) + 5 such that(a) C contains P as a subpath;(b) C contains an adjustable segment A (with end-vertices x , x );(c) every edge of C has a vertex in R , unless it is an edge in A .Now we choose a cycle C in G satisfying (a)(b)(c) with order as large as possible butat most m . If ν ( C ) = m , then we are done. So we assume that ν ( C ) ≤ m −
1. We claimthat V ( R ) ⊂ V ( C ). Assume the contrary. Let v be a vertex in U = V ( R − C ).Using the same analysis in Lemma 8, we can conclude that A = x x ′ x ′′ ux ′′ x ′ x (if X ∩ X = ∅ ) or A = x x ′ xx ′ x (if X ∩ X = ∅ ) and P is a longest path of R . Thus ν ( P ) ≥ p and U ∪ { s, t } is contained in a common component of R . Furthermore, we canassume that every vertex in X \{ x ′ , x ′ , x } is not joined to U ∪ { s, t } .Let W be the union of X and the set of vertices in G − R that are not joined to U ∪ { s, t } . Then | W | ≥ q . If every vertex in W is in V ( C ), then noting that there are at10ost 5 vertices in W each of which has a successor on C such that it is not in R − P , wehave ν ( C ) ≥ ν ( P ) + | W | + ( | W | − ≥ p + 2 q − ≥ m, a contradiction. So we assume that there is a vertex w in W that is not in V ( C ). Let v ′ be the predecessor of x in C . Clearly v ′ ∈ U ∪ { s, t } . Then C ′ = v ′ wvx x ′′ −→ C [ x ′′ , v ′ ](if X ∩ X = ∅ ) or C ′ = v ′ wvx x −→ C [ x, v ′ ] (if X ∩ X = ∅ ) is a required cycle of order ν ( C ) + 1, a contradiction. Thus as we claimed, every vertex in R is in C . This implies C satisfies (b)(c) and(d) V ( R ) ⊂ V ( C ).Now we choose a cycle C in G satisfying (b)(c)(d) with order as large as possible butat most m . If ν ( C ) = m , then we are done. So we assume that ν ( C ) ≤ m −
1. By asimilar argument as above, we can conclude that A = x x ′ x ′′ ux ′′ x ′ x (if X ∩ X = ∅ ) or A = x x ′ xx ′ x (if X ∩ X = ∅ ).We claim that there are two vertices u , u in C such that u , u are in a commoncomponent of R and u +1 , u +2 ∈ V ( R ). Assume the contrary. Note that every componentof R has at least 2 vertices, there is at most one vertex in a component, such that it has asuccessor on C in R , and there are 4 vertices of C (in the adjusted segment) each of whichis not a successor of some vertex in R . Thus ν ( C ) ≥ ν ( R ) + (cid:24) ν ( R )2 (cid:25) + 4 = (cid:24) ν ( R )2 (cid:25) + 4 ≥ m, a contradiction. Thus as we claimed, there are two edges u u +1 , u u +2 such that u , u arein a common component of R and u +1 , u +2 ∈ V ( R ).If there is a vertex y in X \ V ( C ) that is joined to { u , u } , then we use y instead ofthe vertex x ′ , x ′ or x in C . Thus we assume that every vertex in X \ V ( C ) is not joinedto { u , u } . Let W be the union of X and the set of vertices in G − R that are not joinedto { u , u } . Then | W | ≥ q . If every vertex in W is in C , then ν ( C ) ≥ ν ( R ) + | W | ≥ ν ( R ) + q ≥ m, a contradiction. Thus we assume that there is a vertex w in W that is not in C .If u +1 , u +2 are in distinct components of R , then C ′ = u wu ←− C [ u , u +1 ] u +1 u +2 −→ C [ u +2 , u ]is a required cycle with order ν ( C ) + 1. Now we assume that u +1 , u +2 are in a commoncomponent of R .If there is a vertex y ′ in X \{ w } that is joined to { u +1 , u +2 } , then we use y ′ instead ofthe vertex x ′ , x ′ or x in C . Thus we assume that every vertex in X \ V ( C ) \{ w } is notjoined to { u +1 , u +2 } . 11et W ′ be the union of X and the set of vertices in G − R that are not joined to { u +1 , u +2 } . Then | W ′ | ≥ q . If every vertex in W ′ \{ w } is in C , then ν ( C ) ≥ ν ( R ) + | W ′ | − ≥ ν ( R ) + q − ≥ m, a contradiction. Thus we assume that there is a vertex w ′ in W \{ w } that is not in C . Let C ′ = u wu ←− C [ u , u +1 ] u +1 w ′ u +2 −→ C [ u +2 , u ]. Then C ′′ = C ′ − x x ′ x ′′ ∪ x x ′′ (if X ∩ X = ∅ )or C ′′ = C ′ − x x ′ x ∪ x x (if X ∩ X = ∅ ) is a required cycle of order ν ( C ) + 1, acontradiction. The case of n = 2 is trivial. For the case of n = 3 or n = 4, we are done by Theorem 6.Thus in the following we will assume that n ≥ t = t ( n, m ). By Lemma 1, t ( n, m ) = min { t : t / ∈ L [ t − m + 1 , n − } . Thus t − ∈ L [ t − m, n − t − P ki =1 t i , where t i ∈ [ t − m, n − ≤ i ≤ k . Let G bea graph with k components H , . . . , H k such that H i is a clique on t i vertices. Note that G contains no P n since every component of G has less than n vertices; and G containsno W m since every vertex of G has less than m nonadjacent vertices. Thus G is a graphon t − G contains no P n and G contains no W m . This implies that R ( P n , W m ) ≥ t .Now we will prove that R ( P n , W m ) ≤ t . Assume not. Let G be a graph on t verticessuch that G contains no P n and G contains no W m .Let s = m + n − t (i.e., ν ( G ) = m + n − s ). Claim 1. ≤ s ≤ ⌊ ( n + 5) / ⌋ . Proof.
Let t ′ = m + n −
1. Since t ′ − m + 1 = n , [ t ′ − m + 1 , n −
1] = ∅ , and t ′ / ∈ L ( ∅ ) = { } ,we have t ≤ t ′ = m + n −
1, and this implies that s ≥ s ≤ ( n + 5) /
4. By Lemma 1, t / ∈ L [ t − m + 1 , n − t / ∈ [ k ( t − m + 1) , k ( n − k ≥
1. That is, t ∈ [ k ( n −
1) + 1 , ( k + 1)( t − m + 1) − k .If k ≤
2, then by t ≤ ( k + 1)( t − m + 1) −
1, we get that t ≥ k + 1 k m − ≥ m − > n − ≥ t − m + 1) − , a contradiction. Thus we assume that k ≥ m ≤ ( k n − k + 2 k ) / ( k + 1), then s = m + n − t ≤ k n − k + 2 kk + 1 + n − ( k ( n −
1) + 1)= n + 2 k − k + 1 ≤ n + 54 . If m > ( k n − k + 2 k ) / ( k + 1), then s = m + n − t ≤ m + n − ( k + 1 k m − n − mk + 1 < n − k n − k + 2 kk ( k + 1) + 1= n + 2 k − k + 1 ≤ n + 54 . Thus the claim holds.We list the possible values of s for n ≤ n s ≤ Table 1: The possible values of s for n ≤ Claim 2.
Let v be an arbitrary vertex of G and G ′ ⊂ G − v − N ( v ). Then G ′ containsno C m . Proof.
Otherwise, noting that v is nonadjacent to every vertex in the C m , there will be a W m in G (with the hub v ). Claim 3. δ ( G ) ≥ ⌈ n/ ⌉ − s + 1. Proof.
Assume the contrary. Let v be a vertex of G with d ( v ) ≤ ⌈ n/ ⌉ − s . Then G ′ = G − v − N ( v ) has at least m + ⌊ n/ ⌋ − G ′ contains no P n , byTheorem 2, G ′ contains a C m (note that m ≥ n + 1), a contradiction to Claim 2.From Claims 1 and 3, one can see that δ ( G ) ≥ n ≥ Case 1. G is disconnected. Case 1.1.
Every component of G has order less than n .Let H i , 1 ≤ i ≤ k , be the components of G . Since t / ∈ L [ t − m + 1 , n − H , with order at most t − m . Thus P ki =2 ν ( H i ) ≥ m . Since ν ( H i ) ≤ n − ≤ ⌊ m/ ⌋ . By Lemma 3, G − H contains a C m , a contradiction. Case 1.2.
There is a component of G with order at least n .13et H be a component of G with the largest order. Note that ν ( H ) ≥ n . If everyvertex of H has degree at least ⌊ n/ ⌋ , then by Lemma 2, H contains a P n , a contradiction.Thus there is a vertex v in H with d ( v ) ≤ ⌊ n/ ⌋ −
1. Let G ′ = G − v − N ( v ). Then ν ( G ′ ) = ν ( G ) − − d ( v ) ≥ m + n − s − − j n k + 1= m + l n m − s ≥ m. Since ν ( H ) ≥ n > d ( v ), G ′ is disconnected. Let H be the union of ω ( G ′ ) − G ′ . We will prove that ν ( H ) ≥ m + ⌊ n/ ⌋ − ν ( G ′ ).Let H ′ be a component of G other than H . If H ′ ⊂ H , then ν ( H ) ≥ ν ( H ′ ) ≥ δ ( G ),and ν ( H ) + ν ( G ′ ) − m − j n k ≥ δ ( G ) + ν ( G ′ ) − m − j n k ≥ l n m − s + 1 + m + l n m − s − m − j n k = l n m + par( n ) + 2 − s ≥ . If H ′
6⊂ H , then ν ( H ) = ν ( G ′ ) − ν ( H ′ ) ≥ ν ( G ′ ) − ⌊ ν ( G ) / ⌋ , and ν ( H ) + ν ( G ′ ) − m − j n k ≥ ν ( G ′ ) − (cid:22) ν ( G )2 (cid:23) + ν ( G ′ ) − m − j n k ≥ (cid:16) m + l n m − s (cid:17) − (cid:22) m + n − s (cid:23) − m − j n k = l n m + par( n ) + (cid:24) m − n − s (cid:25) ≥ (cid:24) m − s (cid:25) ≥ (cid:24) n + 1 − s (cid:25) ≥ . Now by Lemma 6, G ′ contains a C m , a contradiction. Case 2. G has connectivity 1.Note that δ ( G ) ≥ ⌈ n/ ⌉ − s + 1 ≥
2. Every end-block of G is 2-connected. Case 2.1.
There is an end-block of G with order at least ⌈ m/ ⌉ + 1.Let B be an end-block of G with the maximum order, and x be the cut-vertex of G contained in B . Let x ′ be a cut-vertex of G such that the longest path between x and x ′ is as long as possible. Clearly x ′ is contained in some end-blocks. Let B ′ be an end-blockof G containing x ′ ( B = B ′ ). Let v be a vertex in B − x such that d B − x ( v ) is as small aspossible. 14 laim 4. d B − x ( v ) ≤ ⌈ ( n + 2 s − par( n )) / ⌉ − , if x = x ′ ; ⌊ ( n + 2 s − par( n )) / ⌋ − , if xx ′ is a cut-edge of G ; ⌈ ( n + 2 s − par( n )) / ⌉ − , otherwise . Proof.
We set a parameter a such that a = 0 if x = x ′ , 1 if xx ′ is a cut-edge of G , and 2otherwise. So there is a path between x and x ′ of length at least a .By Claim 3 and Lemma 2, B ′ contains a path from x ′ of order at least ⌈ n/ ⌉ − s + 2,and G − ( B − x ) contains a path from x of order at least ⌈ n/ ⌉ − s + a + 2.Note that ν ( B ) ≥ ⌈ m/ ⌉ + 1 ≥ ⌊ n/ ⌋ + s − a −
1. If δ ( B − x ) ≥ ⌊ ( ⌊ n/ ⌋ + s − a − / ⌋ ,then by Lemma 2, B contains a path from x of order at least ⌊ n/ ⌋ + s − a −
1. Thus G contains a P n , a contradiction. This implies that δ ( B − x ) ≤ (cid:22) ⌊ n/ ⌋ + s − a − (cid:23) − (cid:24) n + 2 s − par( n ) − a (cid:25) − . Thus the claim holds.Note that ν ( B − x − v − N ( v )) = ν ( B ) − − d B − x ( v ) ≥ l m m + 1 − − (cid:24) n + 2 s − par( n )4 (cid:25) + 2 ≥ (cid:24) m − n + 2 s − par( n ) + 24 (cid:25) + 1 ≥ . This implies that V ( B ) \{ x, v }\ N ( v ) = ∅ . Case 2.1.1. x = x ′ .In this case, G has only one cut-vertex x . Let G ′ = G − x − v − N ( v ). Then G ′ isdisconnected and ν ( G ′ ) = ν ( G ) − − d B − x ( v ) ≥ m + n − s − − (cid:24) n + 2 s − par( n )4 (cid:25) + 2= m + (cid:22) n + par( n ) − s (cid:23) ≥ m. Let H be the union of any ω ( G ′ ) − G ′ . We will prove that ν ( H ) ≥ m + ⌊ n/ ⌋ − ν ( G ′ ). 15f B ′ − x
6⊂ H , then ν ( H ) = ν ( G ′ ) − ν ( B ′ − x ) ≥ ν ( G ′ ) − ⌊ ( ν ( G ) − / ⌋ , and ν ( H ) + ν ( G ′ ) − m − j n k ≥ ν ( G ′ ) − (cid:22) ν ( G ) − (cid:23) + ν ( G ′ ) − m − j n k ≥ (cid:18) m + (cid:22) n + par( n ) − s (cid:23)(cid:19) − (cid:22) m + n − s − (cid:23) − m − j n k ≥ (cid:24) m + 2 · n − s − − m + n − s − − n (cid:25) = (cid:24) m + n − s − (cid:25) ≥ (cid:24) n − s (cid:25) ≥ . Now we assume that B ′ − x ⊂ H . In this case ν ( H ) ≥ ν ( B ′ − x ) ≥ δ ( G ), and ν ( H ) + ν ( G ′ ) − m − j n k ≥ δ ( G ) + ν ( G ′ ) − m − j n k ≥ l n m − s + 1 + m + (cid:22) n + par( n ) − s (cid:23) − m − j n k ≥ (cid:22) n + 5par( n ) + 4 − s (cid:23) . Note that 3 n + 5par( n ) + 4 − s ≥ n = 8 and s = 3. Petty Case. n = 8 and s = 3.In this case ν ( B ′ − x ) ≥ d B − x ( v ) ≤
2. If ν ( H ) ≥
3, or if d B − x ( v ) = 1, then it iseasy to see that ν ( H ) ≥ m + ⌊ n/ ⌋ − ν ( G ′ ). Now we assume that ν ( B ′ − x ) = ν ( H ) = 2and d B − x ( v ) = 2. This implies that B ′ is a triangle, there are only two blocks B, B ′ , andevery vertex in B − x has degree at least 2 in B − x . If B − x has a cut-vertex, thennoting that every end-block of B − x has at least three vertices, B contains a path from x of order at least 6, and G contains a P , a contradiction. So we assume that B − x is2-connected.Note that B − x contains a cycle of order at least 4. Let C be a longest cycle of B − x .If ν ( C ) ≥
5, then there is also an path from x in B of order at least 6, a contradiction.Thus we assume that ν ( C ) = 4. If there is a component of B − x − C with order at least 2,or if there is a vertex in B − x − C adjacent to two consecutive vertices on C , then it is easyto find a cycle longer than C . Thus B − x − C consists of isolated vertices and every vertexis adjacent to two nonconsecutive vertices on C . If there are two vertices in B − x − C adjacent to different vertices on C , we can also find a longer cycle. Thus all the vertices of B − x − C have the same neighbors on C . This implies that B − x − v − N ( v ) is disconnectedand then ν ( H ) ≥ ν ( B ′ − x ) + 1 = 3. Thus we also have ν ( H ) ≥ m + ⌊ n/ ⌋ − ν ( G ′ ).By Lemma 6, G ′ contains a C m , a contradiction. Case 2.1.2. xx ′ is a cut-edge of G and there is only one end-block containing x ′ .16y Claim 4, d B − x ( v ) ≤ ⌊ ( n + 2 s − par( n )) / ⌋ −
2. Let G ′ = G − x − v − N ( v ). Then ν ( G ′ ) = ν ( G ) − − d B − x ( v ) ≥ m + n − s − − (cid:22) n + 2 s − par( n )4 (cid:23) + 2= m + (cid:24) n + par( n ) − s (cid:25) ≥ m. Now let H be the union of any ω ( G ′ ) − G ′ . If B ′ ⊂ H , then ν ( H ) ≥ ν ( B ′ ) ≥ δ ( G ) + 1, and ν ( H ) + ν ( G ′ ) − m − j n k ≥ δ ( G ) + 1 + ν ( G ′ ) − m − j n k ≥ l n m − s + 2 + m + (cid:24) n + par( n ) − s (cid:25) − m − j n k = (cid:24) n + 5par( n ) + 8 − s (cid:25) ≥ . If B ′
6⊂ H , then ν ( H ) = ν ( G ′ ) − ν ( B ′ ) ≥ ν ( G ′ ) − ⌊ ν ( G ) / ⌋ , and ν ( H ) + ν ( G ′ ) − j n k − m ≥ ν ( G ′ ) − (cid:22) ν ( G )2 (cid:23) + ν ( G ′ ) − m − j n k ≥ (cid:18) m + (cid:24) n + par( n ) − s (cid:25)(cid:19) − (cid:22) m + n − s (cid:23) − m − j n k ≥ (cid:24) m + 2 · n + par( n ) − s − m + n − s − n (cid:25) = (cid:24) m + n + par( n ) − s (cid:25) ≥ (cid:24) n + par( n ) + 1 − s (cid:25) ≥ . By Lemma 6, G ′ contains a C m , a contradiction. Case 2.1.3. xx ′ / ∈ E ( G ), or xx ′ is not a cut-edge of G , or there are at least two end-blockscontaining x ′ .Let G ′ = G − x − x ′ − v − N ( v ). Note that in this case ω ( G ′ ) ≥
3, and we have ν ( G ′ ) = ν ( G ) − − d B − x ( v ) ≥ m + n − s − − (cid:22) n + 2 s − par( n )4 (cid:23) + 2= m + (cid:24) n + par( n ) − s − (cid:25) ≥ m. Now let H be the union of any ω ( G ′ ) − G ′ . If B ′ − x ′ ⊂ H , thennoting that ω ( G ′ ) ≥ ν ( H ) ≥ ν ( B ′ − x ′ ) + 1 ≥ δ ( G ) + 1, and ν ( H ) + ν ( G ′ ) − m − j n k ≥ δ ( G ) + 1 + ν ( G ′ ) − m − j n k ≥ l n m − s + 2 + m + (cid:24) n + par( n ) − s − (cid:25) − m − j n k = (cid:24) n + 5par( n ) + 4 − s (cid:25) ≥ . B ′ − x ′
6⊂ H , then ν ( H ) = ν ( G ′ ) − ν ( B ′ − x ′ ) ≥ ν ( G ′ ) − ⌊ ν ( G ) / ⌋ + 1, and ν ( H ) + ν ( G ′ ) − m − j n k ≥ ν ( G ′ ) − (cid:22) ν ( G )2 (cid:23) + 1 + ν ( G ′ ) − m − j n k ≥ (cid:18) m + (cid:24) n + par( n ) − s − (cid:25)(cid:19) − (cid:22) m + n − s (cid:23) + 1 − m − j n k ≥ (cid:24) m + 2 · n − s − − m + n − s − n (cid:25) = (cid:24) m + n − s − (cid:25) ≥ (cid:24) n − s − (cid:25) ≥ . By Lemma 6, G ′ contains a C m , a contradiction. Case 2.2.
Every end-block of G has order at most ⌈ m/ ⌉ . Claim 5.
Let G ′ be a disconnected subgraph of G . If(a) ν ( G ′ ) ≥ m ; and(b) there are two components of G ′ , each of which is an end-block removing a cut-vertexof G contained in the end-block,then the order sum of every ω ( G ′ ) − G ′ is m + ⌊ n/ ⌋ − ν ( G ′ ). Proof.
Let B − x and B ′ − x ′ be two components of G ′ , where B, B ′ are two end-blocksof G and x, x ′ are two cut-vertices of G contained in B and B ′ , respectively.Let H be the union of any ω ( G ′ ) − G ′ . We first assume that both B − x and B ′ − x ′ ⊂ H . Then ν ( H ) ≥ ν ( B − x ) + ν ( B ′ − x ′ ) ≥ δ ( G ), and ν ( H ) + ν ( G ′ ) − m − j n k ≥ δ ( G ) + ν ( G ′ ) − m − j n k ≥ (cid:16)l n m − s + 1 (cid:17) + m − m − j n k = l n m + par( n ) + 2 − s ≥ . Now we assume that H does not contain B − x or B ′ − x ′ . Without loss of generality, weassume that H does not contain B − x . Then ν ( H ) = ν ( G ′ ) − ν ( B − x ) ≥ ν ( G ′ ) −⌈ m/ ⌉ +1,and ν ( H ) + ν ( G ′ ) − m − j n k ≥ ν ( G ′ ) − l m m + 1 + ν ( G ′ ) − m − l n m ≥ m − l m m + 1 − m − j n k = j m k − j n k + 1 ≥ . Thus the claim holds.
Case 2.2.1. G has only two end-blocks. 18et B and B ′ be the two end-blocks of G , and let x and x ′ be the cut-vertices of G contained in B and B ′ , respectively. Note that ν ( G ) − ν ( B ) − ν ( B ′ ) ≥ m + n − s − · l m m = n − s − par( m ) ≥ . This implies that V ( G ) \ V ( B ) \ V ( B ′ ) = ∅ .Note that in this case G − ( B − x ) − ( B ′ − x ′ ) + xx ′ is 2-connected. If every vertex in G − B − B ′ has degree at least 2 s − par( n ) −
3, then by Lemma 2, there is a path from x to x ′ of order at least 2 s − par( n ) −
2. Note that B contains a path from x of order atleast ⌈ n/ ⌉ − s + 2, and B ′ contains a path from x ′ of order at least ⌈ n/ ⌉ − s + 2. Thus G contains a P n , a contradiction. This implies that there is a vertex v in G − B − B ′ with d ( v ) ≤ s − par( n ) − G ′ = G − x − x ′ − v − N ( v ). Then ν ( G ′ ) ≥ ν ( G ) − − d ( v ) ≥ m + n − s − − s + par( n ) + 4= m + n + par( n ) + 1 − s ≥ m. By Claim 5, the order sum of every ω ( G ′ ) − G ′ is at least m + ⌊ n/ ⌋ − ν ( G ′ ). By Lemma 6, G ′ contains a C m , a contradiction. Case 2.2.2. G has at least three end-blocks.Let x and x ′ be two cut-vertices of G such that the longest path between x and x ′ in G is as long as possible. Clearly x and x ′ are both contained in some end-blocks. Let B and B ′ be two end-blocks of G containing x and x ′ , respectively. Let v be a vertex in V ( B − x ) ∪ V ( B ′ − x ′ ) such that d G − x − x ′ ( v ) is as small as possible. We assume withoutloss of generality that v ∈ V ( B − x ). Claim 6. d B − x ( v ) ≤ ⌊ n/ ⌋ − , if x = x ′ ; ⌈ n/ ⌉ − , if xx ′ is a cut-edge of G ; ⌊ n/ ⌋ − , otherwise . Proof.
We set a parameter a such that a = 0 if x = x ′ , 1 if xx ′ is a cut-edge of G , and 2otherwise. So there is a path between x and x ′ of length at least a .If δ ( B − x ) ≥ ⌊ ( n − a ) / ⌋−
1, then δ ( B ′ − x ′ ) ≥ ⌊ ( n − a ) / ⌋−
1. By Lemma 2, B containsa path from x of order at least ⌊ ( n − a ) / ⌋ + 1 and B ′ contains a path from x ′ of order atleast ⌊ ( n − a ) / ⌋ + 1. Thus G contains a path of order at least n + 1 − par( n − a ) ≥ n , acontradiction. Now we obtain that δ ( B − x ) ≤ ⌊ ( n − a ) / ⌋ − ase 2.2.2.1. x = x ′ .Let G ′ = G − x − v − N ( v ). Then ν ( G ′ ) = ν ( G ) − − d B − x ( v ) ≥ m + n − s − − j n k + 2= m + l n m − s ≥ m. Note that every end-block of G other than B removing x is a component of G ′ . ByClaim 5 and Lemma 6, G ′ contains a C m , a contradiction. Case 2.2.2.2. xx ′ is a cut-edge of G .In this case, G has the only two cut-vertices x and x ′ . Let G ′ = G − x − x ′ − v − N ( v ).Then ν ( G ′ ) = ν ( G ) − − d B − x ( v ) ≥ m + n − s − − l n m + 3= m + j n k − s ≥ m. Note that every end-block of G other than B removing x or x ′ is a component of G ′ .By Claim 5 and Lemma 6, G ′ contains a C m , a contradiction. Case 2.2.2.3. xy / ∈ E ( G ) or xy is not a cut-edge of G .Let B ′′ be an end-block of G other than B and B ′ , and let x ′′ be the cut-vertex of G contained in B ′′ (possibly x ′′ = x or x ′ ). Let G ′ = G − x − x ′ − x ′′ − v − N ( v ). Then ν ( G ′ ) ≥ ν ( G ) − − d B − x ( v ) ≥ m + n − s − − j n k + 3= m + l n m − s − ≥ m. Note that B ′ − x ′ and B ′′ − x ′′ are two components of G ′ . By Claim 6 and Lemma 6, G ′ contains a C m , a contradiction. Case 3. G is 2-connected.By Claim 3 and Lemma 2, G contains a cycle of order at least 2( ⌈ n/ ⌉ − s + 1) = n − s + par( n ) + 2. Let C be a longest cycle of G (with a given orientation). Supposethat ν ( C ) = n − r , where r ≤ s − par( n ) − . H be a subgraph of a component of G − C , and let N C ( H ) = { z , z , . . . z k } , where k = d C ( H ), and z i , 1 ≤ i ≤ k , are in order along C . We call the subpath −→ C [ z i , z i +1 ](the indices are taken modulo k ) a good segment of C (with respect to H ); moreover, if z i and z i +1 are joined to two distinct vertices x, y in H , then we call −→ C [ z i , z i +1 ] a bettersegment of C (with respect to H ); moreover, if there is a path from x to y in G − C oforder at least 3, then we call −→ C [ z i , z i +1 ] a best segment of C (with respect to H ). Since G is 2-connected, we conclude that for any component H of G − C , there are at least twogood (better, best) segments of C with respect to H if ν ( H ) ≥ ν ( H ) ≥ ν ( H ) ≥ H is not a star, respectively). Note that every good (better, best) segment has orderat least 3 (4, 5, respectively).For a vertex x of C , we use x + to denote the successor, and x − the predecessor, of x on C . For a subset X of V ( C ), we set X + = { x + : x ∈ X } and X − = { x − : x ∈ X } .Now we consider a component H of G − C . If H is non-separable, then H is a K , a K or 2-connected; if H is separable, then H has at least two end-blocks. In the later case,we call an end-block of H removing the cut-vertex contained in the end-block a branch of H (also, of G − C ). Claim 7.
Let H be a component of G − C and u ∈ V ( H ).(1) If H is non-separable, then H contains a path from u of order at least min { ν ( H ) , ⌈ r/ ⌉} .(2) If H is separable and D is a branch of H not containing u , then H contains a pathfrom u of order at least min { ν ( D ) + 1 , ⌈ r/ ⌉} . Proof.
We first claim that for any two vertices u, v ∈ V ( H ), d H ( u )+ d H ( v ) ≥ ⌈ r/ ⌉ , unless uv is a cut-edge of H . Assume that uv is not a cut-edge of H . Then H contains a pathfrom u to v of order at least 3. Let N C ( { u, v } ) = { z , z , . . . , z k } , where z i , 1 ≤ i ≤ k , arein order along C . If z i is joined to exactly one vertex of u, v , then −→ C [ z i , z i +1 ] is a goodsegment of C with respect to { u, v } ; if z i is adjacent to both u and v , then −→ C [ z i , z i +1 ] is abest segment with respect to { u, v } . This implies that d C ( u ) + d C ( v ) ≤ ⌊ ( n − r ) / ⌋ and d H ( u ) + d H ( v ) = d ( u ) + d ( v ) − d C ( u ) − d C ( v ) ≥ · (cid:16)l n m − s + 1 (cid:17) − (cid:22) n − r (cid:23) = (cid:24) n + r (cid:25) + par( n ) + 2 − s ≥ l n m + l r m + 2 − s ≥ l r m . Now we prove the claim.(1) If H contains only one or two vertices, then the assertion is trivially true. So weassume that ν ( H ) ≥
3. Let u ′ be a vertex in H such that d H ( u ′ ) is as small as possible.Thus d H ( v ) ≥ ⌈⌈ r/ ⌉ / ⌉ for any vertex v ′ ∈ V ( H ) \{ u ′ } . By Lemma 2, H contains a pathfrom u of order at least min { ν ( H ) , ⌈ r/ ⌉} . 212) Let B be the end-block of H containing D and b be the cut-vertex of H containedin B . If B contains only two vertices, then the assertion is trivially true. So we assumethat ν ( D ) ≥
3, from which we can see that B is 2-connected. Let u ′ be a vertex in B − b such that d H ( u ) is as small as possible. Thus every vertex in V ( B ) \{ b, u ′ } hasdegree at least ⌈⌈ r/ ⌉ / ⌉ in B . By Lemma 2, B contains a path from b of order at leastmin { ν ( B ) , ⌈ r/ ⌉} , and H contains a path from u of order at least min { ν ( B ) , ⌈ r/ ⌉} =min { ν ( D ) + 1 , ⌈ r/ ⌉} .Now we choose D among all the non-separable components and branches of G − C such that the order of D is as small as possible. We set a parameter a such that a = 0 if D is a non-separable component, and a = 1 if D is a branch of G − C .If D is a branch of G − C , then let H be the component of G − C , and B the end-blockof G − C , containing D ; if D is a component of G − C , then let H = B = D . Case 3.1. ν ( D ) = 1.Let v be the vertex in D . If D = H , then let R = G − C − H , X = N + C ( H ). If D = H ,then let y be a vertex in H − B , R = G − C − B − y and X = N + C ( H ) ∪ { y } . Thus everycomponent of R is joined to at most one vertex in X . Moreover, we have ν ( R ) = ν ( G ) − ν ( C ) − − a = m + n − s − n + r − − a = m + r − s − a − , and | X | = d C ( H ) + a ≥ d C ( v ) + a = d ( v ) ≥ l n m − s + 1 . Let G ′ = G [ V ( R ) ∪ X ]. Note that there is a path of order at least 2+2 a with an end-vertexin C and all other vertices in H . We have r ≥ a , and ν ( G ′ ) = ν ( R ) + | X | ≥ m + r − s − a − l n m − s + 1= m + l n m + r − s − a ≥ m + l n m + 2 − s ≥ m. Claim 8. D = H or d C ( H ) ≥ Proof.
Assume that D = H and d C ( H ) = 2. Since d C ( H ) = d ( v ) ≥ ⌈ n/ ⌉ − s + 1, wehave n ≤
8. We claim that every component of G − C is an isolated vertex. Supposeon the contrary that there is a component H ′ of G − C with order at least 2. Note thatthere are at least two better segments of C with respect to H ′ . We have ν ( C ) ≥
6, and G [ V ( C ) ∪ V ( H ′ )] contains a P , a contradiction. Thus as we claimed, every component of G − C is an isolated vertex. 22ote that ν ( R ) = m + r − s −
1. Since s ≤ n ≤
8) and r ≥
2, we have ν ( R ) ≥ m −
2. If ν ( R ) ≥ m , then there is a C m in G ′ ; if ν ( R ) = m −
1, then r = s ≤ N + C ( H ) is nonadjacent to every vertex in R , and thereis a C m in G ′ ; if ν ( R ) = m −
2, then r = s − ≤
2, and the two vertices in N + C ( H )are nonadjacent to every vertex in R , and there is a C m in G ′ . In any case we get acontradiction. So we conclude that D = H or d C ( H ) ≥ | X | ≥ C ′ in R with order r + par( r ), then let P be a path between C and C ′ , and C ∪ P ∪ C ′ will contain a P n , a contradiction. Thus we assume that R containsno cycle of order r + par( r ). Since ν ( R ) + 1 −
32 ( r + par( r )) = m + r − s − a − −
32 ( r + par( r )) ≥ m − s − a − l r m − par( r ) ≥ n − s − ≥ , by Lemma 4, there is a path in R of order at least p = ν ( R ) + 1 − r + par( r )2 = m + r − s − a − − l r m = m + j r k − s − a. Note that p + 2 | X | − ≥ m + j r k − s − a + 2 (cid:16)l n m − s + 1 (cid:17) − m + n + par( n ) + j r k − s − a − ≥ m + n + par( n ) − s − a. We can see that p + 2 | X | − ≥ m , when n ≥
9, unless n = 11 or 12 and a = 1. If n ≤ | X | ≥
3, we also have p + 2 | X | − ≥ m + j r k − s − a + 3 ≥ m − s + 3 ≥ m. By Lemma 7, G ′ contains a C m , a contradiction. Petty Case. n = 11 or 12 and a = 1.We claim that every component of G − C is a K , K , K or a star K ,k . Suppose thecontrary that there is a component H ′ of order at least 4 which is not a star. Since thereare at least two best segments of C with respect to H ′ , we can see that ν ( C ) ≥
8. Notethat there is a path P of order at least 5 with one end-vertex in C and all other verticesin H ′ . This implies that G [ V ( C ) ∪ V ( H ′ )] contains a P , a contradiction. Thus as weclaimed, every component of G − C is a K , K , K or a star K ,k .Since H is not a K , K or K , we conclude that H is a star. Now we choose acomponent H ′ of G − C that is a maximum star of G − C , and let u ′ be the center of H ′ ,23 ′ and y ′ be two end-vertices of H ′ . Let R ′ = G − C − { u ′ , v ′ , y ′ } , X ′ = N + C ( H ′ ) ∪ { y ′ } and G ′′ = G [ V ( R ′ ) ∪ X ′ ]. By the analysis above, we have ν ( R ′ ) ≥ m + r − s − | X ′ | ≥ l n m − s + 1 . Since ν ( R ′ ) ≥ m + r − s − ≥ n + 2 − s ≥
20. If G − C has at least three components,then R ′ is disconnected; if G − C has exactly two components, then H ′ is a star with atleast 4 vertices, and R ′ is connected; if G − C consists of only one component H ′ , then R ′ = H ′ − { u ′ , v ′ , y ′ } is empty, and thus disconnected. Thus in any case, R ′ is connected.Let H ′′ be a component of R ′ with the maximum order. If ν ( H ′′ ) ≤ ⌈ ν ( R ′ ) / ⌉ , thenevery vertex of R ′ has degree at least ⌊ ν ( R ′ ) / ⌋ in R ′ . By Lemma 2, R ′ contains aHamilton path. If ν ( H ′′ ) ≥ ⌈ ν ( R ′ ) / ⌉ + 1, then H ′′ is a star with at least 4 vertices.Let u ′′ be the center of the star. Then every vertex in V ( R ′ ) \{ u ′′ } has degree at least ⌈ ν ( R ′ ) / ⌉ in R ′ − u ′′ . By Lemma 2, R ′ − u ′′ contains a Hamilton cycle and R ′ contains aHamilton path. In any case R ′ contains a path of order at least p ′ = ν ( R ′ ). Thus we have p ′ + 2 | X ′ | − ≥ ν ( R ′ ) + | X ′ | ≥ m. By Lemma 7, G ′′ contains a C m , a contradiction. Case 3.2. ν ( D ) = 2.Let v, v ′ be the two vertices in D . If D = H , then let R = G − C − H , X = N + C ( H ), X = N − C ( H ). If D = H , then let y be a vertex in H − B , let R = G − C − B − y , X = N + C ( H ) ∪ { y } , X = N − C ( H ) ∪ { y } . Thus every component of R is joined to at mostone vertex in X i , i = 1 ,
2, and ν ( R ) = ν ( G ) − ν ( C ) − − a = m + n − s − n + r − − a = m + r − s − a − . Let X = X ∪ X and G ′ = G [ V ( R ) ∪ X ]. Note that there is a path of order at least 3 + 3 a with an end-vertex in C and all other vertices in H . We have that r ≥ a .Let N C ( H ) = { z , z , . . . , z k } , where z i , 1 ≤ i ≤ k , are in order along C . Sincethere are at least two better segments, we have | X \ X | = | X \ X | ≥
2. For any vertex z i ∈ N C ( H ): if z i is adjacent to exactly one vertex in { v, v ′ } , then −→ C [ z i , z i +1 ] is a goodsegment; if z i is adjacent to both v and v ′ , then −→ C [ z i , z i +1 ] is a better segment. This24mplies that | X | ≥ d C ( v ) + d C ( v ′ ) + a ≥ (cid:16)l n m − s + 1 − − a (cid:17) + a = n + par( n ) − s − a, and ν ( G ′ ) = ν ( R ) + | X |≥ m + r − s − a − n + par( n ) − s − a ≥ m + 3 + 3 a − s − a − n + par( n ) − s − a ≥ m + n + par( n ) + 1 − s ≥ m. Since there are at least two better segments of C with respect to H , ν ( C ) ≥
6. Thusthere is a path in G [ V ( C ) ∪ V ( H )] of order at least 8, which implies that n ≥ Claim 9. D = H or d C ( H ) ≥ Proof.
Assume that D = H and d C ( H ) = 2. Note that the two segments of C with respectto H are both better. Since d C ( H ) ≥ d ( v ) − ≥ ⌈ n/ ⌉ − s , we have n ≤
12. We claimthat every component of R has order at most 3. Suppose on the contrary that there isa component H ′ of G − C that has order at least 4. Note that H ′ is not a star. Thereare at least two best segments of C with respect to H ′ , which implies that ν ( C ) ≥
8. ByClaim 7, there is a path of order at least 5 with one end-vertex in C and all other verticesin H ′ . Thus G [ V ( C ) ∪ V ( H ′ )] contains a P , a contradiction. Thus as we claimed, everycomponent of R has order 2 or 3.Note that ν ( R ) = m + r − s −
2. Since s ≤ n ≤
12) and r ≥
3, we have ν ( R ) ≥ m −
3. If ν ( R ) ≥ m , then by Lemma 3 there is a C m in G ′ ; if ν ( R ) = m − m −
2, then we have r ≤ s + 1 ≤
5, and one of the two vertices in N + C ( H ) ( N − C ( H ))is nonadjacent to every vertex in R , and there is a C m in G ′ ; if ν ( R ) = m −
3, then r = s − ≤
3, and every vertex in N + C ( H ) and N − C ( H ) is nonadjacent to every vertexin R , and there is a C m in G ′ . In any case, we get a contradiction. So we conclude that D = H or d C ( H ) ≥ | X | = | X | ≥ R of order r + par( r ), then there will be a path of order at least n in G . Thus we assume that R contains no cycle of order r + par( r ). Since ν ( R ) + 1 −
32 ( r + par( r )) = m + r − s − − a + 1 −
32 ( r + par( r )) ≥ m − s − a − l r m − par( r ) − ≥ n − s − ≥ ,
25y Lemma 4, there is a path in R of order at least p = ν ( R ) + 1 − r + par( r )2 = m + r − s − − a + 1 − l r m = m + j r k − s − a − . Note that p + 2 | X | − ≥ m + j r k − s − a − n + par( n ) − s − a ) − m + 2 n + 2par( n ) + j r k − s − a − ≥ m + 2 n + 2par( n ) − s − . We can see that p + 2 | X | − ≥ m , when n ≥
13. If n ≤
12, then noting that d C ( H ) + a ≥ | X | ≥
5, we also have p + 2 | X | − ≥ m + j r k − s − a − ≥ m − s + 5 ≥ m. By Lemma 8, G ′ contains a C m , a contradiction. Case 3.3. ≤ ν ( D ) ≤ ⌈ r/ ⌉ − r ≥
7. If D = H , then let R = G − C − H , X = N + C ( H ) and X = N − C ( H ).If D = H , then let y be a vertex in H − B which is not a cut-vertex of H − B , let R = G − C − B − y , X = N + C ( H ) ∪ { y } and X = N − C ( H ) ∪ { y } . Thus every componentof R is joined to at most one vertex in X i , i = 1 ,
2, and ν ( R ) = ν ( G ) − ν ( C ) − ν ( D ) − a ≥ m + n − s − n + r − l r m + 1 − a = m + j r k + 1 − s − a. Clearly, every component of R has order at least 2, and (cid:24) ν ( R )2 (cid:25) + 4 ≥ (cid:24) m + ⌊ r/ ⌋ + 1 − s − a )2 (cid:25) + 4 ≥ m + (cid:24) m + 3(3 + 1 − s − a )2 (cid:25) + 4 ≥ m + (cid:24) n + 15 − s (cid:25) ≥ m. Let X = X ∪ X and G ′ = G [ V ( G − B ) \ N C ( H )]. Since there are at least two bestsegments with respect to H , we have | X \ X | = | X \ X | ≥
2. Let v be a vertex in D .26ince R contains no cycle of length r + par( r ) and ν ( R ) + 1 −
32 ( r + par( r ))= m + j r k + 1 − s − a + 1 − · l r m ≥ m + 2 − r − r ) − s − a ≥ n + 3 − s + par( n ) + 2 − r ) − s − a ≥ n + par( n ) + 1 − s ≥ ,R contains a path of order at least p = ν ( R ) + 1 − r + par( r )2 ≥ m + j r k + 1 − s − a + 1 − l r m = m + 2 − s − par( r ) − a. Claim 10. D = H or d C ( H ) ≥ Proof.
Assume that D = H and d C ( H ) = 2. Thus d D ( v ) ≥ l n m − s + 1 − l n m − s − , and ν ( D ) ≥ d D ( v ) ≥ l n m − s ≥ s − ≥ l r m − . This implies that ⌈ r/ ⌉ = s − ν ( D ) = ⌈ r/ ⌉ −
1, and d D ( v ) = ν ( D ) −
1. Note that inthis case every vertex in D has degree ν ( D ) −
1, and thus D is a clique.If every vertex in N + C ( H ) is joined to some component of G − C , then by Claim 7, wecan find a path from the cycle C , component H and the two components joined to thetwo vertices in N + C ( H ), of order at least ν ( C ) + 3 ν ( D ) = ν ( C ) + 3 · (cid:16)l r m − (cid:17) = n − r + r + par( r ) + l r m − ≥ n, a contradiction. Thus there is a vertex v ′ in N + C ( H ) that is not joined to every componentof G − C . Let G ′′ = G − C .Since ⌈ r/ ⌉ = s − r ≥
7, we can see that r ≥ s + 1. Thus ν ( G ′′ ) = ν ( G ) − ν ( C ) ≥ m + n − s − n + r = m + r − s ≥ m. Note that in this case, G ′′ − H = R contains a path of order at least p ≥ m +2 − s − par( r ) ≥ m + 3 − r − par( r ) and p + 2 ν ( H ) − ≥ m + 3 − r − par( r ) + 2 · (cid:16)l r m − (cid:17) − m + 3 − r − par( r ) + r + par( r ) − − m. ν ( R ) ≥ m + j r k + 1 − s = m − par( r ) ≥ l m m , by Lemma 5, G ′′ contains a C m , and G contains a W m with the hub v ′ , a contradiction.By Claim 10, | X | = | X | ≥
3. If D = H , then since there are at least two bestsegments with respect to H , we can see that ν ( C ) ≥
8. By Claim 7, there is a pathof order at least 6 with one end-vertex in C and all other vertices in H , which impliesthat G [ V ( C ) ∪ V ( H )] contains a P ; if D = H and d C ( H ) ≥
3, noting that at least twosegments of C with respect to H are best, we have ν ( C ) ≥
10. By Claim 7, there is a pathof order at least 4 with an end-vertex in C and all internal vertices in H , G [ V ( C ) ∪ V ( H )]contains a P as well. Thus we conclude that n ≥ H ′ be a component of R , and let W be the union of X and the set of vertices in V ( C ) \ N C ( H ) not joined to H ′ . For any two vertices x, y with xy ∈ E ( C ): if one of x, y is in N C ( H ), then the other one will be in X ⊂ W ; if none of them is in N C ( H ), then atleast one of them will not be joined to H ′ , otherwise there will be a cycle longer than C .This implies that | W | ≥ ⌈ ( n − r ) / ⌉ + a = q .Since ν ( R ) + q − ≥ m + j r k + 1 − s − a + (cid:24) n − r (cid:25) + a − ≥ m + j n k − s − a ≥ m, ( n ≥
14) and p + 2 q − ≥ m + 2 − s − par( r ) − a + 2 · (cid:18)(cid:24) n − r (cid:25) + a (cid:19) − m + n − r − s − n − r ) − par( r )= m + n − s − n ) + par( n − r ) − par( r ) ≥ m + n − s − , we can see that p + 2 q − ≥ m , unless n = 15, s = 5 and r = 7. Petty Case. n = 15, s = 5 and r = 7.In this case, ν ( C ) = 8 which implies that D = H . It is easy to find a path with twoend-vertices in C and all internal vertices in H of order at least 7. Thus ν ( C ) ≥
12, acontradiction.By Lemma 9, G ′ contains a C m , a contradiction.28 ase 3.4. ν ( D ) ≥ max {⌈ r/ ⌉ , } .By Claim 7, there is a path of order at least 4 with an end-vertex in C and all othervertices in H . Thus we have r ≥
4. Let H ′ be an arbitrary component of R and u ∈ V ( H ′ ).By Claim 7, H ′ contains a path from u of order at least ⌈ r/ ⌉ . Thus for any edge xy ∈ E ( C ), either x or y is not joined to any components of G − C , otherwise there willbe a P n in G . Moreover, if r is odd and x is joined to some component, say H ′ , of G − C ,then x ++ will not be joined to any component of G − C other than H ′ as well. Case 3.4.1.
Every component of G − C has order at most r − v be a vertex in N + C ( H ), and let G ′ = G [ V ( G − C ) ∪ N + C ( H ) \{ v } ]. Note that v isnonadjacent to every vertex in G ′ , and every component of G ′ has order at most r − ≤ s − par( n ) − − ≤ l n m ≤ j m k . Let u be a vertex in H . Since d C ( H ) ≥ d C ( u ) ≥ d ( u ) − ν ( H ) + 1 ≥ d ( u ) + 2 − r, and ν ( G ′ ) = ν ( G ) − ν ( C ) + d C ( H ) − ≥ m + n − s − n + r + d ( v ) + 1 − r ≥ m − s + l n m − s + 1 + 1= m + l n m + 2 − s ≥ m, by Lemma 3, there is a C m in G ′ , a contradiction. Case 3.4.2.
There is a component of G − C of order at least r .Let H ′ be a component of G − C with order at least r . We claim that there is a vertex u in H ′ with d H ′ ( u ) ≤ ⌈ r/ ⌉ −
1. Suppose the contrary that every vertex of H ′ has degreeat least ⌈ r/ ⌉ in H ′ . If H ′ is 2-connected, then by Lemma 2, there is a cycle of order atleast r in H ′ , and G will contain a P n ; if G is separable, letting B ′ be any end-block of H ′ , b ′ be the cut-vertex of H ′ contained in B ′ , and u ′ be any vertex in V ( B ) \{ b ′ } , thenthere is a path from b ′ to u ′ of order at least ⌈ r/ ⌉ + 1. Thus G will contain a P n as well.So we assume that there is a vertex u in H ′ with d H ′ ( u ) ≤ ⌈ r/ ⌉ − v be a vertex in N + C ( H ′ ), X = N + C ( H ′ ) \{ v } . If r is odd, then let −→ C [ z, z ′ ] be abetter segment of C with respect to H ′ not containing v , and we add z ++ to X . Let G ′ = G [ V ( G − C ) ∪ X ]. Note that v is nonadjacent to every vertex in G ′ , and there areno edges between G − C and X . 29ince d C ( H ′ ) ≥ d C ( u ) = d ( u ) − d H ′ ( u ) ≥ l n m − s + 1 − l r m + 1 = l n m + 2 − l r m − s, we have | X | = d C ( H ) − r ) ≥ l n m + 1 − j r k − s and ν ( G ′ ) = ν ( G ) − ν ( C ) + | X |≥ m + n − s − n + r + l n m + 1 − j r k − s ≥ m − s + l r m + l n m − s + 1 ≥ m + l n m + 3 − s ≥ m. Since G − C contains no cycle of length r + par( r ) and ν ( G − C ) + 1 −
32 ( r + par( r ))= m + r − s + 1 − · l r m ≥ m − l r m − par( r ) − s ≥ n − s ≥ ,G − C contains a path of order at least p = ν ( G − C ) + 1 − r + par( r )2= m + r − s + 1 − l r m = m + j r k + 1 − s. Clearly | X | ≥
1. If ⌊ r/ ⌋ ≥ s −
2, then p + 2 | X | − ≥ m + j r k + 1 − s + 2 − ≥ m + s − − s + 2 − m. If ⌊ r/ ⌋ ≤ s −
3, then p + 2 | X | − ≥ m + j r k + 1 − s + 2 · (cid:16)l n m + 1 − j r k − s (cid:17) − m + n + par( n ) + 2 − j r k − s ≥ m + n + par( n ) + 5 − s ≥ m. ν ( G − C ) = m + r − s ≥ l m m , by Lemma 5, there is a C m in G ′ , a contradiction.The proof is complete. (cid:3) A linear forest is a forest such that every component of it is a path. From our main resultof the paper, we can conclude the following result. Corollary 1.
Let n ≥ , m ≥ n + 1 and F be a linear forest on m vertices. Then R ( P n , K ∨ F ) = t ( n, m ) . Proof.
Note that the graph constructed in the beginning of Section 3 contains no P n andits complement contains no K ∨ F . We conclude that R ( P n , K ∨ F ) ≥ t ( n, m ). On theother hand, since K ∨ F is a subgraph of W m , we have R ( P n , K ∨ F ) ≤ R ( P n , W m ) ≤ t ( n, m ).For the case F is an empty graph, the above formula gives the Ramsey numbers ofpaths versus stars when m ≥ n + 1. In fact, Parsons [10] gave all the values of thepath-star Ramsey numbers by a recursive formula. The interested readers can compareour formula with the recursive one in [10]. Acknowledgments
The authors are very grateful to Professor Yunqing Zhang for providing them the paper[12].
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