aa r X i v : . [ m a t h . C O ] F e b The anti-Ramsey number for paths
Long-Tu Yuan ∗ Abstract
We determine the exactly anti-Ramsey number for paths. This confirms a conjec-ture posed by Erd˝os, Simonovits and S´os in 1970s.
Key words:
Anti-Ramsey numbers; paths.
AMS Classifications:
A subgraph of an edge-colored graph is rainbow if all of its edges have different colors.For a given graph H , the anti-Ramsey number AR( n, H ) of H is the maximum numberof colors in an edge-colored K n such that K n does not contain a copy of rainbow H .The anti-Ramsey number was introduced by Erd˝os, Simonovits and S´os [3]. In thesame paper, they observed that if we color a copy of K k − in K n with different colors andcolor the remaining edges a new color, then K n contains no rainbow paths on k vertices.Let X ⊆ V ( K n ) with size ⌊ ( k − / ⌋ and let i = 1 if k is odd and i = 2 if k is even. If wecolor the edges incident X with different colors and color the remaining edges with i newcolors, then we can easily check that K n does not contain rainbow paths on k vertices.Hence, they asked whether those two configurations are best possible.Denote by P k the path on k vertices. Let t = ⌊ ( k − / ⌋ . In [18], Simonovits andS´os determined AR( n, P k ) for n ≥ c t , where c is a constant. They also claimed thattheir result held for n ≥ t/ c , where c is a constant (without proof). The exactlyanti-Ramsey number for paths is still not know. For other related results on this topic werefer the interested readers to a survey of Fujita, Magnant, and Ozeki [6] and some newresults as [4, 9, 10, 11, 12, 14, 15, 19, 21].The main result of this paper is the following theorem which settles an old conjectureposed by Erd˝os, Simonovits and S´os [3] almost fifty years ago. Theorem 1 .
Let P k be a path on k vertices and ℓ = ⌊ ( k − / ⌋ . If n ≥ k ≥ , then AR( n, P k ) = max (cid:26)(cid:18) k − (cid:19) + 1 , (cid:18) ℓ − (cid:19) + ( ℓ − n − ℓ + 1) + ǫ (cid:27) , where ǫ = 1 if k is odd and ǫ = 2 otherwise. The extremal results of an n -vertex graph proved by stability results usually need n to be sufficiently large. Our proof of Theorem 1 bases on a recently result of F¨uredi,Kostochka, Luo and Verstra¨ete [7, 8] holding for graphs with arbitrary number of vertices.Hence, we can apply the stability results to determine the exactly anti-Ramsey number forpaths. It is very interesting to obtain other exactly extremal result by stability method. ∗ School of Mathematical Sciences and Shanghai Key Laboratory of PMMP, East China Normal Univer-sity, 500 Dongchuan Road, Shanghai 200240, P.R. China. Email: [email protected]. Supportedin part by National Natural Science Foundation of China grant 11901554 and Science and TechnologyCommission of Shanghai Municipality (No. 18dz2271000, 19jc1420100). Notation and basic lemmas
Given two graphs G and H , we say that G is H -free if G does not contain a copy of H as a subgraph. For a given graph H , the Tur´an number ex( n, H ) of H is the maximumnumber of edges in an n -vertex H -free graph. Similarly, the connected Tur´an number of agiven graph H , denoted by excon( n, H ), is the maximum number of edges of an n -vertex H -free connected graph. The anti-Ramsey problem is strongly connected with the Tur´anproblem. So we introduce some results about Tur´an problem first.Erd˝os and Gallai [2] first studied the Tur´an numbers of paths. Later, Faudree andSchelp [5] and independently Kopylov [13] improved Erd˝os and Gallai’s result to the fol-lowing. Theorem 2 (Faudree, Schelp [5] and Kopylov [13])
Let n ≥ k . Then ex( n, P k ) = s (cid:0) k − (cid:1) + (cid:0) r (cid:1) , where n = s ( k −
1) + r and ≤ r ≤ k − . Moreover, the extremal graphsare characterized. For connected graphs without containing a copy of P k , Balister, Gy˝ori, Lehel andSchelp [1] and independently Kopylov [13] proved the following theorem.We first introduce the following graphs which play an important role in extremalproblems for paths and cycles. For integers n ≥ k ≥ a , let H ( n, k, a ) be the n -vertex graphwhose vertex set is partitioned into three sets A, B, C such that | A | = a, | B | = n − k + a and | C | = k − a and the edge set consists of all edges between A and B together with alledges in A ∪ C . Let h ( n, k, a ) = e ( H ( n, k, a )). Theorem 3 (Balister, Gy˝ori, Lehel, Schelp [1] and Kopylov [13])
Let n ≥ k and s = ⌊ ( k − / ⌋ . Then ex con ( n, P k ) = max { h ( n, k − , , h ( n, k − , s ) } . Moreover, theextremal graph is either H ( n, k − , or H ( n, k − , s ) . Let n ≥ k and ℓ = ⌊ ( k − / ⌋ be positive integers. Define ar ( n, k ) = max { h ( k, k − , − , h ( n, k − , ℓ − − i } , where i = 0 if k is odd and i = 1 if k is even. Then Theorem 1 states that if the numberof colors in an edge-colored K n is at least ar( n, k ) + 1, then K n contains a rainbow copyof P k . The following lemma only needs Theorems 2, 3 and some basic calculations. Wemove its proof to Appendix A. Lemma 4
Let k ≥ k ≥ and t ≥ . Let n ≥ k − , n ≥ n ≥ . . . ≥ n t ≥ and n = P ti =0 n i ≥ k + k − ≥ . Then ex con ( n , P k ) + t X i =1 ex con ( n i , P k ) + t − ≤ ar ( n, k + k − . Very recently, F¨uredi, Kostochka, Luo and Verstra¨ete [7, 8] considered the stabilityresults of the well-known Erd˝os-Gallai theorems on cycles and paths. Let C k be the set ofcycles of length at least k . Let G be an n -vertex connected P k -free ( n -vertex 2-connected C k -free) graph. Their results state that if e ( G ) is close to the maximum value of numberof edges of n -vertex connected P k -free (2-connected C k -free) graphs, then G must be asubgraph of some well-specified graphs. We will use the following two corollaries of themain theorems in [7, 8] (Theorem 1.6 in [7] and Theorem 2.3 in [8], also see results in Theorem 2.3 in [8] states the stability result for 2-connected C k -free graphs. Note that if we add newa vertex and join it to all vertices of a connected P k -free graph, then the obtained graph is 2-connected C k -free. Theorem 2.3 in [8] can be extended to stability result for connected P k -free graphs as Theorem1.6 was extended by Theorem 1.4 in [7]. k for the purpose of proving ourmain result. Corollary 5 (F¨uredi, Kostochka, Luo and Verstra¨ete [7, 8])
Let k ≥ be odd and ℓ = ( k − / . Let G be an n -vertex connected graph without containing a path on k vertices. Then e ( G ) ≤ max { h ( n, k − , , h ( n, k − , ℓ − } unless G is a subgraph of H ( n, k − , . Corollary 6 (F¨uredi, Kostochka, Luo and Verstra¨ete [7, 8])
Let k ≥ be even and ℓ = ⌊ ( k − / ⌋ . Let G be an n -vertex connected graph without containing path on k ver-tices. Then e ( G ) < max { h ( n, k − ,
2) + 1 , h ( n, k − , ℓ − } unless(a) G is a subgraph of H ( n, k − , , or(b) G is a subgraph of H ( n, k − , ℓ ) , or(c) G is an acyclic P -free graph when k = 6 , or(d) G = H ( n, k − , ℓ − . Remark.
For k = 6 in Corollary 6, see Theorem 5.1(4) in [7]. Actually, the result weused here is extended by Theorem 5.1(4) in [7] as Theorem 1.6 was extended by Theorem1.4 in [7]. Corollary 6( d ) ( e ( G ) = h ( n, k − , ℓ − d ).We also need the following simple lemma (see Lemma 2.2 in [20]). Lemma 7
Let G be a bipartite graph with classes A and B . Let | A | = | B | = ℓ . If e ( G ) ≥ ( ℓ − ℓ + 2 , then G contains a cycle of length ℓ . The representing graph of a graph G with an edge coloring c is a spanning subgraph of G obtained by taking one edge of each color of c . For a set of edges E of G , we use c ( E )to denote the colors of edges in E . For a set of colors C , when an edge e is colored by acolor in C , we say e is colored by C for short. Given a graph G , we use T ( G ) to denotethe set of its cut edges. Definition 8
Given a graph G with an edge coloring c , we say that the pair ( G, c ) is a good edge coloring if there is a connected representing graph L n of G with a non-emptyset of cut edges X ⊆ T ( L n ) such that each e ∈ E ( G ) between components of L n − X arecolored by c ( X ). Lemma 9
Let G be a graph with an edge coloring c and let L n be the set of connectedrepresenting graphs of c . If C = T L n ∈L n { c ( e ) : e ∈ T ( L n ) } is not empty, then ( G, c ) is agood edge coloring with a representing graph L ∗ n and a set of cut edges X ⊆ T ( L ∗ n ) suchthat C ⊆ c ( X ) . Proof.
Let L n ∈ L n be a representing graph of G . Since C = T L n ∈L n { c ( e ) : e ∈ T ( L n ) } is not empty, L n is connected with at least one cut edge. We obtain a subgraph of L n bythe following procedure. Delete the edges of L n colored by C and denote the obtainedgraph by L n . Let C be the colors of the edges of G between any two components of L n . We say a graph is acyclic if it is connected without containing a cycle. The lemma maybe appears in some old paper, but I do not find it. L n colored by C and denote the obtained graph by L n . We go on thisprocedure and finally obtain a minimal spanning subgraph L tn of L n for some integer t .It is enough to show that the edges of G between any two components of L tn are coloredby c ( X ) with X ⊆ T ( L n ). We will show that L in is obtained from L n by deleting edges of T ( L n ) for i = 1 , . . . , t . This will complete our proof of the lemma. Suppose for contrarythat there is an edge e ′ ℓ ∈ E ( G ) between two components of L ℓ n not colored by c ( T ( L n )).We choose ℓ as small as possible. Clearly, we have ℓ > L in for i < ℓ isobtained from L n by deleting edges from T ( L n ). Then we may add e ′ ℓ to L n and deletethe edge e ℓ of L n colored by c ( e ′ ℓ ). Since e ℓ is not a cut edge of L n , the obtained graph e L n ∈ L n is connected. Moreover, by the minimality of ℓ , e L n contains a cycle containingan edge e ℓ of T ( L n ) colored by C ℓ with 0 ≤ ℓ ≤ ℓ −
1. We choose ℓ as small aspossible. If ℓ = 0, then e ℓ is colored by C and e ℓ is not a cut edge of e L n . This is acontradiction to definition of C . Let ℓ ≥
1. Then, from e L n , after deleting the edge e ℓ and adding an edge e ′ ℓ colored by c ( e ℓ ) between the components of L ℓ n , the obtainedgraph e L n ∈ L n is connected. Moreover, e L n contains a cycle containing one edge e ℓ of T ( L n ) colored by C ℓ with ℓ < ℓ . If ℓ = 0, then get a contradiction to definition of C .Otherwise, we may go on the above procedure until we have ℓ s = 0 for some s . Thus thereis an edge e ℓ s ∈ T ( L n ) colored by C and this edge is not a cut edge for e L sn ∈ L n . Thisfinal contradiction completes the proof of Lemma 9.Let c be an edge coloring of K n with maximum number of colors such that K n containsno copy of rainbow P k . Taking a representing graph L n of K n with a maximum component.The configurations before Theorem 1 show that e ( L n ) ≥ ar ( n, k ). Suppose that e ( L n ) ≥ ar ( n, k ) + 1 . (1)We will finish our proof of Theorem 1 by contradictions. Claim. L n is connected. Proof.
Suppose for contrary that L n is not connected. Let C , C , . . . , C ℓ be the compo-nents of L n with | C | ≥ | C | ≥ . . . ≥ | C ℓ | . We choose L n as following. For i = 1 , . . . , ℓ ,subject to the choice of C i , we choose C i +1 with | V ( C i +1 ) | = s i +1 maximum first and then | E ( C i +1 ) | = t i +1 maximum.Let T ( C ) be the set of cut edges of C . Then T ( C ) is not empty. Otherwise, C is2-connected. Let X = V ( C ) and Y = V ( L n ) \ V ( C ). Choose an edge xy in K n with x ∈ X and y ∈ Y . Let L ′ n be the representing graph obtained from L n by adding the edge xy and deleting the edge in L n with color c ( xy ). Then L ′ n contains a component with size s + 1, a contradiction to our choice of L n . Moreover, each edge between X and Y arecolored by T ( C ).Let L ∗ n be the set of representing graphs of K n containing a component with s verticesand t edges. Let C ∗ = \ L ′ n ∈L ∗ n { c ( e ) : e ∈ T ( L ′ n [ C ]) } . Then each edge of K n between X and Y is colored by a color in C ∗ . Otherwise, as theabove argument (for some L ′ n ∈ L ∗ n ), there is a representing graph containing a componentwith size s + 1, a contradiction. In particular, we show that C ∗ is not empty. Hence, itfollows from Lemma 9 that the pair ( c, K n [ C ]) is a good coloring with a representing graph L n [ C ] and a set of cut edges X ⊆ T ( L n [ C ]) such that C ∗ ⊆ c ( X ). Let P be a longestpath in L n [ C ] − X on k vertices. Without loss of generality, let e C be the component of L n [ C ] − X containing a copy of P k . Assume L n − X − V ( e C ) contains a path e P on k − k vertices. Since each edge between P and e P are colored by c ( X ), K n contains a rainbowcopy of P k , a contradiction. Thus L n − X − V ( e C ) is P k − k -free. If ( k − / ≤ k ≤ k − L n − X is at least | c ( X ) | + 2, by Lemma 4, wehave e ( L n ) ≤ ar ( n, k ), a contradiction to (1). If k = k −
2, then L n − X − V ( e C ) is anindependent set. In this case we can consider K n [ C ] since it contains ar ( n, k ) + 1 colors.Suppose that k < ( k − /
2. Then we consider the subgraph G of G = K n obtainedby deleting V ( C ) and the edges colored by c ( E ( C )). Then as the previous argument,by Lemma 9, the pair ( c, G [ C ]) is a good coloring with a representing graph L n [ C ] anda set of cut edges X ⊆ T ( L n [ C ]). Let e C be the component of L n [ C ] − X containinga longest path P on k vertices. If ( k − / ≤ k ≤ k −
3, then by Lemma 4, we have e ( L n ) ≤ ar ( n, k ), a contradiction (other components of L n − X − X do not contain a pathon k − k vertices). If k = k −
2, then L n − X − X − V ( e C ) is an independent set. Hencewe can consider K n [ C ∪ C ] since K n [ C ∪ C ] contains ar ( n, k ) + 1 colors. Suppose that k < ( k − /
2. We may go on this procedure and obtain X ℓ such that L ℓn [ C ℓ ] − X ℓ ( X ℓ canbe an empty set) contains a longest path P ℓ on k ℓ < ( k − / k − P ℓ has less than ( k − / L n − S ℓi =1 X i dose not containa path on at least ( k − / L n = S ℓi =1 L in [ C i ]. Moreover, the number ofcomponents of L n − S ℓi =1 X i is P ℓi =1 ( c ( X i ) + 1). Thus, since ⌈ ( k − / ⌉ − ≤ k −
3, byLemma 4, we have e ( L n ) ≤ ar ( n, k ), a contradiction to (1). If P ℓ has k − C i is a tree for 1 ≤ i ≤ ℓ −
1. Let x x be a pendent edge of C , where x is aleaf of C . If there is an edge color by c ( x x ) between V ( C ) \ { x } and L n − V ( C ),then K n − { x } contains ar ( n, k ) + 1 colors. Thus we may consider the graph K n − { x } .Hence the edges between V ( C ) \ { x } and L n − V ( C ) are not colored by c ( x x ). Inparticular, the edges between x and L n − V ( C ) are not colored by c ( x x ). Thus thereis a path on k vertices containing the edge x x , a contradiction. The proof of the claimis complete.By the claim, L n is a connected graph. We divide the proof basing on the parity of k . Case 1. k is odd, i.e., k = 2 ℓ + 1 . For k = 5 ,
7, we have ar ( n, k ) + 1 > excon( n, P k ). Hence L n contains a copy of P k , acontradiction. Let k ≥
9, i.e., ℓ ≥
4. Note that ar ( n, k ) + 1 > max { h ( n, k − , , h ( n, k − , ℓ − } . By Corollary 5 and (1), each connected representing subgraph of K n is asubgraph of H ( n, k − , A ∪ B ∪ C be the partition of H ( n, k − ,
1) in definition. Thus, since L n is connected,each vertex in B has degree one in L n . Basic calculation shows that if n > (5 ℓ − / > (5 ℓ − ℓ ) / (2 ℓ − ar ( n, k )+1 > h ( n, k − , n ≤ (5 ℓ − /
2. By (1), there are at most n − ℓ − ≤ (5 ℓ − / − ℓ < ℓ − L n inside A ∪ C . Let L ∗ n be the representing graph obtained from L n by adding an edge e inside B and deleting the edge of L n colored by c ( e ). Thus L ∗ n contains at most n − k + 1with degree one and hence L ∗ n is not a subgraph of H ( n, k − , L ∗ n is connected. Suppose that L ∗ n is not connected. Then L ∗ n contains a unique isolatedvertex, say x . Let L ∗ n − = L ∗ n −{ x } . Then we have e ( L ∗ n − ) ≥ ar ( n, k )+1 ≥ ar ( n − , k )+1.Go on the previous arguments repeatedly we can finally get a contradiction. The proof ofCase 1 is complete. Case 2. k is even, i.e., k = 2 ℓ + 2 . Note that ar ( n, k ) + 1 ≥ max { h ( n, k − ,
2) + 1 , h ( n, k − , ℓ − } . By Corollary 6 and(1), for each connected representing graph L n of K n , we have the following: • ( a ) L n is a subgraph of H ( n, k − , • ( b ) L n is a subgraph of H ( n, k − , ℓ ), or • ( c ) L n is an acyclic P -free graph when k = 6, or5 ( d ) L n = H ( n, k − , ℓ − b ), let A ∪ B ∪ C be the partition of H ( n, k − , ℓ ) in definition, i.e, | A | = ℓ , | C | = 1 and | B | = n − ℓ −
1. Let X = A and Y = B ∪ C . By (1) there are at least( ℓ − | B ∪ C | + 3 edges between X and Y . Hence, there exists a vertex y in Y with degree ℓ . Moreover, since L n [ X, Y \ { y } ] contains a subgraph on 2 ℓ vertices containing X with( ℓ − ℓ + 2 edges, by Lemma 7, L n [ X, Y \ { y } ] contains a cycle, e C , of length 2 ℓ . Let yy ′ be an edge in K n [ Y ] with y ′ / ∈ V ( e C ). Then one can find a rainbow path of length 2 ℓ + 2 in K n [ V ( e C ) ∪ { y, y ′ } ], a contradiction. For ( c ), we have e ( L n ) ≤ n −
1, contradicts (1). For( d ), let L n = H ( n, k − , ℓ − L ∗ n be representing graph obtained from L n by addingan edge e not in L n and deleting the edge colored by c ( e ) in L n . It is obviously that L ∗ n contains a copy of P k , a contradiction. Finally, let L n be a subgraph of H ( n, k − , e ( L n ) ≤ h ( n, k − , n ≤ ⌊ (5 ℓ + 3) / ⌋ for ℓ ≥
3. If k ≥
8, i.e., ℓ ≥
3, then by (1), there are at most n − ℓ − ≤ (5 ℓ +3) / − ℓ ≤ ℓ − L n inside A ∪ C . We get a contradiction similarly as Case 1. If k = 6, thenby (1) we have L n = H ( n, , K n contains a rainbowcopy of P . This final contradiction completes our proof of Theorem 1. References [1] P.N. Balister, E. Gy˝ori, J. Lehel, and R.H. Schelp, Connected graphs without longpaths,
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Proof.
The proof of Lemma 4 bases on Theorems 2 and 3 and some basic calculations.Let k ≥ k ≥ t ≥
1. Let n ≥ k − n ≥ n ≥ . . . ≥ n t ≥ n = P ti =0 n i ≥ k + k − ≥
5. Since excon( n i , P k ) ≤ n i − k ≤ n i ≥ k , the lemma holdseasily for k ≤
4. Hence, we may suppose that k ≥ Claim.
Let m = s ( k −
1) + r with s ≥ k ≥ ≤ r ≤ k −
1. Let m ≥ m ≥ . . . ≥ m t ≥ m = P ti =1 m i . Then P ti =1 ex( m i , P k ) + t − ≤ ex( m, P k ) + s. Proof.
Clearly, we have P ti =1 ex( m i , P k ) ≤ ex( m, P k ) . Thus the claim holds trivially for t − ≤ s . Now suppose that t − > s . Then we have 1 ≤ m t ≤ m t − ≤ k − . Note thatex( n, P k ) = (cid:0) n (cid:1) for n ≤ k −
1. By Theorem 2, we have ex( m t − , P k ) + ex( m t , P k ) + 1 = (cid:0) m t − (cid:1) + (cid:0) m t (cid:1) + 1 ≤ ex( m t − + m t , P k ). Thus we have t X i =1 ex( m i , P k ) + t − ≤ t − X i =1 ex( m i , P k ) + ex( m t − + m t , P k ) + t − . If t − ≤ s , then we are done. Suppose that t − > s . Repeating the above argument t − s − m , m , . . . , m t − , m t − + m t ), we have t X i =1 ex( m i , P k ) + t − ≤ s X i =1 ex( m ′ i , P k ) + ex s +2 X i = s +1 m ′ i , P k ! + s ≤ ex( m, P k ) + s, where 1 ≤ m ′ s +1 ≤ m ′ s +2 ≤ k − P s +2 i =1 m ′ i = m . The proof of the claim is complete.Let n − n = s ′ ( k −
1) + r ′ with 1 ≤ r ′ ≤ k −
1. By the claim, we haveexcon( n , P k ) + t X i =1 excon( n i , P k ) + t − ≤ excon( n , P k ) + ex( n − n , P k ) + s ′ . s = ⌊ ( k − / ⌋ . Since k ≥
3, we have s ≤ ℓ −
1. We will finish our proof in thefollowing two cases.
Case 1. k + k − k + k − k = 2 ℓ + 1. Basic calculation shows that ar ( n, k ) = h ( k, k − , − n ≤ (5 ℓ − / ar ( n, k ) = h ( n, k − , ℓ −
1) for n ≥ (5 ℓ − /
2. Let n ≤ ⌊ (5 ℓ − / ⌋ .We divide the proof into the following three subcases:(a.1) excon( n , P k ) = (cid:0) k − (cid:1) , i.e, n = k −
1. By Theorem 2 and a detailed calculation,we haveexcon( n , P k ) + ex( n − n , P k ) + s ′ = (cid:18) k − (cid:19) + ex( n − k + 1 , P k ) + s ′ ≤ (cid:18) k − (cid:19) + ex (cid:18)(cid:22) ℓ − (cid:23) − k + 1 , P k (cid:19) + s ′ ≤ (cid:18) k + k − (cid:19) + 1 = h ( k, k − , − . (a.2) excon( n , P k ) = h ( n , k , k ≥
5, calculations in [1] show that if k iseven, then k ≤ n ≤ (5 k − / k is odd, then k ≤ n ≤ (5 k − /
4. Hence,we haveexcon( n , P k ) + ex( n − n , P k ) + s ′ = (cid:18) k − (cid:19) + ( n − k + 2) + ex( n − n , P k ) + s ′ < (cid:18) k − (cid:19) + ex( n − k + 1 , P k ) + s ′ < (cid:18) k + k − (cid:19) + 1 = h ( k, k − , − , where the last strict inequality holds similarly as before.(a.3) excon( n , P k ) = h ( n , k , s ), i.e, n ≥ (5 k − / k and n ≥ (5 k − / k . Let i = 1 when k is odd and i = 0 when k is even. Thenexcon( n , P k ) + ex( n − n , P k ) + s ′ ≤ (cid:18) s (cid:19) + s ( n − s ) + i + ex (cid:18) ℓ − − n , P k (cid:19) + s ′ 1, where s ′ = ⌈ n ′ / ( k − ⌉ − s ′ = ⌈ ( n ′ − / ( k − ⌉ − 1. Wedivide the proof into the following three subcases:(b.1) excon( n , P k ) = (cid:0) k − (cid:1) , i.e, n = k − 1. Thenexcon( n , P k ) + ex( n − n , P k ) + s ′ = (cid:18) k − (cid:19) + s ′ (cid:18) k − (cid:19) + (cid:18) r ′ (cid:19) + s ′ 1, we have excon( n , P k ) + ex( n − n , P k ) + s ′ = (cid:18) s (cid:19) + s ( n − s ) + i + s ′ (cid:18) k − (cid:19) + (cid:18) r ′ (cid:19) + s ′ 2. Let n ≤ ⌊ (5 ℓ + 2) / ⌋ .We divide the proof into the following three subcases:(a.1) excon( n , P k ) = (cid:0) k − (cid:1) , i.e, n = k − 1. Then we haveexcon( n , P k ) + ex( n − n , P k ) + s ′ = (cid:18) k − (cid:19) + ex( n − k + 1 , P k ) + s ′ ≤ (cid:18) k − (cid:19) + ex (cid:18)(cid:22) ℓ + 22 (cid:23) − k + 1 , P k (cid:19) + s ′ < (cid:18) k + k − (cid:19) + 1 = h ( k, k − , − . (a.2) excon( n , P k ) = h ( n , k , n ≤ (5 k − / k and n ≤ (5 k − / k . Thenexcon( n , P k ) + ex( n − n , P k ) + s ′ = (cid:18) k − (cid:19) + ( n − k + 2) + ex( n − n , P k ) + s ′ < (cid:18) k + k − (cid:19) + 1 = h ( k, k − , − . (a.3) excon( n , P k ) = h ( n , k , s ), i.e, n ≥ (5 k − / k and n ≥ (5 k − / k . Let i = 1 when k is odd and i = 0 when k is even. Thenexcon( n , P k ) + ex( n − n , P k ) + s ′ ≤ (cid:18) s (cid:19) + s ( n − s ) + i + ex (cid:18)(cid:22) ℓ + 22 (cid:23) − n , P k (cid:19) + s ′ 1. Thenexcon( n , P k ) + ex( n − n , P k ) + s ′ = (cid:18) k − (cid:19) + ex( n − k + 1 , P k ) + s ′ = (cid:18) k − (cid:19) + s ′ (cid:18) k − (cid:19) + (cid:18) r ′ (cid:19) + s ′ < h ( n, k − , ℓ − − , where the strict inequality holds by n ≥ ⌈ (5 ℓ + 2) / ⌉ .9b.2) excon( n , P k ) = h ( n , k , n ≤ (5 k − / k and n ≤ (5 k − / k . Thenexcon( n , P k ) + ex( n − n , P k ) + s ′ = (cid:18) k − (cid:19) + ( n − k + 2) + ex( n − n , P k ) + s ′ < (cid:18) ℓ − (cid:19) + ( ℓ − n − ℓ + 1) + 1 = h ( n, k − , ℓ − − , where the strict inequality holds by n ≥ ⌈ (5 ℓ + 2) / ⌉ .(b.3) excon( n , P k ) = h ( n , k , s ), i.e, n ≥ (5 k − / k and n ≥ (5 k − / k . Let i = 1 when k is odd and i = 0 when k is even. Recall that s ≤ ℓ − 1, we haveexcon( n , P k ) + ex( n − n , P k ) + s ′ = (cid:18) s (cid:19) + s ( n − s ) + i + ex( n − n , P k ) + s ′ < (cid:18) ℓ − (cid:19) + ( ℓ − n − ℓ + 1) + 1 = h ( n, k − , ℓ − − , where the strict inequality holds by n ≥ ⌈ (5 ℓ + 2) / ⌉⌉