On the Edit Distance of Powers of Cycles
aa r X i v : . [ m a t h . C O ] S e p On the Edit Distance of Powers of Cycles
Zhanar Berikkyzy Ryan R. Martin Chelsea Peck
Department of MathematicsIowa State UniversityAmes, Iowa, U.S.A { zhanarb,rymartin } @iastate.edu August 30, 2015
Mathematics Subject Classifications: Primary 05C35; Secondary 05C80
Abstract
The edit distance between two graphs on the same labeled vertex set is definedto be the size of the symmetric difference of the edge sets. The edit distance functionof a hereditary property H is a function of p ∈ [0 ,
1] that measures, in the limit, themaximum normalized edit distance between a graph of density p and H .In this paper, we address the edit distance function for Forb( H ), where H = C th ,the t th power of the cycle of length h . For h ≥ t ( t + 1) + 1 and h not divisibleby t + 1, we determine the function for all values of p . For h ≥ t ( t + 1) + 1 and h divisible by t + 1, the function is obtained for all but small values of p . We alsoobtain some results for smaller values of h . Keywords: edit distance, colored regularity graphs, powers of cycles
The edit distance in graphs was introduced independently by Axenovich, K´ezdy, andMartin [2] and by Alon and Stav [1]. The question considered is “Given a class of graphs H what is the minimum number m = m ( n ) such that for every graph on n vertices, thereis a set of m edge-additions and edge-deletions that ensure the resultant graph is a memberof H ?” For every hereditary property H , there is a p = p ( H ) such that the Erd˝os-R´enyirandom graph G ( n, p ) is asymptotically extremal [1]. A hereditary property is a family ofgraphs that is closed under isomorphism and the taking of induced subgraphs.The edit distance function of a hereditary property H is a function of p ∈ [0 ,
1] thatmeasures, in the limit, the maximum normalized edit distance between a graph of density p and H . A principal hereditary property , denoted Forb( H ), is a hereditary property thatconsists of the graphs with no induced copy of a single graph H . Most of the known editdistance functions are of the form Forb( H ). These include the cases where H is a split1raph [7] (including cliques and independent sets), complete bipartite graphs K ,t [9] and K , [5] and cycles C h where h is small [6]. In this paper, we compute the edit distancefunction for powers of cycles.For positive integers t and h , the t th power of a cycle of length h is denoted C th and hasvertex set { , . . . , h } , where two vertices are adjacent in C th if and only if their distanceis at most t in C h .The notation in this paper primarily comes from [8]. The edit distance between graphs G and G ′ on the same labeled vertex set is denoted dist( G, G ′ ) and satisfies dist( G, G ′ ) = | E ( G ) △ E ( G ′ ) | . The edit distance between a graph G and a hereditary property H isdist( G, H ) = min { dist( G, G ′ ) : V ( G ) = V ( G ′ ) , G ′ ∈ H} . The edit distance function of a hereditary property H measures the maximum distanceof a density p graph from H , i.e.ed H ( p ) = lim n →∞ max { dist( G, H ) : | V ( G ) | = n, | E ( G ) | = ⌊ p (cid:0) n (cid:1) ⌋} / (cid:0) n (cid:1) . Balogh and Martin [3] showed that this limit exists and is equal to lim n →∞ E [dist( G ( n, p ) , H )],using the result of Alon and Stav in [1]. The function has a number of interesting prop-erties: Proposition 1 (Balogh-Martin [3]) . If H is a hereditary property, then ed H ( p ) is contin-uous and concave down over p ∈ [0 , . By the proposition above, the function ed H achieves its maximum in [0 , d ∗H , and the the set of all values of p for which the maximum isachieved by p ∗H .A colored regularity graph (CRG) , K , is a complete graph with a partition of thevertices into white VW( K ) and black VB( K ), and a partition of the edges into whiteEW( K ), gray EG( K ), and black EB( K ). We say that a graph H embeds in K , denoted H K , if there is a function ϕ : V ( H ) → V ( K ) so that if h h ∈ E ( H ), then either ϕ ( h ) = ϕ ( h ) ∈ VB( K ) or ϕ ( h ) ϕ ( h ) ∈ EB( K ) ∪ EG( K ), and if h h / ∈ E ( H ), theneither ϕ ( h ) = ϕ ( h ) ∈ VW( K ) or ϕ ( h ) ϕ ( h ) ∈ EW( K ) ∪ EG( K ).Given a hereditary property H , it is easy to see that it can be expressed as H = T { Forb( H ) : H ∈ F ( H ) } for some family of graphs F ( H ). We denote K ( H ) to be thesubset of CRGs such that no forbidden graph embeds into them, i.e. K ( H ) = { K : H K, ∀ H ∈ F ( H ) } . In our case, K ( H ) = { K : H K } for H = Forb( H ) and so a CRG K is a sub-CRG of ˜ K if K can be obtained by deleting vertices of ˜ K .For every CRG K we associate a function g on [0 ,
1] defined by g K ( p ) = min { x T M K ( p ) x : x T = 1 , x ≥ } , (1)where [ M K ( p )] ij = p, if v i v j ∈ EW( K ) or v i = v j ∈ VW( K );1 − p, if v i v j ∈ EB( K ) or v i = v j ∈ VB( K );0 , if v i v j ∈ EG( K ). (2)2he g function of CRGs can be used to compute the edit distance function. Balogh andMartin [3] proved that ed H ( p ) = inf K ∈K ( H ) g K ( p ) and Marchant and Thomason [5] furtherproved that the infimum is achieved by some K , i.e. ed H ( p ) = min K ∈K ( H ) g K ( p ). So, for every p ∈ [0 , K ∈ K ( H ) such that ed H ( p ) = g K ( p ). It is also shown in [5]that in order to find such CRG we only need to look at so called p -core CRGs. A CRG˜ K is p -core if g ˜ K ( p ) < g K ( p ) for every sub-CRG K of ˜ K .The CRG with r white vertices, s black vertices and all edges gray is denoted K ( r, s ).The clique spectrum of the hereditary property H = Forb( H ), denoted Γ( H ), is the set ofall pairs ( r, s ) such that H K ( r, s ). It is easy to see that, for any hereditary property H its clique spectrum Γ = Γ( H ) can be expressed as a Ferrers diagram. That is, if r ≥ r, s ) ∈ Γ, then ( r − , s ) ∈ Γ and if s ≥ r, s ) ∈ Γ, then ( r, s − ∈ Γ. An extreme point of a clique spectrum Γ is a pair ( r, s ) ∈ Γ such that ( r + 1 , s ) and ( r, s + 1)do not belong to Γ. The set of all extreme points of Γ is denoted by Γ ∗ .Define the function γ H ( p ) = min { g K ( r,s ) ( p ) : ( r, s ) ∈ Γ( H ) } . Clearly, ed H ( p ) ≤ γ H ( p ).Moreover, one only need consider the extreme points rather than whole Γ itself, that is, γ H ( p ) = min { g K ( r,s ) ( p ) : ( r, s ) ∈ Γ ∗ ( H ) } .In this paper, the hereditary properties we consider are of the form H = Forb( C th ).Since Theorem 20 in [8] gives ed Forb( K h ) ( p ) = p/ ( h − h ≥ t + 2.For convenience, we denote ℓ a = (cid:6) ht + a +1 (cid:7) , for a ∈ { , . . . , t } . We also denote p = ℓ t − .The motivation for these values will be discussed in Section 5.The main results of this paper are Theorems 2 and 3. Theorem 2.
Let t ≥ and h ≥ max { t ( t + 1) , } be integers. For all p ∈ [0 , , γ H ( p ) = min a ∈{ , ,...,t } (cid:26) p (1 − p ) a (1 − p ) + ( ℓ a − p (cid:27) , if ( t + 1) | h ; γ H ( p ) = min a ∈{ , ,...,t } (cid:26) pt + 1 , p (1 − p ) a (1 − p ) + ( ℓ a − p (cid:27) , if ( t + 1) |6 h . Note: If a = 0, then p (1 − p ) a (1 − p )+( ℓ a − p = p (1 − p )( ℓ − p , which we define to be − pℓ − at p = 0. Theorem 3.
Let t ≥ and h ≥ t ( t + 1) + 1 be positive integers and let H = Forb( C th ) .If ( t + 1) |6 h , then for ≤ p ≤ , and if ( t + 1) | h then for p ≤ p ≤ , we have that ed H ( p ) = γ H ( p ) . (3) Corollary 4.
Let h ≥ be a positive integer and H = Forb( C h ) . • If h is even, then for ⌈ h/ ⌉ − ≤ p ≤ , ed H ( p ) = min (cid:26) p (1 − p )1 − p + ( ⌈ h/ ⌉ − p , − p ⌈ h/ ⌉ − (cid:27) . If h is odd, then for ≤ p ≤ , ed H ( p ) = min (cid:26) p , p (1 − p )1 − p + ( ⌈ h/ ⌉ − p , − p ⌈ h/ ⌉ − (cid:27) . It was shown in [6] and in [5], respectively, thated
Forb( C ) ( p ) = p/ Forb( C ) ( p ) = p (1 − p ) . It follows from this and the above corollary that when t = 1, the furthest graph fromForb( C h ) is a graph which has density p ∗ = 1 / ( ⌈ h/ ⌉ − ⌈ h/ ⌉ + 1) when h ≥ h
6∈ { , , , , } , and has density p ∗ = 1 / (1 + p ⌈ h/ ⌉ −
1) when h ∈ { , , , , } .Also, observe that the maximum value of the edit distance function can be an irrationalnumber.Our proof techniques often require us to compare the g function of a CRG to one ofthe individual functions that are given in Theorem 2. However, when h is large enoughat most 3 of these functions are necessary to define γ H . Corollary 5.
Let t ≥ and h ≥ t + 10 t + 24 be positive integers. Recall ℓ t = (cid:6) h t +1 (cid:7) and p = ℓ − t . Then • If ( t + 1) | h , then for p ≤ p ≤ , ed H ( p ) = min (cid:26) p (1 − p ) t (1 − p ) + ( ℓ t − p , − pℓ − (cid:27) . • If ( t + 1) |6 h , then for ≤ p ≤ , ed H ( p ) = min (cid:26) pt + 1 , p (1 − p ) t (1 − p ) + ( ℓ t − p , − pℓ − (cid:27) . The rest of the paper is organized as follows: Section 2 provides some definitions andbasic results, Section 3 gives the proof of Theorem 2, Section 4 gives Lemma 13 whichis the key lemma for the proof of Theorem 3, Section 5 gives the proof of Theorem 3,Section 6 gives the proofs of some helpful lemmas and facts, and Section 7 gives someconcluding remarks.
All graphs considered in this paper are simple. For standard graph theory notation pleasesee [12], for the edit distance notation please see [8]. A sub-CRG K ′ of a CRG K is a component if it is maximal with respect to the property that, for all v, w ∈ V ( K ′ ), thereexists a path consisting of white and black edges entirely within K ′ . It is easy to computethe g function of a CRG given the g function of its components:4 roposition 6 ([6]) . Let K be a CRG with components K (1) , . . . , K ( r ) and p ∈ [0 , .Then ( g K ( p )) − = P ri =1 ( g K ( i ) ( p )) − . Note that by Proposition 6, g K ( r,s ) ( p ) = (cid:16) rp + s − p (cid:17) − .Let K be a CRG, v ∈ V ( K ), and let x be an optimal solution to the quadraticprogram (1). The weight of v , denoted x ( v ), is the entry corresponding to v of the vector x . We say that w ∈ V ( K ) is a gray neighbor of v ∈ V ( K ) if w is adjacent to v via a grayedge. White and black neighbors are defined analogously. The set of all gray neighbors of v is denoted by N G ( v ) and the number of vertices adjacent to v via gray edges is denotedby deg G ( v ), i.e. deg G ( v ) = | N G ( v ) | .In contrast, the gray degree of v , denoted d G ( v ), is the sum of the weights of grayneighbors of v , i.e. d G ( v ) = P { x ( w ) : w ∈ N G ( v ) } . Similarly, the white degree of v ,denoted d W ( v ), is the sum of the weights of the white neighbors of v plus the weight of v if and only if it is a white vertex. The black degree of v , denoted d B ( v ), is the sum of theweights of the black neighbors of v plus the weight of v if and only if it is a black vertex.So, d G ( v ) + d W ( v ) + d B ( v ) = 1 for all v ∈ V ( K ).The number of common gray neighbors of vertices v and w is denoted by deg G ( v, w ).The gray codegree of vertices v and w , denoted d G ( v, w ), is the sum of the weights of thecommon gray neighbors of v and w . For a set of vertices { v , v , . . . , v ℓ } , we say v v · · · v ℓ is a gray path if v i v i +1 ∈ EG( K ) for i = 1 , . . . , ℓ −
1. Analogously, we say v v · · · v ℓ v is a gray cycle if v v ℓ ∈ EG( K ) and v i v i +1 ∈ EG( K ) for i = 1 , . . . , ℓ −
1. Proposition 7 givesa structural classification of p -core CRGs. Proposition 7 (Marchant-Thomason [5]) . Let K be a p -core CRG. • If p = 1 / , then all of the edges of K are gray. • If p < / , then EB( K ) = ∅ and there are no white edges incident to white vertices. • If p > / , then EW( K ) = ∅ and there are no black edges incident to black vertices. Proposition 8 gives a formula for d G ( v ) for all v ∈ V ( K ) and Proposition 9 uses thisto give a bound on the weight of each v . Proposition 8 ([6]) . Let p ∈ (0 , and K be a p -core CRG with optimal weight function x . • If p ≤ / , then x ( v ) = g K ( p ) / (1 − p ) for all v ∈ VW( K ) , and d G ( v ) = p − g K ( p ) p + 1 − pp x ( v ) , for all v ∈ VB( K ) . • If p ≥ / , then x ( v ) = g K ( p ) /p for all v ∈ VB( K ) , and d G ( v ) = 1 − p − g K ( p )1 − p + 2 p − − p x ( v ) , for all v ∈ VW( K ) . roposition 9 ([6]) . Let p ∈ (0 , and K be a p -core CRG with optimal weight func-tion x . • If p ≤ / , then x ( v ) ≤ g K ( p ) / (1 − p ) for all v ∈ VB( K ) . • If p ≥ / , then x ( v ) ≤ g K ( p ) /p for all v ∈ VW( K ) . γ H func-tion In this section we compute the γ H function, which gives an upper bound for the editdistance function. Recall that for any t ≥ h ≥ t + 2 and a ∈ { , . . . , t } , we denote ℓ a = (cid:6) ht + a +1 (cid:7) . Proof of Theorem 2.
Our first observation is the value of the chromatic number of C th ,denoted χ ( C th ). Proposition 10 (Prowse-Woodall [11]) . Let t ≥ and h ≥ max { t + 1 , } be positiveintegers. Let h = q ( t + 1) + r , where r ∈ { , . . . , t } . Then, χ ( C th ) = t + ⌈ r/q ⌉ + 1 . Inparticular, if h ≥ max { t ( t + 1) , } , then χ ( C th ) = (cid:26) t + 1 , if ( t + 1) | h ; t + 2 , if ( t + 1) |6 h . Let h ≥ max { t ( t + 1) , t + 2 } and χ = χ ( C th ). Denote the vertices of C th by { , . . . , h } such that distinct i and j are adjacent if and only if | i − j | ≤ t (mod h ). For each a ∈ { , . . . , t } , we first show that ( a, ℓ a − ∈ Γ = Γ(Forb( C th )) and then show that( a, ℓ a ) Γ. We will also show that if χ > t + 1 then { ( t + 1 , , . . . , ( χ − , } ⊂ Γ butthat ( t + 1 , Γ.This will imply that Γ ∗ ⊆ { ( a, ℓ a −
1) : a = 0 , , . . . , t }∪{ ( χ − , } , which is a strongerresult than we need. Case 1: a ∈ { , . . . , t } . First, we show that ( a, ℓ a − ∈ Γ. By contradiction, assume there is a partition of V ( C th ) into a independent sets and ℓ a − k = ℓ a −
1, and let C , . . . , C k bethe cliques. We may assume that the vertices in each C i are consecutive. This is becauseif j and j are in the same clique, then by the nature of adjacency in the power of a cycle,every vertex between j and j is adjacent to every member of the clique, and hence canbe added to the clique. Thus, | C i | ≤ t + 1 for i = 1 , . . . , k .For i = 1 , . . . , k −
1, let B i be the set of vertices between C i and C i +1 , and let B k be the set of vertices between C k and C . The sets B i might or might not be empty. Ifsome | B i | ≥ a + 1, then the first a + 1 ≤ t + 1 vertices form a clique and so must be indifferent independent sets, which is not possible since there are only a independent sets.Therefore, | B i | ≤ a for i = 1 , . . . , k . 6onsequently, we need k ( t + a + 1) ≥ h in order to cover C th with a independent setsand k cliques. Hence, k ≥ ℓ a , a contradiction to our choice of k . Thus ( a, ℓ a − ∈ Γ for a = 0 , . . . , t .Next, we show that ( a, ℓ a ) Γ. Again, let k = ℓ a −
1. For i = 1 , . . . , k , let S i = { ( i − t + a + 1) + 1 , . . . , i ( t + a + 1) } and let S k +1 = { , . . . , h } − ∪ ki =1 S i . For i = 1 , . . . , k ,let C i be the first t + 1 vertices of S i and let C k +1 be the first min { t + 1 , | S k +1 |} verticesof S k +1 . For j = 1 , . . . , a , let A j consist of the ( t + 1 + j ) th vertex of S , . . . , S k and the( t + 1 + j ) th vertex of S k +1 if | S k +1 | ≥ t + 1 + j .The sets ( A , . . . , A a , C , . . . , C k +1 ) form a partition of V ( C th ). Clearly each C i , i =1 , . . . , k , is a clique of size t + 1 and since there is a clique of size t + 1 between pairs ofvertices in each A j , each A j is an independent set. Thus ( a, ℓ a ) Γ for a = 0 , . . . , t . Case 2: a ≥ t + 1 . If ( t +1) | h , then Proposition 10 gives that C th can be partitioned into t +1 independentsets and so ( t +1 , Γ. If ( t +1) |6 h , then Proposition 10 gives that χ ≥ t +2 and since C th cannot be partitioned into fewer than χ independent sets, we have ( t +1 , , . . . , ( χ − , ∈ Γ. Since C th can be partitioned into χ independent sets, ( χ, Γ.Finally, let k = ⌈ h/ ( t + 1) ⌉ −
1. For j = 1 , . . . , t + 1, let A j = { ( i − t + 1) + j : i =1 , . . . , k } . Let C = { k ( t + 1) + 1 , . . . , h } . The sets ( A , . . . , A t +1 , C ) form a partition of V ( C th ). Clearly, C is a clique of size at most t + 1 and since there are at least t verticesbetween pairs of vertices in each A j , each A j is an independent set. Thus ( t + 1 , Γ.Using Proposition 6, if h = q ( t + 1) + r where r ∈ { , . . . , t } , then γ H ( p ) = min a ∈{ , ,...,t } (cid:26) p (1 − p ) a (1 − p ) + ( ℓ a − p (cid:27) , if r = 0; γ H ( p ) = min a ∈{ , ,...,t } (cid:26) pt + ⌈ r/q ⌉ , p (1 − p ) a (1 − p ) + ( ℓ a − p (cid:27) , if r = 0.Restricting ourselves to h ≥ min { t ( t + 1) , } , we have the result in the statement ofthe theorem. Before we can prove Theorem 3, we need to study the properties of the CRGs into which C th does not embed. Recall that we may assume h ≥ t + 2. An important property ofsuch CRGs is that the set of lengths of gray cycles on black vertices is restricted, as isshown in Lemma 13. Its proof needs the technical inequalities in Facts 11 and 12. Forcompleteness, we give their proofs in Section 6. Fact 11.
Let h, x, y be positive integers. Then(a) ⌊ h/x ⌋ ≥ y if and only if ⌊ h/y ⌋ ≥ x .(b) ⌈ h/x ⌉ ≤ y if and only if ⌈ h/y ⌉ ≤ x . act 12. Let t ≥ , h ≥ max { t ( t − , t + 2 } , and a ∈ { , . . . , t − } be positive integers.Then (cid:6) ht + a +1 (cid:7) ≤ (cid:4) ht (cid:5) . Lemma 13 is a key lemma in proving our main result of Theorem 3.
Lemma 13.
Let p ∈ (0 , / and let t ≥ and h ≥ t + 2 be integers. Let ˜ K be a p -core CRG with exactly a white vertices such that C th ˜ K . Let K be the sub-CRG of ˜ K induced by the set of all black vertices of ˜ K . Then, the following occurs:(a) If a ∈ { , . . . , t − } and h ≥ t − t , then K has no gray cycle which has length in (cid:8)(cid:6) ht + a +1 (cid:7) , . . . , (cid:4) ht (cid:5)(cid:9) .(b) If a = t , then | V ( K ) | ≤ ℓ t − .(c) If a ≥ t + 1 , then ( t + 1) |6 h and V ( K ) = ∅ . Note:
We interpret a gray cycle of length 2 to be a gray edge.
Proof of Lemma 13.
Denote the vertices of C th by { , . . . , h } such that distinct i and j are adjacent if and only if | i − j | ≤ t (mod h ). Partition:
Let K have a gray cycle on vertex set { v , . . . , v k } such that v i v i +1 is a grayedge, where the indices are taken modulo k . We describe a partition of V ( C th ), whichgives an interval of forbidden gray cycle lengths. We will construct at most a independentsets and k cliques C , . . . , C k such that there is no edge between nonconsecutive cliques.Partition V ( C th ) into k sets of consecutive vertices S , . . . , S k , with each set S i of sizeeither ⌈ h/k ⌉ or ⌊ h/k ⌋ . We will eventually construct the at most a independent sets and k cliques C , . . . , C k with C i ⊆ S i such that there is no edge between C i and C i ′ unless | i − i ′ | = 1 (mod k ).If a = 0, then simply let C i = S i for i = 1 , . . . , k . Using Fact 14, each C i has sizeat least t and so nonconsecutive sets have no edge between them. Fact 14 is a simpleobservation of number theory. Fact 14.
A set of size h can be partitioned into sets of size t or t + 1 if and only if h ≥ t ( t − . Moreover, for any k ∈ {⌈ h/ ( t + 1) ⌉ , . . . , ⌊ h/t ⌋} , such a partition exists withexactly k parts. So, we assume a ≥ a ′ ∈ {⌊ h/k ⌋ − t, ⌈ h/k ⌉ − ( t + 1) } such that 0 ≤ a ′ ≤ a .This is possible as long as both (a) 0 ≤ ⌊ h/k ⌋ − t and (b) ⌈ h/k ⌉ − ( t + 1) ≤ a . (Thisis only nontrivial if k | h , in which case at least one of the two choices of a ′ will be in0 , . . . , a .)If a ′ = 0, again let C i = S i for i = 1 , . . . , k . If a ′ ≥
1, let A j consist of the j th vertex ofeach of S , . . . , S k and let C i = S i − ∪ a ′ j =1 A j . Observe that if a ′ ≥
1, then | S i | ≥ t + 1 andso there are at least t vertices between each pair of vertices in every A j . Therefore, A j isan independent set for j = 1 , . . . , a ′ . We have | C i | ≤ t + 1 so C i is a clique for i = 1 , . . . , k .In addition, | C i | ≥ t and so there are no edges between C i and C i ′ unless | i − i ′ | (mod k ).8he mapping, for all a ≥
0, is as follows: Map each A j to a different white vertexand C i to v i for i = 1 , . . . , k . If a = 0, Fact 14 gives that K has no cycle with length in {⌈ h/ ( t + 1) ⌉ , . . . , ⌊ h/t ⌋} . If a ≥
1, Fact 11, gives that K has no cycle with length in (cid:26)(cid:24) ht + a + 1 (cid:25) , . . . , (cid:22) ht (cid:23)(cid:27) , (4)and (4) is valid in the case of a = 0 also. Case (a) . The result is given by (4). It suffices to show that (cid:6) ht + a +1 (cid:7) ≤ (cid:4) ht (cid:5) . Fact 12 gives thatthis holds if h ≥ t − t . Case (b) . In this case, we use a second partition. Partition V ( C th ) into k + 1 consecutive parts, S , . . . , S k +1 , where k = ⌈ h/ (2 t + 1) ⌉ − r = h − ( k − t + 1). Since h ≥ t + 2, k ≥
1. Let | S | = · · · = | S k − | = 2 t + 1, | S k | = ⌈ r/ ⌉ and | S k +1 | = ⌊ r/ ⌋ . Note that t + 1 ≤ | S k +1 | ≤ | S k | ≤ t + 1.For j = 1 , . . . , t , let A j consist of the j th vertex in each part and let C i = S i − S tj =1 A j .Each of A , . . . , A t is an independent set. Furthermore, there are no edges between C i and C i ′ if i = i ′ . Therefore, K has at most k = ⌈ h/ (2 t + 1) ⌉ − A , . . . , A t can be mapped arbitrarily to each of the t white vertices and C , . . . , C k +1 canbe mapped arbitrarily to k + 1 different black vertices in K . Case (c) . If ( t + 1) | h , then χ ( C th ) = t + 1 and ˜ K having at least t + 1 white vertices meansthat C th embeds in ˜ K , a contradiction. If ( t + 1) |6 h , then partition V ( C th ) into k = ⌊ h/ ( t + 1) ⌋ + 1 parts S , . . . , S k of consecutive vertices, each of S , . . . , S k − of size t + 1.For j = 1 , . . . , t + 1, let A j consist of the j th vertex in each S i for i = 1 , . . . , k −
1. Thegraph induced by V ( C th ) − S t +1 j =1 A j forms a clique of size at most t in S k . Since all verticesin K are black, this clique will embed into any vertex of V ( K ). Thus V ( K ) = ∅ . ed H = γ H We will use Lemma 13 to prove Theorem 3. Recall that h ≥ t ( t + 1) + 1 ≥ t ( t + 1). ByProposition 10, this means χ ( C th ) = t + 1 if ( t + 1) | h and χ ( C th ) = t + 2 if ( t + 1) |6 h . Proof of Theorem 3.
By definition ed H ( p ) ≤ γ H ( p ) for all p ∈ [0 , Case 1: p ∈ [1 / , γ H ( p ) = − pℓ − for p ∈ [1 / , act 15. Let h and t be positive integers. If h ≥ ( t + 1) + 1 , then − pℓ − ≤ pt + 1 . For a ∈ { , . . . , t } if h ≥ ( t + 1)( t + a ) + 1 , then for all p ∈ [1 / , , − pℓ − ≤ p (1 − p ) a (1 − p ) + ( ℓ a − p . Note:
The condition h ≥ t ( t + 1) + 1 suffices to achieve all of the conclusions in Fact 15.By Proposition 16 below, ed H ( p ) = γ H ( p ) for the two values of p ∈ { / , } . Proposition 16 (Balogh-Martin [3]) . If H is a hereditary property, then ed H (1 /
2) = γ H (1 / . Moreover, if K ℓ ∈ H for all positive integers ℓ , then ed H (1) = γ H (1) = 0 and if K ℓ ∈ H for all positive integers ℓ , then ed H (0) = γ H (0) = 0 . We have ed H ( p ) ≤ γ H ( p ) and the two functions are equal at p = 1 / p = 1.The function γ H ( p ) is linear over p ∈ [1 / ,
1] for h ≥ t ( t + 1) + 1. By Proposition 1,ed H ( p ) is continuous and concave down, so we may conclude that ed H ( p ) = γ H ( p ) = − pℓ − for p ∈ [1 / , H (0) = γ H (0) = 0. Let p ∈ (0 , /
2) and ed H ( p ) = g ˜ K ( p ) for some p -core CRG ˜ K . Assume by contradiction that g ˜ K ( p ) < γ H ( p ). Suppose ˜ K has a white vertices. Recall that for any t ≥ h ≥ t + 2 and a ∈ { , . . . , t } , we denote ℓ a = (cid:6) ht + a +1 (cid:7) . We consider several cases and show that we arrive at a contradiction ineach case. Case 2: a ≥ t and p ∈ (0 , / a ≥ t + 1, then by Lemma 13 (c) , V ( K ) = ∅ . As long as h ≥ max { t ( t + 1) , } ,Proposition 10 gives that χ ( C th ) ≤ t + 2 with equality only if ( t + 1) |6 h . Thus, a = t + 1and Proposition 6 gives that g ˜ K ( p ) = p/ ( t + 1), a contradiction to the assumption that g ˜ K ( p ) < γ ˜ K ( p ).If a = t , then Case (b) of Lemma 13 gives that | V ( K ) | ≤ ℓ t −
1. Consequently, g K ( p ) ≥ − pℓ t − . We can partition ˜ K into t + 1 sub-CRGs, K and t white vertices, and useProposition 6 to conclude that( g ˜ K ( p )) − ≤ tp − + (cid:18) − pℓ t − (cid:19) − g ˜ K ( p ) ≥ p (1 − p ) t (1 − p ) + ( ℓ t − p Hence, ed H ( p ) ≥ γ H ( p ), again a contradiction. This concludes Case 2.10 ase 3: a ≤ t − p ∈ (0 , / K is a CRG with a white vertices, with 0 ≤ a ≤ t −
2. By Proposition 6, g ˜ K ( p ) − = ap − + g − K ( p ). Therefore, g K ( p ) < (cid:18) max a ′ ∈{ , ,...,t } (cid:26) a ′ − ap + ℓ a ′ − − p (cid:27)(cid:19) − =: g ( a, t ; p ) . (5)Given our assumptions on g K ( p ), Lemma 17 gives lower bounds on the gray degree ofvertices and the codegree of pairs of vertices. Recall that deg G ( v ) denotes the number ofgray neighbors of v ∈ V ( K ). Lemma 17.
Let p ∈ (0 , / , t ≥ be an integer and a ∈ { , . . . , t − } . Let p = ℓ − t = (cid:6) h t +1 (cid:7) − . Let K be a p -core CRG with all black vertices such that g K ( p ) < g ( a, t ; p ) .Then(a) for every v ∈ V ( K ) , we have deg G ( v ) ≥ ℓ a +1 , and(b) for every v, w ∈ V ( K ) , deg G ( v, w ) ≥ (cid:26) ℓ a +2 , if a ≤ t − ; , if a = t − and p ≥ p . Note:
Since h ≥ t + 2, it is the case that ℓ a +1 ≥ a ≤ t − ℓ a +2 ≥ a ≤ t − F with vertex set V ( K ) and edge set EG( K ).Using Lemma 17, the lower bound on the number of common gray neighbors of v and w gives a structural restriction on this graph. Note that the length of a path is defined tobe the number of vertices in said path. Lemma 18.
Fix integers t ≥ , h ≥ max { t ( t − , t + 2 } and a ∈ { , . . . , t − } . Recallthat ℓ a = ⌈ h/ ( t + a + 1) ⌉ and let L = ⌊ h/t ⌋ .Let F be a graph with no cycle with length in { ℓ a , . . . , L } and every pair of verticeseither has at least ℓ a +2 ≥ common neighbors if a ≤ t − or has at least commonneighbor if a = t − .Then F has no cycle of length more than ℓ a − . Now we consider a maximum-length path in the graph F . If such a path can be madeinto a cycle, then Proposition 19 gives that F must be Hamiltonian. By Lemma 18, thismeans that | V ( K ) | ≤ ℓ a − g K ( p ) ≥ − pℓ a − , which is the g function for theCRG on ℓ a − a ′ = a . Proposition 19 is a common argument in proofs of Hamiltoniancycle results, including the classical theorems of Dirac [4] and Ore [10]. Proposition 19.
Let F be a connected graph. If some path of maximum length forms acycle, then F is Hamiltonian.
11o we may assume that every maximum-length path in F is not a cycle. Let v · · · v ℓ be such a maximum length path. The common neighbors of v and v ℓ in F must be onthis path, otherwise F has a longer path. From Lemma 17, it follows that v and v ℓ haveat least ℓ a +2 ≥ Lemma 20.
Fix integers t ≥ , h ≥ t + 2 and a ∈ { , . . . , t − } . Recall that ℓ a = ⌈ h/ ( t + a + 1) ⌉ . Let F be a graph with no cycle of length longer than ℓ a − , with everyvertex having degree at least ℓ a +1 ≥ and with every pair of vertices having at least onecommon neighbor. Furthermore, let F have the property that no maximum length pathforms a cycle.Let v · · · v ℓ be a path of maximum length in F . Then v and v ℓ have exactly onecommon neighbor v c on this path. Furthermore, N ( v ) ⊆ { v , . . . , v c } and N ( v ℓ ) ⊆{ v c , . . . , v ℓ } . This concludes Case 3.
Case 4: a = t − p ∈ [ p , / K is a CRG with a = t − g − K ( p ) =( t − p − + g − K ( p ). Therefore, g K ( p ) < g ( t − , t ; p ) = (cid:18) max a ′ ∈{ , ,...,t } (cid:26) a ′ − ( t − p + ℓ a ′ − − p (cid:27)(cid:19) − ≤ − pℓ t − − . Again, we consider the graph F with vertex set V ( K ) and edge set EG( K ). ByLemma 17, every vertex in F has degree at least ℓ t and every pair of vertices has atleast one common neighbor. By Lemma 18, F has no cycle of length more than ℓ t − − F isHamiltonian, which means | V ( K ) | ≤ ℓ t − −
1. As a result, g K ( p ) ≥ − pℓ t − − , a contradiction.So we may assume that every maximum-length path in F is not a cycle. Let v . . . v ℓ be such a maximum-length path such that, in K , the sum x ( v ) + x ( v ℓ ) is the largestamong such paths. Let v c be the unique common neighbor of v and v ℓ .Let v have d neighbors in F . Since v cannot have neighbors outside of this path,the sum of the weights, in K , of the neighbors of v satisfy d G ( v ) ≤ x ( v ) + · · · + x ( x c ).Notice that if v i ∈ { v , . . . , v c − } is a predecessor of a neighbor of v , then it is an endpointof a path containing the same ℓ vertices, namely v i v i − · · · v v i +1 v i +2 · · · v c · · · v ℓ . Henceall d predecessors of gray neighbors of v (including v itself) have weight at most x ( v ).All other vertices have weight at most g K ( p )1 − p . Proposition 8 gives p − g K ( p ) p + 1 − pp x ( v ) = x ( v ) + d G ( v ) ≤ x ( v ) + · · · + x ( v c ) ≤ d x ( v ) + ( c − d ) g K ( p )1 − p . Rearranging the terms, we obtain g K ( p ) (cid:18) c − d − p + 1 p (cid:19) ≥ − x ( v ) (cid:18) d − − pp (cid:19) . p − ≤ p − = ℓ t and ℓ t < d +1, we may, by Lemma 17, lower bound the right-handside by using x ( v ) ≤ g K ( p )1 − p from Proposition 9, g K ( p ) (cid:18) c − d − p + 1 p (cid:19) ≥ − g K ( p )1 − p (cid:18) d − − pp (cid:19) g K ( p ) (cid:18) c − p (cid:19) ≥ . Lemma 18 bounds the size of the longest cycle, so c ≤ ℓ t − −
1. Thus, g K ( p ) ≥ − pc ≥ − pℓ t − − ≥ g ( t − , t ; p ), a contradiction. This concludes Case 4. Case 5: a = t − p ∈ (0 , p ).It remains to prove the theorem for 0 < p < p = ℓ − t in the case where ( t + 1) |6 h and a = t − Fact 21.
Let h and t be positive integers such that h ≥ t + 2 . Let p = ℓ − t = (cid:6) h t +1 (cid:7) − and recall that γ H ( p ) = min a ∈{ ,...,t } (cid:26) pt + 1 , p (1 − p ) a (1 − p ) + ( ℓ a − p (cid:27) . Then γ H ( p ) = p/ ( t + 1) for p ∈ [0 , p ] . We have ed H ( p ) ≤ γ H ( p ) and the previous case gives that the two functions are equalat p = p . They are also equal at p = 0. By Fact 21, the function γ H ( p ) is linear over p ∈ [0 , p ] for h ≥ t + 2. By Proposition 1, ed H ( p ) is continuous and concave down, sowe may conclude that ed H ( p ) = γ H ( p ) = pt +1 for p ∈ [0 , p ].This concludes Case 5 and completes the proof of Theorem 3. Proof of Corollary 5.
The case of t = 1 is covered by Corollary 4.Let a ∈ { , . . . , t − } .If p ≥ aa + ℓ − ℓ a , then p (1 − p ) a (1 − p ) + ( ℓ a − p ≥ − pℓ − . If p ≤ t − at − a + ℓ a − ℓ t , then p (1 − p ) a (1 − p ) + ( ℓ a − p ≥ p (1 − p ) t (1 − p ) + ( ℓ t − p . Therefore, it suffices to show t − at − a + ℓ a − ℓ t ≥ aa + ℓ − ℓ a ( ℓ − ℓ a )( t − a ) ≥ ( ℓ a − ℓ t ) a. (6)13o that end,( ℓ − ℓ a )( t − a ) − ( ℓ a − ℓ t ) a = ( t − a ) ℓ + aℓ t − tℓ a > ( t − a ) ht + 1 + ah t + 1 − tht + a + 1 − t = at ( t − a ) h ( t + 1)( t + a + 1)(2 t + 1) − t ≥ t ( t − h ( t + 1)(2 t )(2 t + 1) − t. If h ≥ t + 10 t + 12 + t − , then (6) is satisfied and the corollary follows. Proof of Fact 11.
We only need to prove one direction because x and y are arbitrary. Inboth cases, we will prove the forward implication. (a) Let ⌊ h/x ⌋ ≥ y and h = qx + r , where r ∈ { , . . . , x − } . Then y ≤ ⌊ h/x ⌋ = q , so h ≥ xy + r . Thus ⌊ h/y ⌋ ≥ x + ⌊ r/y ⌋ ≥ x . (b) Let ⌈ h/x ⌉ ≤ y and h = qx − r , where r ∈ { , . . . , x − } . Then y ≥ ⌈ h/x ⌉ = q , so h ≤ yx − r . Thus ⌈ h/y ⌉ ≤ x − ⌊ r/y ⌋ ≤ x . Proof of Fact 12.
Clearly, if a ∈ { , . . . , t − } , then (cid:6) ht + a +1 (cid:7) ≤ (cid:6) ht +1 (cid:7) so it suffices to provethis fact for a = 0. Let h = qt + r with r ∈ { , . . . , t − } . Since h ≥ t ( t − q ≥ t − ≥ r . Then (cid:24) ht + 1 (cid:25) = q + (cid:24) r − qt + 1 (cid:25) ≤ q = (cid:22) ht (cid:23) . Proof of Fact 15. If h ≥ ( t + 1) + 1, then t + 2 ≤ ⌈ h/ ( t + 1) ⌉ = ℓ . Consequently, t + 1 ≤
12 ( ℓ + t ) ≤ p ( ℓ + t )and so − pℓ − ≤ pt +1 .For a ∈ { , . . . , t } , let h = q ( t + 1) + r , where r ∈ { , . . . , t + 1 } . The bound h ≥ ( t + 1)( t + a ) + 1 ensures q ≥ t + a . Then, a + (cid:24) ht + a + 1 (cid:25) = a + (cid:24) q ( t + a + 1) + r − qat + a + 1 (cid:25) = q + (cid:24) a ( t + a + 1) + r − qat + a + 1 (cid:25) ≤ q + (cid:24) a ( t + a + 1) + t + 1 − ( t + a ) at + a + 1 (cid:25) ≤ q + 1 = (cid:24) ht + 1 (cid:25) − pℓ − ≤ p (1 − p ) a (1 − p )+( ℓ a − p . Proof of Lemma 17.(a)
Let v ∈ V ( K ). Using Proposition 8,deg G ( v ) ≥ (cid:24) d G ( v )max { x ( u ) } (cid:25) ≥ & p − g K ( p ) p + − pp x ( v ) g K ( p )1 − p ' ≥ ( p − g K ( p ))(1 − p ) pg K ( p ) = 1 − pg K ( p ) − − pp> max a ′ ∈{ , ,...,t } (cid:26) ( a ′ − a )(1 − p ) + ( ℓ a ′ − pp − − pp (cid:27) = max a ′ ∈{ , ,...,t } (cid:26) ( a ′ − a − − p ) p + ℓ a ′ − (cid:27) ≥ ℓ a +1 − . The last inequality is obtained by choosing a ′ = a + 1. (b) By inclusion-exclusion, 1 ≥ d G ( v ) + d G ( w ) − d G ( v, w ), we have that d G ( v, w ) ≥ p − g K ( p ) p + − pp ( x ( v ) + x ( w )) − > p − g K ( p ) p . Therefore,deg G ( v, w ) ≥ (cid:24) d G ( v, w )max { x ( u ) } (cid:25) ≥ & p − g K ( p ) pg K ( p )1 − p ' = 1 − pg K ( p ) − − p ) p> max a ′ ∈{ , ,...,t } (cid:26) ( a ′ − a )(1 − p ) + ( ℓ a ′ − pp − − p ) p (cid:27) = max a ′ ∈{ , ,...,t } (cid:26) ( a ′ − a − − p ) p + ℓ a ′ − (cid:27) . If a ≤ t −
2, then we choose a ′ = a + 2. Then deg G ( v, w ) > ℓ a +2 −
1, and becausedeg G ( v, w ) is an integer, deg G ( v, w ) ≥ ℓ a +2 .If a = t −
1, then we choose a ′ = t . Then deg G ( v, w ) > − − pp + ℓ t − ℓ t − p − ≥ p ≥ p = ℓ − t . Because deg G ( v, w ) is an integer, deg G ( v, w ) ≥ Proof of Lemma 18.
We say that a long cycle is a cycle of length at least L + 1 and will show that there areno long cycles. Let v · · · v ℓ be a smallest cycle in G among all those length greater than L .15 ase 1: ≤ a ≤ t − t ≥
2. Consider the path v · · · v ℓ a − on the cycle v · · · v ℓ v . There is no cycle of length ℓ a and so the common neighbors of v and v ℓ a − are all in { v , . . . , v ℓ a − } . Note that Lemma 17 establishes that v and v ℓ a − have at least ℓ a +2 ≥ v and v ℓ a − are in { v , . . . , v ℓ a − } , we have ℓ a − ≥ ℓ a +2 .Hence, ht + a + 3 ≤ (cid:24) ht + a + 3 (cid:25) ≤ (cid:24) ht + a + 1 (cid:25) − < ht + a + 1 − h > ( t + a + 1)( t + a + 3).This gives that the number of common neighbors of v and v ℓ a − is at least ℓ a +2 = (cid:6) ht + a +3 (cid:7) ≥ t + a + 2 ≥ v and v ℓ a − has at least two common neighbors in { v , . . . , v ℓ a − } . Let i > j < ℓ a − { v , . . . , v ℓ a − } that are common neighbors of v and v ℓ a − . That is, 3 ≤ i ≤ j ≤ ℓ a − v v i v i +1 · · · v ℓ − v ℓ has length ℓ − i + 2. The cycle v v · · · v j − v j v ℓ a − v ℓ a · · · v ℓ − v ℓ has length j + ℓ − ℓ a + 2.Since these two cycles have length less than ℓ , they cannot be long cycles. Hence, theirlength is at most ℓ a −
1, giving us ℓ − i + 2 ≤ ℓ a − ℓ + j − ℓ a + 2 ≤ ℓ a − . We can add these inequalities and use the fact that ℓ ≥ L + 1. Rearranging the terms,we conclude the following:3 ℓ a − L − ≥ ℓ a − ℓ − ≥ j − i + 1 ≥ ℓ a +2 − . (7)To verify there are no long cycles, we must show that (7) produces a contradiction.Since 0 ≤ a ≤ t − ℓ a − L − (cid:24) ht + a + 1 (cid:25) − (cid:22) ht (cid:23) − < (cid:18) ht + a + 1 + 1 (cid:19) − (cid:18) ht − (cid:19) − ht + a + 3 − − h ( at + a + 4 a + 3) t ( t + a + 1)( t + a + 3) < (cid:24) ht + a + 3 (cid:25) − ℓ a +2 − , a contradiction for all t ≥ h ≥ t + 2. Therefore, for 0 ≤ a ≤ t − G has no cycleof length longer than ℓ a −
1. 16 ase 2: a = t − v and v ℓ t − − are in { v , . . . , v ℓ t − − } , we have ℓ t − − ≥
1. Hence, 1 ≤ ℓ t − − (cid:24) h t (cid:25) − < h t − h > t , which means ℓ t − ≥ v · · · v ℓ t − on the cycle v · · · v ℓ v . To see there is no cycle of length ℓ t − + 1, we set h = q (2 t ) − r with q ≥ r ∈ { , . . . , t − } and have ℓ t − + 1 = (cid:24) h t (cid:25) + 1 = q + 1 ≤ q − ≤ q + (cid:22) − rt (cid:23) ≤ (cid:22) ht (cid:23) = L. Since v and v ℓ t − have a common neighbor v i , either v v i v i +1 · · · v ℓ v or v · · · v i v ℓ t − v ℓ t − +1 · · · v ℓ v has length less than ℓ . Without loss of generality, we will assume that it is the former.This gives ℓ t − − ≥ ℓ − i + 2 ≥ ℓ − ( ℓ t − −
1) + 2 ≥ L + 1 − ( ℓ t − −
1) + 2 . Consequently, 2 ℓ t − − L − ≥ . (8)To see that (8) is contradicted,2 ℓ t − − L − (cid:24) h t (cid:25) − (cid:22) ht (cid:23) − < (cid:18) h t + 1 (cid:19) − (cid:18) ht − (cid:19) − − < . Therefore, for a = t − G has no cycle of length longer than ℓ t − − Proof of Proposition 19.
Let v · · · v ℓ be a longest path in G such that v v ℓ ∈ E ( G ). If G is not Hamiltonian, there exists a w ∈ V ( G ) − { v , . . . , v ℓ } . Because G is connected, thereexists i ∈ { , . . . , ℓ } and w ′ ∈ V ( G ) − { v , . . . , v ℓ } such that v i is adjacent to w ′ . Thenthere is a longer path: v i +1 · · · v ℓ v · · · v i w ′ , a contradiction. Proof of Lemma 20.
Because v · · · v ℓ is a longest path in F , neither v nor v k can have neighbors off this path,as that would yield a longer path. Thus N ( v ) ∪ N ( v ℓ ) ⊆ { v , . . . , v k } in F . Case 1: ℓ ≤ ℓ a .If v i is adjacent to v , then v i − cannot be adjacent to v ℓ . Thus, the predecessors of17 ( v ) and the neighbors of v ℓ are disjoint subsets in { v , . . . , v ℓ − } . Since both v and v ℓ have degree at least ℓ a +1 , hence 2 ℓ a +1 ≤ ℓ − ≤ ℓ a − . However, ℓ a − ℓ a +1 − (cid:24) ht + a + 1 (cid:25) − (cid:24) ht + a + 2 (cid:25) − < ht + a + 1 − ht + a + 2 = − h ( t + a )( t + a + 1)( t + a + 2) < . (9) Case 2: ℓ ≥ ℓ a + 1.Partition the vertices of this path into 2 s + 1 consecutive sets A , B , A , . . . , A s , B s with s ≥
0, constructed so that, in each set A i , neighbors of v appear before neighborsof v ℓ as follows:We let neighbors of v be denoted with v p i and neighbors of v ℓ be denoted with v q i in this construction. Let A contain v and add consecutive vertices of this path untilwe arrive at a neighbor of v ℓ . From this point forward we do not allow another neighborof v to be in A , i.e. we continue adding consecutive vertices until we reach the lastneighbor v q of v ℓ before another neighbor v p of v . Then A = { v , . . . , v q } , and wedefine B = { v q +1 , . . . , v p − } . Note that this definition does not preclude B beingan empty set. Continuing with this algorithm, we define sets A = { v p , . . . , v q } and B = { v q +1 , . . . , v p − } , where v p is a neighbor of v on this path, v q is the last neighborof v ℓ in A before another neighbor v p of v as shown in Figure 1. We continue in thisway and define sets A i = { v p i , . . . , v q i } and B i = { v q i − +1 , . . . , v p i − } for i ∈ { , . . . , s } ,adding the last vertex v ℓ into the set A s . v · · · v q v q +1 · · · v p − v p · · · v q v q +1 · · · v ℓ A B A Figure 1: Partition of vertices of the path. Sets A i are iteratively constructed so thatthey contain consecutive vertices of this path starting with a neighbor of v and endingwith the last neighbor of v ℓ so that no neighbor of v appears after neighbors of v ℓ in eachset. Sets B i contain consecutive vertices between sets A i − and A i , if there are any. Thefirst vertex is placed in A and the last vertex v ℓ in A s .Now we analyze this partition: 18 We call the sets B i , i ∈ { , . . . , s } , gaps as they do not contain any neighbors ofeither v or v ℓ , but only contain vertices that succeed a given neighbor of v ℓ andprecede a given neighbor of v . According to the definition, gaps may be empty, butwe will see below that this is not possible in this case. • Each set A i , i ∈ { , . . . , s } , contains at most one common neighbor of v and v ℓ . • By construction, neighbors of v (other than a common neighbor, if exists) precedeneighbors of v ℓ in each A i , i ∈ { , . . . , s } .It will suffice to show that s = 0. This will imply that no neighbor of v follows thefirst neighbor of v ℓ on this path, which further implies that N ( v ) entirely precedes N ( v ℓ ),except possibly for a single common vertex. Since v and v ℓ have at least one commonneighbor, the lemma will follow.Notice that v · · · v q v ℓ v ℓ − · · · v p v is a cycle as seen in Figure 1. In fact, for any i ≥
1, removing the gap B i from vertices { v , . . . , v ℓ } forms a cycle, so by assumption, ℓ − | B i | ≤ ℓ a − P si =1 | B i | ≥ s ( ℓ − ℓ a + 1).On the other hand, by the degree assumption and since each set A i contains at mostone common neighbor of v and v ℓ , we obtain 2 ℓ a +1 ≤ | N ( v ) | + | N ( v ℓ ) | ≤ ( P si =0 | A i | ) +( s + 1) −
2. Combining these two inequalities we have ℓ = s X i =0 | A i | + s X i =1 | B i | ≥ ℓ a +1 − ( s + 1) + 2 + s ( ℓ − ℓ a + 1)= s ( ℓ − ℓ a ) + 2 ℓ a +1 + 1 . If s ≥
1, then we have ℓ ≥ ℓ − ℓ a + 2 ℓ a +1 + 1 which simplifies to ℓ a − ℓ a +1 − ≥ s = 0 and the lemma follows. Proof of Fact 21.
We need to show that γ H ( p ) = p / ( t + 1). Since γ H ( p ) = p · min a ∈{ ,...,t } (cid:26) t + 1 , − p a (1 − p ) + ( ℓ a − p (cid:27) , we need to show that ℓ a − ℓ t − ≤ t − a + 1 for all a ∈ { , . . . , t − } .To do this, let h = q (2 t + 1) − r where r ∈ { , . . . , t } and q ≥ h ≥ t + 2).Then, ℓ a − ℓ t − q − (cid:18) q − (cid:24) q ( t − a ) − rt + a + 1 (cid:25)(cid:19) ≤ q − (cid:18) q − q ( t − a ) + t + at + a + 1 (cid:19) = t − a + 1 + t − a + 2 t − q ( t − a )( q − t + a + 1) , which is at most t − a + 1 if q ≥ a ≤ t − q = 2. In the case where a = t − q = 2, then ℓ a − ℓ t − = 1 + (cid:6) − r t (cid:7) ≤ t − a + 1.19 Conclusion and open questions
We have obtained the edit distance function over all of its domain for C th when t + 1 doesnot divide h and h ≥ t ( t + 1) + 1. When t + 1 divides h and h ≥ t ( t + 1) + 1, we haveobtained the function for p ∈ [ p , p = (cid:6) h t +1 (cid:7) − . The function, however, is notknown when t + 1 divides h and p ∈ [0 , p ) or when h ≤ t ( t + 1).As to the case of p < p (and h sufficiently large), we showed that if K ∈ K (Forb( C th ))is a p -core CRG with p < / a = t − g K ( p ) = γ Forb( C th ) ( p ).Therefore, to solve the problem for the remaining case when t + 1 divides h , and p is small,one only needs to consider CRGs with exactly t − p ≥ p to ensure that the graph induced by the blackvertices and gray edges of the CRG has the property that any two vertices have at leastone common neighbor. Such a condition need not hold for small p .As to reducing the lower bound required of h , we note that in the proof of Theorem 3,we required h ≥ t ( t + 1) + 1 in Fact 15. This ensured that the γ H function for p ∈ [1 / ,
1] was linear and by the concavity and continuity of the edit distance function (seeProposition 1), this ensures that ed H ( p ) = γ H ( p ) in that interval. So, more carefulanalysis of the case p ≥ / h , but thesearguments are very different from the case where p < /
2. Elsewhere, we only require h ≥ max { t ( t − , t + 2 } in order to complete the proof of Theorem 3. This bound isrequired in several places. See Fact 12, Lemma 13, Lemma 18 but especially the basicFact 14 which says that a set of size h can be partitioned into sets of size t or t + 1 ifand only if h ≥ t ( t − h smaller than max { t ( t − , t + 2 } in general. Acknowledgements
Berikkyzy was supported through Wolfe Research Fellowship of Department of Mathe-matics at Iowa State University. Martin’s research was partially supported by the Na-tional Security Agency (NSA) via grant H98230-13-1-0226. Martin’s contribution wascompleted in part while he was a long-term visitor at the Institute for Mathematics andits Applications. He is grateful to the IMA for its support and for fostering such a vi-brant research community. Peck’s research is based upon work supported by the NationalScience Foundation Graduate Research Fellowship under Grant No. DGE0751279.
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