Hamilton cycles, minimum degree and bipartite holes
aa r X i v : . [ m a t h . C O ] A p r Hamilton cycles, minimum degree and bipartiteholes
Colin McDiarmid [email protected]
Nikola Yolov [email protected]
April 20, 2016
Abstract
We present a tight extremal threshold for the existence of Hamiltoncycles in graphs with large minimum degree and without a large “bipartitehole“ (two disjoint sets of vertices with no edges between them). Thisresult extends Dirac’s classical theorem, and is related to a theorem ofChv´atal and Erd˝os.In detail, an ( s, t ) -bipartite-hole in a graph G consists of two disjointsets of vertices S and T with | S | = s and | T | = t such that there are noedges between S and T ; and e α ( G ) is the maximum integer r such that G contains an ( s, t )-bipartite-hole for every pair of non-negative integers s and t with s + t = r . Our central theorem is that a graph G with at least3 vertices is Hamiltonian if its minimum degree is at least e α ( G ).From the proof we obtain a polynomial time algorithm that eitherfinds a Hamilton cycle or a large bipartite hole. The theorem also yieldsa condition for the existence of k edge-disjoint Hamilton cycles. We seethat for dense random graphs G ( n, p ), the probability of failing to containmany edge-disjoint Hamilton cycles is (1 − p ) (1+ o (1)) n . Finally, we discussthe complexity of calculating and approximating e α ( G ). Hamilton cycles are one of the central topics in graph theory, see for exam-ple [BM08]. The problem of recognising the existence of a Hamilton cycle in agraph is included in Karp’s 21 NP-complete problems [Kar72]. Recall that δ ( G )denotes the minimum degree d ( v ) of a vertex v in G . An early result by Dirac[Dir52] states: Theorem 1 (Dirac’s Theorem) . A graph G with n ≥ vertices is Hamiltonianif δ ( G ) ≥ n/ . The theorem is sharp, since the disjoint union of two complete n -vertexgraphs has minimum degree n − S and T of vertices in a graph, we let E ( S, T ) denote theset of edges with one end in S and one in T . Definition 1.1. An ( s, t )-bipartite-hole in a graph G consists of two disjointsets of vertices S and T with | S | = s and | T | = t such that E ( S, T ) = ∅ . Wedefine the bipartite-hole-number e α ( G ) to be the least integer r which may bewritten as r = s + t − for some positive integers s and t such that G does notcontain an ( s, t ) -bipartite-hole. An equivalent definition of e α ( G ) is the maximum integer r such that G contains an ( s, t )-bipartite-hole for every pair of non-negative integers s and t with s + t = r . Observe that e α ( G ) = 1 if and only if G is complete, and e α ( G ) ≥ α ( G ), where α ( G ) is the stability number of G . Also note that for1 ≤ a ≤ b , we have e α ( K a,b ) = b and e α ( K a,b ) = min { b + 1 , a + 1 } . (Here K a,b denotes the complete bipartite graph with parts of sizes a and b , and G denotesthe complement of G .)The following is our main theorem. It arose from our investigations of therandom perfect graph P n , where we wished to show that P n is Hamiltonian withfailure probability e − Ω( n ) , see [MY16]. Theorem 2.
A graph G with at least 3 vertices is Hamiltonian if δ ( G ) ≥ e α ( G ) . This result is sharp in the sense that for every positive integer r there is anon-Hamiltonian graph with δ ( G ) = r = e α ( G ) −
1. An example is G = K r,r +1 ,where δ ( G ) = r and e α ( G ) = r + 1. Theorem 2 generalises Theorem 1 of Dirac.Indeed, a graph G with δ ( G ) ≥ n/ , ⌊ n/ ⌋ )-bipartite-hole, and hence δ ( G ) ≥ n/ ≥ e α ( G ). Also, Theorem 2 can be extended to provide a sufficientcondition for the existence of many edge-disjoint Hamilton cycles; and in factthe next result will be deduced quickly from Theorem 2. Theorem 3.
Let r ≥ be an integer, and let G be a graph with at least 3vertices such that δ ( G ) ≥ ( r + 1) e α ( G ) + 3 r . Then G contains r + 1 edge-disjointHamilton cycles. Note that by setting r = 0 in Theorem 3 we regain Theorem 2.It is perhaps not surprising that determining e α ( G ) is NP-hard and that itis hard to approximate, see Section 5 below. However, Theorem 2 can be madealgorithmic. Theorem 4.
There is an algorithm which, on input a graph G with n ≥ vertices, in O ( n ) time outputs either a Hamilton cycle or a certificate that e α ( G ) > δ ( G ) . Theorem 3 can also be made algorithmic. One can repeatedly use the al-gorithm in Theorem 4 to find a Hamilton cycle, remove its edges from G andrepeat, or if no cycle is found, output a certificate that e α ( G ) is large. Thisyields: 2 heorem 5. There is an algorithm that, on input a graph G with n ≥ vertices,in O ( n ) time outputs a non-negative integer r , a collection of r edge-disjointHamilton cycles of G , and a certificate that e α ( G ) > δ ( G ) − rr +1 . Containing a large bipartite hole is not a certificate for the absence of Hamil-ton cycles; there are Hamiltonian graphs for which the algorithm will stop beforeoutputting a Hamilton cycle, which is to be expected, since deciding whetheror not a graph is Hamiltonian is NP-complete.We conclude the paper by applying Theorem 3 to show quickly that fora sufficiently dense random graph G , the probability of G failing to containmany edge-disjoint Hamilton cycles is well-estimated by the probability that G contains a vertex with too small degree ( < r ), or indeed contains an isolatedvertex. Theorem 6.
Let < ǫ < , let ≤ p = p ( n ) ≤ − ǫ , and let r = r ( n ) be apositive integer. If p ( n ) √ nr ( n ) log n → ∞ as n → ∞ , then the probability that G ( n, p ) fails to contain at least r edge-disjoint Hamilton cycles is (1 − p ) (1+ o (1)) n . Setting r = 1 we obtain: Corollary 7. If p ( n ) √ n/ log n → ∞ as n → ∞ , then the probability that G ( n, p ) fails to be Hamiltonian is (1 − p ) (1+ o (1)) n . Finding sufficient conditions for the existence of Hamilton cycles has been anactive area of research for more than sixty years. Among the most well-knownconditions are Dirac’s Theorem [Dir52], Theorem 1; and a generalisation by Ore[Ore60], which states that an n -vertex graph G is Hamiltonian if d ( u )+ d ( v ) ≥ n for any pair of non-adjacent vertices u and v . These were further generalised byBondy and Chv´atal, and others, see the book by Bondy and Murty [BM08] andsee [Gou03, Li13] for surveys. Both conditions are further generalised by Fan[Fan84], where he proved that a 2-connected graph G of order n is Hamiltonianif max( d ( u ) , d ( v )) ≥ n/ u, v with distance2. See [DeL00] for a survey.One of these generalisations, by Chv´atal and Erd˝os [CE72], has a sharpcondition close to the one in this paper. We denote the vertex connectivity of G by κ ( G ) and the number of vertices of G by v ( G ). Theorem 8 (Chv´atal-Erd˝os Theorem) . A graph G with at least 3 vertices isHamiltonian if κ ( G ) ≥ α ( G ) . There are interesting connections between Theorems 2 and 8, and betweenthe parameters κ , δ , α and e α . For example, κ ( G ) ≤ δ ( G ) ≤ v ( G ) − α ( G ) and α ( G ) ≤ e α ( G ) ≤ v ( G ) − κ ( G ). Furthermore, we will see in Lemma 3.1 that κ ( G ) ≥ δ ( G ) − e α ( G ) + 2. 3omparing Theorems 2 and 8, neither condition implies the other. Hereis an example of a graph G that meets the conditions of Theorem 2 but notTheorem 8. It has vertex set V ( G ) = { a } ∪ B ∪ C ∪ D , such that | B | = k + ℓ , | C | = k , | D | = ℓ + 1, and all these sets are disjoint. All edges between { a } and B , between B and C , and between C and D are present, B and D are complete,and C is independent. It is easy to see that k = κ ( G ) < α ( G ) = k + 1, and δ ( G ) = k + ℓ ≥ max { k + 1 , ℓ + 3 } = e α ( G ) for ℓ ≥ k ≥ ℓ + 3. In the otherdirection, C satisfies κ = 2 = α but δ = 2 < e α .A more recent related result is by Hefetz, Krivelevich and Szab´o [HKS09].Roughly speaking, the authors prove that expanding graphs without large bipar-tite holes are Hamiltonian. Their results cover a wide range of graphs, includ-ing relatively sparse graphs. Compared to [HKS09], we focus on tight extremalthresholds, simple self-contained proofs and the right conditions for edge-disjointHamilton cycles.Hamilton cycles in random graphs have been well-studied, see for exam-ple [Fri89, Bol01]. In [KS83] Koml´os and Szemer´edi prove that if p = p ( n ) = k ( n ) (cid:0) n (cid:1) ; k ( n ) = 12 n log n + 12 n log log n + c n n, then lim n →∞ P ( G ( n, p ) is Hamiltonian) = c n → −∞ e − e − c if c n → c c n → ∞ . Frieze proves in [Fri85] a similar result for random bipartite graphs. The evolu-tionary process G n,t is defined as follows: G n, is the empty graph on n verticesand G n,k +1 is obtained from G n,k by adding an edge uniformly at random.Ajtai, Koml´os and Szemer´edi [AKS85] and Bollob´as [Bol84] showed that withhigh probability the hitting time for Hamiltonicity equals the hitting time forminimal degree at least two. A necessary condition for a graph to be Hamiltonian is to be 2-connected, soTheorem 2 implies that every graph G with δ ( G ) ≥ e α ( G ) is 2-connected. Wegive one preliminary lemma before proving Theorems 2 and 4. Lemma 3.1.
The following holds for every graph G : κ ( G ) ≥ δ ( G ) + 2 − e α ( G ) . Proof.
Suppose for a contradiction the vertices v and v are separated by a set S of size less than δ ( G ) + 2 − e α ( G ). Let s and t be positive integers such that e α ( G ) + 1 = s + t and G has no ( s, t )-bipartite-hole. Then e α ( G )+12 ≤ max( s, t ) ≤ e α ( G ). Now the closed neighbourhoods N [ v i ] satisfy | N [ v i ] \ S | ≥ δ ( G )+1 −| S | ≥ α ( G ) ≥ max( s, t ). The sets N [ v ] \ S and N [ v ] \ S are disjoint because S is aseparator, but | N [ v ] \ S | ≥ s and | N [ v ] \ S | ≥ t , so there is an edge betweenthem and S does not separate v from v , a contradiction.As an aside before proving Theorem 2, suppose the graph G with at least 3vertices satisfies δ ( G ) ≥ e α ( G ) −
2. Then κ ( G ) ≥ δ ( G ) + 2 − e α ( G ) ≥ e α ( G ) ≥ α ( G ) . Hence the conditions of the Chv´atal-Erd˝os Theorem are met, and so G is Hamil-tonian. Theorem 2.
A graph G with at least 3 vertices is Hamiltonian if δ ( G ) ≥ e α ( G ) .Proof. If e α ( G ) = 1, then G is complete, and so G is Hamiltonian. Thus we maysuppose that e α ( G ) ≥
2. We will show that if P is a maximal length path in G , then G [ V ( P )] is Hamiltonian. This, together with the connectedness of G following from Lemma 3.1, is enough to complete the proof.Indeed, suppose P is a maximal length path in G , n = v ( P ), and label thevertices in V ( P ) with [ n ] := { , . . . , n } in the order they appear in the path,after choosing an arbitrary orientation. We may assume that vertices 1 and n are not adjacent. For a set S ⊆ V ( P ), define S + to be the set of successors x + of elements x in S , and define S − to be the set of predecessors x − . We leave S + undefined if n ∈ S and S − is undefined if 1 ∈ S .We now describe three situations when P can be closed to form a cycle.The first yields a standard proof of Dirac’s and Ore’s theorems, the second in-volves ‘non-crossing’ edges from the end vertices, and the third involves ‘crossingedges’.(a) If for some j ∈ (1 , n ) we have j ∈ N (1) and j − ∈ N ( n ), then 1 j − nj − − V ( P ) (where we follow the path P from j to n andfrom j − to 1). See Figure 1.1 j − j ∈ (1 , n ) n Figure 1: Single flip(b) If for some k ∈ (1 , n ) there exist i ∈ N (1) ∩ (1 , k ] and j ∈ N ( n ) ∩ [ k, n )such that i − is adjacent to j + , then 1 − i − j + − nj − i V ( P ). Here we may have i = j ; see Figure 2.(c) If for some k ∈ (1 , n ) there exist i ∈ N (1) ∩ [ k, n ) and j ∈ N ( n ) ∩ [1 , k )such that i + is adjacent to j + , then 1 − jn − i + j + − i V ( P ). Here we may have j + = i ; see Figure 3.5 i − i ∈ (1 , k ] j ∈ [ k, n ) j + n Figure 2: Double nested flip1 j ∈ [1 , k ) j + i ∈ [ k, n ) i + n Figure 3: Double cross flipWe shall show that at least one of these situations must hold. Suppose fora contradiction that this is not the case. Then for every k ∈ (1 , n ) E [( N (1) ∩ (1 , k ]) − , ( N ( n ) ∩ [ k, n )) + ] = ∅ (1)since (b) does not hold; and E [ { } ∪ ( N (1) ∩ [ k, n )) + , ( N ( n ) ∩ [1 , k )) + ] = ∅ (2)since (a) and (c) do not hold.Let 1 ≤ s ≤ t be such that e α ( G )+1 = s + t and G has no ( s, t )-bipartite-hole.Since e α ( G ) ≥
2, we have s ≤ e α ( G )+12 < e α ( G ), and hence | N (1) ∩ (1 , | = 1 ≤ s ≤ δ ( G ) − < | N (1) ∩ (1 , n ] | = d (1) . Therefore we can choose k ∈ (1 , n ) such that | N (1) ∩ (1 , k ] | = s . Equation (1)implies that | N ( n ) ∩ [ k, n ) | < t . Since | N ( n ) ∩ [1 , k ) | + | N ( n ) ∩ [ k, n ) | ≥ δ ( G ),we have | N ( n ) ∩ [1 , k ) | > δ ( G ) − t ≥ e α ( G ) − t = s −
1, and so | N ( n ) ∩ [1 , k ) | ≥ s .Now from (2) we deduce | N (1) ∩ [ k, n ) | < t −
1, hence | N (1) ∩ [ k, n ) | ≤ t − n , we have δ ( G ) ≤ | N (1) ∩ (1 , k ] | + | N (1) ∩ [ k, n ) | ≤ s + t − ≤ δ ( G ) − , and this contradiction completes the proof.Next we consider edge-disjoint Hamilton cycles. We need a preliminarylemma. For graphs F and G with the same vertex set V , we define F ∪ G =( V, E ( F ) ∪ E ( G )) and F − G = ( V, E ( F ) \ E ( G )). Lemma 3.2.
Suppose H , . . . , H r are r ≥ Hamilton cycles in a graph G andlet H = H ∪ . . . ∪ H r . Then e α ( G − H ) + 1 ≤ ( r + 1)( e α ( G ) + 1) . roof. Let 1 ≤ s ≤ t be such that e α ( G ) + 1 = s + t and G has no ( s, t )-bipartite-hole. Let U, W ⊆ V ( G ) be disjoint sets of size s and 2 rs + t respectively. Now | W \ Γ H ( U ) | ≥ | W | − r X i =1 | Γ H i ( U ) | ≥ rs + t − rs = t. But G has no ( s, t )–bipartite-hole, so G − H has no ( s, rs + t )–bipartite-hole.Finally, we see that e α ( G − H ) + 1 ≤ s + 2 rs + t ≤ ( r + 1)( e α ( G ) + 1), since s ≤ e α ( G )+12 . Theorem 3.
Let r ≥ be an integer, and let G be a graph with at least 3vertices such that δ ( G ) ≥ ( r + 1) e α ( G ) + 3 r . Then G contains r + 1 edge-disjointHamilton cycles.Proof. We sequentially find edge-disjoint Hamilton cycles H , H , . . . . Let 0 ≤ i ≤ r and suppose we have found H , . . . , H i . Let G i = G − ∪ j ≤ i H j . Then byLemma 3.2 e α ( G i ) ≤ ( i + 1)( e α ( G ) + 1) − ≤ ( r + 1) e α ( G ) + r ≤ δ ( G ) − r ≤ δ ( G i ) . Hence by Theorem 2 we can find H i +1 edge-disjoint from H , . . . H i .A certificate that e α ( G ) ≥ k consists of pairs ( S i , T i ) for i = 1 , . . . , ⌊ k/ ⌋ suchthat S i , T i ⊆ V ( G ), S i ∩ T i = ∅ , E ( S i , T i ) = ∅ , and | S i | = i , | T i | = k − i . Theorem 4.
There is an algorithm which, on input a graph G with n ≥ vertices, in O ( n ) time outputs either a Hamilton cycle or a certificate that e α ( G ) > δ ( G ) .Proof. First check if G is connected. If not, pick two connected components andnote that each has size at least δ ( G ) + 1. For each i = 1 , . . . , ⌊ ( δ ( G ) + 1) / ⌋ ,any i vertices from one of these components together with any δ ( G ) + 2 − i from the other form a bipartite hole, and hence we can find a certificate that e α ( G ) ≥ δ ( G ) + 2. So we can assume that G is connected.Maintain a path P with initial length at least two. The algorithm performsat most n steps, and the length of P increases with each one. On each step,check if a terminal vertex of P has a neighbour outside V ( P ), and if so extend P . Otherwise, following the proof of Theorem 2, we can either find a sequenceof bipartite holes forming a certificate as required and halt, or close P to forma cycle. This cycle is either Hamiltonian and then the algorithm halts, or fromthe connectivity of G we can attach an edge xy with x ∈ V ( P ) and y V ( P )to obtain a strictly longer path starting from y and spanning V ( P ) ∪ { y } .Each step takes O ( n ) time, so the total time spent is O ( n ). The following result is phrased to cover the existence of one Hamilton cycle,and of many. 7 emma 4.1.
Fix < ǫ < and let ≤ p = p ( n ) ≤ − ǫ for all n . Given r = r ( n ) ≥ , let A r be the event that G ( n, p ) contains at least r edge-disjointHamilton cycles, and let A cr be the complementary event. Then n log(1 − p ) ≤ log P ( A cr ) ≤ n log(1 − p ) + (2 + o (1)) r √ n log n. Proof.
Let G ∼ G ( n, p ), t = ⌈√ n ⌉ and d = r (2 t ) + 3 r −
3. From Theorem 3 wehave { e α ( G ) ≤ t } ∩ { δ ( G ) ≥ d } ⊆ A r , so { δ ( G ) = 0 } ⊆ A cr ⊆ { e α ( G ) > t } ∪ { δ ( G ) < d } . Clearly P ( δ ( G ) = 0) ≥ (1 − p ) n = exp( n log(1 − p )). Also, the probability thatvertex n has degree at most d − d − n −
1] such that each other vertex is not adjacent to n . Thus P ( δ ( G ) < d ) ≤ n (cid:18) n − d − (cid:19) (1 − p ) ( n − − ( d − ≤ n d (1 − p ) n ǫ − d = exp( n log(1 − p ) + d (log n + log(1 /ǫ ))) . Further P ( e α ( G ) > t ) ≤ P ( G has a ( t, t )-bipartite-hole) ≤ (cid:18) nt (cid:19) (1 − p ) t ≤ (cid:16) ent (cid:17) t (1 − p ) t ≤ e t n t (1 − p ) n = exp( n log(1 − p ) + √ n (log n + O (1)));and the required upper bound on log P ( A cr ) follows since d = (2 + o (1)) r √ n .Theorem 6 and Corollary 7 follow directly from Lemma 4.1. e α ( G ) Computing e α ( G ) is closely related to the following problem: Maximum Balanced Complete Bipartite Subgraph (BCBS):
Instance : A positive integer k and a bipartite graph G with parts A and B where | A | = | B | ; Question : Does G contain a complete bipartite graph with k vertices in eachpart; that is, does G have a subgraph K k,k ?We use lemma 2.2 of [ADL + Bipartite Hole-Number (BHN):
Instance : A positive integer k and a graph G ; Question : Is e α ( G ) ≥ k ?To compare the two problems we introduce the following lemma:8 emma 5.1. Given a graph G and an integer k , let G ′ be formed from G byadding a disjoint copy of K k − , k ; and let G φk be the complement of G ′ . Then K k,k ⊆ G if and only if e α ( G φk ) ≥ k .Proof. We see that G φk has an induced copy of K k − ∪ K k , and so it has an( s, k − s )-bipartite-hole for each s = 1 , . . . , k −
1. Thus e α ( G φk ) ≥ k if and onlyif G φk has a ( k, k )-bipartite-hole; and that happens if and only if the complementof G has a ( k, k )-bipartite-hole, if and only if G has a subgraph K k,k .The next proposition follows as a corollary. Proposition 5.2.
The BHN problem is NP-complete.
In fact, a stronger statement could be given: the BHN problem is hardto approximate. To this end we use a result from [FK04] stating that theBCBS problem cannot be approximated within a factor of 2 (log n ) δ for some δ >
0, unless 3-SAT can be solved in O (cid:16) n / ǫ (cid:17) time for every ǫ >
0. Thewidely believed Exponential Time Hypothesis (ETH) states that 3-SAT cannotbe solved in 2 o ( n ) time, which provides strong evidence for the inapproximabilityof BCBS. Lemma 5.1 allows us to directly translate these results to hardness ofapproximating e α ( G ): Proposition 5.3.
There exists δ > such that e α ( G ) cannot be approximatedwithin a factor of (log n ) δ provided that 3-SAT / ∈ DT IM E (cid:16) n / ǫ (cid:17) for some ǫ > . In this paper we presented a tight sufficient condition for Hamiltonicity, Theo-rem 2, and used that result to prove an extension concerning the existence of r edge-disjoint Hamilton cycles, Theorem 3. As an application of these theorems,we proved results on disjoint Hamilton cycles in dense random graphs. It waspointed out to one of us by Michael Krivelevich that results from [HKS09] shouldallow us to extend Corollary 7 to much lower edge-probabilities p ( n ), down tonear the threshold for the necessary minimum degree; and indeed this is the caseas long as np ( n ) log log log log n log n log log log n → ∞ , see the appendix in arXiv:math.CO/1604.00888.We are not aware of any examples where the inequality in Theorem 3 is sharpfor r ≥
1. It would be interesting to find such examples or relax the condition.9 ppendix: Note on the probability of containinga Hamilton cycle around the threshold
The external neighbourhood of a set S ⊆ V is denoted by N ( S ), that is N ( S ) = { v ∈ V \ S : v is adjacent to some u ∈ S } . We use the following result by Hefetz et al. [HKS09].
Theorem 9.
Suppose ≤ d ≤ e √ log n and m = m ( n, d ) = log n log log log n log d log log n . If G = ( V, E ) is a graph such thatP1 for each S ⊆ V if | S | ≤ ndm , then | N ( S ) | ≥ d | S | , andP2 for every disjoint A, B ⊆ V if | A | , | B | ≥ n m , then E ( A, B ) = ∅ ,then G is Hamiltonian for large n . Lemma 6.1. If p ≥ m log( e m ) n , then G ( n, p ) fails to satisfy P2 withprobability less than e − np for large n .Proof. It is sufficient to show that the claim holds for each | A | and | B | of size k = ⌈ n m ⌉ . The number of choices for A, B is at most (cid:18) nk (cid:19) ≤ (cid:16) enk (cid:17) k = exp (cid:16) k log enk (cid:17) . The probability that a fixed pair (
A, B ) is edgeless is (1 − p ) k ≤ exp( − pk ).Now use the union bound to deduce that the log of the probability that P2 failsis at most 2 k log enk − pk ≤ − pk ≤ − pn, where the first inequality holds since p ≥ e m ) n m ≥ enk k , and the secondholds for large n . Lemma 6.2. If p ≥ (8 d + 12) log nn , then G ( n, p ) fails to satisfy P1 for some | S | ≥ with probability at most e − pn for large n .Proof. The probability that a set S of size s , 2 ≤ s ≤ ndm , is not expanding isat most p = (cid:18) ns (cid:19)(cid:18) n − s ⌊ ds ⌋ (cid:19) (1 − p ) s ( n − s − ds ) ≤ exp (( d + 1) s log n ) exp (cid:18) − nsp (cid:19) , since n − s − ds ≥ n for large n . Note that ax + bcx + d is increasing in x on ( − dc , ∞ ) iff ad − cb > − dc , ∞ ) otherwise, hence ( d +1) s +1 s − is decreasing10n s on ( , ∞ ), and therefore ( d +1) s +1 s − ≤ d + 12, since s ≥
2. Finally, pn ≥ (8 d + 12) log n ≥ ( d + 1) s + 1 s − log n ⇒ log n + ( d + 1) s log n − nsp ≤ − pn ⇒ np ≤ exp (cid:18) − pn (cid:19) . Taking the union bound over all s ≥ Lemma 6.3.
Suppose d log nn << p < . Then P ( δ ( G ( n, p )) ≤ d ) ≤ exp( − pn (1 + o (1))) . Proof. P ( δ ( G ) < d ) ≤ n (cid:18) n − d − (cid:19) (1 − p ) ( n − − ( d − ≤ n d e − pn d = exp( − pn + d log(2 n )) = exp( − pn (1 + o (1))) . For k ≥ ( k ) n stands for log n if k = 1 and log(log ( k − n )otherwise. To wrap things up, note that log m = Θ(log (2) n ), hence m log m =Θ( log n log (3) n log d ). By solving the equation d log n = m log m we get d log d =Θ(log (3) n ) and hence d = Θ( log (3) n log (4) n ). Therefore if p = log n log (3) nn log (4) n ω (1), theprobability of G ( n, p ) failing to contain a Hamilton cycle is e − np (1+ o (1)) . References [ADL +
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