Hamilton Cycles in the Semi-random Graph Process
aa r X i v : . [ m a t h . C O ] J un Hamilton Cycles in the Semi-random Graph Process
Pu Gao ∗ Department of Combinatorics and OptimizationUniversity of Waterloo, Waterloo, Canada [email protected]
Bogumi l Kami´nski
Decision Analysis and Support UnitSGH Warsaw School of Economics, Warsaw, Poland [email protected]
Calum MacRury ∗ Department of Computer ScienceUniversity of Toronto, Toronto, Canada [email protected]
Pawe l Pra lat ∗ Department of MathematicsRyerson University, Toronto, Canada [email protected]
Abstract
The semi-random graph process is a single player game in which the player isinitially presented an empty graph on n vertices. In each round, a vertex u is presentedto the player independently and uniformly at random. The player then adaptivelyselects a vertex v , and adds the edge uv to the graph. For a fixed monotone graphproperty, the objective of the player is to force the graph to satisfy this property withhigh probability in as few rounds as possible.We focus on the problem of constructing a Hamilton cycle in as few rounds aspossible. In particular, we present a novel strategy for the player which achieves aHamiltonian cycle in (2 + 4 e − + 0 .
07 + o (1)) n < . n rounds, assuming that aspecific non-convex optimization problem has a negative solution (a premise we numer-ically support). Assuming that this technical condition holds, this improves upon thepreviously best known upper bound of 3 n rounds. We also show that the previouslybest lower bound of (ln 2 + ln(1 + ln 2) + o (1)) n is not tight. ∗ The first, the third, and the last authors are partially supported by NSERC. Introduction
In this paper, we consider the semi-random process introduced recently in [2] that canbe viewed as a “one player game”. The process starts from G , the empty graph on thevertex set [ n ] = { , . . . , n } . In each step t , a vertex u t is chosen uniformly at random from[ n ]. Then, the player (who is aware of graph G t and vertex u t ) needs to select a vertex v t and add an edge u t v t to G t to form G t +1 . The goal of the player is to build a (multi)graphsatisfying a given monotonely increasing property A as quickly as possible.A strategy in this context is a sequence of functions f , f , . . . , where for each t ∈ N , f t ( u , v , . . . , u t − , v t − , u t ) is a distribution over [ n ], given the history of the process up to andincluding step t −
1, and vertex u t . Then v t is chosen according to this distribution. If f t isan atomic distribution, then v t is determined by u , v , . . . , u t − , v t − , u t . As f t is determinedby u t , and the history of the process up to step t −
1, it means that the player needs to selecther strategy in advance, before the game actually starts. Given f = ( f , f , . . . ) and realnumber 0 < q <
1, let τ q ( f , n ) be the minimum t for which P ( G t ∈ A ) ≥ q , where ( G i ) ti =0 isobtained following strategy f . Define τ q ( A , n ) = min f τ q ( f , n ) , where the minimum is over all strategies. Let τ A = lim q → − lim sup n →∞ τ q ( A , n ) n ;note that the limit above exists, since for every n the function q → τ q ( A ,n ) n is nondecreasing,as A is monotonely increasing.In this paper, we concentrate on property A = HAM that a graph has a Hamilton cycle.It was observed in [2] that1 . ≤ ln 2 + ln(1 + ln 2) ≤ τ HAM ≤ . We improve both upper and lower bounds for τ HAM . For the upper bound, we need toassume some “technical condition” P that claims that some function is negative on itsdomain—the function is defined in Subsection 2.7. Unfortunately, we could not prove that P holds but, in Subsection 2.9, we provide strong numerical evidence for it. For the lowerbound, we do not optimize the argument (as it gives a small improvement anyway) but aimfor a relatively easy proof that shows that the currently existing bound is not tight.Here are our main results. Theorem 1 is proved in Section 2 whereas the proof of Theo-rem 2 can be found in Section 3. Theorem 1.
Suppose that property P holds. Then, τ HAM ≤ e − + 0 . < . . Theorem 2.
There exists a universal constant ǫ > − such that τ HAM ≥ ln 2 + ln(1 + ln 2) + ǫ. n → ∞ . We say that an event holds asymptoti-cally almost surely ( a.a.s. ) if the probability that it holds tends to 1 as n → ∞ . In theproofs we use the standard Landau notation. Given two sequences of real numbers a n and b n , we write a n = O ( b n ) if there exists a constant C > | a n | ≤ C | b n | for all n . Wewrite a n = o ( b n ) if b n > n and lim n →∞ a n /b n = 0. In order to obtain an upper bound for τ HAM , one needs to propose a strategy for the playerto build a graph during the semi-random process, and show that after a certain number ofsteps the resulting graph is a.a.s. Hamiltonian.In order to warm up, let us recall an observation made in [2] that gives τ HAM ≤ n a.a.s.To see it, the following simple strategy can applied: let v t = ( t −
1) (mod n ) + 1 for all1 ≤ t ≤ n . Note that this is a non-adaptive strategy, that is, function f t does not dependon the history of the process nor vertex u t chosen at time t . More importantly, it is easy tosee that the resulting graph has the same distribution as the well-known G process thatis Hamiltonian a.a.s. [4].In general, the m -out process is defined for any natural number m : each vertex v ∈ [ n ]independently chooses m random out-neighbours from [ n ] to create the random digraph D m out . We then obtain G m out by ignoring orientations. Note that G m out is a multi-graph (itmay have loops or multiple edges) with minimum degree m and precisely mn edges. In themodel, we can either allow these multiple edges and loops, replace multiple edges with singleedges and remove loops, or condition on them not occurring. (Since the probability thatthere are no multiple edges is bounded away from zero, any property that holds a.a.s. in themodel that allows multiple edges also holds a.a.s. when we condition on no multiple edges.)For our application, when the strategy creates a multiple edge or a loop in the underlyingundirected graph, we simply “discard” that edge. That is, we will not use that edge for theconstruction of a Hamilton cycle.This section is structured as follows. In Subsection 2.1, we define a strategy for theplayer that creates a random graph G ∗ . The main goal is to prove that G ∗ , together withsome additional o ( n ) semi-random edges, is Hamiltonian a.a.s. Since the argument is quiteinvolved, an overview of the proof is provided in Subsection 2.2. In Subsection 2.3, weintroduce definitions and notation that will be used through the entire paper. Some usefulproperties of the graphs involved in the argument are extracted and proved in Subsection 2.4.In order to achieve our goal, in particular, we need to prove that a.a.s. G ∗ has a 2-matchingwith o ( n ) components (the definition is provided in Subsection 2.2). Subsection 2.6 preparesus for this task. As already mentioned earlier, this part requires property P that is definedat the end of Subsection 2.7. The proof that a.a.s. G ∗ has a 2-matching with few componentsis finished in Subsection 2.8. Now, it is enough to guide the semi-random process such thatafter additional o ( n ) rounds the graph has a Hamiltonian cycle. This last task does notdepend on the argument used to show that G ∗ has the desired 2-matching and so, in fact,we do it earlier, in Subsection 2.5. 3 .1 Our Strategy In this subsection, we define a strategy for the player that creates a random (multi)graph G ∗ . It will be convenient to work with the directed graph D t underlying G t . For each edge u t v t that is added to G t at time t , we put a directed edge from v t to u t in D t . As mentionedbefore, for the construction of a Hamilton cycle we will only use edges from a subgraph of G t . For any multigraph G , let ˆ G denote the simple graph obtained from G by deleting allloops and replacing all multiple edges by single edges. Thus, the sequence of multigraphs( G t ) immediately yields the corresponding sequence of simple graphs ( ˆ G t ).Consider the following strategy that will be defined in four phases. During the first phase,for 1 ≤ t ≤ n , let v t = ( t −
1) (mod n ) + 1. It is clear that G n has the same distributionas G . Let V and V be the sets of vertices in D n of in-degree 0 and, respectively, ofin-degree 1. During the second phase, two out edges are added from every vertex in V andone out edge is added from every vertex in V . During the third phase, we add 0 . n directededges uniformly at random, that is, in each step, v t is uniformly chosen from [ n ] \ { u t } . Wecall v a deficit vertex if after phase 3 its degree is less than 4. Then, in the fourth and lastphase, we repeatedly add a semi-random edge, coloured golden for convenience, coming outof a deficit vertex until its degree in the current underlying undirected simple graph (thatis, in ˆ G t ) becomes at least 4. Let τ i denote the last step of phase i (in particular, τ = 2 n ).Note that a golden semi-random edge is added out of u only if a loop or a multiple edgeincident with u was created in the process ( G t ) during the first two phases. It is easy toshow, by a standard first moment calculation, that G τ has O (1) loops or parallel edges, and o (1) other types of multiple edges in expectation. If v is incident with a loop in G τ then v may send out up to two semi-random edges in phase 4. If v is incident with a parallel edgein G τ then v may send out at most one semi-random edge in the final phase. Hence, a.a.s. G τ and ˆ G τ have the following property.( E) : There are at most ln ln n double edge or loops in G τ and they are all vertex disjoint.There are at most ln ln n golden edges, inducing vertex-disjoint paths of length 1 or 2,and every pair of deficit vertices are at distance at least ln n/ G τ .Thus a.a.s. the total number of semi-random edges added during the last two phases is(0 .
07 + o (1)) n . Note that the addition of the golden edges guarantees that the minimumdegree of ˆ G τ is at least 4, which will be used in the proof later. Finally, let G ∗ = G τ , themultigraph obtained after the last step of phase 4. As we will only use edges in ˆ G τ ⊆ G ∗ ,we will mainly focus on the process ( ˆ G t ). Let us present an overview of the proof of Theorem 1. First, we will investigate how long ittakes to construct graph G ∗ . Lemma 3.
A.a.s. the following holds | E ( G ∗ ) | = (2 + 4 e − + 0 .
07 + o (1)) n.
4n order to state the next lemma, we need one definition. A in a graph G is a simple subgraph H of G with maximum degree at most 2, that is, a collection ofvertex-disjoint paths and cycles. Moreover, we assume that V ( H ) = V ( G ) so some paths in H could be isolated vertices.Now, we are ready to state the lemma. This is the place where property P is needed. Lemma 4.
Suppose that property P holds. Then, a.a.s. G ∗ has a 2-matching with o ( n ) components. The final ingredient does not require property P anymore. Lemma 5.
Suppose G ∗ has a 2-matching with o ( n ) components. Then, there exists anadaptive strategy such that a.a.s. the semi-random process builds a Hamiltonian graph withinan additional o ( n ) steps. Theorem 1 follows immediately from the above three lemmas. Our strategy for construct-ing a Hamilton cycle in Lemma 5 is the same as that in [4] where a Hamilton cycle is foundin G . We start with a 2-matching F of G ∗ which has o ( n ) components. Then, we takean arbitrary component C of F and let P be a path that spans all vertices of C . By applyingPos´a rotations, we use either edges in G ∗ , or additional o ( n ) edges added to G ∗ to repeatedlyabsorb vertices in other components of F into the long path we carefully construct, untilfinally completing the path into a Hamilton cycle. Having less available edges in G ∗ than in G requires some new treatments in the proof of Lemma 5.In order to prove Lemma 4, as it is done in [4], we will apply Tutte and Berge’s formula forthe size of a maximum 2-matching of ˆ G τ ⊆ G ∗ . However, as we have significantly less edgesin ˆ G τ than in G , it becomes much more challenging to verify the Tutte-Berge conditions.Rough bounds that worked in [4] fail to work in our setting and, in order to achieve a tighterbound we end up with an optimization problem involving a high dimensional objectivefunction. That results in the technical property P that we only support numerically. In this subsection, we introduce basic definitions and notation that will be used throughoutthe paper. Let us start from graph theoretic ones. For a given subset of vertices S ⊆ V ( G ),let G [ S ] be the graph induced by set S , that is, V ( G [ S ]) = S and E ( G [ S ]) = { uw ∈ E ( G ) : u, v ∈ S } ⊆ E ( G ) . Let e ( S ) denote the number of edges induced by set S , that is, e ( S ) = | E ( G [ S ]) | = |{ xy ∈ E ( G ) : x, y ∈ S }| . Moreover, let N ( S ) = { v ∈ V ( G ) \ S : ∃ u ∈ S such that uv ∈ E ( G ) } . Finally, we say that S is an independent set if S induces no edge, that is, e ( S ) = 0.5iven subsets of vertices U, W ⊆ V ( G ), let e ( U, W ) denote the number of edges withexactly one end in U and the other end in W , that is, e ( U, W ) = |{ uw ∈ E ( G ) : u ∈ U, w ∈ W }| . For a given vertex v ∈ V ( G ), let deg( v ) be the degree of v , that is, the number of neighboursof v in G . Let δ ( G ) = min { deg( v ) : v ∈ V ( G ) } denote the minimum degree of a graph G .For a directed graph D and a given vertex v ∈ V ( D ), let deg − ( v ) and deg + ( v ) be the in-and out-degree of v , that is, the number of directed edges going to v and, respectively, goingfrom v in D .For sequences of real numbers a n and b n , we say a n = poly ( n ) if there exists a constant C > n − C < a n < n C for every n . We say a n = O ( b n ) if there exists a constant C > | a n | < C | b n | for all n . We say a n = o ( b n ) if eventually b n > n →∞ a n /b n = 0.Finally, let us introduce the binomial random graph G ( n, p ). More precisely, G ( n, p )is a distribution over the class of graphs with vertex set [ n ] in which every pair { i, j } ∈ (cid:0) [ n ]2 (cid:1) appears independently as an edge in G with probability p . Note that p = p ( n ) may (and inour application it does) tend to zero as n tends to infinity. Let us start with the following simple observations.
Observation 6.
Our process can be coupled such that the following properties hold.(a) G n has the same distribution as G and thus ˆ G τ ⊆ G .(b) ˆ G τ is a subgraph of G -out ; in particular, P (cid:16) S ⊆ E ( ˆ G τ ) (cid:17) ≤ (cid:18) n (cid:19) | S | , for any S ⊆ (cid:18) [ n ]2 (cid:19) . (1) (c) δ ( ˆ G τ ) ≥ and P (cid:16) S ⊆ E ( ˆ G τ ) (cid:17) ≤ (cid:18) . n (cid:19) | S | , for any S ⊆ (cid:18) [ n ]2 (cid:19) (2) P (cid:16) S ⊆ E ( ˆ G τ ) (cid:17) ≤ (cid:18) n (cid:19) | S | , for any S ⊆ (cid:18) [ n ]2 (cid:19) . (3) Proof. Part (a) : The property follows immediately from the construction of our process.
Part (b) : Recall that G τ is constructed from G by adding two out edges from everyvertex in V and one out edge from every vertex in V . If, instead, two out edges are addedfrom every vertex in G , we would get a graph with the same distribution as G .6ence, one may easily couple our process such that G τ is a subgraph of G . In order tosee that (1) holds, note first that P (cid:16) e ∈ E ( ˆ G τ ) (cid:17) = P (cid:16) e ∈ E ( G τ ) (cid:17) ≤ P (cid:16) e ∈ E ( G ) (cid:17) = 1 − (cid:18) − n (cid:19) ≤ n . The desired inequality holds after observing that S ′ ⊆ E ( ˆ G τ ) does not increase the proba-bility that an edge e / ∈ S ′ is also in E ( ˆ G τ ). Part (c) : The fact that δ ( ˆ G τ ) ≥ G τ . For (2),we note that part (b) implies that our process can be coupled such that ˆ G τ is a subgraphof G . As a result, ˆ G τ can be viewed as a subgraph of G ∪ G ( n, . n ). Thus, by theunion bound we get that P ( e ∈ E ( ˆ G τ )) ≤ P ( e ∈ E ( ˆ G τ )) + P ( e ∈ E ( G ( n, . n ))) ≤ n + 0 .
14 + o (1) n < . n . The assertion follows by noting that the presence of other edges do not increase the proba-bility that e ∈ E ( ˆ G τ ).In order to see that (3) holds, we apply the same argument after noting that every vertexsends out at most two golden semi-random edges. As a result, ˆ G τ can be viewed as asubgraph of G ∪ G ( n, . n ).The next lemma collects some important properties of the graphs involved in the processthat will be used in various places of this paper. In particular, part (a) immediately impliesLemma 3. Lemma 7.
A.a.s. the following properties hold.(a) D τ has asymptotically e − n vertices of in-degree 0 and e − n vertices of in-degree 1.In other words, | V | = ( e − + o (1)) n and | V | = (2 e − + o (1)) n .(b) For every ǫ > there exists δ = δ ( ǫ ) > such that for all S ⊆ [ n ] with | S | ≤ δn , S induces at most (1 + ǫ ) | S | edges in ˆ G τ .(c) All S ⊆ [ n ] with | S | ≤ . n induce at most . | S | edges in ˆ G τ .Proof. Part (a) : Let v ∈ [ n ] be any vertex of D n = D τ . Clearly, P (deg − ( v ) = 0) = (cid:18) − n (cid:19) n = e − + o (1) P (deg − ( v ) = 1) = (2 n ) · n · (cid:18) − n (cid:19) n − = 2 e − + o (1) . It follows that E ( | V | ) = ( e − + o (1)) n and E ( | V | ) = (2 e − + o (1)) n . It is straightforward toshow the concentration for these random variables (for example, by using the second momentmethod; we omit details) and so part (a) holds.7 art (b) : Let us fix ǫ > s = s ( n ) ∈ N . By (3), the expected number of sets S ⊆ [ n ]with | S | = s that induce at least (1 + ǫ ) s edges in ˆ G τ is at most g ( s ) := (cid:18) ns (cid:19)(cid:18) (cid:0) s (cid:1) (1 + ǫ ) s (cid:19) (cid:18) n (cid:19) (1+ ǫ ) s ≤ (cid:16) ens (cid:17) s (cid:18) es / ǫ ) s (cid:19) (1+ ǫ ) s (cid:18) n (cid:19) (1+ ǫ ) s = (cid:18) e ǫ . ǫ (1 + ǫ ) ǫ (cid:16) sn (cid:17) ǫ (cid:19) s ≤ (cid:18) . e (cid:18) . esn (cid:19) ǫ (cid:19) s . Clearly, g ( s ) ≤ (cid:0) . e (6 . eδ ) ǫ (cid:1) s ≤ (1 / s , provided that s ≤ δn and δ = δ ( ǫ ) > δ is(13 e ) − /ǫ / (6 . e )). On the other hand, if (for example) s ≤ ln n , then g ( s ) ≤ n − ǫs/ ≤ n − ǫ/ .It follows that the expected number of sets S ⊆ [ n ] with | S | ≤ δn that induce at least(1 + ǫ ) | S | edges is at most δn X s =1 g ( s ) ≤ ln n X s =1 n − ǫ/ + δn X s =ln n (1 / s ≤ (ln n ) n − ǫ/ + 2(1 / ln n = o (1) . Part (b) holds by Markov’s inequality.
Part (c) : For a given s = s ( n ) ∈ N , let X s be number of sets S ⊆ [ n ] with | S | = s thatinduce at least 1 . s edges in ˆ G τ , and let Y s be the number of sets S ⊆ [ n ] with | S | = s thatinduce at least t ( s ) edges in ˆ G τ , where t ( s ) = (cid:26) . s if s ≤ ln n . s if s > ln n. As a.a.s. ˆ G τ ∈ E , it follows that a.a.s. X s ≤ Y s for all s , since the number of goldenedges induced by S is at most min { s/ , ln ln n } ≤ min { . s, ln ln n } , given ˆ G τ ∈ E , and1 . s − ln ln n ≥ . s when s > ln n . Let g ( s ) = E ( Y s ). By (2), we get that g ( s ) ≤ (cid:18) ns (cid:19)(cid:18) (cid:0) s (cid:1) t ( s ) (cid:19) (cid:18) . n (cid:19) t ( s ) ≤ (cid:16) ens (cid:17) s (cid:18) es / t ( s ) (cid:19) t ( s ) (cid:18) . n (cid:19) t ( s ) . If s ≤ ln n , then g ( s ) ≤ n − . . On the other hand, if ln n < s ≤ δn with δ = 0 . g ( s ) ≤ e (cid:18) . e . (cid:19) . δ . ! s < . s . It follows that the expected number of sets S ⊆ [ n ] with | S | ≤ δn that induce at least 1 . | S | edges in ˆ G τ is at most δn X s =1 g ( s ) ≤ ln n X s =1 n − . + δn X s =ln n . s = o (1) . Part (c) holds by Markov’s inequality. 8 .5 Proof of Lemma 5
The whole subsection is devoted to prove Lemma 5. In order to achieve it, we will use apowerful proof technique introduced by Pos´a in [9]. Suppose that F is a 2-matching (that is,a collection of vertex-disjoint paths and cycles) of ˆ G τ ⊆ G ∗ with o ( n ) components. We willuse Pos´a rotations to extend a path in ˆ G τ to longer and longer paths, and eventually extenda Hamilton path to a Hamilton cycle, by adding o ( n ) extra semi-random edges. During theprocess of extending the paths, we will use edges in ˆ G τ whenever possible. If no edges inˆ G τ are of help, then we will use semi-random edges where we strategically choose v t to helpus with the extension of the paths.We start from a path P = u u . . . u h in F . If F is a collection of cycles, then wearbitrarily take a cycle and let P be the path obtained by deleting an arbitrary edge in thatcycle. Given a path P and an edge u h u j , 1 < j < h − h , namely, P ′ = u u . . . u j , u h , u h − , . . . , u j +1 with a new endpoint u j +1 . We call thisoperation a Pos´a rotation . Let S be the set of paths in ˆ G τ on the same set of verticesas P obtained by fixing u and performing any sequence of Pos´a rotations on P . Let End denote the union of the end vertices of paths in S other than u .Let us independently consider the following two cases: Case 1: there is x ∈ End and y / ∈ V ( P ) such that xy ∈ E ( ˆ G τ ) . If y is in a cycle C in F , thenwe can extend P to a longer path on V ( P ) ∪ V ( C ). On the other hand, if y is in a path P ′ in F , then without loss of generality we may assume that P ′ = v v . . . v ℓ . . . , v h with v ℓ = y where ℓ > h/
2. We can now extend P to a longer path on vertex set V ( P ) ∪ { v , . . . , v ℓ } .After that operation, the number of vertex-disjoint paths and cycles remains the same ordecreases by one. Case 2: for every x ∈ End , N ( x ) ⊆ V ( P ) . Colour vertices in
End blue or red as follows.If u i ∈ End and none of the two neighbours of u i (or just one neighbour of u i if i = h ) on P are in End , then colour u i red; otherwise, colour it blue. Let us start with the followingobservation about red vertices. Claim 8.
Let U denote the set of red vertices in End . Then U induces an independent setin ˆ G τ .Proof. For a contradiction, suppose that x, y are both red vertices in
End and xy is an edgein ˆ G τ . Without loss of generality, suppose that y was added to End before x and let P ′ bethe path obtained via Pos´a rotation with y being the other end. Let x = u i . Since x is red,neither u i − nor u i +1 is in End . Thus, the two neighbours of x on P ′ must be u i − and u i +1 .But then we can get another path on V ( P ) via Pos´a rotation on P ′ where one of u i − and u i +1 becomes an end vertex. This contradicts with the fact that x is red. It follows that U must be an independent set in G ∗ .By the usual argument of Pos´a rotation, for every u i ∈ N ( End ), we must have { u i − , u i +1 } ∩ End = ∅ . In particular, it implies that | N ( End ) | < | End | . However, using the above claim, we get aslightly stronger bound. Let x and x be the number of red and, respectively, blue vertices9n End . Since the set of red vertices in
End induces an independent set in ˆ G τ , it follows that | N ( End ) | ≤ x + x − | End | + x − . (4)Our next task and the main ingredient of the proof of the lemma is the next claim. Claim 9. | End | = Ω( n ) .Proof. In order to simplify the notation, let S = End . Let ǫ > | S | ≥ ǫ n . For a contradiction,suppose that | S | < ǫ n , and let N i = { x / ∈ S : e ( { x } , S ) = i } , n i = |N i | . Claim 10.
For every < ǫ ≤ , P i ≥ n i ≥ (2 − ǫ ) | S | , provided ǫ = ǫ ( ǫ ) is sufficientlysmall. Indeed, by Lemma 7(b) applied with ǫ ′ = ǫ/ S ∪ N ( S ), we get that a.a.s. e S ∪ [ i ≥ N i ! ≤ (1 + ǫ ′ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) S ∪ [ i ≥ N i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , provided ǫ is sufficiently small. It follows that e ( S ) + X i ≥ in i ≤ (1 + ǫ ′ ) | S | + X i ≥ n i ! . (5)On the other hand, by Observation 6(c), δ ( ˆ G τ ) ≥ e ( S ) + X i ≥ n i ≥ | S | . Substituting 2 e ( S ) ≤ (2 + 2 ǫ ′ ) | S | + P i ≥ (2 + 2 ǫ ′ − i ) n i from (5) into the above yields(2 − ǫ ′ ) | S | ≤ n + X i ≥ (2 + 2 ǫ ′ − i + 1) n i By the definition of ǫ ′ and as ǫ ′ ≤ /
2, we get(2 − ǫ ) | S | = (2 − ǫ ′ ) | S | ≤ n + X i ≥ (2 + ǫ − i + 1) n i (6) ≤ n ≤ X i ≥ n i . (7)This finishes the proof of the claim.We apply the above claim with ǫ = 0 .
05 so we may assume that | N ( S ) | ≥ (2 − ǫ ) | S | if | S | ≤ ǫ n . (8)10y (4) and (8), (2 − ǫ ) | S | ≤ | S | + x − , and hence (1 − ǫ ) | S | ≤ x − . Let X denote the set of red vertices and let X be the set of blue vertices in S . Let X ′ ⊆ X be the set of red vertices with at least 2 blue neighbours. Claim 11. | X ′ | ≤ . ǫ | S | . Indeed, consider the subgraph of G induced by Y = X ′ ∪ X . By the definition of X ′ , Y induces at least 2 | X ′ | edges. Hence, 2 | X ′ | ≤ . | X ′ | + | X | ). As x < ǫ | S | , we have | X ′ | ≤ (1 . / . ǫ | S | < . ǫ | S | , which finishes the proof of the claim.Therefore, every vertex in X \ X ′ has at least 3 neighbours in S . Thus, e ( S, S ) ≥ | X \ X ′ | ≥ − ǫ ) | S | − . ǫ | S | ) ≥ − . ǫ ) | S | . That is, X i ≥ in i ≥ (3 − . ǫ ) | S | . (9)By (6) and noting that 2 i − ≥ i for every i ≥ − ǫ ) | S | ≤ n + (2 + ǫ ) X i ≥ n i − X i ≥ in i . Plugging the lower bound for P i ≥ in i from (9) yields(2 + ǫ ) X i ≥ n i ≥ n + (2 + ǫ ) X i ≥ n i ≥ (2 − ǫ ) | S | + (3 − . ǫ ) | S | = (5 − . ǫ ) | S | . Thus, | N ( S ) | = X i ≥ n i ≥ − . ǫ ǫ | S | > . | S | , as ǫ < .
05. This contradicts with | N ( S ) | ≤ | S | + x − < | S | . It follows then that | S | ≥ ǫ n .Now, it is straightforward to finish the proof of Lemma 5. Proof of Lemma 5.
We extend P whenever possible, and if it is not possible, then | V ( P ) | ≥ ǫ n by Claim 9. The vertices outside of P are in a collection F of o ( n ) paths and cycles. Let v t be an arbitrary vertex outside of P that is either an end vertex of a path, or any vertexin a cycle. If the semi-random process selects a vertex u t ∈ End then, by performing Pos´arotations, we extend P to a longer path by absorbing a path or a cycle in F that was not in P . The number of components in F goes down by 1. Otherwise, we simply ignore u t and v t ,and repeat until eventually u t ∈ End . Since | End | ≥ ǫ n , the probability that u t ∈ End is atleast 1 /ǫ . Hence, it takes O (1) trials on average to absorb a path or a cycle. Since there areonly o ( n ) paths or cycles to be absorbed, it follows immediately from Chernoff bound thata.a.s. an additional o ( n ) edges are enough to be added to G ∗ to make it Hamiltonian.11 .6 Preparation for the Proof of Lemma 4 Our aim now is to prove that G ∗ has a 2-matching with o ( n ) components. In order to achieveit, we will apply the consequence of the Tutte-Berge matching formula [10, Theorem 30.7] toˆ G τ , which is a simple graph and a subgraph of G ∗ . However, we need one more definitionbefore we can state it.Given a simple graph G , let κ ( G ) be the number of edges in a maximum 2-matching of G (that is, κ ( G ) is the size of a maximum 2-matching). The Tutte-Berge matching theoremimplies the following. Theorem 12.
Let G be a simple graph on the vertex set [ n ] . Then, κ ( G ) = min ( n + | U | − | S | + X X (cid:22) e ( X, S )2 (cid:23)) , where U and S are disjoint subsets of [ n ] , S is an independent set, and X ranges over thecomponents of G − U − S . Despite the fact that the above theorem provides the exact value for κ ( G ), it is not soeasy to apply it in the context of random graphs. Fortunately, if G belongs to some family ofgraphs, then we get an easier property to check. We will first define the family, then prove aweaker but more workable statement, and finally show that a.a.s. G ∗ belongs to the family.Let C cyclic be the family of graphs on the vertex set [ n ] which satisfy the followingproperties: there are at most n/ ln n subsets S ⊆ [ n ] with | S | ≤ ln n/
10 such that S inducesa connected subgraph with at least the same number of edges as the number of vertices; thatis, G [ S ] is connected and | E ( G [ S ]) | ≥ | S | . Corollary 13.
Suppose G ∈ C cyclic and all 2-matchings of G have more than γ componentsfor some γ ≥ . Then, G has vertex partition [ n ] = S ∪ T ∪ R ∪ U such that(a) S is an independent set and G [ T ] is a forest;(b) | S | ≥ max {| U | , γ − n/ ln n } ;(c) e ( S ∪ T ) + e ( S ∪ T, R ) ≤ | T | + 2 | S | − | U | − γ + 33 n/ ln n ;(d) e ( R, T ) = 0 .Proof.
The proof is almost identical to that in [4] so we only briefly sketch the argumenthere. Let F be a maximum 2-matching of G . Since G ∈ C cyclic , the number of cycles in F is at most n/ ln n + n/ (ln n/
10) = 11 n/ ln n . Let c ( F ) and e ( F ) denote the numbers ofcomponents and, respectively, edges in F . Then, γ ≤ c ( F ) ≤ n − e ( F ) + 11 n/ ln n = n + 11 n/ ln n − κ ( G ) . Thus, κ ( G ) ≤ n + 11 n/ ln n − γ .Let S and U be a pair of disjoint subsets of [ n ] that minimize n + | U | − | S | + X X ∈C (cid:22) e ( X, S )2 (cid:23) , (10)12here C is the set of components of G − U − S , and S is an independent set. Let T be theunion of components of G − U − S that are trees, and R = [ n ] − U − S − T . By Theorem 12 andour earlier observation, we get that κ ( G ) = n + | U | − | S | + P X ∈C j e ( X,S )2 k ≤ n + 11 n/ ln n − γ ,and so | U | − | S | + X X ∈C (cid:22) e ( X, S )2 (cid:23) ≤ n/ ln n − γ. (11)By our construction, [ n ] = S ∪ T ∪ R ∪ U is a partition of the vertex set, and properties (a)and (d) hold. It remains to show that properties (b) and (c) also hold. It follows immediatelyfrom inequality (11) that | S | ≥ | U | + X X ∈C (cid:22) e ( X, S )2 (cid:23) + γ − n/ ln n ≥ γ − n/ ln n, since | U | ≥ P X ∈C j e ( X,S )2 k ≥
0. On the other hand, by Theorem 12 and the fact that(
S, U ) is chosen such that it minimizes (10), we have n ≥ κ ( G ) ≥ n + | U | − | S | , which implies | S | ≥ | U | . This shows that property (b) holds.For part (c), let p and q denote the number of components in G [ T ] and, respectively, G [ R ]. Since G ∈ C cyclic , there are at most n/ ln n components in G [ R ] of order at mostln n/
10, and at most 10 n/ ln n components in G [ R ] of order greater than ln n/
10. It followsthat q ≤ n/ ln n . Then, X X ∈C (cid:22) e ( X, S )2 (cid:23) ≥ ( e ( S, T ) − p ) + ( e ( S, R ) − q )2 ≥ e ( S, T ) + e ( S, R ) − p − n/ ln n . Hence,11 n/ ln n − γ ≥ | U | − | S | + X X ∈C (cid:22) e ( X, S )2 (cid:23) ≥ | U | − | S | + e ( S, T ) + e ( S, R ) − p − n/ ln n . It follows that e ( S, T ) + e ( S, R ) ≤ n/ ln n − γ + 2 | S | − | U | + p. Now condition (c) follows since e ( S ∪ T ) + e ( S ∪ T, R ) = e ( S, T ) + e ( T ) + e ( S, R ) ≤ e ( S, T ) + e ( S, R ) + | T | − p, as T induces a forest and e ( T, R ) = 0.Let us now show that ˆ G τ belongs to the family C cyclic and so Corollary 13 can be applied. Lemma 14.
A.a.s. ˆ G τ ∈ C cyclic .Proof. Let Z be the family of sets S with | S | ≤ ln n/
10 where S induces a connected subgraphof ˆ G τ with at least | S | edges, and let Z = |Z| . We will show that E [ Z ] = o ( n/ ln n ) whichproves the lemma as it implies that Z ≤ n/ ln n by Markov’s inequality.13or a given S ⊆ [ n ] with | S | ≤ ln n/
10, let X S be the indicator random variable that S induces a connected subgraph of ˆ G τ with at least | S | edges. Let X = P S :3 ≤| S |≤ ln n/ X S . Itfollows that E [ X ] ≤ ln n/ X s =3 (cid:18) ns (cid:19) s s − (cid:18) s (cid:19) (cid:18) . n (cid:19) s ≤ ln n/ X s =3 (8 . e ) s = O (cid:0) (8 . e ) ln n/ (cid:1) = O ( n . ) = o ( n/ log n ) . (Indeed, there are (cid:0) ns (cid:1) sets of cardinality s , s s − spanning trees of K s , and (cid:0) s (cid:1) choices foran additional edge. By (2), the probability that selected edges are present in G ∗ is at most(8 . n ) s .)Note that X counts those sets S ∈ Z that already satisfy the desired property in thesubgraph ˆ G τ . We may assume that ˆ G τ has property E . Hence, it is sufficient to furtherbound the number of sets S ∈ Z that contain exactly one deficit vertex v and induce atleast one golden edge incident with v . Let Y S be the indicator variable that S is such a set.Let Y = P S :3 ≤| S |≤ ln n/ Y S and we immediately have Z ≤ X + Y . Hence, our next task isto upper bound E [ Y ]. There are | S | ways to choose vertex v in S to be the deficit vertex.Then either v is incident with a loop, or a multiple edge in G τ . We will only bound E [ Y S A ]where A denotes the event that the deficit vertex in S is incident with a loop in G τ ; theother case can be dealt with analogously. Let s = | S | . There are s s − (cid:0) s (cid:1) ways to specify aset of s edges that must be induced by S . Given a specification of such s edges, there areat most s ways to specify one of them to be golden. The probability for that specific edgeto be golden is at most 2 /n < . /n (as v sends out 2 golden edges in total). There couldbe another edge among the s edges that is golden, and the conditional probability for thatis at most 2 /n < . /n . Moreover, the probability that v is incident with a loop is O (1 /n ).It follows now that E [ Y A ] ≤ ln n/ X s =3 (cid:18) ns (cid:19) s · s s − (cid:18) s (cid:19) (cid:18) . n (cid:19) s · O (1 /n ) = O (cid:18) ln nn (cid:19) · ln n/ X s =3 (8 . e ) s = o (1) . As we already mentioned, similar calculations show that E [ Y B ] = o (1), where B is theevent that the deficit vertex in S is incident with a parallel edge in G τ . Combining all ofthe above, we have E [ Z ] ≤ E [ X ] + E [ Y A ] + E [ Y B ] = o ( n/ ln n ). The lemma follows byMarkov’s inequality.Let us fix an arbitrarily small ǫ >
0. After combining Lemma 14 and Corollary 13, itremains to show that a.a.s. there is no vertex partition S ∪ T ∪ U ∪ R of ˆ G τ satisfyingproperties (a)–(d) in Corollary 13 with some γ ≥ ǫn . However, the distribution of ˆ G τ iscomplicated. As a result, we will work on G τ instead and use Property E , which impliesthat ˆ G τ misses at most ln ln n edges of G τ .It will be convenient to colour edges of G τ in one of the four colours: blue, green,red, and yellow. Recall that G τ is constructed during the first three phases, and G τ isobtained by adding up to ln ln n golden semi-edges to G τ . During the first phase, G n and the corresponding directed graph D n are created; V and V are the sets of vertices in D n of in-degree 0 and, respectively, of in-degree 1. Let us colour edges of G n green if their14ounterparts in D n are directed into one of the vertices in V ∪ V which we also colour green.The remaining edges are coloured blue. During the second phase graph G τ is created; letus colour edges added during this phase red. Finally, edges added during the third phaseare coloured yellow.Let us consider any partition [ n ] = S ∪ T ∪ U ∪ R . For any i ∈ { S, T, U, R } , let α i bethe fraction of vertices that belong to set i (that is, α i = | i | /n ) and let γ i be the fraction ofvertices of i that are green (that is, γ i = | G i | /α i n where G i is the set of green vertices inset i ). Moreover, let β i be the fraction of vertices in G i that received no incoming edge in D n (that is, β i = | G i ∩ V | / | G i | ). In order to simplify the notation, we define the followingvectors: ααα = ( α i ) i ∈{ S,T,U,R } , βββ = ( β i ) i ∈{ S,T,U,R } , γγγ = ( γ i ) i ∈{ S,T,U,R } . It follows immediatelyfrom the above definitions that the following properties hold: X i ∈{ S,T,U,R } α i = 1 , , ≤ γ i ≤ , ≤ β i ≤ i ∈ { S, T, U, R } . (12)Next, for i, j ∈ { S, T, U, R } , let b ij · (2 α i n ), g ij · (2 α i n ), and r ij · (2 β i + (1 − β i )) γ i α i n denote the numbers of blue, green and, respectively, red edges from set i to set j . Vectors bbb = ( b ij ) i,j ∈{ S,T,U,R } , ggg = ( g ij ) i,j ∈{ S,T,U,R } , and rrr = ( r ij ) i,j ∈{ S,T,U,R } describe the distribution ofedges of a given colour between parts. Let y · . n denote the number of yellow edges thatare either incident to a vertex in U , or are induced by R . Let y · . n denote the number ofyellow edges that are induced by T . Note that there are (1 − y − y ) · . n edges between S and R ∪ T . Hence, the vector ( y , y , − y − y ) describes the distribution of yellow edges.Let us fix uuu = ( ααα, βββ, γγγ, bbb, ggg, rrr, y , y ). Our goal is to upper bound the probability P ( uuu )that there exists a partition [ n ] = S ∪ T ∪ U ∪ R with | i | = α i n for i ∈ { S, T, U, R } , andsubsets i ′ ⊆ i for i ∈ { S, T, U, R } with | i ′ | = γ i α i n such that the following properties hold: • there are exactly b ij · (2 α i n ) blue directed edges from set i to set j ; • there are exactly g ij · (2 α i n ) green directed edges from set i to set j ; • there are exactly r ij · (2 β i + (1 − β i )) γ i α i n red directed edges from set i to set j ; • there are exactly y · . n yellow edges that are either incident to a vertex in U , orare induced by R ; • there are exactly y · . n yellow edges that are induced by T ; • there are no yellow edges inside S , or between R and T ; • all vertices in i ′ received at most 1 incoming green edge; • all vertices in i \ i ′ received at least 2 incoming blue edges.We will show that P ( uuu ) ≤ poly ( n ) exp( f ( uuu ) n ) for some explicit function f ( uuu ). Unfortunately,this function is quite involved so we will define it in the next section.15 .7 Property P Let ǫ = 2 − . Let us start with an observation that, due to Lemma 7, we may assume thatthe parameter uuu is of a specific form, that is, it satisfies the following constraints: − ǫ < X i ∈{ S,T,U,R } γ i α i − e − < ǫ (13) − ǫ < X i ∈{ S,T,U,R } β i γ i α i − e − < ǫ (14) − ǫ < X i ∈{ S,T,U,R } α i X j ∈{ S,T,U,R } g ij − e − < ǫ , (15)Indeed, equations (13) and (14) follow from the fact that a.a.s. | V | + | V | = (3 e − + o (1)) n and, respectively, | V | = ( e − + o (1)) n (Lemma 7(a)); equation (15) follows from the factthat the number of green edges is equal to | V | and so a.a.s. it is asymptotic to 2 e − n . Wealso have the following set of obvious constraints: X j ∈{ S,T,U,R } ( b ij + g ij ) = 1 , for all i ∈ { S, T, U, R } (16) X j ∈{ S,T,U,R } r ij = 1 , for all i ∈ { S, T, U, R } (17) y + y ≤ ααα, βββ, γγγ, bbb, ggg, rrr, y , y ∈ [0 , . (19)(For the ease of notation, we write that a vector is in [0 ,
1] when every component of thevector is in [0 , α S ≥ α U , (20)2 α T b T T + 2 α T g T T + γ T α T (2 β T + (1 − β T )) r T T + 0 . y ≤ α T + min { ǫ , α T } , (21) c SS = c RT = c T R = 0 , for all c ∈ { b, g, r } . (22)The first constraint comes immediately from property (b), and the last constraint followsfrom properties (a) and (d). The second constraint comes from the fact that e ( T ) ≤ | T | inˆ G τ required by property (a), which together with Property E imply that e ( G ) ≤ | T | + ǫ n and e ( G ) ≤ | T | in G τ (note there can be at most | T | loops or double edges induced by T ).Finally, let us note that Properties (c) and E , and Lemma 3 imply that a.a.s. the number ofedges incident with U or induced by R is at least | E ( G τ ) |− e ( S ∪ T ) − e ( S ∪ T, R ) ≥ (2 + 4 e − + 0 .
07 + o (1)) n − | T | − | S | + 2 | U | + 2 γ − n/ ln n − ln ln n ≥ (2 + 4 e − + 0 . n − | T | − | S | + 2 | U | = (4 e − + 0 .
07 + 4 α U + α T + 2 α R ) n. α U + γ U α U (1 + β U ) + 2 α S ( b SU + g SU ) + γ S α S r SU (1 + β S ) + 2 α T ( b T U + g T U )+ γ T α T r T U (1 + β T ) + 2 α R ( b RU + b RR + g RU + g RR )+ γ R α R ( r RU + r RR )(1 + β R ) + 0 . y ≥ e − + 0 .
07 + 4 α U + α T + 2 α R . (23)For i ∈ { S, T, U, R } , the number of blue edges coming into set i must be at least 2 α i (1 − γ i ) n ,as every vertex in i \ ( V ∪ V ) must receive at least 2 blue edges. This yield the followingset of constraints: X j ∈{ S,T,U,R } α j b ji ≥ α i (1 − γ i ) , for all i ∈ { S, T, U, R } . (24)Finally, the number of green edges coming into each set satisfies the following constraints: X j ∈{ S,T,U,R } α j g ji = α i γ i (1 − β i ) , for all i ∈ { S, T, U, R } . (25)Now, we are ready to show that P ( uuu ) ≤ poly ( n ) exp( f ( uuu ) n ) for some explicit function f ( uuu ). Given a vector of non-negative real numbers aaa with P i a i = 1, let H ( aaa ) = − P i a i ln a i .If aaa = ( a , a ), then we simply write H ( a ) for H ( aaa ). By convention, we set 0 log(0) = 0, forany a > a log(0) = −∞ , and we treat −∞ < x for every real number x .Given uuu , there are (cid:0) nα S n,α T n,α U n,α R n (cid:1) = poly ( n ) exp( H ( α S , α T , α U , α R ) n ) choices for sets S , T , U , and R . Given S , T , U , R , there are Y i ∈{ S,T,U,R } (cid:18) α i nγ i α i n (cid:19) = poly ( n ) Y i ∈{ S,T,U,R } exp( H ( γ i ) α i n ) = poly ( n ) exp n X i ∈{ S,T,U,R } H ( γ i ) α i ways to choose G S , G T , G U and G R . The probability that the number of blue and greenedges going out of S into each part of S , T , U , R is precisely as prescribed by uuu is equal to poly ( n ) exp( f S n ), where f S =2 α S (cid:16) H ( b SU , b ST , b SR , g SU , g ST , g SR ) + b SU ln((1 − γ U ) α U ) + b ST ln((1 − γ T ) α T )+ b SR ln((1 − γ R ) α R ) + g SU ln( γ U α U ) + g ST ln( γ T α T ) + g SR ln( γ R α R ) (cid:17) . Indeed, there are 2 α S n edges going out of S that are blue or green. We need to partitionthem into 6 classes depending on their colour and to which part they go to. This gives usthe term 2 α S H ( b SU , b ST , b SR , g SU , g ST , g SR ). For each i ∈ { T, U, R } , there are 2 α S b Si n blueedges that need to go to blue vertices of i (hence terms 2 α S b Si ln((1 − γ i ) α i )) and there are2 α S g Si n green edges that need to go to green vertices of i (hence terms 2 α S g Si ln( γ i α i )).Similarly, the probabilities that the number of blue and green edges going out of T , U , R uuu are poly ( n ) exp( f T n ), poly ( n ) exp( f U n ) and,respectively, poly ( n ) exp( f R n ), where f T =2 α T (cid:16) H ( b T S , b
T T , b
T U , g
T S , g
T T , g
T U ) + b T S ln((1 − γ S ) α S ) + b T T ln((1 − γ T ) α T )+ b T U ln((1 − γ U ) α U ) + g T S ln( γ S α S ) + g T T ln( γ T α T ) + g T U ln( γ U α U ) (cid:17) ,f U =2 α U (cid:16) H ( b US , b UT , b UU , b UR , g US , g UT , g UU , g UR ) + b US ln((1 − γ S ) α S )+ b UT ln((1 − γ T ) α T ) + b UU ln((1 − γ U ) α U ) + b UR ln((1 − γ R ) α R ) + g US ln( γ S α S )+ g UT ln( γ T α T ) + g UU ln( γ U α U ) + g UR ln( γ R α R ) (cid:17) , and f R =2 α R (cid:16) H ( b RS , b RU , b RR , g RS , g RU , g RR ) + b RS ln((1 − γ S ) α S ) + b RU ln((1 − γ U ) α U )+ b RR ln((1 − γ R ) α R ) + g RS ln( γ S α S ) + g RU ln( γ U α U ) + g RR ln( γ R α R ) (cid:17) . Given that, the probabilities that the number of red edges going out of S , T , U , R intoeach part of S , T , U , R exactly as dictated by uuu are poly ( n ) exp( g S n ), poly ( n ) exp( g T n ), poly ( n ) exp( g U n ) and, respectively, poly ( n ) exp( g R n ), where g S = α S γ S (2 β S + (1 − β S )) (cid:16) H ( r SU , r ST , r SR ) + r SU ln α U + r ST ln α T + r SR ln α R (cid:17) ,g T = α T γ T (2 β T + (1 − β T )) (cid:16) H ( r T S , r
T T , r
T U ) + r T S ln α S + r T T ln α T + r T U ln α U (cid:17) ,g U = α U γ U (2 β U + (1 − β U )) (cid:16) H ( r US , r UT , r UU , r UR ) + r US ln α S + r UT ln α T + r UU ln α U + r UR ln α R (cid:17) ,g R = α R γ R (2 β R + (1 − β R )) (cid:16) H ( r RS , r RU , r RR ) + r RS ln α S + r RU ln α U + r RR ln α R (cid:17) . In order to continue our computations, we need the following auxiliary lemma on the“balls into bins” model.
Lemma 15.
Fix α > and suppose that αn balls are thrown independently and uniformlyat random into n bins.(a) If α > , then the probability that every bin receives at least two balls is asymptoticto poly ( n ) exp( t ( α ) n ) with t ( α ) = λ − α + α ln( α/λ ) + ln(1 − e − λ − λe − λ ) , where λ = λ ( α ) > is the unique solution of the following equation: λ (1 − e − λ )1 − e − λ − λe − λ = α. (b) If α ≤ , then the probability that every bin receives at most one ball is asymptotic to poly ( n ) exp( κ ( α ) n ) , where κ ( α ) = − α − (1 − α ) ln(1 − α ) . c) If α = 2 , the probability that every bin receives exactly two balls is asymptotic to poly ( n ) exp((ln 2 − n ) . Before we prove the lemma, let us note that f ( λ ) := λ (1 − e − λ )1 − e − λ − λe − λ = λ (1 − (1 − λ + O ( λ )))1 − (1 − λ + λ / O ( λ ))(1 + λ ) = λ + O ( λ ) λ / O ( λ ) = 2 + O ( λ ) , so lim λ → + f ( λ ) = 2. It is also straightforward to see that lim λ →∞ f ( λ ) = ∞ and f ( λ ) is anincreasing function of λ . Hence, indeed, λ = λ ( α ) is well defined. For convenience, we define λ (2) = 0 and set t (2) = lim α → t ( α ) = ln 2 − . This definition of t : [2 , ∞ ) → R unifies parts (a) and (c) in the lemma above. Proof.
Suppose that α ≥
2. Let K be the truncated Poisson variable with parameter λ = λ ( α ) and truncated at 2, that is, P ( K = j ) = e − λ λ j j !(1 − e − λ − λe − λ ) , for every integer j ≥ E K = X j ≥ j · P ( K = j ) = X j ≥ e − λ λ j ( j − − e − λ − λe − λ )= λ − e − λ − λe − λ X j ≥ e − λ λ j j ! = λ (1 − e − λ )1 − e − λ − λe − λ = α, by the definition of λ .Let k , . . . , k n be n independent copies of K . Then, by Gnedenko’s local limit theorem [6],Θ( n − / ) = P n X i =1 k i = αn ! = X j ≥ ,...,j n ≥ P ni =1 j i = αn n Y i =1 e − λ λ j i j i !(1 − e − λ − λe − λ )= e − λn λ αn (1 − e − λ − λe − λ ) n X ∗ , where X ∗ = X j ≥ ,...,j n ≥ P ni =1 j i = αn n Y i =1 j i ! . Hence, X ∗ = poly ( n ) (1 − e − λ − λe − λ ) n e − λn λ αn . αn balls independently and uniformly at random into n bins.By Stirling’s formula ( x ! = poly ( x )( x/e ) x ), the probability that every bin receives at least 2balls is equal to X j ≥ ,...,j n ≥ P ni =1 j i = αn (cid:18) αnj , . . . , j n (cid:19) n − αn = ( αn )! n αn X ∗ = poly ( n ) e − αn α αn X ∗ = poly ( n ) e − αn α αn (1 − e − λ − λe − λ ) n e − λn λ αn = poly ( n ) exp( t ( α ) n ) . This completes the proof of part (a).To show part (b), suppose that α ≤
1. The probability that every bin receives at mostone ball is equal to ( n ) αn n αn = n !( n − αn )! n αn = poly ( n ) exp( κ ( α ) n ) , where ( x ) j = Q j − i =0 ( x − j ) denotes the j -th falling factorial.To show part (c), note that the probability that every bin receives exactly two balls isequal to (2 n )! / n n n = poly ( n ) exp((ln 2 − n ) . This finishes the proof of the lemma.We are now back to our problem. With Lemma 15 at hand, we will be able to prove thefollowing claims.
Claim 16.
The probability that all vertices in [ n ] \ ( G S ∪ G T ∪ G U ∪ G R ) receive at least twoblue edges is equal to poly ( n ) exp (cid:16) n P i ∈{ S,T,U,R } w i (cid:17) , where w i = (1 − γ i ) α i (cid:16) λ i − d i + d i ln( d i /λ i ) + ln(1 − e − λ i − λ i e − λ i ) (cid:17) ,d i = P j ∈{ S,T,U,R } α j b ji (1 − γ i ) α i , and λ i = λ i ( d i ) > is the unique solution of the following equation: λ i (1 − e − λ i )1 − e − λ i − λ i e − λ i = d i . Before we move to the proof, let us remark that, by (24), for every i ∈ { S, T, U, R } wehave d i ≥ λ i is well defined. Proof.
Note that for each i ∈ { S, T, U, R } , the number of blue edges coming into i \ G i isequal to P j ∈{ S,T,U,R } α j nb ji . Moreover, | i \ ( V ∪ V ) | = (1 − γ i ) α i n . The claim followsimmediately from Lemma 15(a) applied with α = P j ∈{ S,T,U,R } α j b ji / (1 − γ i ) α i = d i and thenumber of balls equal to (1 − γ i ) α i n . 20 laim 17. The probability that all vertices in G S ∪ G T ∪ G U ∪ G R receive at most one greenedge is equal to poly ( n ) exp (cid:16) n P i ∈{ S,T,U,R } ˜ w i (cid:17) , where ˜ w i = α i γ i (cid:16) − β i − β i ln β i (cid:17) . Proof.
Note that for each i ∈ { S, T, U, R } , the number of green edges coming into i ∩ ( V ∪ V )is (1 − β i ) γ i α i n . Moreover, | i ∩ ( V ∪ V ) | = γ i α i n . The claim follows immediately fromLemma 15(b) applied with α = (1 − β i ) γ i α i /γ i α i = 1 − β i and the number of bins equal to γ i α i n . Claim 18.
The probability that there are exactly . y n yellow edges incident with U orinduced by R , and exactly . y n yellow edges induced by T is equal to poly ( n ) exp( hn ) ,where h = 0 .
07 ln(0 .
07) + 0 . y ln (cid:18) α U + 2 α U (1 − α U ) + α R . y (cid:19) + 0 . y ln (cid:18) α T . y (cid:19) + 0 . − y − y ) ln (cid:18) α S ( α T + α R )0 . − y − y ) (cid:19) . Proof.
Recall that there are 0 . n yellow edges in total, so the remaining 0 . − y − y ) n yellow edges are between S and R ∪ T . The probability in the claim is equal to (cid:18)(cid:0) α U n (cid:1) + α U (1 − α U ) n + (cid:0) α R n (cid:1) . y n (cid:19)(cid:18) (cid:0) α T n (cid:1) . y n (cid:19)(cid:18) α S n ( α T + α R ) n . − y − y ) n (cid:19)(cid:18) (cid:0) n (cid:1) . n (cid:19) − , which is equal to poly ( n ) exp( hn ) by Stirling’s formula.Combining everything together, it follows that P ( uuu ) = poly ( n ) exp( f ( uuu ) n ), where f ( uuu ) = H ( α S , α T , α U , α R ) + X i ∈{ S,T,U,R } (cid:0) α i H ( γ i ) + f i + g i + w i + ˜ w i (cid:1) + h. Finally, we are ready to state property P . Definition 19 (Property P ) . Suppose there exists δ > such that f ( uuu ) < − δ for all vectors uuu subject to (12)–(25) and α R ≤ . . Let us remark that the reason to separate α R from 1 in the definition of Property P isthat the probability of a specified vertex partition S ∪ T ∪ U ∪ R satisfying Corollary 13(a)–(d)will not be exponentially small when S , T , and U are all of sub-linear size, and thus f is notbounded away from 0 in the entire region (12)–(25).21 .8 Proof of Lemma 4 Proof of Lemma 4.
Suppose that property P holds, that is, there exists δ > f ( uuu ) < − δ for all vectors uuu subject to (12)–(25) and α R ≤ . G τ has a 2-matching with o ( n ) components. Fix ǫ >
0. As mentioned earlier, aftercombining Lemma 14 and Corollary 13, it remains to show that a.a.s. there is no vertexpartition S ∪ T ∪ U ∪ R of ˆ G τ satisfying properties (a)–(d) in Corollary 13 with some γ ≥ ǫn and | R | ≤ . n .The expected number of partitions S ∪ T ∪ U ∪ R satisfying (a)–(d) with γ ≥ ǫn and | R | ≤ . n is at most X uuu P ( uuu ) = X uuu poly ( n ) exp( f ( uuu ) n ) , (26)where the sum is over all possible values of uuu satisfying constraints (12)–(25) and α R ≤ . f ( uuu ) < − δ/ uuu in the range of summation of (26) restricted to α R ≤ . uuu in the summation is clearly poly ( n ). Hence, the expectednumber of partitions S ∪ T ∪ U ∪ R satisfying (a)–(d) where | R | ≤ . n is X uuu poly ( n ) exp( f ( uuu ) n ) = poly ( n ) exp( − δn/
2) = o (1) . It only remains to consider partitions S ∪ T ∪ U ∪ R satisfying (a)–(d) with | R | > . n .Let x denote the number of edges between S and T ; x denote the number of edges between U and T ; x denote the number of edges between S and U ; x denote the number of edges between S and R. Since the minimum degree of ˆ G τ is at least 4, S induces an independent set, and T inducesa forest, we get that x + x + 2 e ( T ) ≥ | T | , e ( T ) < | T | , and x + x + x ≥ | S | . By property (c) and the fact that γ ≥ ǫn , we get that x + e ( T ) + x ≤ | T | + 2 | S | − | U | .Hence, 2 | T | + 2 | S | − | U | + x + x + x > | T | + 2 | S | − | U | + x + x + x + e ( T ) ≥ x + x + x + x + 2 e ( T ) ≥ | S | + | T | ) . (27)It follows that x + x + x ≥ | S | + | T | + | U | ) = 2( | S ∪ T ∪ U | ), that is, S ∪ T ∪ U inducesat least 2 | S ∪ T ∪ U | edges. However, by Lemma 7(c), this does not happen a.a.s. for anypartition with | S ∪ T ∪ U | ≤ . n . The goal of this section is to provide a numerical evidence that property P holds. The opti-mization problem was carefully investigated using the code written in the Julia language [3],22 uMP.jl package [5] with Ipopt solver [11]. The optimization problem we needed to face ischallenging for the following reasons.Firstly of all, it involves a non-convex optimization problem which potentially has manylocal optima (we numerically confirmed that this is the case in our problem). In orderto overcome this challenge, we used a standard multi-start [7] approach for solving globaloptimization problems. However, due to a stochastic nature of the heuristic search procedureused in this process, it means that the results we obtained are only heuristic in nature. Inother words, the numerical results we obtained strongly suggest that the desired propertyholds but this is, unfortunately, not a formal proof of this.Second of all, the objective function contains terms of the form x ln( x ) which have deriva-tives tending to ∞ as x →
0. This creates a challenge when solving the problem usingnumerical methods. More importantly, in the problem there are some local optima for whichsome variables are equal to zero. In order to overcome this problem, we relaxed the originalproblem by replacing x ln( x ) with some other function f ( x ) ≤ x ln( x ) (we need this prop-erty as we deal with a maximization problem and terms of the form x ln( x ) appear with anegative sign in the objective function). Function f ( x ) should be a quadratic function near0, its value and the values of its first and second derivatives should match in the point ofchange of the formula. The exact function we ended up using as a relaxation of x ln( x ) is: f ( x ) = ( x + ln(2 − ) x − − if 0 ≤ x < − x ln( x ) if x ≥ − . The third challenge is that the optimization problem for most of the variables allows thedomain to be [0 ,
1] and we have ln( x ) occurring in multiple places of the formulation of theobjective function (and also other than x ln( x ) which is handled by the relaxation describedabove). This poses another challenge when the solver performs a local search in the pointsnear the boundary of the admissible set. In such cases a logarithm of negative value mightbe considered (note that the solver evaluates the objective function for points contained insome small neighbourhood of a current potential solution before ensuring that the constraintsare satisfied; as a result, if points close to 0 are considered, such neighbourhood couldcontain negative values), which leads to errors when performing the computation. In orderto overcome this problem, we apply the transformation given by the formula g ( x ) = 12 (cid:18) sin (cid:18) π (cid:18) x − (cid:19)(cid:19) + 1 (cid:19) to every variable that is constrained to the interval [0 , ,
1] into the interval [0 ,
1] but it guarantees that if some decision variable istested outside the [0 ,
1] interval it is transformed back to [0 ,
1] interval (such values arerejected later anyway due to the constraints but are tested during the optimization processwhich causes no error). Also note that the transformation we use is an analytic function,which means that it does not introduce additional problems when calculating the first or thesecond derivatives of the objective functions or constraints.In order to explore the solution space thoroughly, we have performed two optimizationprocesses. In the first one, we tested the interior of the solution space, that is, all deci-sion variables that are restricted to [0 ,
1] were in fact constrained even further to be in the230 . , . − . ,
1] actually liedin the [0 . , . ǫ ,various relaxation functions f and space transformation functions g , and many separationmargins from the boundary. In all cases we consistently obtained that the best local optimumfound was below zero. Therefore, it provides a strong numerical support for the conjecturethat the objective value of our optimization problem is negative, that is, property P holds.We independently tested if the third phase (where 0 . n semi-random edges are sprinkled)is required for f ( uuu ) to be negative and bounded away from zero. Denote by ˆ uuu the bestsolution for our original problem we found; it satisfies f (ˆ uuu ) <
0. However, if 0 . n edgesare added instead of 0 . n , then the best solution that solver is able to find is a point uuu ∗ with f ( uuu ∗ ) >
0. This time, all [0 , uuu ∗ turned out to be in theinterval [0 . , . uuu and uuu ∗ . Bothpoints are very close to each other ( k uuu ∗ − ˆ uuu k ∞ = 0 . uuu ∗ is not feasible for the process involving adding 0 . n random edges. As it was done in the argument for an upper bound, it will also be convenient to work withthe directed graph D t underlying G t . For each edge u t v t that is added to G t at time t , weput a directed edge from v t to u t in D t (recall that u t is a random vertex selected by thesemi-random graph process and v t is a vertex selected by the player). The existing lowerbound for τ HAM that was observed in [2] follows from the fact that in order to construct aHamilton cycle, the player has to create a graph with minimum degree at least 2. However,this trivial necessary condition alone requires (ln 2 + ln(1 + ln 2) + o (1)) n steps. Indeed,in order to reach a graph with minimum degree 2, the player has to play greedily duringthe first part of the game by selecting vertices of G t that are of degree 0. This part of thegame ends at step (ln 2 + o (1)) n a.a.s. From that point on, she continues playing greedilyby selecting vertices of degree 1 which requires additional (ln(1 + ln 2) + o (1)) n steps a.a.s.In order to improve the lower bound (unfortunately, only by a hair) we will use anothertrivial observation. We will call a vertex x in D t problematic if it is of in-degree at least 3(out-degree of x is not important) with the in-neighbours y , y , y (if x has in-degree largerthan 3, then these are the first three in-neighbours sorted by the time when they were addedto the graph), each of them of out-degree 1 and in-degree 1. Since y i ’s are of degree 2 in theunderlying graph G t , the three edges y i x must be included in a potential Hamilton cycle but24hen, indeed, vertex x creates a problem. It gives us another trivial necessary condition: if G t has a Hamilton cycle, then there are no problematic vertices. Indeed, if G t has a vertex v adjacent to three vertices, all of which are of degree 2, then G t cannot be Hamiltonian. Thisresults in various types of “problematic” vertices. Our definition focuses only on a particulartype for the purpose of simplifying the proof.The numerical improvement is tiny and the bound we prove is certainly not tight. Hence,we only provide sketches of the proofs. The computations presented in the paper wereperformed by using Maple [8]. The worksheets can be found at the following address [13].For convenience, we will distinguish a few phases in the semi-random graph process. Thefirst phase lasts exactly n ln 2 steps. Our first goal is to show that if the player plays greedily,then a.a.s. there will be linearly many problematic vertices at the end of first phase. Claim 20.
Suppose that the player plays greedily during the first phase of the process. Then,a.a.s. there are ( ξ + o (1)) n problematic vertices at the end of this phase, where ξ = 1128 (cid:0) + 20(ln 2) + 54(ln 2) −
18 ln 2 − (cid:1) ≈ . . Proof.
It is fairly easy to show that the number of problematic vertices is a.a.s. at least ξn for some positive constant ξ . By the standard first and second moment calculations, afterthe first (ln 2 / n steps there will be at least ( e − c c / n vertices of in-degree at least 3 in D t where c = ln 2 /
2. Then, a.a.s. a positive fraction of these vertices turns out problematicduring the next (ln 2 / n steps. Of course, in order to get larger constant ξ it is best to trackthe process and apply the differential equation’s method (see [12] for more information onthe DE’s method). We briefly sketch the argument.For a, b, c ∈ { , } and a ≥ b ≥ c , we will say that a vertex x in D t is of type ( a, b, c ) if it isof in-degree at least 3, with the first three in-neighbours y , y and y (order is not important),each of which has out-degree 1 and in-degree a , b , and c , respectively. In particular, vertexof type (1 , ,
1) is simply a problematic vertex. Similarly, vertices of in-degree 2 could beof type ( a, b ) and vertices of in-degree 1 could be of type ( a ). The remaining vertices ofin-degree at least 1 are called neglected . (Note that neglected vertices can still preventHamilton cycle to be constructed but we simply neglect them.)In order to analyze the process, we need to keep track of 9 random variables associatedwith vertices of different types, random variables X abc , X ab , and X a . In particular, X ( t )is the number of problematic vertices (type (1 , , t . Moreover, let Y ( t )be the number of neglected vertices at the end of step t . It is straightforward to computethe conditional expectations; for example, E (cid:16) X ( t + 1) − X ( t ) | D t (cid:17) = X ( t ) n − X ( t ) n . Indeed, the only chance to create a problematic vertex is when the semi-random processselects the in-neighbour of a vertex of type (1 , ,
0) that is of in-degree 0. On the otherhand, if the process selects any of the first three in-neighbours of a problematic vertex, thisvertex becomes neglected. The other expectations can be computed in a similar way. Thissuggests the following system of differential equations that should reflect the behaviour of25he corresponding random variables: x ′ ( x ) = 1 − x ( x ) − x ( x ) − x ( x ) − x ( x ) − x ( x ) − x ( x ) − x ( x ) − x ( x ) − x ( x ) − y ( x ) − x ( x ) ,x ′ ( x ) = x ( x ) − x ( x ) ,x ′ ( x ) = x ( x ) − x ( x ) ,x ′ ( x ) = x ( x ) − x ( x ) ,x ′ ( x ) = 2 x ( x ) + x ( x ) − x ( x ) ,x ′ ( x ) = 3 x ( x ) + x ( x ) − x ( x ) ,x ′ ( x ) = x ( x ) − x ( x ) ,x ′ ( x ) = 2 x ( x ) + x ( x ) − x ( x ) ,x ′ ( x ) = x ( x ) − x ( x ) ,y ′ ( x ) = x ( x ) + x ( x ) + x ( x ) + 2 x ( x ) + 2 x ( x ) + 3 x ( x ) , with the initial condition that all functions at x = 0 are equal to zero. This system ofequations can be explicitly solved. In particular, we get that x ( x ) = e − x x e − x x e − x x e − x x e − x − e − x x − e − x + 9 e − x . It follows from the DE’s method that a.a.s. X ( t ) = (1 + o (1)) x ( t/n ) n for any 0 ≤ t ≤ n ln 2. Hence, a.a.s. the number of problematic vertices at the end of the first phase is equalto (1 + o (1)) x (ln 2) and the claim holds.The above claim implies that if the player concentrates on achieving minimum degree 2as soon as possible (that is, play greedily until the graph has minimum degree equal to 2),then a.a.s. there will be ( ξ + o (1)) n problematic vertices at the end of the first phase. If shecontinues playing greedily, then a.a.s. some positive fraction of these problematic verticeswill remain present in the graph. Making them negligible will take linearly many steps. Asa result, the player might want to adjust her strategy and not play greedily but start payingattention to problematic vertices instead. We now argue that this will also slow her down.For a given δ ∈ [0 ,
1] ( δ = δ ( n ) could be a function of n ), let F δ be a family of strategiesin which (1 − δ ) n ln 2 steps in the first phase are greedy (that is, the player selects someisolated vertex) but δn ln 2 steps are non-greedy (that is, the player selects some vertex ofdegree at least 1). We will show that playing non-greedily has a penalty in the form ofreaching minimum degree 2 later in comparison to the minimum degree 2 process. Claim 21.
Fix any δ ∈ [0 , . For any strategy from family F δ , a.a.s. it takes at least (ln 2 + ln(1 + ln 2) + ǫ ( δ ) + o (1)) n steps for G t to reach minimum degree 2, where ǫ ( δ ) = ln (cid:0) (2 δ −
1) ln(2 δ − − δ δ ln 2 + (1 + ln 2)2 δ (cid:1) − δ ln 2 − ln(1 + ln 2) , for δ ∈ [0 , / and ǫ ( δ ) = ǫ (1 / for δ ∈ (1 / , . ǫ ( δ ) is an increasing function of δ on [0 , /
2] and ǫ (0) = 0 (which correspondsto the original minimum degree 2 process). Proof.
It is important to notice that the objective here is only to eliminate all vertices ofdegree below 2, and thus the player does not need to worry about problematic vertices.First consider δ ∈ [0 , / δ = 0), it is straightforward to see (for example, by a simple couplingargument) that it is always beneficial to play a greedy move instead of a non-greedy one .Hence, in order to achieve our goal, the best strategy from the family F δ is to play on verticesof degree 0 during the first (1 − δ ) n ln 2 steps. After that, the player should select verticesof degree 1 until the end of the first phase , that is, during the following δn ln 2 steps. Asthere are no restrictions on the game after that (in particular, no restrictions on the numberof non-greedy moves), she should play greedily until the end of the game; that is, play onvertices of degree 0 until they disappear and then play on vertices of degree 1 until theend of the game. Hence, both the first and the second phase are split into two sub-phases,depending on which type of vertices are selected.In order to analyze how long it takes to finish this process, we need to keep track oftwo random variables: Y ( t ) and Z ( t ), the number of vertices at time t of degree 0 and 1,respectively. We say that a move is of type i (where i ∈ { , } ) if the player plays on avertex of degree i . It is not difficult to see that E (cid:16) Y ( t + 1) − Y ( t ) | G t and type i (cid:17) = − δ i =0 − Y ( t ) n E (cid:16) Z ( t + 1) − Z ( t ) | G t and type i (cid:17) = δ i =0 − δ i =1 + Y ( t ) n − Z ( t ) n . where δ A is the Kronecker delta function ( δ A = 1 if A is true and δ A = 0 otherwise). Thecorresponding system of DEs is y ′ ( x ) = − δ i =0 − y ( x ) z ′ ( x ) = δ i =0 − δ i =1 + y ( x ) − z ( x ) . The initial condition is y (0) = 1 and z (0) = 0. Moreover, the final values of y ( x ) and z ( x ) after one of the sub-phases are used as the initial values for the next sub-phase. Theconclusion follows from the DE’s method. We skip the details and refer the interested readerto the Maple worksheets available on-line.It is easy to see that if 1 / < δ ≤ F δ a.a.s. takes at least(ln 2 + ln(1 + ln 2) + ǫ (1 /
2) + o (1)) n steps to build a graph with minimum degree at least2. During the second sub-phase of phase 1, the player may select any non-isolated vertexif there are no vertices of degree 1 left. These moves are not helping with building a graphwith minimum degree 2 and thus it takes even longer to complete the process.Our next task is to estimate the number of problematic vertices at the end of the firstphase, provided that the player uses a strategy from family F δ . For any strategy f of F δ which does not prioritize greedy moves first, there exists another strategy within F δ which does prioritize greedy moves first, and whose completion time is stochastically dominated by thecompletion time of f . laim 22. Fix any δ ∈ [0 , ξ/ (2 ln 2)] , where ξ is defined in Claim 20. For any strategy fromfamily F δ , a.a.s. there are at least ( ξ − δ ln 2 + o (1)) n problematic vertices at the end of thefirst phase.Proof. It is not clear what the best strategy for minimizing the number of problematicvertices is. So, in order to keep the argument as simple as possible, we will help the playerand propose to play the following auxiliary game, a mixture of on-line and off-line variantsof the game. We simply run the greedy algorithm by selecting an isolated vertex in eachstep of the process. It follows from Claim 20 that a.a.s. there are ( ξ + o (1)) n problematicvertices at the end of the first phase. After that, we ask the player to ‘rewind’ the processand carefully ‘rewire’ δ fraction of moves in any way she wants keeping the remaining 1 − δ fraction of moves greedy, as required. Each modified move affects at most two problematicvertices so the number of problematic vertices decreases by at most 2 · δn ln 2. Since thistask clearly is much easier for the player than the original one, the lower bound follows.Our final task is to combine all results together. Claim 23.
Fix any δ ∈ [0 , ξ/ (2 ln 2)] , where ξ is defined in Claim 20. For any strategy fromfamily F δ , a.a.s. it takes at least (ln 2 + ln(1 + ln 2) + ǫ ( δ ) + ǫ ( δ ) + o (1)) n steps for G t to reach minimum degree 2 and remove all problematic vertices that were createdduring the first phase. Function ǫ ( δ ) is defined in Claim 21 and ǫ ( δ ) = ln (cid:0) τ ( δ ) + 1 (cid:1) ,τ ( δ ) = ( ξ − δ ln 2) exp( − − ǫ ( δ )) . Proof.
As in the proof of the previous claim, it is not clear what the best strategy is. Sincewe aim for an easy argument without optimizing the constants, we propose the player to playthe following auxiliary game. We let her play the degree-greedy algorithm from the family F δ which optimizes the time needed to achieve minimum degree 2 (without worrying aboutproblematic vertices). At the end of the first phase we artificially ‘destroy’ some problematicvertices (if needed), leaving only ( ξ − δ ln 2 + o (1)) n of them in the graph. Clearly, this isan easier game for the player to play. Indeed, by Claim 22 any strategy from F δ creates atleast that many problematic vertices and so this is certainly a sweet deal for her.The player continues the game trying to reach minimum degree at least 2 and to destroythe remaining problematic vertices. It is straightforward to see that the best strategy isto continue playing the degree-greedy algorithm, destroying the remaining isolated verticesbefore playing vertices of degree 1. That part is taking (ln(1 + ln 2) + ǫ ( δ ) + o (1)) n stepsby Claim 21. In the meantime, vertices selected by the random graph process land on theneighbours of problematic vertices. The probability that a given problematic vertex is notdestroyed is equal to (cid:18) − n (cid:19) (ln(1+ln 2)+ ǫ ( δ )+ o (1)) n = exp (cid:16) − (cid:0) ln(1 + ln 2) + ǫ ( δ ) (cid:1)(cid:17) + o (1) . τ ( δ ) + o (1)) n problematic vertices at this point.After that, the player has to destroy the remaining problematic vertices. Obviously, thebest strategy is to choose v t to be one of the first three neighbours of a problematic vertex.A problematic vertex x can also be destroyed if u t happens to be one of these neighbours.Let Y ( t ) be the number of problematic vertices at the end of step t (for simplicity countingfrom t = 0). It is straightforward to see that E (cid:16) Y ( t + 1) − Y ( t ) | G t (cid:17) = − − Y ( t ) n . The corresponding DE is y ′ ( x ) = − − y ( x ) with the initial condition y (0) = τ ( δ ). It followsthat y ( x ) = − / τ ( δ ) + 1 / e − x and so we get that a.a.s. it takes another ( ǫ ( δ ) + o (1)) n steps to finish the game, and the claim holds.Theorem 2 follows immediately from Claim 23. Let us first extend ǫ ( δ ) to [0 ,
1] bysetting ǫ ( δ ) = 0 for δ ∈ ( ξ/ (2 ln 2) , δ ∈ [0 , F δ a.a.s. takes at least (ln 2 + ln(1 + ln 2) + ǫ ( δ ) + ǫ ( δ ) + o (1)) n steps to build aHamilton cycle. Note that ǫ ( δ ) is an increasing function of δ ; the more non-greedy movesthe player needs to play, the longer the game is. On the other hand, ǫ ( δ ) is a decreasingfunction on [0 , ξ/ (2 ln 2)] with ǫ ( ξ/ (2 ln 2)) = 0; the non-greedy moves can be spent ondestroying problematic vertices and so the number of them decreases with δ . After morecareful investigation we get that ǫ ( δ ) + ǫ ( δ ) is a decreasing function on [0 , ξ/ (2 ln 2)] andthen it is equal to ǫ ( δ ) and so it starts increasing. Therefore we get that ǫ = min δ (cid:16) ǫ ( δ ) + ǫ ( δ ) (cid:17) = ǫ (cid:18) ξ (cid:19) + ǫ (cid:18) ξ (cid:19) = ǫ (cid:18) ξ (cid:19) ≈ . · − . References [1] O. Ben-Eliezer, L. Gishboliner, D. Hefetz, M. Krivelevich, Very fast construction ofbounded degree spanning graphs via the semi-random graph process. Proceedings ofthe 31st Symposium on Discrete Algorithms (SODA’20), 728–737 (2020).[2] O. Ben-Eliezer, D. Hefetz, G. Kronenberg, O. Parczyk, C. Shikhelman, M. Stojakovic,Semi-random graph process, to appear in
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