Spanning trees at the connectivity threshold

Yahav Alon Michael Krivelevich Peleg Michaeli

October 29, 2020

Abstract

We present an explicit connected spanning structure that appears in a random graphjust above the connectivity threshold with high probability.

1 Introduction

The binomial random graph G(n, p) is a graph on n vertices, in which every pair of verticesis connected independently with probability p. It is a well known and thoroughly studiedmodel (see, e.g., [6, 8, 12]). A fundamental result, due to Erdős and Rényi [7], it that G(n, p)exhibits a sharp threshold for connectivity at p = p(n) = log n/n1. More precisely, settingp = (log n + f(n))/n and G ∼ G(n, p), if f(n) → ∞ then G is with high probability2 (whp)connected, and if f(n)→ −∞ then G is whp not connected. This threshold coincides with thethreshold for the disappearance of isolated vertices (vertices of degree 0), and, in fact, isolatedvertices are the bottleneck for connectivity in a stronger sense.

Evidently, a connected graph has a spanning tree. This raises the natural question of whichspanning trees can we expect to find in a connected random graph. The standard proof for theconnectivity threshold is obtained using the dual definition of connectivity, namely, by showingthat whp there is an edge in every cut; this seems to provide no hint of which trees appearabove the threshold. One can check, however, that just above the connectivity threshold arandom graphs contains Θ(log n) vertices of degree 1. Obviously, these vertices must all beleaves in any spanning tree. In particular, one cannot expect to find a Hamilton path (or anyspanning tree with a constant number of leaves). In this work, we present an explicit spanningtree that appears in a random graph just above the connectivity threshold whp. In fact, weprove something stronger, by presenting a concrete unicyclic connected spanning subgraph.Let n, t, ` be integers such that t · (` + 1) ≤ n. A KeyChain with parameters n, t, `, denotedKC(n, t, `), is a cycle on n− t vertices with additional t vertices of degree 1 (“keys”) which havedistinct neighbours in that cycle, where the distance between two consecutive such neighboursis `. Formally, it is the graph H = (V,E) with V = [n] and

E = {{i, i+ 1} | i ∈ [n− t− 1]} ∪ {{n− t, 1}} ∪ {{i`, n− t+ i} | i ∈ [t]}.

See Fig. 1 for an example. Consider the following sequence: a1 = 1, aj+1 = daj · log n/100e,and set j0 to be the minimum index j for which aj ≥ 10n/ log n. Let further t := blog nc and` = 2j0.

1Here and later the logarithms have natural base.2With probability tending to 1 as n tends to infinity.

1

Figure 1: KC(24, 5, 3).

Theorem 1.1. Let p be such that np− log n→∞. Then, G ∼ G(n, p) whp contains KC(n, t, `)as a subgraph.

As KC(n, t, `) is connected and spanning, and in fact contains⌈n−t

2

⌉distinct (unlabelled)

spanning trees when n is sufficiently big, Theorem 1.1 gives a set of distinct trees, all of whichappear in G ∼ G(n, p) whp. Our proof provides, however, further spanning (but not necessarilyconnected) graphs that appear whp in random graphs above the connectivity threshold (suchas KeyChains on linearly many vertices alongside a disjoint cycle spanning the remaining setof vertices).

1.1 Discussion

Hamiltonicity Komlós and Szemerédi [14] and independently Bollobás [5] showed that thethreshold for the appearance of a Hamilton cycle in random graphs is p = (log n+ log log n)/n.This coincides with the threshold for the disappearance of vertices of degree 1, an obviousobstacle in obtaining a Hamilton cycle. In particular, if np − log n − log logn → ∞ andG ∼ G(n, p), thenG contains a Hamilton path as a spanning tree. Thus, our result is interesting,and perhaps surprising, for values of p which satisfy np = logn + f(n), where f(n) → ∞ butf(n)− log logn 6→ ∞. Our result states that in this intermediate regime, we can still constructa specific tree which contains a path that spans almost all of the vertices of the graph.

Spanning trees in random graphs Theorem 1.1 gives an explicit set of trees, each ofwhich spans a binomial random graph just above the connectivity threshold whp. Which otherspanning trees can we expect to find around the same edge density? Recently Montgomery [16]solved a conjecture by Kahn (see [13]) according to which every n-vertex tree of maximumdegree at most ∆ appears in G(n, p) whp if np ≥ C log n for C = C(∆). Earlier, Hefetz,Krivelevich and Szabó [11] showed that every bounded degree n-vertex tree with either linearlymany leaves or a linearly long bare path appears in G(n, p) whp if np ≥ (1 + ε) log n forsome ε > 0. The question of which trees are likely to appear in G(n, p) if one only requiresnp− log n→∞ remains open.

The maximum common subgraph problem Finding a maximum common subgraph oftwo graphs (sometimes called maximum common edge subgraph, or MCES) is an NP-hardproblem (for example, there is a trivial polynomial reduction from the Hamiltonicity problemto MCES) with real-world applications in computer science and in chemistry (see, e.g., [4,18]).Theorem 1.1 can be thought of as an explicit (connected) subgraph which is typically commonto random graphs past the connectivity threshold. In particular, writing M(G1, G2) for thesize of a maximum common subgraph of G1, G2, Theorem 1.1 gives M(G1, G2) ≥ n whp for

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independently sampled G1, G2 ∼ G(n, p), where np − log n → ∞. The following proposition,whose proof appears in Section 5, shows that this bound is asymptotically tight.

Proposition 1.2. For every ε > 0 there exists δ > 0 for which the following holds. Suppose p ≤n−1+δ, and let G1, G2 ∼ G(n, p) be two independent random graphs. Then, whp, M(G1, G2) ≤(1 + ε)n.

1.2 Proof outline

We begin by listing useful properties of random graphs just above the connectivity threshold.With these properties in hand, we continue as follows. First, we take care of all “keys” of theKeyChain: these consist of all vertices of degree 1 in the graph, in addition to some verticesof degree 2. We aim to connect neighbours of these keys by equal-length paths, constructinga comb-like graph. Eventually, we would want to connect the ends of the comb by a pathwhich spans the remaining vertices. This cannot be done naively, however, since some of theremaining vertices may have most (or all) of their neighbours inside the comb. To overcomethis difficulty we make a preparatory step in which we put aside small degree vertices with theirneighbours. This is stated precisely in Section 4. The “comb” is then found in the set of theremaining vertices.

Organisation of the paper We start by reviewing some preliminaries in Section 2. InSection 3 we list and prove useful (and mostly standard) properties of random graphs justabove the connectivity threshold. In Section 4 we prepare our graph, handling future “keys”and small degree vertices, construct the “comb” and close it to a KeyChain, concluding theproof of Theorem 1.1. We finish by a quick proof of Proposition 1.2 in Section 5.

2 Preliminaries

In this section we provide several definitions and results to be used in this paper.

2.1 Notation

The following graph theoretic notation is used. For a graph G = (V,E) and two disjoint vertexsubsets U,W ⊆ V , we let EG(U,W ) denote the set of edges of G adjacent to exactly one vertexfrom U and one vertex from V , and let eG(U,W ) = |EG(U,W )|. Similarly, EG(U) denotesthe set of edges spanned by a subset U of V , and eG(U) stands for |EG(U)|. The (external)neighbourhood of a vertex subset U , denoted by NG(U), is the set of vertices in V \U adjacentto a vertex of U , and for a vertex v ∈ V we set NG(v) = NG({v}). The degree of a vertexv ∈ V , denoted by dG(v), is its number of incident edges. For an integer 0 ≤ i < |V |, we letDi = Di(G) be the set of vertices of degree i in G, and let D≤i = D≤i(G) :=

⋃ij=0Dj . Finally,

we let ni = ni(G) := |Di| and n≤i = n≤i(G) := |D≤i|. In the above notation, we sometimesomit the subscript G if the graph G is clear from the context. We occasionally suppress therounding notation to simplify the presentation. For two functions f = f(n), g = g(n) we writef ∼ g to indicate that f = (1 + o(1))g.

2.2 Probabilistic bounds

We will make use of the following useful bound (see, e.g., in, [12, Chapter 2]).

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Theorem 2.1 (Chernoff bounds). Let X =∑n

i=1Xi, where Xi ∼ Bernoulli(pi) are independent,and let µ = EX =

∑ni=1 pi. Let 0 < α < 1 < β. Then

P(X ≤ αµ) ≤ exp(−µ(α logα− α+ 1)),P(X ≥ βµ) ≤ exp(−µ(β log β − β + 1)).

The following are trivial yet useful bounds.

Claim 2.2. Let X ∼ Bin(n, p) with µ = np and let 1 ≤ k ≤ n. Then

P(X ≥ k) ≤(enpk

)k.

For a proof see, e.g., [9].

Claim 2.3. Let X ∼ Bin(n, p) with µ = np, write q = 1− p and let 1 ≤ k ≤ np/q. Then

P(X ≤ k) ≤(enp

kq

)ke−np.

2.3 Pósa’s lemma and corollaries

For an overview of the rotation–extension technique, we refer the reader to [15].

Lemma 2.4 (Pósa’s lemma [17]). Let G be a graph, let P = v0, . . . , vt be a longest path in G,and let R be the set of all v ∈ V (P ) such that there exists a path P ′ in G with V (P ′) = V (P )and with endpoints v0 and v. Then |N(R)| ≤ 2|R| − 1.

Recall that a non-edge of G is called a booster if adding it to G creates a graph whichis either Hamiltonian or whose longest path is longer than that of G. For a positive integer kand a positive real α we say that a graph G = (V,E) is a (k, α)-expander if |N(U)| ≥ α|U |for every set U ⊆ V of at most k vertices. The following is a widely-used fact stating that(k, 2)-expanders have many boosters. For a proof see, e.g., [15].

Lemma 2.5. Let G be a connected (k, 2)-expander which contains no Hamilton cycle. Then Ghas at least (k + 1)2/2 boosters.

It is not hard to see that an (n/4, 2)-expander on n vertices is connected, a fact that we willuse later on. We will also use the following deterministic lemma from [10].

Lemma 2.6 ([10, Lemma 4.6]). Let m, d ≥ 1 be integers, and let H be a graph on h ≥ 4mvertices satisfying the following properties:

1. δ(H) ≥ 2;

2. No vertex v ∈ V (H) with d(v) < d is contained in a 3- or a 4-cycle, and every two distinctvertices u, v ∈ V (H) with d(u), d(v) < d are at distance at least 5 apart;

3. Every set F ⊆ V (H) of size at most 5m spans at most d|F |/10 edges;

4. There is an edge between every pair of disjoint sets F1, F2 ⊆ V (H) of size m each.

Then H is a (h/4, 2)-expander (and, in particular, it is connected).

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3 Properties of random graphs

We now present and prove some typical properties of random graphs above the connectivitythreshold, to be used in the proof of Theorem 1.1.

DenoteSmall = Small(G) := D≤logn/10(G).

Additionally, setγ = 10−4.

Lemma 3.1. Let 1� f(n)� log logn, p = (log n+ f(n))/n and G ∼ G(n, p). Then, whp, Ghas the following properties:

(P1) ∆(G) ≤ 10 log n;

(P2) |D1| ≤ log n ≤ |D2|;

(P3) There is no path of length at most 0.2 log n/ log logn in G whose (possibly identical)endpoints lie in Small;

(P4) |Small ∪N(Small)| ≤ n0.6;

(P5) There in no U ⊆ V (G) with |U | ≤ 10n/ log n, e(U, V (G) \ U) ≥ |U | log n/11 such that|N(U)| ≤ |U | log n/18;

(P6) Every set U ⊆ V (G) of size at most γn/5000 spans at most γ log n · |U |/1000 edges;

(P7) For every U,W ⊆ V (G) disjoint with 10n/ log n ≤ |U |, |W | ≤ n/9, |N(U)∩N(W )| ≥ n/9;

(P8) For every U,W ⊆ V (G) disjoint with |U |, |W | ≥ γn25000 , |E(U,W )| ≥12 |U ||W | log n/n;

Proof of (P1). Since d(v) ∼ Bin(n− 1, p) for every v ∈ V (G), we have

P(d(v) ≥ 10 log n) ≤(

n

10 log n

)p10 logn ≤

(enp

10 log n

)10 logn= o(1/n),

and the statement follows by the union bound.

Proof of (P2). The probability that a given vertex is of degree 1 is at most np(1 − p)n−2 =o(log n/n). Thus E[n1] = o(log n) and by Markov’s inequality P(n1 > log n) = o(1). Theexpectation of n2 is n ·

(n−1

2

)p2(1− p)n−3 = ω(log n). On the other hand, the variance is

Var[n2] = E[n2

2]− E[n2]2

= −E[n2]2 + E[n2] +∑

(u,v)∈V (G)2u6=v

P(dG(u) = dG(v) = 2)

= (−1 + o(1)) · E[n2]2 + n(n− 1) ·

((n− 2

2

)2p4(1− p)2n−7 + np3(1− p)2n−6

)= o(E[n2]2

),

which, by Chebyshev’s inequality, implies that P(n2 < log n) = o(1).

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Proof of (P3). Write L := 0.2 log n/ log log n. Let 1 ≤ ` ≤ L and let P = (v0, . . . , v`) be asequence of `+ 1 distinct vertices from V (G), with the one possible exception v0 = v`.

Suppose first that v0 6= v`. Let S := V (G) \ {v0, v1, v`−1, v`}, let AP be the event that P iscontained in G as a path, and let Ad be the event that d({v0, v`}, S) ≤ log n/5. By Theorem 2.1with α = logn5·(2n−5)p = 1/10 + o(1), we obtain that

P(Ad) ≤ exp(−(2n− 5)p · (α logα− α+ 1)) ≤ n−1.3.

The events AP ,Ad are independent, hence P(AP ∧Ad) ≤ p`n−1.3. Let A be the event that thereexists a path P = v0, . . . , v` with 1 ≤ ` ≤ L in G such that AP and d(v0), d(v`) ≤ log n/10, theprobability of which is at most P(AP ∧ Ad). By the union bound, summing over all sequencelengths and all sequences, we get

P(A) ≤L∑`=1

n`+1p`n−1.3 ≤L∑`=1

nlog (np)logn

·`−0.3 ≤ L · n1.1 log logn

logn·L−0.3 ≤ L · n−1/20 = o(1).

The case v0 = v` is similar. Let S := V (G) \ {v1, v`−1} and let Ad be the event d(v0, S) ≤log n/10. Once again by Theorem 2.1, P(Ad) ≤ n−0.6, and the events AP ,Ad are independent,hence P(Ap∧Ad) ≤ p`n−0.6. Let A′ be the event that there exists a cycle P of length 3 ≤ ` ≤ Lsuch that AP and d(v0) ≤ log n/10. By the union bound, P(A′) ≤

∑L`=3 n

`p`n−0.6 = o(1).Finally, observe that P(G ∈ (P3)) ≥ 1−P(A)−P(A′), and so the statement is obtained.

Proof of (P4). Thus, by Theorem 2.1 we have P(d(v) ≤ log n/10) ≤ n−0.6, and therefore byMarkov’s inequality |Small| ≤ n0.5 whp. By the definition of Small we have that |Small ∪N(Small)| ≤ n0.6 whp.

Proof of (P5). For U ⊆ V (G) with |U | = k, letAU be the event that e(U, V (G)\U) ≥ k log n/11and |N(U)| ≤ k log n/18. On this event, there exists W ⊆ V (G) of size k log n/18, disjoint ofU , which contains N(U). Denote this event by AU,W . Evidently, using Claim 2.2,

P(AU,W ) ≤ P(Bin(k2 log n/18, p) ≥ k log n/11

)· (1− p)k(n−k−k logn/18)

≤(

11ekp

18

)k logn/11· e−pk(n−k−k logn/18) ≤

(11ekp

18

)k logn/11· e−0.4npk.

For k ≤ 10n/ log n let Ak denote the event that there exists U with |U | = k for which AUoccurs. By the union bound over all possible choices of U,W we have that

P(Ak) ≤(

n

k + k log n/18

)(k + k log n/18

k

)· P(AU,W )

≤(

en

k + k log n/18

)(1/18+o(1))k logn· logk n · P(AU,W )

≤(

50n

k log n

)(1/18+o(1))k logn·(

11ekp

18

)k logn/11· e−0.4npk

≤

[(n

k log n

)1/18−1/11+o(1)· 501/17 · 21/11 · e−0.4

]k logn≤ exp(k log n · (log(10)/25 + log(50)/17 + log(2)/11− 0.4)) = e−Ω(k logn).

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Finally, let A denote the event that there exists U with 1 ≤ |U | ≤ 10n/ log n for which AUoccurs. By the union bound over all possible cardinalities k = |U | we have

P(A) ≤10n/ logn∑k=1

P(Ak) ≤∞∑k=1

e−Ω(k logn) = o(1).

Proof of (P6). Fix U ⊆ V (G) and set γ′ = γ/5000. By Claim 2.2, the probability that |E(U)| ≥m for some m ≥ 1 is at most (e|U |2p/m)m. Hence, by the union bound, the probability thatthere is a subset U ⊆ V (G) negating (P6) is at most

γ′n∑k=1

(n

k

)(ekp

5γ′ log n

)5γ′ logn·k

≤γ′n∑k=1

((enk

)·(

0.6k

γ′n

)5γ′ logn)k

≤γ′n∑k=1

(e

γ′· 0.6Ω(logn)

)k=

γ′n∑k=1

o(1)k = o(1),

and the statement follows.

Proof of (P7). If U,W ⊆ V (G) are disjoint with 10n/ log n ≤ |U |, |W | ≤ n/9, and |N(U) ∩N(W )| < n/9, then there is a set X ⊆ V (G) \ (U ∪W ) of size 0.6n such that every vertexx ∈ X is connected to no vertex in U , or connected to no vertex in W . By the union bound,the probability of this is at most

n/9∑k= 10n

logn

n/9∑s= 10n

logn

(n

k

)(n

s

)(n

0.6n

)·(

(1− p)k + (1− p)s)0.6n

≤ n2 · (9e)2n/9 · (2e)0.6n · 20.6n · exp(− 10n

log np · 0.6n

)≤ exp((o(1) + log(9e)/2 + 2 log 2 + 1− 10) · 0.6n) = o(1),

and the statement follows.

Proof of (P8). If U,W ⊆ V (G) are disjoint, and |U |, |W | ≥ γn/25000, then by Theorem 2.1,the probability that e(U,W ) < 12 |U ||W | log n/n ≤

23E(e(U,W )) is of order

exp(−Ω(E(e(U,W )))) = exp(−Ω(|U ||W | log n/n)) = exp(−Ω(n log n)).

By the union bound, the probability that such U,W exist is at most

3n · exp(−Ω(n log n)) = o(1).

The following lemma describes yet another property of G(n, p), which might need furtherexplanation. Recall that G ∼ G(n, p) is not assumed to be Hamiltonian, and in the relevantregime typically does not contain a Hamilton path. We will need, however, to find a path (and,in fact, many such paths) which spans a large predetermined portion of its vertices. It is nothard to see that subgraphs spanned by carefully chosen (large) sets of vertices are (very) good

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expanders. Our way to argue that they are Hamiltonian, however, will be to show that theycontain sparser expanders. To this end, we will use the method of “random sparsification”which has become a fairly standard tool in the study of Hamiltonicity (see, e.g., in [2,3,9,15]).The main idea behind this, which is the essence of the next lemma, is that we can show thatwhp every sparse expander will have a relative booster.

Lemma 3.2. Let 1 � f(n) � log log n, p = (log n + f(n))/n and G ∼ G(n, p). Then, whp,for every W ⊆ V (G) of size |W | = h ≥ n/10 and for every (h/4, 2)-expander H on W which isa subgraph of G with at most γn log n/100 edges, G contains a booster with respect to H.

For later references we will name the property described above (P9).

Proof. Fix W ⊆ V (G) such that h := |W | ≥ n/10, and let H be an (h/4, 2)-expander on Wwith m edges, that is a subgraph of G. Recall that H is connected. Hence, by Lemma 2.5 weknow that G[W ] has at least n2/3200 boosters with respect to H. Thus, the probability thatG[W ] contains H but no booster thereof is at most

pm · (1− p)n2/3200 ≤ pm ·(

1− log nn

)n2/3200≤(

2 log n

n

)m· exp(−n log n/3200).

Write β = γ/100. As there are at most 2n choices for W and at most(n2

m

)≤(en2/m

)mchoices

for H for each 1 ≤ m ≤ βn log n, we have, by the union bound, that the probability that thereexist such W and H for which G[W ] does not contain a booster with respect to H is at most

2n · exp(−n log n/3200) ·βn logn∑m=1

(2en log n

m

)m. (1)

Set g(m) = (2en log n/m)m and observe that g′(m) = g(m) · (log(2en log n/m) − 1), which ispositive for 1 ≤ m ≤ βn log n. Thus, the sum in (1) can be bounded from above by

βn log n · (2e/β)βn logn = exp((β log(2e/β) + o(1))n log n),

which, recalling that β = γ/100 = 10−6, is smaller than exp(n log n/3201), and thus (1) tendsto 0 as n→∞.

4 Constructing a KeyChain

In Section 3 we have identified useful properties which are satisfied by random graphs whp. Inthis section we will assume our graph possesses these properties, and show that this determin-istically implies that the graph contains a KeyChain.

For convenience, let us repeat some definitions from Sections 1 and 3. Consider the followingsequence: a1 = 1, aj+1 = daj · log n/100e, and set j0 to be the minimum index j for whichaj ≥ 10n/ log n. Let further

t := blog nc and ` = 2j0.

Finally, recall that γ = 10−4. In this section we prove the following lemma, which, when puttogether with Lemmas 3.1 and 3.2, completes the proof of Theorem 1.1.

Lemma 4.1. Any n-vertex graph G satisfying Properties (P1)–(P9) contains KC(n, t, `) as asubgraph.

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Our plan is as follows. We want to construct a “comb”, which consists of t keys (all verticesof degree 1 in addition to some vertices of degree 2), and equal-length paths between neighboursof consecutive keys. We will then want to connect the endpoints of the comb by a path whichspans the remaining set of vertices. As hinted in the introduction, we cannot carelessly do so,as some vertices outside the comb might have most or all of their neighbours inside the comb.Instead, we have to make a preparatory step, in which we put aside vertices of small degree(except the future “keys” of the KeyChain) along with their neighbourhoods. This preparatorystep is Lemma 4.2. Given the partition in Lemma 4.2, we construct a comb in its large part(in Lemma 4.5), and then connect the endpoints with a path that spans the remaining set ofvertices.

In the next lemmas we assume G is an n-vertex graph satisfying Properties (P1)–(P9).

Lemma 4.2. There exist a partition V (G) = V ? ∪ V ′ with |V ?| ∼ 2γn and a set K ⊆ V ′ with|K| = t for which the following holds:

(a) D1(G) ⊆ K ⊆ D≤2(G);

(b) N(K) ⊆ V ′;

(c) d(v, V ?) ≤ 200γ log n and d(v, V ′) ≥ log n/20 for every v ∈ V ′ \K;

(d) If K ⊆ X ⊆ V ′ satisfies |X| ≤ n/2 and w1, w2 ∈ X \ K then there exist z1 ∼ w1 andz2 ∼ w2 in V (G) \X, and a Hamilton path from z1 to z2 in G[V (G) \X].

The proof of Lemma 4.2 is based on two ingredients. The first ingredient (Lemma 4.3) takescare of the actual partition obtained by Lemma 4.2. The second ingredient (Lemma 4.4), whichis the core of the proof, gives (d), by showing that inside “well-prepared” sets, one can findHamilton paths with linearly many distinct endpoints, emerging from a given vertex.

Lemma 4.3. There exist disjoint sets K,U1, U2 ⊆ V (G) with |K| = t and |U1|, |U2| ∼ γn forwhich the following holds. Write V ′ = V (G) \ (U1 ∪ U2). Then

(a) D1(G) ⊆ K ⊆ D≤2(G);

(b) For every v ∈ K, N(v) ⊆ V ′ \K;

(c) If v /∈ Small then γ log n/100 ≤ d(v, U1), d(v, U2) ≤ 100γ log n and d(v, V ′) ≥ log n/20;

(d) If v ∈ Small \K then v and all of its neighbours are in U1.

Proof. The proof involves an application of a the symmetric form of the Local Lemma (see,e.g., [1, Chapter 5]; a similar application appears in [11] and in [10]). Write V = V (G),α = 1/10, s = 1/γ. Let r = bn/sc ∼ γn and let A1, . . . , Ar, Z be a partitioning of the verticesof G into r “blobs” Ai of size s and an extra set Z with 0 ≤ |Z| < s. For j ∈ [r] let (x1j , x2j )be a uniformly chosen pair of distinct vertices from Aj . For i = 1, 2 define U

′i = {xij}rj=1.

Clearly, |U ′1| = |U ′2| = r and U ′1 ∩ U ′2 = ∅. For every v /∈ Small let B−v be the event thatd(v, U ′i) < 2γα

2 log n for some i = 1, 2, and let B+v be the event that d(v, U ′i) > γα−2 log n/2for some i = 1, 2. For such v, let L(v) be the set of blobs that contain neighbours of v,namely, L(v) = {Aj : N(v) ∩ Aj 6= ∅}. For j ∈ [r] write nj(v) = |N(v) ∩ Aj |, and note that∑

j nj(v) ≥ d(v) − s ≥ α log n/2. For i = 1, 2 and j ∈ [r] let χij(v) be the indicator of theevent that xij is a neighbour of v, and note that Eχij(v) = γnj(v). Observe that for i = 1, 2,

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d(v, U ′i) =∑

j χij(v), hence E[d(v, U ′i)] = γ

∑j nj(v) ≥ γα log n/2. Thus, by Theorem 2.1,

P(B−v ) ≤ n−c− for some c− > 0. Similarly, by (P1), E[d(v, U ′i)] = γ∑

j nj(v) ≤ γα−1 log n.Thus, by Theorem 2.1, P(B+v ) ≤ n−c+ for some c+ > 0. We conclude that for Bv = B−v ∪ B+vwe have P(Bv) ≤ n−c for some c > 0.

For two distinct vertices u, v /∈ Small say that u, v are related if L(u) ∩ L(v) 6= ∅. Fora vertex u /∈ Small let R(u) be the set of vertices in V \ Small which are related to u, andnote that |R(u)| ≤ s∆(G)2, which is, by (P1), at most C log2 n for some C > 0. Note that Buis mutually independent of the set of events {Bv | v ∈ (V \ Small) \R(u)}. We now apply thesymmetric case of the Local Lemma: observing that en−c · C log2 n < 1 (for large enough n),we get that with positive probability, none of the events Bv occur. We choose U ′1, U ′2 to satisfythis.

Choose a set K of size t arbitrarily to satisfy (a); this is possible due to (P2). WriteK+ = K ∪N(K) and S+ = Small ∪N(Small) ⊇ K+. Note that |K+| ≤ 3 log n by (P2) andthat |S+| ≤ n0.6 by (P4). Define U1 = (U ′1 ∪ S+) \ K+ and U2 = U ′2 \ S+ and observe that|U1|, |U2| ∼ γn. Due to (P3), N(Small \K) ⊆ S+ \K+, hence the construction satisfies (d).Let v /∈ Small. The fact that G satisfies (P3) implies that v has at most 1 neighbour in S+.Thus, for every v /∈ Small and i = 1, 2 it holds that |d(v, Ui)− d(v, U ′i)| ≤ 1, and, in addition,d(v, V ′) ≥ d(v, V \ (U ′1 ∪ U ′2)) − 1 ≥ α log n/2, hence (c) is satisfied. By the discussion aboveand by (P3), (b) is also satisfied. We have thus proved the statement.

Lemma 4.4. If W ⊆ V (G) is a vertex subset such that |W | ≥ n/10, δ(G[W ]) ≥ 2, and suchthat for every v ∈ W we have d(v,W ) ≥ min{d(v), γ log n/100}, then for every w ∈ W thereexists R ⊆ W with |R| ≥ n/40 such that for each y ∈ R, there is a Hamilton path in G[W ]whose endpoints are w and y.

Proof. Write β = γ/100 and set d0 = β log n. Let W ⊆ V (G) satisfy h := |W | ≥ n/10and δ(G[W ]) ≥ 2, and assume that for every v ∈ W we have d(v,W ) ≥ min{d(v), β log n}.We select a random edge subgraph H of G[W ] as follows. For each v ∈ W , if d(v,W ) ≤ d0set E(v) = EG(v,W ); otherwise, namely if d(v,W ) > d0, then set E(v) to be a (uniformly)selected set of d0 random edges of G[W ] which are incident to v. Let H = (W,EH) withEH =

⋃v∈W E(v). Observe that |E(H)| ≤ h · d0 ≤ βn log n/10.

We now show that H is, with positive probability, a (connected) (h/4, 2)-expander. Takingd = d0 and m = βn/250, and noting that h ≥ n/10 ≥ 4m, it is enough to show that Hsatisfies, with positive probability, Conditions 1–4 in Lemma 2.6. For the first condition, notethat δ(H) ≥ min{d0, δ(G[W ])} ≥ 2. The second condition holds as it holds for G by (P3) (sinced ≤ log n/10), and clearly also for every subgraph thereof. Similarly, noting that 5m = βn/50,the third condition holds as it holds for G by (P6).

We move on the prove the fourth condition of Lemma 2.6. Let F1, F2 ⊆W with |F1|, |F2| ≥m. By (P8) we know that |E(F1, F2)| ≥ cn log n for c = 10−6β2. For u ∈ F1 for whichdG(u, F2) ≥ 1 let Au be the event that none of the edges of D(u) is incident to a vertex ofF2. By the construction of H, if dG(u,W ) ≤ d0 then P(Au) = 0. On the other hand, ifdG(u,W ) > d0 then, using (P1),

P(Au) ≤(dG(u,W )− dG(u, F2)

d0

)/

(dG(u,W )

d0

)=

d0−1∏i=0

dG(u,W )− dG(u, F2)− idG(u,W )− i

≤(

1− dG(u, F2)dG(u,W )

)d0≤(

1− dG(u, F2)10 log n

)d0≤ exp(−dG(u, F2) · β/10).

10

Note also that Au are independent for different u ∈ F1. Thus,

P(EH(F1, F2) = ∅) ≤ exp

− β10

∑u∈F1

dG(u, F2)

= exp(− β10|EG(F1, F2)|

)≤ e−c′n logn

for c′ = 10−7β3. By taking the union bound over all at most 22n choices of F1, F2, we see thatCondition 4 of Lemma 2.6 holds whp.

Our next step is to show that G[W ] is Hamiltonian. Fix a subgraph H of G[W ] which isa (h/4, 2)-expander. To find a Hamilton cycle in G[W ] we define a sequence H0, H1, . . . ,Hhof subgraphs of G[W ] as follows. Set H0 = H. For each i ≥ 0, if Hi is Hamiltonian then setHi+1 = Hi; otherwise, let ei be a booster of Hi which is contained in G[W ]. Note that such abooster is guaranteed to exist by (P9), as |E(Hi)| ≤ |E(H)| + i ≤ βn log n/10 + h ≤ βn log n.Evidently, one cannot add h boosters to a graph on h vertices sequentially without making itHamiltonian, hence Hh is a Hamiltonian subgraph of G[W ].

Now, let w ∈W , and let P be a Hamilton path in G[W ] with w being one of its endpoints.Let R be the set of endpoints y of Hamilton paths of G[W ] with endpoints w and y. Evidently, asG[W ] is Hamiltonian, R is not empty. Moreover, by Lemma 2.4 we have |NG[W ](R)| ≤ 2|R|−1.Since G[W ] is a (h/4, 2)-expander (since H is such), it must be the case that |R| > h/4 ≥ n/40,so the assertion of the lemma holds.

We are now ready to prove Lemma 4.2.

Proof of Lemma 4.2. Let K,U1, U2 be the disjoint subsets of V = V (G) obtained in Lemma 4.3.Set V ? = U1 ∪ U2 and V ′ = V \ V ? (so K ⊆ V ′ is of size t and D1 ⊆ K ⊆ D≤2, hence (a) issatisfied). Note that (b) and (c) are also satisfied by Lemma 4.3. Let K ⊆ X ⊆ V ′ satisfy|X| ≤ n/2 and let w1, w2 ∈ X \K. Write V ′′ = V ′ \X and partition V ′′ = V ′′1 ∪ V ′′2 as equallyas possible. For i = 1, 2, let Wi = V

′′i ∪ Ui, and choose a neighbour zi of wi in Wi; this is

possible since d(wi, Ui) ≥ γ log n/100 by the condition in Lemma 4.3. Note that |Wi| ≥ n/5and for every v ∈ Wi it holds that d(v,Wi) ≥ min{d(v), γ log n/100}, hence by Lemma 4.4there exists a set Ri ⊆ Wi with |Ri| ≥ n/40 such that for every y ∈ Ri there is a Hamiltonpath spanning Wi from zi to y. In view of (P8), there exists an edge e between R1 and R2 withendpoints yi ∈ Ri, say. For i = 1, 2, denote by Qyi the Hamilton path between wi and yi. Wenow construct a Hamilton path on G[V (G) \X] as follows (as depicted in Fig. 2):

z1Qy1−−→ y1

e−→ y2Qy2−−→ z2,

hence (d) is satisfied.

Let V (G) = V ? ∪ V ′ be the partition obtained by Lemma 4.2, and let K ⊆ V ′ be the setof size t obtained by it. Write K = {x1, . . . , xt}, and for each i ∈ [t] let wi be an arbitraryneighbour of xi in V

′ (there exist such neighbours due to the properties of K,V ′, and they aredistinct due to (P3)). Set Q = {w1, . . . , wt}. Recall the definitions of aj and j0 from Section 1.

Lemma 4.5. There is a sequence of paths P1, ..., Pt−1 ⊆ G[V ′] for which the following holds:

1. The endpoints of Pi are {wi, wi+1} for all 1 ≤ i ≤ t− 1;

2. The length of Pi is exactly ` for all 1 ≤ i ≤ t− 1;

11

V ′U1

U2

K

Xw1

w2

W1

W2

z1

z2

y1

R1

y2

R2

Figure 2: Visualisation of the proof of Lemma 4.2.

3. V (Pi)∩ V (Pi+1) = {wi+1} for all 1 ≤ i < t− 1, and V (Pi)∩ V (Pj) = ∅ for all 1 ≤ i, j ≤t− 1 such that |i− j| > 1.

Proof. For U ⊆ V ′, set N ′(U) := NG(U) ∩ V ′ (and similarly N ′(v) := NG(v) ∩ V ′ for v ∈ V ′).By (P3), Q ∩ Small = ∅, and N ′(wi) ∩ N ′(wj) = ∅ for all i 6= j. For each 1 ≤ i ≤ t letYi, Zi ⊆ N ′(wi) be arbitrary disjoint subsets of size |Yi| = |Zi| = a2 (such sets exist, by theconstruction of V ′ in Lemma 4.2). We now construct the required paths sequentially. For1 ≤ i < t we assume that P1, P2, ..., Pi−1 have already been constructed, and construct a pathPi with the desired properties. Additionally, we construct Pi to be such that its internal vertices

do not belong to K ′ := K ∪Q∪(⋃t−1

k=1(Yk ∪ Zk+1))

(and, accordingly, assume that the internal

vertices of P1, P2, ..., Pi−1 do not belong to Yi, Zi+1).Set S2 := Yi, T2 := Zi+1. Now, for j ≤ `/2, given the sets S2, ..., Sj−1, T2, ..., Tj−1 ⊆ V ′ we

construct sets Sj , Tj ⊆ V ′ with the following properties:

• Sj ⊆ N ′(Sj−1), Tj ⊆ N ′(Tj−1);

• |Sj | = |Tj | = aj ;

• Sj ∩(i−1⋃k=1

V (Pk)

)= Tj ∩

(i−1⋃k=1

V (Pk)

)= ∅;

• Sj ∩

(j−1⋃k=2

(Sk ∪ Tk)

)= Tj ∩

(j−1⋃k=2

(Sk ∪ Tk)

)= ∅;

• Sj ∩K ′ = Tj ∩K ′ = Sj ∩ Tj = ∅.

We make the following observation, obtained from properties (P5),(P6) and from the con-struction of K,V ′, V ∗ in Lemma 4.2: if U ⊆ V ′ \ K is of size at most 10n/ log n, then|N ′(U)| ≥ |U | · log n/30. Indeed, assume otherwise, then

|N(U)| ≤ |N ′(U)|+ 200γ|U | log n ≤(

1

30+

1

50

)|U | log n ≤ |U | log n/18.

12

On the other hand, since |U | ≤ 10nlogn ≤ γn/5000, by (P6), U spans at most γ|U | log n/1000edges, and since U ∩ Small = ∅, we have

e(U, V (G) \ U) ≥∑u∈U

d(u)− 2 · e(U) ≥ |U | log n/11.

a contradiction to (P5). Therefore, since |Sj−1| = |Tj−1| = aj−1 ≤ a`/2−1 < 10n/ log n, we have

|N ′(Sj−1)|, |N ′(Tj−1)| ≥ aj−1 · log n/30.

This inequality implies the existence of two disjoint subsets S′, T ′ of N ′(Sj−1), N′(Tj−1), re-

spectively, of size at least aj−1 · log n/60. In addition, recalling that ` = o(log n), we get∣∣∣∣∣(i−1⋃k=1

V (Pi)

)∪

(j−1⋃k=2

(Sk ∪ Tk)

)∣∣∣∣∣ ≤ i · `+ 2 · j · aj−1 = o(aj−1 · log n).We now wish to make sure that we can choose large enough subsets of S′, T ′ which do notintersect K ′. To this end, note that by (P3) N ′(S2)∩K ′, N ′(T2)∩K ′ ⊆ Q and |Q| = o(a2 log n),and for j > 3, |K ′| = O(log2 n) = o(aj−1 log n), so for every 3 ≤ j ≤ `/2 we have

|N ′(Sj−1) ∩K ′|, |N ′(Tj−1) ∩K ′| = o(aj−1 · log n).

So overall we get∣∣∣∣∣(S′ ∪ T ′) ∩(K ′ ∪

(i−1⋃k=1

V (Pk)

)∪

(j−1⋃k=2

(Sk ∪ Tk)

))∣∣∣∣∣ = o(aj−1 · log n),which implies that there are subsets Sj , Tj of S

′, T ′ with all the listed properties. Finally,observe that |S`/2| = |T`/2| = a`/2, and therefore

10n/ log n ≤ |S`/2|, |T`/2| ≤10n

log n·⌈

log n

100

⌉≤ n

9,

and therefore, by (P7),

|N ′(S`/2) ∩N ′(T`/2)| ≥ |NG(S`/2) ∩NG(T`/2)| − |V ∗| ≥(

1

9− 3γ

)· n ≥ n/10,

which implies that N ′(S`/2)∩N ′(T`/2) contains a vertex that is not a member of(⋃i−1

k=1 V (Pk))∪(⋃`/2−1

k=2 (Sk ∪ Tk))

. By the definitions of S2, ..., S`/2, T2, ..., T`/2, this proves that there is a path

Pi of length ` between wi and wi+1 with all our desired properties.

This concludes the proof of Lemma 4.1. Indeed, let P =⋃ti=1 Pi be the union of the paths we

have found in Lemma 4.5, and let X = P ∪ {{xi, wi}}ti=1 be the “comb”. By Lemma 4.2, thereexist neighbours z1 ∼ w1 and zt ∼ wt outside the comb, and a Hamilton path in G[V (G) \X]between z1 and zt. The union of the comb, the edges {w1, z1} and {wt, zt} and the Hamiltonpath constitutes a copy of KC(n, t, `) in G (see Fig. 3).

13

V ′U1

U2

Kx1

...

xt

Xw1

...

wt

W1

W2

z1

zt

R1

R2

Figure 3: Visualisation of the proof of Lemma 4.1.

5 Maximum common subgraph

In this short section we prove Proposition 1.2.

Proof of Proposition 1.2. We may assume that ε > 0 is small enough. Let δ = δ(ε) > 0 to bechosen later and p = n−1+δ. Let m = (1 + ε)n, and let Am be the event that there exists asubgraph H of G1 with m edges which is also a subgraph of G2. By the union bound over thepossible choices of H and the permutations of the vertices of G2, we obtain

P(Am) ≤((n

2

)m

)· n! · p2m ≤

((enp2

2(1 + ε)

)1+ε· n

)n≤(

2 · n(−1+2δ)(1+ε)+1)n.

Taking δ = δ(ε) > 0 small enough, the last term is vanishing.

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Yahav AlonSchool of Mathematical Sciences, Tel Aviv University, Tel Aviv 6997801, IsraelEmail: yahavalo@tauex.tau.ac.il

Michael KrivelevichSchool of Mathematical Sciences, Tel Aviv University, Tel Aviv 6997801, IsraelEmail: krivelev@tauex.tau.ac.ilResearch supported in part by USA-Israel BSF grant 2018267 and by ISF grant 1261/17.

Peleg MichaeliSchool of Mathematical Sciences, Tel Aviv University, Tel Aviv 6997801, IsraelEmail: peleg.michaeli@math.tau.ac.ilThis research is supported by ERC starting grant 676970 RANDGEOM and by ISF grant 1207/15.

15

http://www.ams.org/mathscinet-getitem?mr=120167http://www.ams.org/mathscinet-getitem?mr=3675279http://www.ams.org/mathscinet-getitem?mr=3789834https://arxiv.org/abs/2007.12111http://www.ams.org/mathscinet-getitem?mr=2993127http://www.ams.org/mathscinet-getitem?mr=1782847http://www.ams.org/mathscinet-getitem?mr=3508728http://www.ams.org/mathscinet-getitem?mr=680304http://www.ams.org/mathscinet-getitem?mr=3998769http://www.ams.org/mathscinet-getitem?mr=389666mailto:yahavalo@tauex.tau.ac.ilmailto:krivelev@tauex.tau.ac.ilmailto:peleg.michaeli@math.tau.ac.il

IntroductionDiscussionProof outline

PreliminariesNotationProbabilistic boundsPósa's lemma and corollaries

Properties of random graphsConstructing a KeyChainMaximum common subgraph