k-regular subgraphs near the k-core threshold of a random graph

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1 -regular subgraphs near the -core threshold of a random graph Dieter Mitsche Michael Molloy Pawe l Pra lat April 5, 208 Abstract We prove that G n,p=c/n w.h.p. has a -regular subgraph if c is at least e Θ above the threshold for the appearance of a subgraph with minimum degree at least ; i.e. an non-empty -core. In particular, this pins down the threshold for the appearance of a -regular subgraph to a window of size e Θ. In this paper, we study the threshold for the Erdős-Rényi random graph model G n,p=c/n to have a -regular subgraph where is fixed. This problem was first studied by Bollobás, Kim and Verstraëte [5] who proved, amongst other things, that G n,p=c/n w.h.p. has a -regular subgraph when c is at least roughly 4. Letzter [0] proved that this threshold is sharp 2. This problem is reminiscent of the -core, a maximal subgraph with minimum degree at least. Pittel, Spencer and Wormald [4] established the threshold for G n,p to have a non-empty -core to be a specific constant c = + o we specify c more precisely in 4 below. This provides a lower bound on the threshold for a -regular subgraph, and it is natural to as: Question: Is the threshold for a -regular subgraph equal to the -core threshold? Bollobás, Kim and Verstraëte [5] proved that the answer is No for = 3 and conjectured that it is No for all 4. On the other hand, Pretti and Weigt [6] provided a non-rigorous analysis and claimed that it indicates the answer is Yes for 4. Pra lat, Verstraëte, and Wormald [5] proved that w.h.p. the + 2-core of G n,p if it is nonempty contains a -regular spanning subgraph. Chan and Molloy [6] proved the same for the + -core. So the -regular subgraph threshold is at most c + c +. We will reduce this bound to within an exponentially small distance as a function of from c : Theorem. For a sufficiently large constant, and for any c c + e /300, G n,p=c/n w.h.p. contains a -regular subgraph. Universite de Nice Sophia-Antipolis, Nice, France. Department of Computer Science, University of Toronto, Toronto, ON, Canada. Department of Mathematics, Ryerson University, Toronto, ON, Canada. A property is said to hold with high probability w.h.p. if it holds with probability tending to one as n. 2 Meaning that there is a function ρ n such that for any ɛ > 0, G n,p=c/n w.h.p. has no -regular subgraph for c = ρ n ɛ and w.h.p. has a -regular subgraph for c = ρ n + ɛ.

2 It is not hard to see that the -core cannot have a -regular spanning subgraph; for example w.h.p. it has many vertices of degree + whose neighbours all have degree. Our approach is to start with the -core and repeatedly remove such vertices, along with other problematic vertices. We will then apply a classic theorem of Tutte to show that what remains has a spanning -regular subgraph. The aforementioned papers [5, 6] applied Tutte s theorem to the +2- and +-core. The new arguments required in this wor are i stripping the -core down to something to which Tutte s theorem can be applied and ii applying Tutte s theorem to it. The first part requires a delicate variant of the configuration model see the discussion at the beginning of Section 3., whereas the presence of degree vertices brings new challenges to the second part. The number of problematic vertices, as described above, is linear in n. Furthermore, removing them from the -core will cause a linear number of vertices to have their degrees drop below. It is not surprising that if c is too close to c, then w.h.p. what remains will have no -core, and so this argument will not wor unless c is bounded away from c. Fortunately, when is large, the number of problematic vertices is very small but linear in n: e Θ n. So we only need c to be bounded away from c by an exponentially small distance in terms of. Furthermore, the subgraph that we show to have a -factor consists of all but e Θ n vertices of the -core. This is consistent with a result of Gao [8] who proved that any -regular subgraph must contain all but at most ɛ n vertices of the -core where ɛ 0 as grows. Organization of the paper: We begin, in section, by presenting Tutte s condition. Section 2 contains some brief probabilistic tools. The stripping procedure to find our subgraph is given and analyzed in section 3; this is most of the wor. Finally, in section 4, we show how to prove that the subgraph satisfies Tutte s condition and thus has a spanning -regular subgraph. Tutte s condition We begin by presenting Tutte s theorem for establishing that a graph has a -regular spanning subgraph. Recall that a -regular spanning subgraph is called a -factor. Let Γ be a graph with minimum degree at least. Definition 2. L = LΓ is the set of low vertices of Γ, i.e. the vertices v with d Γ v =, and H = HΓ is the set of high vertices of G, i.e. the vertices v with d Γ v +. For any set of vertices Z, we use Z L, Z H to denote Z L, respectively Z H. For any S V Γ we use es to denote the number of edges of Γ with both endpoints in S. For any disjoint S, T V H, we use es, T to denote the number of edges of G from S to T and qs, T to denote the number of components Q of H \ S T such that Q and eq, T have different parity. Theorem 3 [8]. A graph Γ with minimum degree at least has a -factor if and only if for every pair of disjoint sets S, T V Γ, S qs, T + T v T d Γ\S v. 2

3 Corollary 4. A graph Γ with minimum degree at least has a -factor if for every pair of disjoint sets S, T V Γ, S + v T H d Γ v qs, T + es, T. Proof Rearranging the terms in Theorem 3, we obtain that the condition given there for the existence of a -factor is equivalent to every pair of disjoint sets S, T V Γ satisfying: d Γ v + S qs, T + T + es, T i.e. v T d Γ v + S qs, T + es, T v T which is equivalent to since d Γ v = for all v T L. In all but one case, we will in fact show that S, T satisfy the stronger condition, which implies since d Γ v + for all v T H : S + T H qs, T + es, T. 2 Remar: In previous papers [6, 5] Theorem 3 was applied to subgraphs with minimum degree at least +, specifically the + -core [6] and the + 2-core [5]. So it sufficed to prove the weaer bound S + T qs, T + es, T. We begin by showing that in Corollary 4 we may assume that S H and every component counted by qs, T has a high vertex: Lemma 5. A graph Γ with minimum degree has a -factor if holds for every pair of disjoint sets S, T V Γ satisfying: M S H; and M2 every component Q counted by qs, T satisfies Q H. Proof We will prove that if holds for every S, T satisfying M and M2 then holds for every S, T. So Corollary 4 implies that Γ has a -factor. To do this, we show that if S, T violates and violates either M or M2 then we can modify S, T so that M and M2 both hold but is still violated, thus contradicting the hypothesis of our lemma. Suppose there exist two disjoint sets S, T V Γ that violate both and M. So there exists some u S L. We will show that after moving u to V Γ \ S T, will still fail for the new pair of sets. By applying this procedure iteratively for every u S L, we obtain two sets violating and satisfying M. Let S := S \ {u}. Note first that S = S and es, T = es, T d T u. Now, since d Γ u =, there are at most d T u neighbours of u in V Γ \ S T observe that some neighbours of u might be in S. In the worst case, all neighbours of u in V Γ \ S T belong to different components all contributing to qs, T ; moreover, after moving u to V Γ \ S T, they 3

4 form one connected component of V Γ \ S T, and so do not contribute to qs, T anymore. In any case, qs, T qs, T d T u. Since T H is left unchanged by switching from S to S, we have S + d Γ v = S + d Γ v v T H v T H and hence still fails. < qs, T + es, T qs, T + d T u + es, T + d T u = qs, T + es, T, Now suppose there exist two disjoint S, T V Γ violating and which satisfy M but not M2. Let Q be a component of Γ\S T such that Q and eq, T have different parity and with Q H =. Note that eq, V Γ\Q = eq, S T has the same parity as u Q du = Q since Q H =. So eq, V Γ\Q has a different parity than eq, T and thus eq, S 0. Now we move Q into T ; i.e. we set T = T Q. Since Q L, we have T H = T H. Since Q is a component of Γ\S T, this move does not affect whether any other component counts towards q; i.e. qs, T = qs, T. Moreover, as we argued above, eq, S > 0 and so es, T es, T +. Combining this with S + v T H d Γ v < qs, T +es, T yields S + v T H d Γ v < qs, T +es, T and hence still fails. Clearly, S has not changed and hence M still holds. Repeated applications result in a pair S, T that violates and satisfies M and M2. 2 Probabilistic preliminaries Pittel, Spencer and Wormald [4] established the -core threshold to be: c = min x>0 x e x 2 i=0 xi i!. 3 In [5] the asymptotic value of c is determined up to an additive O/ log =o term. Setting q = log log2π, we have /2 c = + q /2 + + q + O. 4 q 3 log We will use the following well-nown bounds on tail probabilities nown as Chernoff s bound see, for example, [9], Theorem 2.. Let X Binl, p be distributed as a binomial random variable with l trials and success probability p, so E[X] = µ = pl. Then, Pr X µ t exp t2 2µ and Pr X µ + t exp t 2 2µ + t/3. 4

5 In addition, all of the above bounds hold for the general case in which X = l i= X i and X i is the Bernoulli random variable with parameter p i with possibly different p i s. The Hoeffding-Azuma inequality can be generalized to include random variables close to martingales. One of our proofs, proof of Lemma 9, will use the supermartingale method of Pittel et al. [4], as described in [20, Corollary 4.]. Let G 0, G,..., G l be a random process and let X i be a random variable determined by G 0, G,..., G i, 0 i l. Suppose that for some real b and constants c i, EX i X i G 0, G,..., G i < b and X i X i c i for each i l. Then, for every α > 0, Pr [For some i with 0 i l : X i X 0 ib + α] exp α2 2 j c2 j. 5 3 Finding the subgraph Let N and set β = e / We begin with a random graph G = G n,p with p = c/n for some constant c satisfying c + 0 β =: c min c c max := c + /2, 7 where c is the threshold of the emergence of the -core. Our goal is to find for sufficiently large a subgraph K of the -core with certain properties, which will ensure that it has a -factor. Our first property is simply a degree requirement. Of course, K must have minimum degree at least ; for technical reasons, it will help if the maximum degree is bounded by a constant; we arbitrarily chose this to be 2. In the introduction, we noted that K cannot have any vertex of degree greater than whose neighbours all have degree. It is not hard to build similar problematic local structures; it turns out that we can eliminate all of them by not allowing any vertex of degree greater than that has many neighbours of degree ; our second property enforces this. Our third property simply says that K has linear size. Our final property is a trivial necessary condition for having a -factor. K for every vertex v K, d K v 2; K2 for every vertex v K with d K v +, we have {w N K v : d K w = } 9 0 ; K3 K n 3 ; K4 K is even. In fact, we will be able to find an induced subgraph K of G satisfying these properties. 5

6 3. The stripping procedure In order to achieve our goal, we are going to use a carefully designed stripping procedure during which one vertex is removed in each step until a subgraph K satisfying properties K, K2, and K3 remains. It will be easy to modify K to enforce the final property K4, if necessary, at the end. We found property K2 to be particularly challenging to enforce. The typical approach to this sort of problem is to repeatedly remove a vertex if it violates one of K-3. Often one can argue that at every step, the remaining graph is uniformly random conditional on its degree sequence for example, this happens when analyzing the -core stripping process. In some situations, the vertex set is initially partitioned into a fixed number of parts, and one must condition on the number of remaining neighbours each vertex has in each part; this is more complicated but in principle not much more difficult than conditioning on the degree sequence. In the present situation, enforcing K3 requires conditioning on the number of remaining neighbours each vertex has in W, the set of vertices of degree. However, this is not an initial partition; W changes during the process. This made our analysis more difficult. In dealing with this problem, it helps to partition W into W 0, the vertices that initially have degree, and W, the vertices whose degrees change to during the process. W is much smaller than W 0 and so we can afford to delete vertices if they have at least two neighbours in W rather than at least 9 0. This simpler deletion rule helps us deal with the fact that W is changing throughout our stripping process. We begin with the -core of G, as any subgraph K satisfying K must be a subgraph of the -core. The -core of a graph can be found by repeatedly deleting vertices of degree less than from the graph so this can be viewed as an initial phase of our stripping procedure. We continue stripping the graph as explained below and, throughout our procedure, we partition the vertex set as follows: W 0 = the vertices in the remaining graph that had degree in the -core of G; W = the vertices of degree at most in the remaining graph that are not in W 0 ; R = the vertices of degree greater than in the remaining graph. Note that vertices may move from R to W during our procedure, but no vertex leaves W 0 unless it is deleted. The following definition governs the stripping procedure. Definition 6. We say a vertex v is deletable if in the initial -core: D degv > 2; D2 v / W 0 that is, degv + and v has at least 2 neighbours in W 0; or if in the remaining graph: D3 degv < ; D4 v R and v has at least two neighbours that are in W ; or 6

7 D5 v W and v has a neighbour that is either i in R and deletable, or ii in W. Furthermore, D6 once a vertex becomes deletable it remains deletable. Remars: Let us mae three remars. a Deleting vertices in W with non-deletable neighbours in W is not required for properties K, K2, and K3; we only delete them because it helps with our analysis. b In many common stripping processes for example, the -core process, we have the property that the subgraph we eventually obtain does not depend on the order in which vertices are deleted. That is not true for our procedure whether a vertex ever becomes deletable can depend on the deletion order. However, our goal is only to obtain a subgraph with the desired properties K, K2, and K3 and so this wors for our purpose. c Deleting a vertex that is deletable may cause some non-deletable vertices to become deletable which, in turn, might force more non-deletable vertices to change their status. However, this domino effect will eventually stop possibly with all remaining vertices mared as deletable because of D6. At any point of the algorithm, we use Q to denote the set of deletable vertices. Recall from 6 that β = e /200. We will show that w.h.p. we reach Q = within βn steps. It will be convenient to force our stripping procedure to halt after βn steps regardless of whether it has reached the desired property that is, Q =. Now, we are ready to introduce our stripping procedure. STRIP. Begin with the -core of G, and initialize Q to be all vertices v with degv > 2 or v / W 0 and v has at least 2 neighbours in W Until Q = or until we have run βn iterations, let v be the next vertex in Q, according to a specific fixed vertex ordering. Let N be the set of neighbours of v. a Remove v from the graph and from Q. b If any u N that is in R now has degree at most, then move u from R to W. c If any vertex w / Q is now deletable, place w into Q. Clarification: In step 2c, w does not leave whichever of W 0, W, R it was in. So it is now in, for example, Q W 0. Remar: Note that initially no vertex has degree less than and W = so, indeed, all initially deletable vertices are added to Q in step. The following observation is an immediate consequence of Definition 6. Indeed, parts a and b follow from D5; part c follows from D4. 7

8 Observation 7. The following properties hold. a If u W then u has no neighbours in W \Q. b If u R Q then u has no neighbours in W \Q. c If u R\Q then u has at most one neighbour in W. Here is another straightforward but useful observation. Observation 8. If the stripping procedure terminates with Q =, then in the remaining subgraph: the degree of every vertex is in [, +,..., 2] and each vertex in R has at most neighbours of degree provided that 3. Thus, the remaining subgraph satisfies properties K and K Configuration models We model the -core of G = G n,p=c/n with the configuration model introduced by Bollobás [4], and inspired by Bender and Canfield [2]. We are given the degree sequence of a graph that is, the degree of each vertex. We tae degv copies of each vertex v and then choose a uniform pairing of those vertex-copies. Treating each pair as an edge gives a multigraph. It is well-nown see, for example, the result of McKay [] that the multigraph for a degree sequence meeting some mild conditions; these conditions are met in our application is simple with probability tending to a positive constant, and it follows that if a property holds w.h.p. for a uniformly random configuration then holds w.h.p. for a uniformly random simple graph on the same degree sequence; see the survey of Wormald [9] for more on this, and for a history of the configuration model and other related models. Throughout our analysis, we often refer to a pair in the configuration as an edge in the corresponding multigraph. A sub-configuration is simply a subset of the pairs in a configuration, and thus yields a subgraph of the multigraph. We define C to be the -core of G n,p=c/n. We expose the degree sequence D of C and then define Λ to be a uniform configuration with degree sequence D. We will prove: Lemma 9. W.h.p. STRIP terminates with Q = when run on Λ. As described above, we can transfer our results on random configurations to random simple graphs thus obtaining: Lemma 0. W.h.p. STRIP terminates with Q = when run on C core properties We will require the following properties of the configuration Λ. Setup for Lemma : is a sufficiently large constant, and c < c c max = c + /2. C is the -core of G n,p=c/n. Λ is a uniform configuration with the same degree sequence as C. Finally, W 0, R are as defined in Section 3.. 8

9 Lemma. Before the first iteration of STRIP: a Λ > 0.99n; b 0.99 n < W 0 <.0 n ; c the total degree of all vertices with degree greater than 2 is at most e /6 n; d there are at least n 5 edges with both endpoints in W 0; e there are at least 2 n edges from W 0 to R; f there are at least 3n edges with both endpoints in R; g C has at most e /3 n vertices of degree at most 2 and with at least 2 neighbours in W 0; h at least n 200 vertices in R have no neighbours in W 0. Proof First recall from Section 3. that, before the first iteration of STRIP, W 0 is the set of vertices with degree in Λ, and R is the set of vertices with degree greater than in Λ. Part a is well-nown; the size of the -core approaches n as grows see, for example, the results of Molloy [2] or Gao [8]. Indeed, in [2] it is proved that w.h.p. the -core has size ζn+on, with ζ = ζc = e x where x = xc is the greatest solution to i=0 c = fx := x i i! = x i e x i! i = Pr P ox, x e x 2 i=0 xi /i!. 8 Recall from 3 that c is the minimum value of f over all x > 0. Simple analysis of f shows that there is exactly one value of x for which fx = c ; we denote that value by x. For every c > c there are exactly two solutions for x. It is straightforward to verify that f x 0 for x x and so for c c we have x xc xc max. Recall from 4 that q = log log2π. [5] shows that note that x is denoted as λ in [5]. It follows that as. Part a holds for large enough. x = + q /2 + q 3 + o, 9 ζ = Pr P ox Pr P ox, Corollary 3 of [7] establishes that for any constant i, the number of vertices of degree i in the -core is w.h.p. λ i n + on where λ i = Pr P ox = i = e x x i. 0 i! 9

10 In particular, w.h.p. W 0 = e x x! n + on, and so to prove part b we will estimate e x x!. Below, setting δ = δc = xc x, we will prove that c c max implies δ log. So for now, we will restrict our attention to x = x + δ with 0 δ log. Using + y = exp y + Oy 2, and / + y = y + Oy 2 we get that e x x! = e x δ x! = e x x! = e x x! = e x x! + δ = e x x exp δ + δ + O x! x exp δ + δ q / /2 + Oq / + Oδ/ exp δq / /2 + Oδ log / δq / /2 + Olog 3 / δ 2 = e x x + o, for δ log.! x 2 by 9 since q = log + O Using Stirling s formula, 9 and the fact that + y = exp y y 2 /2 + y 3 /3 + Oy 4, we get for some ɛ = ɛ = o 2 e x x! = e x ex + O 2π = e q /2 q 3 + ɛ + q /2 + q 3 + ɛ + O 2π = exp q 2 + ɛq / /2 + Oq 2 / 2π Therefore = + ɛq / /2 + Olog 2 / = + o. e x x! = + o since q = log/2π and e y = + y + Oy 2 3 for x = x + δ with 0 δ log. 4 Next we prove that our upper bound c c max = c + /2 implies and so 4 establishes part b. To do this we estimate the derivative of fx over the range 0 δ log to show that at δ = log and x = x + δ we have cx = fx > c max. Rewriting 8 as: fx = x e x i xi i!, 0

11 the derivative is f x = e x i xi i! x e x i xi i! + e x i xi e x i xi i! = e x i xi i! xe x x 2 2! e x 2. i xi i! Recalling q = log log2π, we have for 0 δ log xe x x 2 = e x x x = e x x! 2 i! δq / /2 + Olog 3 / x 2!! = + ɛ δq / /2 + Olog 3 / x = + ɛ δq / /2 + Olog 3 / q / /2 + Oq / by 9 and since + y = y + Oy2 by 2 by 3 = ɛ + δq / /2 + Olog 3 /. 5 Standard bounds for the tail probabilities of a Poisson random variable see eg. [, Theorem A..5], along with 4 yield that for 0 δ log we have Pr P ox 2 = O /2, and so e x i x i i! = Pr P ox 2 = O /2. 6 Substituting 5 and 6 into 5 and recalling q = log + O yields f x = ɛ + δq / /2 + O = ɛ + δlog / /2 + O /2 /2, for x = x + δ and 0 δ log. 7 Therefore, recalling that fx = c and since ɛ = o = oδ we have for sufficiently large fx + log = fx + Θlog 2 log / /2 + log O /2 > c + /2 = c max. Since fx is monotone increasing for x > x, it follows that xc x + log for all c c c max, thus proving. Therefore 4 and 0 yield part b. For part c it is easier to show the desired property for G instead of C; the conclusion for C will trivially follow. The degree of each vertex in G is a random variable X with the binomial distribution Binn, c/n with E[X] c <., provided that is large enough. Hence, by

12 Chernoff s bound, we get that the expected total degree of all vertices with degree greater than 2 is at most n 2 + l Pr X 2 + l n l2 + l exp l/3 l l 2 n + l exp l 2 + l = Oe /4 n < e /5 n, 4 provided is sufficiently large. Since the concentration can be proved with a straightforward concentration argument using, for example, Azuma s Inequality or an easy second moment argument, we omit the details. This establishes part c. For the remaining parts, let us first observe that the total degree of all vertices in G is 2Bin n 2, c/n with expectation cn <.0n, provided is sufficiently large. Hence, by Chernoff s bound, w.h.p. it is at most.02n, and this upper bound clearly holds for the total degree of the vertices of C. On the other hand, part a implies that it is at least 0.99n. For parts d,e,f, we focus on the partners of the vertex-copies in W 0. Part b implies that the total degree of the vertices in W 0 is between 0.99n and.0n. Expose the partners in Λ of the vertex-copies of W 0, one at a time. At step i.0n, the probability that the partner chosen is in W 0 is between p i and q i, where p i = max { } 0.99n 2i.02n, 0 { } { }.0n i.0n i and q i = max 0.99n 2i, 0 max 0.98n, 0, 8 provided is sufficiently large. Hence, the number of vertex-copies in W 0 whose partner is in W 0 can be stochastically lower/upper bounded by the two sums of independent Bernoulli random variables: the first one with parameters p i and the second one with q i. The expected value of the first sum is more than 0.24n/, and the expectation of the second one is at less than 0.53n/. The concentration follows immediately from Chernoff s bound and we get that w.h.p. the number of edges with both endpoints in W 0 is at least n 5 and at most n, which finishes part d. Now, parts e and f follow deterministically. The number of edges from W 0 to C\W 0 is at least 0.99n 2 n 2n for large enough. Finally, there are at least 0.99n 2.0n/2 edges with both endpoints in C\W 0 which is more than 3n for large enough. Part g is slightly more complicated. For a contradiction, suppose that there are e /3 n vertices of degree at most 2 and with at least 2 neighbours in W 0. Note that some of them might be from W 0. This implies deterministically that there exists a set S of 2 e /3 n vertices of degree at most 2 with at least 8 e /3 n edges between S and W 0 \ S. We will show that w.h.p. this is not possible. Indeed, let us fix set S and expose partners of all vertex-copies from W 0 \ S in Λ. Arguing similarly as for 8, the number of edges from W 0 \ S to S can be stochastically upper bounded by the binomial random variable X Bin.0n, 2 S 0.98n, with E[X] < 3e /3 n. 2

13 It follows from Chernoff s bound that the expected number of sets S with a large number of edges to W 0 \ S is at most n Pr X e /3 n 2e +/3 2 e /3 n exp.4 e /3 n = o, e /3 n provided that is sufficiently large. Part g follows from Marov s inequality. Finally, let us move to part h. We showed earlier that the total degree of the vertices of C is at least 0.99n and the total degree of the vertices in W 0 is at most.0n. It follows from parts a, b, and c that there are at least 0.98n vertices in R that are of degree at most 2 and, of course, at least +. We pic arbitrarily 0.3n of them and expose partners of all corresponding vertex-copies. Note that, regardless of the history of the process, the probability that a given vertex of degree l 2 has no neighbour in W 0 is at least.0n l.4 2 e 3, 0.99n 20.3n provided that is sufficiently large. Hence, the number of vertices in R that have no neighbours in W 0 is bounded from below by the random variable X Bin0.3n, e 3 with E[X] = 0.3e 3 n > 0.006n. Part h holds by Chernoff s bound and the proof of the lemma is finished. Corollary 2. When STRIP terminates, the remaining subgraph has size at least n 3 ; that is, it satisfies K3. Proof This follows from Lemma a, since 0.99n βn n 3, for sufficiently large. 3.4 Stripping a configuration We will apply STRIP to the random configuration Λ. We restate STRIP here, adding details about how we expose the pairs of the configuration. STRIP2. Begin with Λ and initialize Q to be all vertices v with degv > 2 or v / W 0 and v has at least 2 neighbours in W Until Q = or until we have run βn iterations, let v be the next vertex in Q, according to a specific fixed vertex ordering. a Expose the partners of every vertex-copy of v of course, if they are not already exposed. Let N be the set of neighbours of v. b Remove v from Λ and from Q, along with all vertex-copies of v and their partners. c If any u N that is in R has its degree decreased to at most, then i. move u from R to W ; 3

14 ii. for each vertex-copy of u that has a partner in W R, expose that partner. d If any vertex w / Q is now deletable, place w into Q. We must carefully trac the information that is exposed while running STRIP2. If the partition W 0, W, R were fixed throughout our procedure, then we would simply expose the number of remaining neighbours that each vertex has in each part. But because W and R change during the process, our exposure is more delicate. We found that the best way to deal with this complication includes exposing many of the remaining edges involving W and R. Definition 3. Suppose the vertices of a configuration are partitioned into R, W 0, W. We define the RW-information to be: for each vertex v W 0, deg W Rv, deg W0 v; for each vertex v W R, deg R v, deg W0 v, deg W v; all vertex-copy pairs that have one vertex-copy in W and the other in W R. Given a particular partition into R, W 0, W, and RW-information, Ψ, let Ω Ψ be the set of all configurations on vertex set R W 0 W with RW-information Ψ; that is, all configurations in which each v has deg R v, deg W0 v, deg W v, deg W Rv equal to the values prescribed in Ψ, and where the set of pairs with one copy in W and the other in W R is as prescribed in Ψ. Suppose that at a certain step during STRIP2, we set R = R, W 0 = W 0, and W = W. Then the RW-information will capture everything that has been exposed thus far. However, it is important to note that in the definition of RW-information, R, W 0, W are not necessarily set in this way; they can be any partition of the vertices. This arises in the proof of Lemma 4 below; in particular, it is the reason that we need to prove Claim 2. We define W 0 i, W i, Ri, and Qi to be those vertex sets after applying i iterations of STRIP2 to Λ. We define Ψi to be the RW-information of the remaining configuration with partition R = Ri, W 0 = W 0 i, and W = W i. The next lemma will allow us to analyze STRIP2. Lemma 4. For any t 0: the configuration that remains after t iterations of STRIP2 is distributed uniformly from Ω Ψt. Proof Let H be what remains after t iterations of STRIP2. Consider any H Ω Ψt, and form Λ from Λ by replacing the pairs of H with the pairs of H. Claim : Each vertex v has the same degree in both Λ and Λ. Proof: Indeed, G, G differ only on the pairs of H, H and, by construction, v has the same degree in H as in H. The claim holds. Claim says that Λ, Λ have the same degree sequence D and so both are equally liely to be chosen as our initial configuration. The remainder of this proof will establish that if we apply t iterations of STRIP to Λ we will obtain H. This will imply that H, H are both equally liely 4

15 to be what remains after applying t iterations of STRIP2 to the our initial random configuration. This will finish the proof of the lemma. We define Hi, H i to be the sub-configurations remaining after applying i iterations of STRIP2 to Λ, Λ, respectively. So H = Ht and we wish to show that H = H t. Note that this does not follow from the fact that H Ω Ψt ; Ψt specifies, for example, that each vertex v has the same number of neighbours from R = Rt, but it does not imply that the vertices in R all have degree greater than in what remains after running t iterations of STRIP2 on Λ ; in fact it does not even imply that the vertices of R remain after t iterations of STRIP2 on Λ. We let W 0 i, W i, R i, Q i denote the vertex sets W 0, W, R, Q after applying i iterations of STRIP2 to Λ. In what follows, we use degv, deg R v, etc. to denote degrees in Λ and deg v, deg R v, etc. to denote degrees in Λ. Claim 2: for every 0 i t, W 0 i = W 0 i, W i = W i, Ri = R i, and Qi = Q i. Proof: By definition, W 0 = W 0 =. Claim implies that W 00 = W 0 0 and R0 = R 0. To prove that Q0 = Q 0 we apply Claim and we also need to argue that each vertex v has the same number of neighbours in W 0 0= W 0 0 in both Λ and Λ. Note that W 0 t W 0 0. The number of neighbours v has in W 0 t is specified by Ψt to be the same in both Λ, Λ, and the set of edges from v to W 0 0\W 0 t is identical in both Λ, Λ as W 0 0\W 0 t is not in H. Having established the base case, we proceed by induction on i. Suppose that Claim 2 holds for some i < t. Since Qi = Q i, the i + st vertex deleted is the same for both Λ, Λ, since we choose the next deletable vertex according to the same ordering in both procedures; let v be that vertex. So the vertex set of the subgraph after i + steps is the same in both procedures. Furthermore, v / H as it is removed during the first t iterations on Λ and thus the set of pairs including a copy of v is the same in both Λ, Λ and hence in both remaining configurations. The ey observation is that all decisions made in STRIP2 are based on sets of pairs that are equal in Hi, H i. This implies that we will mae the same changes to the various vertex sets in both configurations. Indeed, the decisions made are determined entirely by: Pairs containing copies of v: these are the same in Hi, H i since v / H. degu in step 2c: this is the same in Hi, H i by Claim and the fact that all pairs removed during the first i t steps of STRIP2 are the same in both Λ, Λ, as they each include at least one vertex not in H. The decision as to whether w enters Q in Step 2d is determined by the degree of w, and by pairs of vertex-copies where one is in W i = W i and the other is in W i = W i or Ri = R i. Any such pair will either a not be in H and hence will be the same in Λ, Λ or b will have one copy in W t and the other in either W t or Rt, as vertices may move from R to W but do not leave W unless they are deleted; all such pairs are specified by Ψt to be the same in both Λ, Λ. Thus all such pairs are the same in Hi and H i. Note: The choice of pairs to expose in step 2c,ii is not relevant here as it does not affect whether vertices become deletable. However, it is easy to chec that those choices also are the same in both Hi and H i. 5

16 So STRIP2 maes the same decisions, and carries out the same steps during iteration i + on both Λ and Λ. This yields the claim for iteration i + and so the proof of the claim is finished. Therefore STRIP2 removes the same sequence of vertices for the first t iterations on Λ and on Λ, and so H is what remains after t iterations on C. This finishes the proof of the lemma as explained above. Lemma 4 allows us to analyze the configuration remaining after any iteration t of STRIP2 by taing a uniform member of Ω Ψt. This is fairly simple as the members of Ω Ψt all decompose into the union of a few configurations which can be analyzed independently using the configuration model. Given any three disjoint vertex sets W 0, W, R, a set of RW-information Ψ, and a configuration Λ Ω Ψ we define: Λ W0 Λ Ψ - the sub-configuration induced by W 0 ; Λ W Λ Ψ - the sub-configuration induced by W ; Λ R Λ Ψ - the sub-configuration induced by R; Λ W0,W R Λ Ψ - the bipartite sub-configuration induced by W 0, W R; Λ W,R Λ Ψ - the bipartite sub-configuration induced by W, R. Note that Ψ specifies the vertex-copies and pairs of Λ W, Λ W,R; thus it also specifies which vertex-copies in W are in Λ W0,W R. To select a uniform member of Λ Ψ, we can select the other three configurations independently; that is,. For each vertex v R, choose a uniform partition of the vertex-copies of v not paired with copies in W into those that will be paired with copies in W 0, R, according to deg W0 v, deg R v. 2. For each vertex v W 0, choose a uniform partition of the vertex-copies of v into those that will be paired with copies in W 0, W R, according to deg W0 v, deg W Rv. 3. Choose Λ W0 by taing a uniform matching on the appropriately selected vertex-copies in W Choose Λ R by taing a uniform matching on the appropriately selected vertex-copies in R. 5. Choose Λ W0,W R by taing a uniform bipartite matching on the appropriately selected vertex-copies in W 0, R and the vertex-copies of W that are implicitly specified by Ψ to be paired with W 0. To see that this yields a uniform member of Λ Ψ, note that there is a bijection between the union of these choices and the members of Λ Ψ ; note also that the number of choices for steps 3,4,5 is independent of the partitions chosen in steps,2. In every use of this model, we will use it to analyze the configuration remaining at a particular point during STRIP2. We will set W 0 = W 0, W = W, R = R and so we will use the notation, for example, Λ W,R. 6

17 The crucial parameter in our analysis is the branching parameter that comes from exploring W 0 0 with a branching process. It is well-nown that for any c > c, W 0 0 is subcritical and so this parameter is less than one see eg. [3, 7]. We need to show how far away it is from one when c = c min. This bound is the only reason that we require c c min rather than c > c in 7. Lemma 5. Let N be a sufficiently large constant and set α = 9 β. Let G = G n,p be a random graph with p = c/n for some constant c + α = c min c c max = c + /2. Then, w.h.p. in the -core of G, i.e. at step i = 0 of STRIP: u W 0 deg W0 udeg W0 u < α. 9 u W 0 deg W0 u Proof Recall from 6 that β = e /200 and so α is very small. We are going to use the notation and observations used in the proof of Lemma ; in particular, we set x = x + δ, and recall from that c c max implies δ log. See 8 for the definition of fx and x, 9 for the definition of x, and 0 for the degree distribution of the -core. In particular, recall that c = fx and c = fx. Thus, since c is bounded away from c, x is bounded away from x. Indeed, since α < log /, and since ɛ = o and so ɛ + α = o, it follows from 7 that over the range [x, x + α], f does not exceed ɛ + α log / + O/ < log / for sufficiently large. Therefore, fx + α < fx + α log / < c + α = c min c. Since c = fx = fx + δ and f is increasing, this implies δ > α. 20 We consider the configuration Λ with the same degree sequence as the -core of G. Recall that W 0 is the set of vertices with degree exactly in Λ. We could determine the degree sequence of the subconfiguration induced by W 0 and then bound the LHS of 9, but we obtain a simpler calculation by considering the following experiment: choose a uniformly random vertex-copy ι from W 0 conditional on ι being paired in Λ with another copy from W 0 ; let u W 0 be the vertex of which ι is a copy. Set Z to be the number of other copies of u that are paired with vertex-copies in W 0. Note that EZ is the LHS of 9. One way to choose ι is to repeatedly tae a uniform vertex-copy from W 0 and expose its partner; halt the first time that the partner is also in W 0. By Lemma, a linear proportion of the copies are in W 0 and so w.h.p. we only expose on copies before halting. Now, having found u, we expose the partners of the remaining copies of u. So EZ is simply the expected number of these partners that are in W 0, which by 0 and the fact that all but on vertex-copies of Λ remain unexposed is + o gx, where gx := e x x! i i e x x i i! = e x x! x e x i xi i! = e x x! x Pr P ox. Therefore w.h.p. the LHS of 9 is equal to gx + o. As mentioned in the proof of Lemma, x minimizes the function fx. Setting f x = 0, it is a simple exercise to show that gx = ; 7

18 see [2] for the details. Hence, using the fact that Pr P ox Pr P ox for x > x, it follows from 2 that gx e x x! δlog / /2 + o x + δ Pr P ox = gx δlog / /2 + o 2 δlog //2 < α, by 20 provided is large enough. 3.5 The procedure terminates quicly In this section, we will prove that w.h.p. STRIP2 terminates with Q = when run on Λ. In order to show that Q reaches within βn iterations, we will eep trac of a weighted sum of the total degree of the vertices in Q, and we will show that this parameter drifts towards zero. Within this parameter, the change in the number of edges from Q to W 0 is the most delicate. In particular, the most sensitive part of our process is avoiding cascades that could be formed when the deletion of vertices in W 0 Q causes too many other vertices in W 0 to be added to Q. Roughly speaing, Lemma 5 ensures that such cascades do not occur. So we place a high weight on the number of edges from Q W 0 to W 0. We place an even higher weight on the edges from Q\W 0 to W 0 ; these edges play a different role in the analysis because they initiate the potential cascades. A i = deg W0 v v Q W 0 B i = deg W0 v v Q\W 0 D i = v Q deg W Rv X i = A i + B i + 6 βd i Note that, indeed, since β = e /200 < 0 for large enough, the edges counted by A i and B i have much higher weights in X i than those counted by D i. We will prove that X i has a negative drift, and that w.h.p. it drifts to zero before βn iterations of STRIP2. At that point, if any vertices remain in Q then they all have degree zero and so will be removed without any new vertices being added to Q. Observation 6. Throughout STRIP2, we always have: a W 3βn; b W 0 n 2 ; c ER, W 0 n 3 ; 8

19 d EW 0, W 0 n 6 ; e ER, R n 4 ; f Q 7 2 βn. Proof Note that in STRIP2 we carry out at most βn iterations, where β = e /200 from 6, and that in each iteration we delete one vertex. The total degree of the vertices from the initial set Q is at most e /6 n + 2e /3 n < βn by Lemma c,g. Every other deleted vertex has degree at most 2 and so the total degree of all deleted vertices is at most 2βn + βn = 3βn. Every vertex that moves to W is the neighbour of a deleted vertex; this proves part a. Part b follows from Lemma b since 0.99 β > 2. The remaining parts all follow from Lemma using similar arguments, each time noting that these values have only changed by some small polynomial in times the number of deleted vertices. Indeed, part c follows from Lemma e since 2 3β > 3. From Lemma d we get part d as 5 3β > 6. Part e follows from Lemma f since 3 3β > 4. Finally, for part f observe that every vertex that moves to Q is the neighbour of a deleted vertex or belongs to the second neighbourhood of a deleted vertex. Lemma c,g implies that the total degree of the initial set Q is at most e /6 + 2 e /3 n and every vertex outside of this initial set has degree at most 2. So the deletion of all initial members of Q can cause at most 2 + e /6 + 2 e /3 n < βn other vertices to be moved to Q. Any vertex of Q that was not part of the initial set has degree at most 2 and so its removal causes at most 22 + vertices to be added to Q. So after at most βn deletions, at most βn βn vertices have been added to the initial set Q. This implies part f and so the proof is finished. The ey parameter in bounding the drift of X is controlled by the following lemmas. Lemma 7. At any iteration i βn of STRIP2: if a vertex-copy σ is chosen uniformly at random from the remaining W 0 -copies in Λ W0, then E deg W0 uσ < 2 α, 2 where α = 9 β is from Lemma 5. Proof The expectation we are bounding is equal to: deg W0 u 2 / deg W0 u. 2 u W 0 u W 0 By Lemma 5, at iteration i = 0 this is u W 0 deg W0 udeg W0 u + u W 0 deg W0 u < α + = 2 α. u W 0 deg W0 u At iteration i βn, each of the i vertices that have been deleted decreases the denominator of 2 by at most 2 the largest effect is when we remove a vertex of W 0 with neighbours in 9

20 W 0 and so the denominator has decreased by at most 2βn. The numerator has not increased. Since the denominator was initially at least n/5 by Lemma d, the value of 2 is at most 2 α n/5 n/5 2βn = 2 α 2βn n/5 < 2 α β < 2 α 2, as α = 9 β. This brings up to our ey lemma: Lemma 8. At every iteration i βn of STRIP2, if the vertex v Q that is deleted has degree at least one, then EX i X i 2 6 β. Remar: If v has degree zero then X i = X i deterministically as during that iteration of STRIP2, no vertex-copies will be removed from the configuration, and no vertices will join Q. Proof Vertex v is incident to d 0 = deg W0 v edges to W 0, d = deg W v edges to W and d R = deg R v edges to R. We consider the effect on X i X i of deleting each of these pairs. First note that X i can only increase through vertices joining Q. So most of this analysis focuses on the expected number of vertices that are added to Q. Case : v W. When removing an edge to W : The removed pair is not random; it is specified by Ψ. By Observation 7a, the other endpoint of the edge is already in Q. So no new vertices are added to Q and D i decreases by two. Thus X i = X i 2 6 β. When removing an edge to R: The removed pair is not random; it is specified by Ψ. By Observation 7c, the other endpoint u of the edge is either already in Q or has no other neighbours in W ; either way, any neighbours that u has in W are already in Q. If u has degree + then u moves to W and we expose its at most neighbours in W R. Neighbours of u in W are not random, they are specified by Ψ; and as we said earlier, they are already in Q. To expose a neighbour of u in R, we choose the partner of a copy of u in Λ R ; that is, we choose a uniform vertex-copy in Λ R. The only way for the choice of partners of u in Λ R to cause vertices to be added to Q is if we choose a copy of a vertex with another neighbour in W so that neighbour is added to Q or if we choose a copy of a vertex in Q so u is added to Q. By Observation 6a,e,f: i there are at least n/2 vertex-copies in Λ R to choose from, ii at most 6 3 βn of them are copies of the at most 3 2 βn neighbours of W not already in Q as each such neighbour has degree at most 2, and iii at most 27 2 βn + βn < 5 3 βn of them are copies of vertices in R Q as the total degree of vertices with degree greater than 2 is at most e /6 n <βn by Lemma c. So the expected number of vertices added to Q is at most deg R u 2 3 βn/n/ β. Each vertex added to Q will increase X i by at most 2 2 the extreme case is if it has 2 neighbours in W 0. In addition, the removal of the pair from v to R causes D i to decrease by at least one. So provided is large enough. EX i X i 6 β β < 2 6 β, 20

21 When removing an edge to W 0 : We remove a copy of v and its partner in Λ W0,W R; that partner is selected by choosing a uniform vertex-copy from the W 0 -copies in Λ W0,W R, and the vertex that partner is a copy of is added to Q. No other vertices are added to Q in this step. A i + D i can increase by at most, as the vertex of W 0 added to Q has remaining degree at most. On the other hand, the removal of the pair causes B i to decrease by one. So X i X i + =. Case 2: v R. When removing an edge to W : The removed pair is not random; it is specified by Ψ. By Observation 7b, the other endpoint of the edge is already in Q. So X i = X i 2 6 β as in Case. When removing an edge to R: The edge is a pair from Λ R and so we choose the other vertexcopy, u, uniformly from those in Λ R. If u has degree + then, regardless of whether u NW, we move u to W and expose all of its neighbours in W R. The neighbours of u in R are selected uniformly from the vertex-copies of Λ R ; the neighbours of u in W are specified by Ψ. Any selected neighbour that is in NW \u is added to Q and if we select an neighbour in R Q then u is added to Q; the same analysis as in Case shows that the expected number of such vertices added to Q is at most 42 3 β. If u moves to W and u has a neighbour in W then both u and that neighbour go into Q; by Observation 7b,c, u has at most one neighbour in W, and as we argued in Case, the probability that u NW is at most 6 3 βn/n/2 = 2 2 β. So the expected number of vertices that move to Q because we chose u NW is at most 24 2 β. Every addition to Q increases X i by at most 2 2, and the removal of the edge from v to R decreases D i by at least one. Putting this together, we have for sufficiently large. EX i X i 6 β β β < 2 6 β, When removing an edge to W 0, the same analysis as in Case shows that X i X i <. Case 3: v W 0. When removing an edge to W R: The edge is a pair from Λ W0,W R and so we choose the other vertex-copy uniformly from the W R-copies in Λ W0,W R. By Observation 6a,c there are at least n/3 R-copies to choose from and at most 3 2 βn W -copies. So the probability that we choose a copy from W is at most 3 2 βn/n/3 = 9 2 β. If this happens then the selected neighbour is added to Q. If we choose a copy from R, then we apply essentially the same analysis as in Case 2 to bound the expected number of additions to Q; the only difference is that the number of R-copies to choose the partner of v from is at least n/3 rather than at least n/2, as we apply Observation 6c rather than Observation 6e. This leads to an expectation of at most 8 3 β+42 3 β. So in total, the expected number of additions to Q is at most 9 2 β+8 3 β+42 3 β < 70 3 β. Each such addition increases X i by at most 2 2, and the removal of the edge from v to W R decreases D i by at least one. So for sufficiently large. EX i X i 6 β β < 2 6 β, When removing an edge to W 0 : The edge is a pair from Λ W0 and so we choose a uniform vertex-copy from those in Λ W0. By Lemma 7, the expected increase in A i is at most 2 α 2 where 2

22 α = 9 β. The increase in D i is at most, and deleting the pair decreases A i by two. Putting this together, EX i X i α + 6 β < α 2 4 < 2 6 β. So in every case, the deletion of a copy of v and its partner results in an expected change in X i of less than 2 6 β. Since v has degree at least one, this yields the lemma. Our bounds on the drift of X i and the initial size of Q imply that our procedure stops quicly. Lemma 9. W.h.p. STRIP2 halts with Q = within βn iterations. Proof We begin by showing that w.h.p. we reach X i = 0 long before step i = βn. As we argued in the proof of Lemma 6, Lemma c,g imply that at iteration i = 0, the total degree of the vertices in Q is w.h.p. at most e /6 + 2 e /3 n and so we have X 0 e /6 + 2 e /3 n < e /0 n. Recall from 6 that β = e /200. By Lemma 8, for every i βn, EX i+ X i b, where b := 2 6 β. Note also that deterministically, for every i βn, X i+ X i 24 2 = 8 3, since the degree of a vertex not belonging to Q is at most 2, and we add at most 4 2 vertices to Q in each iteration the extreme case is when 2 vertices move from R to W and each has 2 neighbours in R, each with another neighbour in W. We will use the Martingale inequality from the end of Section 2. We cannot apply this directly to X i since X i stops changing when it reaches zero. So instead, we couple X i to a process which is allowed to drop below zero. We define X i+ = X i+ whenever X i > 0, and if X i = 0 then X i+ = X i with probability b and X i+ = X i otherwise. So for all i we have EX i+ X i b and X i+ X i 83. Setting i := β n, we have 4 EX i X 0 i b < e /0 n 2 2 β 2 n < e /50 n. Applying 5 with x = e /0 n, c = 8 3 and l = i yields PrX i > 0 PrX i > 0 exp e /50 n 2 2β 4 n8 3 2 = exp Ωn = o. At this point, Q e /0 n i < 5β n, since as mentioned above, we add at most 4 2 vertices 2 to Q in each iteration. If X i = 0 then there are no edges from Q to the remaining vertices outside of Q. From that point on, no vertices will be added to Q and so STRIP2 halts after Q further steps. So the total number of steps is w.h.p. at most i + 5β n < βn. 2 This proves Lemma 9 since carrying out STRIP2 performs the same steps as carrying out STRIP on Λ the only difference is that STRIP2 also exposes some information. Since properties that hold w.h.p. on Λ also hold w.h.p. on C, the -core of G n,p=c/n, see the discussion in Section 3.2, this yields Lemma 0. 22