Robust hamiltonicity of random directed graphs

Transcription

1 Robust hamiltonicity of random directed grahs Asaf Ferber Rajko Nenadov Andreas Noever Ueli Peter Nemanja Škorić Institute of Theoretical Comuter Science ETH Zurich, 8092 Zurich, Switzerland Abstract In his seminal aer from 1952 Dirac showed that the comlete grah on n 3 vertices remains Hamiltonian even if we allow an adversary to remove n/2 edges touching each vertex. In 1960 Ghouila-Houri obtained an analogue statement for digrahs by showing that every directed grah on n 3 vertices with minimum in- and out-degree at least n/2 contains a directed Hamilton cycle. Both statements quantify the robustness of comlete grahs (digrahs) with resect to the roerty of containing a Hamilton cycle. A natural way to generalize such results to arbitrary grahs (digrahs) is using the notion of local resilience. The local resilience of a grah (digrah) G with resect to a roerty P is the maximum number r such that G has the roerty P even if we allow an adversary to remove an r-fraction of (in- and out-going) edges touching each vertex. The theorems of Dirac and Ghouila-Houri state that the local resilience of the comlete grah and digrah with resect to Hamiltonicity is 1/2. Recently, this statements have been generalized to random settings. Lee and Sudakov (2012) roved that the local resilience of a random grah with edge robability = ω (log n/n) with resect to Hamiltonicity is 1/2 ± o(1). For random directed grahs, Hefetz, Steger and Sudakov (2014+) roved an analogue statement, but only for edge robability = ω (log n/ n). In this aer we significantly imrove their result to = ω ( log 8 n/n ), which is otimal u to the olylogarithmic factor.

2 1 Introduction A Hamilton cycle in a grah or a directed grah is a cycle that asses through all the vertices of the grah exactly once, and a grah is Hamiltonian if it contains a Hamilton cycle. Hamiltonicity is one of the central notions in grah theory, and has been intensively studied by numerous researchers. It is well-known that the roblem of whether a given grah contains a Hamilton cycle is N P-comlete. In fact, Hamiltonicity was one of Kar s 21 N P-comlete roblems [12]. Since one can not hoe for a general classification of Hamiltonian grahs, as a consequence of Kar s result, there is a large interest in deriving roerties that are sufficient for Hamiltonicity. A classic result by Dirac from 1952 [7] states that every grah on n 3 vertices with minimum degree at least n/2 is Hamiltonian. This result is tight as the comlete biartite grah with arts of sizes that differ by one, K m,m+1, is not Hamiltonian. Note that it also answers the following question: Starting with the comlete grah on n vertices K n, what is the maximal integer such that for any subgrah H of K n with maximum degree, the grah K n H obtained by deleting the edges of H from K n is Hamiltonian? This question not only asks for a sufficient condition for a grah to be Hamiltonian, it also asks for a quantification for the local robustness of the comlete grah with resect to Hamiltonicity. A natural generalization of this question is to relace the comlete grah with some other base grah. Recently, questions of this tye have drawn a lot of attention under the notion of resilience. Roughly seaking, given a monotone increasing grah roerty P and a grah or a digrah G which satisfies P, the resilience of G with resect to P measures how much one must change G in order to destroy P. Since one can destroy many natural roerties by small changes (for examle, by isolating a vertex), it is natural to limit the number of edges touching any vertex that one is allowed to delete. This leads to the following definition of local resilience. Definition 1.1 (Local resilience). Let P be a monotone increasing grah roerty. For a grah G, the local resilience is while for a digrah G the local resilience is r(g, P) := min{r : H G such that v V (G) d H (v) r d G (v) and G H does not have P}, r(g, P) := min{r : H G such that v V (G) d + H (v) r d+ G (v) and d H (v) r d G (v) and G H does not have P}. Sudakov and Vu initiated the systematic study of resilience of random and seudorandom grahs in [18], and since then this field has attracted substantial research interest (see e.g. [2, 3, 4, 6, 8, 13, 14]). Let us denote with HAM the grah roerty of containing a Hamilton cycle and the directed grah roerty of containing a directed Hamilton cycle. Lee and Sudakov [14] roved that for = ω (log n/n), a tyical G G(n, ) satisfies r(g, HAM) (1/2 ± o(1)). Note that this result is asymtotically otimal with resect to the constant 1/2 as, for any ε > 0, a tyical G G(n, ) can be disconnected by the adversary without removing more than a fraction of (1/2 + ε) of the edges around any vertex. Namely, the adversary can take artition of the vertex set into two equal sets 1

3 (u to divisibility conditions) and delete all the edges between the sets. This is ossible since w.h.. a vertex from G(n, ) has roughly half of its edges in both arts. Furthermore, the aforementioned result is also otimal with resect to the arameter as it is well known that a tyical grah G G(n, ) is not Hamiltonian for = o(log n/n) (see [5]). For a ositive integer n and 0 = (n) 1, let D(n, ) denote the binomial robability sace of digrahs on the set of vertices [n] = {1,..., n}. That is, an element D D(n, ) is generated by including each of the n(n 1) ossible ordered airs of [n] with robability, indeendently of all other airs. For this model, Frieze [9] showed that a tyical digrah D D(n, ) is Hamiltonian for (log n + ω(1))/n. As a next ste, it is natural to ask for an analogue to the result of Lee and Sudakov [14] for random digrahs. To this end, Hefetz, Steger and Sudakov [10] roved the following theorem. Note that it is asymtotically otimal with resect to the resilience for the same reasons as in the case of undirected grahs. However, it is far from otimal with resect to the bound on the edge robability. Theorem 1.2 ([10]). For any constant β > 0, if = ω (log n/ n) then w.h.. a digrah G D(n, ) satisfies r(g, HAM) (1/2 ± β). In the roof of Theorem 1.2, Hefetz, Steger and Sudakov extensively used the Regularity Lemma and the fact that for = ω (log n/ n) a tyical digrah G D(n, ) contains many transitive triangles touching each vertex. Generalizing it to smaller values of would at least require to relace triangles with some sarser gadgets. However, any gadget of constant size could in rincile only allow to be of a form = ω(n 1+ε ), where the constant ε deends on the gadget. In general, roblems related to Hamilton cycles in digrahs are known to be much harder than their counterarts in the undirected setting, mainly since the Posá rotation-extension technique (see [16]) is, in its simlest form, not alicable to directed grahs. In this aer we use the absorbing method, initiated by Rödl, Ruciński and Szemerédi [17], combined with an elegant embedding argument of Montgomery [15] to rove the following theorem. Note that bound on is otimal u to a olylogarithmic factor and, as mentioned earlier, the constant 1/2 can not be imroved. Theorem 1.3. For any constant β > 0, if = ω ( log 8 n/n ) then w.h.. a digrah G D(n, ) satisfies r(g, HAM) (1/2 ± β). We want to remark that our roof can easily be turned into a simle and efficient randomized algorithm which finds a Hamilton cycle in a digrah with certain seudorandom roerties. The aer is organized as follows. In Section 2 we resent auxiliary lemmas which are used throughout the aer. In Section 3 we give the definition of (n, α, )-seudorandom digrahs and state our main result (Theorem 3.2) concerning the Hamiltonicity of such digrahs. We then show how it imlies Theorem 1.3 and furthermore derive the roof of Theorem 3.2 using two key lemmas, which we call the Connecting Lemma and the Absorbing lemma. In Section 4 we then give a roof of the Connecting Lemma, and finally in Section 5 we use it to rove the Absorbing Lemma. Outline of the roof The roof relies on the alication of the absorbing method which heavily uses the above mentioned Connecting Lemma. Given a random subset X and a list of t airs of vertices avoiding X, the Connecting Lemma enables us to connect the airs by vertex disjoint aths of any rescribed orientation and sufficiently large length, with all internal vertices being in X. With this in mind, the roof roceeds as follows. 2

4 First, we take out a random subset of vertices X and in the remaining digrah create an absorbing (directed) ath P. This ath the roerty that for any subset of vertices X X there exist a directed ath P X which has the same endoints as P and V (P X ) = V (P ) X. In other words, the ath P can absorb X. The existence of such a P is roven by first constructing a suitable gadget consisting of many intertwined aths, before such gadgets are found by reeatedly alying the Connecting Lemma. Next, we cover the set of unused vertices V (G) \ (X V (P )) with roughly n/ log 5 n vertex disjoint directed aths. This is achieved by artitioning the grah into a sequence of sets of size roughly n/ log 5 n and then iteratively finding a erfect matching between any two consecutive arts. Finally, we join the aths obtained in the revious ste together with the absorbing ath P into one directed cycle by connecting the endoints of the aths using the vertices from X. This is again done by alying the Connecting Lemma. 1.1 Notation and definitions For an integer n, let [n] = {1,..., n}. For a, b, c R let (a±b)c denote the interval ((a b)c, (a+b)c). Our grah theoretic notation is standard and follows that of West [19]. In articular we use the following: Given a digrah D we denote by V (D) and E(D) the sets of vertices and arcs of D, resectively, and let v(d) = V (D) and e(d) = E(D). For a subset S V (D), we denote with D[S] the subgrah of D induced by S. For two (not necessarily disjoint) subsets X, Y V (D), set E D (X, Y ) := {(x, y) E(D) : x X and y Y } and let e D (X, Y ) = E D (X, Y ). Furthermore, let N + D (X, Y ) = {y Y : x X and (x, y) E(D)} denote the set of all out-neighbors of X in Y and let N D (X, Y ) = {y Y : x X and (y, x) E(D)} denote the set of all in-neighbors of X in Y. Given a vertex x V (D) and τ {+, }, we abbreviate ND τ ({x}, Y ) to N D τ (x, Y ) and define d τ D (x, Y ) = N D τ (x, Y ) and d± D (x, Y ) = min{d+ D (x, Y ), d D (x, Y )}. We omit the subscrit D whenever there is no risk of confusion. For τ {+, } we denote with τ the oosite sign. Furthermore, for σ {+, } l and i [l], let σ(i) denote i-th member of the l-tule σ, let σ i = (σ(1),..., σ(i)) and let σ denote ( σ(l),..., σ(1)). We call a sequence of vertices P = v 1,..., v l+1 a σ-walk if all the vertices are different, excet that v 1 and v l+1 can be the same vertex, and if v i+1 N σ(i) (v i ) for all 1 i l. Moreover, we say that P connects v 1 to v l+1 and call v 1 and v l+1 its left and right endoint, resectively. The σ-walk P is additionally called an v 1 v l+1 -ath if v 1 v l+1 and σ(i) = +, for all 1 i l. We also use the standard asymtotic notation o, O, ω, and Ω following [11]. Furthermore, for two functions a and b we write a b if a = o(b) and a b if a = ω(b). 2 Tools and reliminaries We introduce the tools used in the roofs of our results. 2.1 Probabilistic tools We need to emloy standard bounds on large deviations of random variables. We mostly use the following well-known bound on the lower and the uer tails of the Binomial distribution and the hyergeometric distribution due to Chernoff and Hoeffding (see [1, 11]). Lemma 2.1. Let X Bin(n, ) and let µ = E(X). Then Pr [X < (1 a)µ] < e a2 µ/2 for every a > 0; 3

5 Pr [X > (1 + a)µ] < e a2 µ/3 for every 0 < a < 3/2. Moreover, the inequalities above also hold if X has the hyergeometric distribution with the same mean. The following is a trivial yet useful bound. Lemma 2.2. Let X Bin(n, ) and k N.Then the following holds: Proof. Pr(X k) ( ) n k k ( en) k. k Pr(X k) ( en ) k. k 2.2 Grah Partitioning The next lemma states that one can artition a digrah into subsets which, roortionally, inherit the lower bound on the in- and out-degree. Lemma 2.3. Let c, ε > 0 be constants, n a sufficiently large integer and 0 < := (n) < 1. Suose that: (i) D is a digrah on n vertices, (ii) U V (D), (iii) k, s 1,..., s k [n] are integers such that s i log1.1 n and s i U. i Then, there exist disjoint subsets S 1,..., S k U such that the following holds for every 1 i k: (a) S i = s i, and (b) for every v V (D), if d ± D (v, U) c U then d ± D (v, S i) (1 ε)cs i. (1) Proof. We rove the lemma only for d + (v, S i ) (1 ε)cs i as the roof for d (v, S i ) (1 ε)cs i follows in a similar fashion. Let U = S 1... S k Z be a artition of U taken uniformly at random from all artitions for which S i = s i for every 1 i k and the leftover set Z is of size Z = U k i=1 s i. Let W := {v V (D) d + (v, U) c U } and let v W be an arbitrary vertex from W. The number of out-neighbors of v in S i is hyergeometrically distributed, thus we have E[d + (v, S i )] = d + (v, U) s i U c log1.1 n. Using this and Lemma 2.1 we obtain the following uer bound [ Pr d + (v, S i ) (1 ε)d + (v, U) s ] i e ε2 c log 1.1 n/2. U 4

6 The set W has at most n vertices and the size of each art S i is a ositive integer, thus the number of arts k is at most n. Taking the union bound over all arts S 1,... S k and all vertices in W we get [ Pr v W i [k], d + (v, S i ) (1 ε)d + (v, U) s ] i n 2 e ε2 c log 1.1 n/2 = o(1), U which comletes the roof. 3 Proof of Theorem 1.3 In this section we introduce the definition of an (n, α, )-seudorandom digrah, which will be the main object of study throughout the aer. In fact, we rove that for = ω( log8 n n ) and any ositive constant α, any sufficiently large (n, α, )-seudorandom digrah contains a directed Hamilton cycle. This will imly the main theorem as we show that w.h.. D(n, ) has the roerty that after deleting at most a (1/2 β)-fraction of the edges from each vertex the resulting digrah is (n, α, )-seudorandom, for some ositive constant α < β. Definition 3.1. A directed grah D on n vertices is called (n, α, )-seudorandom if the following holds: (P1) for every v V (D) we have d ± D (v, V (D)) (1/2 + 2α)n, (P2) for every subset X V (D) of size X log2 n, we have e D (X) X log 2.1 n, (P3) for every two disjoint subsets X, Y V (D) of sizes X, Y log1.1 n, we have e D (X, Y ) (1 + α/2) X Y. Intuitively, we require from an (n, α, )-seudorandom digrah a certain lower bound on the minimum degree and that it does not contain a dense subgrah. As it turns out, these roerties are sufficient for containing a directed Hamilton cycle. Theorem 3.2. Let n be an integer and let = (n) (0, 1) such that = ω( log8 n n ). There exists a constant α 0 > 0 such that for any constant 0 < α α 0 and n sufficiently large, every (n, α, )-seudorandom digrah is Hamiltonian. Before we give the roof of Theorem 3.2, we first show how it imlies Theorem 1.3. Proof of Theorem 1.3. Let β, n and be as stated in the theorem and let α = min{α 0, β/4}. By Theorem 3.2, it is sufficient to rove that G D(n, ) w.h.. satisfies that D = G H is (n, α, )- seudorandom, for every H G as given in Definition 1.1. Using the fact that w.h.. we have d + G (v), d G (v) (1 ± β/2)n for each v V (G), Definition 1.1 imlies that we have to show this for all subgrahs H G which satisfy d + H (v), d H (v) (1/2 β)n, for every v V (H). Let us consider one such subgrah H. 5

7 First, observe that that for every vertex v D we have d ± D (v) (1 β/2 1/2 + β)n (1/2 + 2α)n, (2) thus the roerty (P1) holds. For (P2), let X V (G) be an arbitrary subset of size at most log2 n. Since e G (X) Bin(2 ( ) X 2, ), by Lemma 2.2 we have x log2 n Pr [ e G (X) X log 2.1 n ] ( e X 2 ) X log 2.1 n X log 2.1. n A union bound over the choices for X shows that the robability that there exists a subset X V (G) of size x log2 n such that e G (X) X log 2.1 n is at most ( ) ( n ex 2 ) x log 2.1 n x x log 2.1 ( ( ) ) ex log 2.1 n x n n log 2.1 n x log2 n x log2 n ( n ( ) ) e log 2.1 n x log 0.1 = o(1). n Hence, (P2) holds in G and therefore in D G as well. The roof of (P3) goes similarly. Consider disjoint subsets X, Y V (G) of size at least log 1.1 n/. Then by Lemma 2.1 we have Pr [e G (X, Y ) > (1 + α/2) X Y ] < e Ω( X Y ). A union bound over the choices for X and Y yields that the robability that (P3) fails is bounded above by n x,y= log1.1 n ( n x )( n y ) e Ω(xy) n x,y= log1.1 n n x n y e Ω(max{x,y} log1.1 n) = o(1). This shows G is w.h. such that D is an (n, α, )-seudorandom digrah regardless of the choice of H and thus comletes the roof. We ick α 0 sufficiently small such that (1 + α 0 /2)(1/2 + α 0 /20) < (1/2 + α 0 /3). Throughout the aer, unless stated otherwise, we always assume that D is a (n, α, )-seudorandom digrah, where α is a ositive constant such that α α 0, = ω( log8 n n ) and n is sufficiently large. Moreover, by {V 1, V 2, V 3, V 4, V 5 } we denote the artition of V (D) given by the following claim. Claim 3.3. There exists a artition V (D) = 5 i=1 V i of the vertices of D, such that the following holds: (Q1) for every v V (D) and every i {1, 2, 3, 4, 5}, we have d ± D (v, V i) (1/2 + α) V i, ( (Q2) V 1 = (1 + o(1))n/ log 3 α n and V 2, V 3, V 4 5(1+2α) n, α 4(1+2α) ). n 6

8 Proof. Let s 1 = (1 + o(1))n/ log 3 n, s 2, s 3, s 4 ( α 5(1+2α) n, α 4(1+2α) n) be arbitrarily chosen integers and s 5 = n 4 i=1 s i. As s i log1.1 n for all i {1, 2, 3, 4, 5}, by alying Lemma 2.3 with c = 1/2 + 2α, ε = α/3, V (D) (as U), k = 5, s 1, s 2, s 3, s 4 and s 5 and by using (P1) we obtain sets V 1, V 2, V 3, V 4 and V 5 (S 1, S 2, S 3, S 4 and S 5 in Lemma 2.3) such that V (D) = V 1 V 2 V 3 V 4 V 5 and d ± (v, S i ) (1 ε)cs i = (1 ε) (1/2 + 2α)s i (1/2 + α)s i, for every v V (D) and i {1, 2, 3, 4, 5}, as required. 3.1 Proof of Theorem 3.2 The following two lemmas will serve as our main tool for roving the Hamiltonicity of D. Lemma 3.4 (The Absorbing Lemma). There exists a directed ath P in D with V (P ) V 2 V 3 V 4 such that for every W V 1 there is a directed ath P W with V (P W ) = V (P ) W and such that P W and P have the same endoints. The following lemma states that, under certain assumtions, one can find disjoint σ-walks connecting secified airs of vertices, for arbitrary σ of length Ω(log n) (see Section 1.1 for the definition of a σ-walk). The roof of the Connecting Lemma is a modification of an argument by Montgomery [15]. Lemma 3.5 (The Connecting Lemma). Let l and t be integers such that l 10 log n and t 4 log2 n and let {(a i, b i )} t i=1 be a family of airs of vertices from V (D) with a i a j and b i b j for every distinct i, j [t]. Assume that K V (D) \ t i=1 {a i, b i } is such that (i) K = ω(tl), (ii) for every v K we have d ± (v, K) ( α) K and (iii) for every i [t] we have d ± (a i, K), d ± (b i, K) (1/2 + α) K. Then for every σ {, +} l there exist t internally disjoint σ-walks P 1,..., P t such that for each i, P i connects a i to b i and V (P i ) \ {a i, b i } K. With these two lemmas at hand, we are ready to give a roof of Theorem 3.2. Proof of Theorem 3.2. Let P be a ath obtained from Lemma 3.4 and let U := (V 2 V 3 V 4 V 5 ) \ V (P ). We first show that there exists a family {Q 1,..., Q t } of t [n/ log 5 n, 2n/ log 5 n] vertex-disjoint directed aths such that U = t i=1 V (Q i). It follows from roerty (Q2) that V 5 (1 fact that V 5 U imly α 1+2α )n. Furthermore, roerty (Q1) and the d ± (v, U) d ± (v, V 5 ) (1/2 + α) V 5 (1/2 + α) 1 + α U = (1/2 + α/2) U, (3) 1 + 2α 7

9 for every v U. Alying Lemma 2.3 with c = 1/2 + α/2, ε such that (1 ε)c > 1/2 + α/4, U, s := s 1 =... = s k = n/ log 5 n and k = U /s, together with (3), we obtain disjoint subsets S 1,..., S k U such that d ± (v, S i ) (1 ε)c S i (1/2 + α ) S i, (4) for some constant α > α/4, every i [k] and v U. We now use the following claim, whose roof we defer to the end of the subsection. Claim 3.6. For every i [k 1], there exists a erfect matching from S i to S i+1. Observe that such matchings induce s vertex-disjoint directed aths {Q 1,..., Q s }, each of length k, such that s i=1 V (Q i) = k i=1 S i. On the other hand, by taking each vertex in U \ k i=1 S i to be a 0-length ath, we obtain at most s additional aths {Q s+1,..., Q t }. Note that the family {Q 1,..., Q t } satisfies the desired roerties. As a final ste, we find a cycle C in D which contains aths Q 1,..., Q t, P and maybe some vertices from V 1. Using Lemma 3.4, we can absorb the remaining vertices from V 1 and obtain a Hamilton cycle. We now make this more recise. For i [t ], let us denote with a i and b i the first and the last vertex on the ath Q i. Furthermore, let a t +1 and b t +1 be the first and the last vertex of the ath P. Alying the Connecting Lemma (Lemma 3.5) with l = 10 log n, t = t + 1, the family of airs {(b i, a i+1 )} t i=1 {(b t +1, a 1 )} and V 1 (as K), we obtain vertex disjoint directed aths P 1,..., P t +1. This is indeed ossible since the condition (i) of the Connecting Lemma follows from V 1 = (1±o(1))n/ log 3 n and tl = O(n/ log 4 n), and (ii) and (iii) follow from (Q1). Observe that Q 1, P 1, Q 2,..., Q t, P t, P, P t +1 forms a directed cycle C with V 2 V 3 V 4 V 5 V (C). By Lemma 3.4 there is a ath PV 1 \V (C) with the same endoints as P and such that V (PV 1 \V (C) ) = V (P ) (V 1 \ V (C)). As C contains the ath P, we can relace P with PV 1 \V (C) thus obtaining a Hamilton cycle. Proof of Claim 3.6. We use the following version of Hall s condition (see [19]): there exists a erfect matching from S i to S i+1 if and only if for every subset X S i of size X S i /2 we have N + (X, S i+1 ) X and for every subset Y S i+1 of size Y S i+1 /2 we have N (Y, S i ) Y. Let us assume, towards a contradiction, that there exists a subset X S i of size X S i /2 such that N + (X, S i+1 ) is contained in a set Y of size exactly X 1. We distinguish between two cases: (a) X log 2 n/(2). In this case we have that e D (X, S i+1 \ Y ) = 0, and therefore, using (4) we obtain that e D (X Y ) e D (X, Y ) = e D (X, S i+1 ) (1/2 + α ) X S i+1 = ω( X log 2.1 n), which contradicts (P2) (here we use the fact that S i+1 = ω(log 2.1 n)). (b) log 1.1 n/ < X S i /2. In this case we have, using (4), e D (X, Y ) (1/2 + α ) X S i+1 = 1 + 2α X S i+1 > (1 + α/2) X Y, 2 which contradicts (P3) (here we use the fact that α > α/4). The same argument is alied to a subset Y S i+1 of size Y S i+1 /2. This comletes the roof. 8

10 4 Proof of the Connecting Lemma In this section we rove the Connecting Lemma (Lemma 3.5). The lemma states that if we have a list of t airs of vertices and a set K of size ω(tl) (where l 10 log n) which behaves like a random subset of V (D) then for any σ {+, } l we can connect each air of vertices via t disjoint σ-walks of length l by using only vertices from K. The roof is obtained by adoting an elegant argument of Montgomery [15] into the setting of resilience. 4.1 Exansion roerties and σ-neighborhoods We start with a lemma which says that for any two (not too small) sets X, Y V (D), such that all vertices x X have a large degree in Y, X exands to more than a half of the vertices in Y. Lemma 4.1. Let X, Y V (D) be two (not necessarily disjoint) subsets such that X = log2 n Y 6 log2.1 n α and for each x X d ± (x, Y ) (1/2 + α/2) Y. Then N + (X, Y ), N (X, Y ) (1/2 + α/20) Y. 2, Proof. We only show that N + (X, Y ) (1/2 + α/20) Y as the bound on N (X, Y ) is roven analogously. We show a slightly stronger statement, namely that S X := N + (X, Y ) \ X is of size at least (1/2 + α/20) Y. From the roerty (P2) we have e(x) X log 2.1 n. Together with d ± (x, Y ) (1/2+α/2) Y for every x X and Y 6 log 2.1 n/α, this imlies the following bound on e(x, S X ), e(x, S X ) e(x, Y ) e(x) (1/2 + α/2) X Y X log 2.1 n (1/2 + α/3) X Y. (5) Let us assume towards a contradiction that S X < (1/2 + α/20) Y. We distinguish between two cases: S X < log2 n 2. In this case we have X S X < log2 n. Therefore alying roerty (P2) to the set X S X we conclude e(x S X ) X S X log 2.1 n 2 X log 2.1 n. On the other hand, from (5) and the bound on the size of Y we have e(x S X ) e(x, S X ) (1/2 + α/3) X Y (2 + 3/α) X log 2.1 n, which is a contradiction. log2 n 2 S X (1/2 + α/20) Y. Using roerty (P3) we obtain e(x, S X ) (1 + α/2) X S X (1 + α/2)(1/2 + α/20) X Y < (1/2 + α/3) X Y, which gives a contradiction with (5). Therefore we have N + (X, Y ) S X (1/2 + α/20) Y, as required. 9

11 Next, we introduce the notion of a σ-neighborhood. For given sets A, B V (D), integer l and σ {+, } l, we define N σ (A, B) as follows, N σ (A, B) := {x B a x A and a σ-walk P connecting a x to x and V (P ) \ {a x } B}. In the following lemma we show that given any two subsets X, Y V (D) with good exansion roerties, we can find a vertex x X which can reach more than a half of the vertices in Y via σ-walks. Lemma 4.2. Let γ (0, 1) be a constant and let l be an integer such that l 2 log n. Suose that X, Y V (D) are two disjoint subsets of vertices such that the following holds: (i) Y 6l γ log2 n, (ii) N + (X, Y ), N (X, Y ) 2 log2 n and (iii) for every subset S Y of size S log2 n we have N + (S, Y ), N (S, Y ) (1/2 + γ) Y. Then for any σ {+, } l there exists a vertex x X such that N σ (x, Y ) (1/2 + γ/2) Y. Proof. Recall that σ i = (σ(1),..., σ(i)). We first show that there exists a vertex x X such that N σl 1 (x, Y ) 2 log2 n. In order to do so, we make use of the following claim. Claim 4.3. Let i < l be an integer and A X such that N σi (A, Y ) 2 log2 n. Then there exists a subset A A such that A A /2 and N σi+1 (A, Y ) 2 log2 n. Using Claim 4.3 we rove the lemma as follows. By assumtion (ii) we have N σ(1) (X, Y ) 2 log 2 n/. Alying Claim 4.3 l 2 times, we thus obtain a set X X such that X X /2 l 2 and N σl 1 (X, Y ) 2 log2 n. Since X n and l 2 log n, it follows that X /2 l 2 1 and therefore X = 1. Hence, there exists x X such that N σl 1 (x, Y ) 2 log2 n. Note that, by definition, for each w N σl 1 (x, Y ) there exists a σ l 1 -walk P w connecting x to w with V (P w ) \ {x} Y. Let now M N σl 1 (x, Y ) be a subset of size M = log2 n and let V := ( w M V (P w) ) \{x}. Note that V l M. Using assumtions (i) and (iii) we have N σ(l) (M, Y \ V ) N σ(l) (M, Y ) V (1/2 + γ) Y l M (1/2 + γ/2) Y. Observe that N σ(l) (M, Y \ V ) N σ (x, Y ) and therefore N σ (x, Y ) (1/2 + γ/2) Y. In order to comlete the roof it remains to rove Claim

12 Proof of Claim 4.3. First, note that there exists a subset A A such that A A /2 and N σi (A, Y ) log2 n. Indeed, this is true as otherwise taking an arbitrary artition of the set A = S T, such that S, T A /2, yields N σi (A, Y ) N σi (S, Y ) + N σi (T, Y ) < 2 log2 n, which contradicts the assumtion that N σi (A, Y ) 2 log2 n. Let H N σi (A, Y ) be an arbitrary subset of size H = log2 n. Using assumtion (iii) we have N σ(i+1) (H, Y ) (1/2 + γ) Y. We know that for each v H there exist a σ i -walk P v connecting a vertex from A to the vertex v. Let us denote V := v H V (P v ). Using the uer bound on i we have V l H and thus N σ(i+1) (H, Y \ V ) (1/2 + γ) Y l H 2 log2 n, where the second inequality follows from assumtion (i). Finally, observe that N σ(i+1) (H, Y \V ) N σi+1 (A, Y ) and hence we have N σi+1 (A, Y ) 2 log2 n. This comletes the roof of the lemma. 4.2 The roof The following lemma is an aroximate version of the Connecting Lemma and it is used as the main building block in the roof of the Connecting Lemma. Namely, the lemma states that for a given set of airs {(a i, b i )} t i=1 and sets R A, R B with good exansion roerties we can connect half of the airs via long σ-walks using only vertices from R A R B. Lemma 4.4. Let γ (0, 1) be a constant, let l and t be integers such that l 5 log n and t 4 log2 n and let {(a i, b i )} t i=1 be a family of airs of vertices from V (D) with a i a j and b i b j for every distinct i, j [t]. Furthermore, let R A, R B V (D) \ ( t i=1 {a i, b i }) be disjoint subsets such that the following holds: (i) R A, R B 12tl/γ and (ii) for X {A, B} and for every set S R A R B t i=1 {a i, b i } of size at least log2 n N + (S, R X ), N (S, R X ) (1/2 + γ) R X. we have Then for any σ {+, } l there exists a subset of indices I [t] of size s := t/2 and s internally disjoint σ-walks P i which connect a i to b i (where i I), and are such that V (P i )\{a i, b i } R A R B. Proof. Let I = {i 1,..., i s } [t] be a largest subset of indices with the desired roerty and assume, towards a contradiction, that I = s < t/2. Let us define R A := R A \ i I V (P i ), R B := R B \ i I V (P i ) and I := [t] \ I. Observe that in order to reach the contradiction it suffices to find a σ-walk P connecting some a i to b i, where i I, such that V (P ) \ {a i, b i } R A R B. To this end, let h A and h B be two integers such that h A, h B 2 log n and h A + h B + 1 = l, and consider σ h A and σ h B (recall that σ h A = (σ(1),..., σ(h A )) and σ h B = ( σ(l),..., σ(l h B + 1))). We make use of the following claim. 11

13 Claim 4.5. There exists an index i I for which the following holds: N σha (a i, R A) (1/2 + γ/4) R A and N σhb (b i, R B) (1/2 + γ/4) R B. Before we rove this claim we first finish the roof of the lemma. Let (a i, b i ) be a air of vertices with the index i I obtained by Claim 4.5. For S := N σh A (a i, R A ) we have S (1/2 + γ/4) R A. As R A is obtained by removing the vertices of I t/2 many σ-walks of length l we have R A R A tl 11tl, and thus S log 2 n/. Similarly, by assumtion (ii) we have that N σ(ha+1) (S, R B ) (1/2 + γ) R B and consequently N σ(h A+1) (S, R B) N σ(h A+1) (S, R B ) i I V (P i ) (1/2 + γ) R B tl (i) (1/2 + γ/2) R B (1/2 + γ/2) R B. (6) On the other hand we know from Claim 4.5 that N σh B (b i, R B ) (1/2 + γ/4) R B. Together with (6), this imlies that there exist v N σh A (a i, R A ) and w N σh B (b i, R B ) such that w N σ(h A+1) (v). Therefore we can construct a σ-walk P connecting a i to b i such that P is vertex disjoint from all the revious σ-walks, which contradicts the maximality of I. It remains to rove Claim 4.5. Proof of Claim 4.5. The idea of the roof is to reeatedly aly Lemma 4.2. First, we aly Lemma 4.2 to X := i I {a i}, Y := R A, γ/2 (as γ) and obtain a vertex v 1 X such that N σh A (a, R A ) (1/2 + γ/4) R A. Next, we aly the lemma again but now to X := X \ {v 1} instead (with the other arameters unchanged) and obtain v 2 X \ {v 1 }. After k stes of this rocedure we obtain vertices {v 1,..., v k } with the roerty N σh A (v i, R A ) (1/2 + γ/4) R A, for all 1 i k. Let us now argue that we can indeed aly Lemma 4.2 and estimate the number of stes k. Note that the condition (i) from Lemma 4.2 is satisfied as R A 12tl/γ tl 11tl/γ 6l/γ log2 n. On the other hand, using roerty (ii) of Lemma 4.4 and the fact that R A R A tl we get N (S, R A), N + (S, R A) (1/2 + γ) R A tl (1/2 + γ/2) R A, for any S R A i I {a i} of size at least log2 n. Therefore, X = i I {a i} \ {v 1,..., v i } satisfies assumtion (ii) of Lemma 4.2 as long as I i > log 2 n/. This imlies that we can iterate the rocess for at least k I /2 + 1 stes as I > t/2 2 log2 n. Thus, we obtain V A := {v 1,..., v k } with v j {a i } i I and N σha (v j, R A) (1/2 + γ/4) R A for all j [k]. 12

14 By using the analogous argument with {b i } i I and R B we obtain V B := {w 1,..., w k } such that k > I /2 and w j {b i } i I with the roerty N σhb (w j, R B) (1/2 + γ/4) R B for all j [k]. Therefore, there must exist i I such that a i V A and b i V B, as required by the claim. This finishes the roof of Lemma 4.4. Before roving the Connecting Lemma we need to introduce the following definitions. Definition 4.6. Let T be a rooted tree with edges oriented arbitrarily. Let L(T ) denote the set of leaves of T and let σ {+, } l for some integer l. We say that T is a σ-tree if for each v L(T ) the unique ath from the root of T to v is a σ-walk. Definition 4.7. Let τ {+, } and let X, Y V (D) be two disjoint sets. We say that there is a (2, τ)-matching between X and Y that saturates X if for each x X there are two distinct vertices y 1 x, y 2 x Y such that {y 1 x, y 2 x} N τ (x) and {y 1 x, y 2 x} {y 1 x, y 2 x } = for x x. We are finally ready to rove the main lemma of this section. Proof of Lemma 3.5. Given σ {, +} l, a set of airs {(a i, b i )} i [t] and a set K such that certain roerties are satisfied, we aim to construct t internally disjoint σ-walks P 1,..., P t such that for each i, P i connects a i to b i and V (P i ) \ {a i, b i } K. Let σ be an arbitrary element of {+, } l and let ε > 0 be a sufficiently small constant, in articular such that (1 ε)(1/2 + α) (1/2 + α/2). Throughout the roof we make use of the following arameters: { 6 log 2.2 } n h = max, 2t, m = log 2 t + 1, s i = h for 1 i 2m, s 2m+1 = s 2m+2 = K 4, k = 2m + 2. Alying Lemma 2.3 to s 1, s 2,..., s k, ε, 1/2 + α (as γ),, K (as U) and D we obtain disjoint subsets S 1,..., S k K, such that for every 1 i k the following holds: (a) S i = s i, and (b) for every v V (D), if d ± D (v, K) (1/2 + α) K, then d ± D (v, S i) (1 ε)(1/2 + α)s i. (7) Using (7), roerties (ii) and (iii) and the fact that ε is sufficiently small, we obtain that for any v {a i b i } t i=1 K and for any set S i the following holds: d ± D (v, S i) (1/2 + α/2)s i. (8) 13

15 For simlicity of resentation, let us denote A 0 := t i=1 {a i}, B 0 := t i=1 {b i}, A i := S i for each 1 i m, B i := S m+i for each 1 i m, R A := S 2m+1 and R B := S 2m+2. We first describe, informally, the strategy for finding σ-walks. In a first ste, we aly Lemma 4.4 to find t/2 σ-walks between vertices in A 0 and B 0. Then we find a (2, σ(1))-matching between the leftovers in A 0 and the vertices in A 1 and a (2, σ(l))-matching between the leftovers in B 0 and the vertices in B 1. Let A 1 denote the set of vertices that are matched to a leftover of A 0 and analogously define B 1. Observe that A 1 = B 1 t and therefore one can aly Lemma 4.4 to find A 1 /2 vertex disjoint κ-walks between vertices in A 1 and B 1, where κ := (σ(2),..., σ(l 1)). Note that extending the walks with the matchings yields σ-walks between at least t/4 leftovers of A 0 and the corresonding leftovers of B 0. By iteratively continuing this rocess for roughly log t stes, we construct all the desired walks. Note that the internal vertices of the σ-walks constructed in the described way are contained in sets m i=1 A i m i=1 B i R A R B K. Before roceeding with the descrition of the rocedure, we describe a structure (I s, P s, TA s, T B s), for 0 s m, whose existence is later roved by induction on s: (X1) I s [t] for which (a) if s < m then I s = t t/2 s (b) if s = m then I s = t (X2) P s = {P i } i Is is a collection of vertex-disjoint σ-walks such that P i connects a i to b i and V (P i ) \ {a i, b i } ( s k=1 (A k B k )) R A R B, for all i I s (X3) T s A = {T i A } i [t]\i s is a collection of σ s -trees and T s B = {T i B } i [t]\i s is a collection of σ s -trees such that for each i [t] \ I s (a) T i A is rooted at a i, L(T i A ) A s, L(T i A ) = 2s and V (T i A ) \ {a i} s j=1 A j (b) T i B is rooted at b i, L(T i B ) B s, L(T i B ) = 2s and V (T i B ) \ {b i} s j=1 B j (X4) for any two T 1, T 2 T s A T s B rooted at v 1 and v 2 we have (V (T 1 ) \ v 1 ) (V (T 2 ) \ v 2 ) = (X5) for any two T T s A T s B and any P P s we have V (T ) V (P ) =. The set I s reresents a set of indices of airs which are connected by a σ-walk u to ste s. The collection P s contains σ-walks created u to ste s between airs with indices in I s and the TA s and T B s are collections of trees for each element of a air not connected by a σ-walk u to ste s. First, the roerties (X1) (X5) trivially hold for s = 0, I 0 =, TA 0 = A 0, TB 0 = B 0 and P 0 =. Suose that the roerties hold for some s such that s < m, we will construct (I s+1, P s+1, T s+1 A, T s+1 B ) such that (X1) (X5) still aly. Denote A s = T TA s L(T ) and B s = T TB s L(T ). Let {a i, b i }r i=1 be a erfect matching between vertices of A s and B s with the following roerty: for every 1 i r there is j [t] such that a i and b i are leaves of trees rooted at a j and b j. Using (X1) and (X3) we obtain that r = 2 s (t I s ) 2 s t/2 s t. Next, let R A = R A\ i Is V (P i ) and let R B = R B \ i Is V (P i ). We use the following claim, whose roof is delayed for later. Claim 4.8. For every 1 s m 1 and every subset S K such that S log2 n and X {A, B} the following holds: where R X := R X \ P Ps V (P ). N + (S, R X), N (S, R X) (1/2 + α/40) R X, (9) 14

16 By Claim 4.8 it follows that for every X {A, B} and every subset S K such that S log2 n we have N + (S, R X), N (S, R X) (1/2 + α/40) R X. Therefore, we can aly Lemma 4.4 to α/40 (as γ), the family of airs {(a i, b i )}r i=1, R A (as R A), R B (as R B) and κ := (σ(s + 1),..., σ(l s)), and obtain the following: a set of indices J [r] of size J = r/2 = 2 s 1 (t I s ) and a collection of vertex-disjoint κ-walks W i, such that a ath W i connects a i to b i and V (W i) \ {a i, b i } R A R B, for each i J. Let us ick a subset J J of size J /2 s = (t I s )/2 such that each tree from TB s T A s has at most one leaf indexed with some i J. Note that this is ossible since each tree has 2 s leaves. For technical reasons, when s = m 1 we ick J of size J = 1. Now, for any j J let Q 1 j be the unique σs -walk in the tree T 1 TA s containing a j which connects the root of T 1 to a j. Similarly, let Q 2 j be the unique σs -walk in the tree T 2 TB s containing b j which connects the root of T 2 to b j. We know that Q1 j and Q2 j start in vertices with the same index from A 0 and B 0, by the definition of the matching between A s and B s. Combining the aths Q 1 j, Q2 j and W j we obtain a σ-walk P ij which connects a ij to b ij for some i j [t] \ I s. We define P s+1 = P s ( j J P ij ) and I s+1 = I s {i j j J }. Using the fact that J = (t I s )/2 when s < m 1 we obtain I s+1 = I s + (t I s )/2 = t (t I s (t I s )/2 )) = t (t I s )/2 (X1) = t t/2 s+1. (10) If s = m 1 it follows by assumtion that I s = t 1 and therefore that I s+1 = t. Note that the roerty (X2) holds directly by the construction of the new σ-walk P ij. Thus, we showed that (X1) and (X2) hold for I s+1 and P s+1. In order to construct T s+1 A and T s+1 B with the roerties from (X3) let T A T A s and T B T B s be the subsets which contain all trees that are rooted at vertices from i Is+1 {a i, b i }. For L A = T T A L(T ) and L B = T T B L(T ) it follows from (10) that L A = L B = 2 s t/2 s+1. Let us now state the next claim without the roof, which is given at the end of the section. Claim 4.9. For every 0 i m 1, every τ {+, } and X {A, B} the following holds. For every subset S X i of size S X i+1 /8 there is a (2, τ)-matching from S to X i+1 that saturates S. Using Claim 4.9 we conclude that there exist an L A -saturating (2, σ(s + 1))-matching M A from L A to A s+1, and a L B -saturating (2, σ(l s))-matching M B from L B to B s+1. For every σ j -tree T TA s we denote by T + the σ j -tree obtained by extending T with the arcs of the matching M A incident to L(T ). Similarly, for every σ j -tree T TB s we denote by T + the σ j+1 tree obtained by extending T with the arcs of the matching M B incident to L(T ). Finally, let T s+1 A = T T s T + and A let T s+1 B = T T s T +. It follows from our construction that T s+1 B A and T s+1 B satisfy (X3) (X5). Finally, we are left with roving Claims 4.9 and 4.8 to comlete the roof of the lemma. Proof of Claim 4.9. We rove the existence of a (2, +)-matching from S to X i+1 that saturates S and the roof for such a (2, )-matching follows similarly. Using Hall s Theorem (see e.g. [19]), it is sufficient to rove that for every S S we have N + (S, X i+1 ) 2 S. Assume the existence of a subset S S that violates Hall s condition, i.e. N + (S, X i+1 ) < 2 S. If S log2 n 3 then, since S N + (S, X i+1 ) log2 n, by roerty (P2) we obtain that e D (S N + (S, X i+1 )) 3 S log 2.1 n. (11) 15

17 Moreover, it follows from (8) that d ± (s, X i+1 ) (1/2 + α/2) X i+1, for all s S. All in all, we get e D (S, N + (S, X i+1 )) (1/2 + α/2) S X i+1 > 3 S log 2.2 n, (12) where the second inequality follow from X i+1 6 log2.2 n. The last inequality together with (11) leads to a contradiction. If on the other hand S > log2 n 3 and N + (S, X i+1 ) < 2 S, it follows from (P3) and the assumtion on S that e D (S, N + (S, X i+1 )) (1 + α/2)2 S 2. However by (12) and the assumtion X i+1 4 S we have e D (S, N + (S, X i+1 )) (2 + 2α) S 2. Therefore, by combining the revious inequalities we obtain which is a contradiction. (2 + α) S 2 e D (S, N + (S, X i+1 )) (2 + 2α) S 2, Proof of Claim 4.8. We rove the claim for N + (S, R X ) as the roof for N (S, R X ) follows analogously. By (8) we know that for X {A, B} and every v {a i, b i } t i=1 K the following holds at each ste s: d ± (v, R X ) (1/2 + α/2) R X. Alying Lemma 4.1 to S (as X), R X (as Y ) we get N + (S, R X ) (1/2 + α/20) R X. Since P Ps V (P ) tl and R X = ω(tl) we conclude that N + (S, R X) (1/2 + α/20) R X tl (1/2 + α/40) R X. 5 Proof of the Absorbing Lemma In this section we rove Lemma 3.4. A main ingredient in our roof is the concet of an absorber. Roughly seaking, in our setting an absorber A x for a vertex x is a digrah which contains x and two designated vertices x s and x t, such that A x contains two x s x t -aths: one which consists of all the vertices in V (A x ) and the other which consists of all the vertices in V (A x ) \ {x}. Definition 5.1. Let l x be an integer and A x a digrah of size l x + 1. Then for some distinct vertices x, x s, x t V (A x ), the digrah A x is called an absorber for a vertex x with starting vertex x s and terminal vertex x t if the following holds: there exists an x s x t -ath P x A x of length l x 1 that does not contain x (the non-absorbing ath), and there exists an x s x t -ath P x A x of length l x (the absorbing ath) In the following lemma we describe the structure of our absorber. Lemma 5.2. Let k and l be integers and consider a digrah A x of size 3 + 2k(l + 1) constructed as follows: 16

18 x s x x t s x 1 t x 1 s x 2 t x 2 s x 3 t x 3 s x 4 t x 4 s x 5 t x 5 s x 6 t x 6 Figure 1: The absorber A x for k = 3. The cycle C is drawn with solid arrows. The dashed arrows reresent directed aths of arbitrary length. The art inside the rectangle can be reeated to obtain absorbers for larger k. (i) A x consists of a cycle C of length 4k + 3 with an orientation of the edges and labeling of the vertices as shown in Figure 1, and (ii) A x contains 2k airwise disjoint directed s x i tx i -aths P i (for each i [2k]), each of which is of length l. Then A x is an absorber for the vertex x. Proof. It is easy to see that P x := x s, x, s x 1, P 1, t x 1, s x 2, P 2, t x 2,..., s x 2k, P 2k, t x 2k, x t is an absorbing ath. On the other hand, the ath P x := x s, s x 2, P 2, t x 2, s x 4, P 4, t x 4,..., s x 2k, P 2k, t x 2k, sx 1, P 1, t x 1, s x 3, P 3, t x 3,..., s x 2k 1, P 2k 1, t x 2k 1, x t uses all vertices excet x, thus it is a non-absorbing ath. We refer the reader to Figure 1 for clarification. The roof of the Absorbing Lemma consists of two main stes. First, we construct an absorber A x for each x V 1 such that the non-absorbing ath of A x is contained in V 2 V 3 and V (A x ) V (A x) = for x x. Second, using the Connecting Lemma (Lemma 3.5) we connect all the non-absorbing aths of these absorbers into one long ath using vertices from V 4. We build the absorbers A x in D by first finding the cycle of the absorber and then connecting all the designated airs of vertices via directed aths. To do so we again use the Connecting Lemma (note that a cycle is a σ-walk, for some σ). Proof of Lemma 3.4. Let A x be an absorber given by Lemma 5.2 for k = 3 log n and l = 10 log n. Recall that A x consists of a cycle C x of length 4k + 3 = 12 log n + 3 with a rescribed orientation σ and 2k = 6 log n disjoint directed aths P 1,..., P 2k, each of length l, connecting the airs of designated vertices (s x 1, tx 1 ),..., (sx 2k, tx 2k ) on the cycle. In order to find such a cycle C x for each x V 1, we aly the Connecting Lemma (Lemma 3.5) to the set V 2 (as K), l = 12 log n + 3, t = V 1 and a family of airs {(x, x)} x V1. This gives us a σ-walk of length 4k + 3 from x to itself, for every x V 1, which corresonds to the cycle C x as required for the absorber. Moreover, all obtained cycles {C x } x V1 are disjoint and contain (aart from the vertices to be absorbed) only vertices in V 2. Note that we can aly the Connecting 17

19 Lemma as V 2 = ω( V 1 log n), t = V 1 4 log2 n and by roerty (Q1) we have that (ii) and (iii) from the Connecting Lemma are satisfied. Next, using the vertices in V 3, for each x V 1 we connect the air of designated vertices (s x i, tx i ) on a cycle C x by a directed ath. To do this we aly the Connecting Lemma to V 3 (as K) with l := 10 log n, t = 2k V 1, and {(s x i, tx i ) x V 1, 1 i 2k} to find all the required aths to comlete the absorbers. We can indeed do that as V 3 = ω( V 1 log 2 n), t = 2k V 1 4 log2 n and by roerty (Q1) we have that (ii) and (iii) from the Connecting Lemma are satisfied. Finally, we build a directed ath which contains all the non-absorbing aths of A x s. To do so, recall that by Definition 5.1 every absorber A x has a start vertex x s and a terminal vertex x t. Let us arbitrarily enumerate vertices from V 1 as V 1 = {x 1,..., x h }, where h = V 1. Aly the Connecting Lemma to V 4 (as K), l = 10 log n, t = V 1 1 and a family of airs {(t xi, s xi+1 )} i [h 1] to find the required aths of length l that connect all non-absorbing aths of the absorbers into one directed ath P. Again, we are allowed to aly the lemma as V 4 = ω( V 1 log n) and by roerty (Q1) we have that (ii) and (iii) from the Connecting Lemma are satisfied. It is now easy to see that the ath P has the required roerties. Let W V 1 be an arbitrary subset of V 1 and let {A w w W } be the set of absorbers for vertices in W. By the definition of absorber for each A w there is an absorbing ath starting and ending at the same vertices as the non-absorbing ath, but which contains the vertex w as well. By relacing the non-absorbing aths of {A w w W } in P with the corresonding absorbing aths we obtain a ath PW which has the same endoints as P and V (PW ) = V (P ) W. References [1] N. Alon and J. H. Sencer. The robabilistic method. John Wiley & Sons, [2] J. Balogh, B. Csaba, and W. Samotij. Local resilience of almost sanning trees in random grahs. Random Structures & Algorithms, 38(1-2): , [3] S. Ben-Shimon, M. Krivelevich, and B. Sudakov. Local resilience and Hamiltonicity Maker Breaker games in random regular grahs. Combinatorics, Probability and Comuting, 20(02): , [4] S. Ben-Shimon, M. Krivelevich, and B. Sudakov. On the resilience of Hamiltonicity and otimal acking of Hamilton cycles in random grahs. SIAM Journal on Discrete Mathematics, 25(3): , [5] B. Bollobás. Random grahs. Sringer, [6] J. Böttcher, Y. Kohayakawa, and A. Taraz. Almost sanning subgrahs of random grahs after adversarial edge removal. Electronic Notes in Discrete Mathematics, 35: , [7] G. A. Dirac. Some theorems on abstract grahs. Proceedings of the London Mathematical Society, 3(1):69 81, [8] A. Frieze and M. Krivelevich. On two Hamilton cycle roblems in random grahs. Israel Journal of Mathematics, 166(1): , [9] A. M. Frieze. An algorithm for finding Hamilton cycles in random directed grahs. Journal of Algorithms, 9(2): ,

20 [10] D. Hefetz, A. Steger, and B. Sudakov. Random directed grahs are robustly Hamiltonian. arxiv rerint arxiv: , [11] S. Janson, T. Luczak, and A. Ruciński. Random grahs, volume 45. John Wiley & Sons, [12] R. M. Kar. Reducibility among combinatorial roblems. Sringer, [13] M. Krivelevich, C. Lee, and B. Sudakov. Resilient ancyclicity of random and seudorandom grahs. SIAM Journal on Discrete Mathematics, 24(1):1 16, [14] C. Lee and B. Sudakov. Dirac s theorem for random grahs. Random Structures & Algorithms, 41(3): , [15] R. Montgomery. Embedding bounded degree sanning trees in random grahs. arxiv rerint arxiv: , [16] L. Pósa. Hamiltonian circuits in random grahs. Discrete Mathematics, 14(4): , [17] V. Rödl, A. Ruciński, and E. Szemerédi. An aroximate Dirac-tye theorem for k-uniform hyergrahs. Combinatorica, 28(2): , [18] B. Sudakov and V. H. Vu. Local resilience of grahs. Random Structures & Algorithms, 33(4): , [19] D. B. West. Introduction to grah theory. Prentice Hall Uer Saddle River,