A probabilistic approach to consecutive pattern avoiding in permutations

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<ul><li><p>Journal of Combinatorial Theory, Series A 120 (2013) 9981011</p><p>Contents lists available at SciVerse ScienceDirect</p><p>Aa</p><p>G</p><p>De</p><p>a</p><p>ArReAv</p><p>KePeCoMCMPr</p><p>1.</p><p>(</p><p>of{1qu</p><p>stproncape</p><p>ta</p><p>00htJournal of Combinatorial Theory,Series A</p><p>www.elsevier.com/locate/jcta</p><p>probabilistic approach to consecutive patternvoiding in permutations</p><p>uillem Perarnau</p><p>partament de Matemtica Aplicada IV, Universitat Politcnica de Catalunya, BarcelonaTech, Spain</p><p>r t i c l e i n f o a b s t r a c t</p><p>ticle history:ceived 30 August 2012ailable online 21 February 2013</p><p>ywords:rmutationsnsecutive pattern avoidingonotone patternsP conjecture</p><p>obabilistic method</p><p>We present a new approach to the problem of enumeratingpermutations of length n that avoid a xed consecutive patternof length m. We use this approach to give explicit upper and lowerbounds on the number of permutations avoiding a consecutivepattern of length m. As a corollary, we obtain a simple proof of theCMP conjecture from Elizalde and Noy, regarding the most avoidedpattern, recently proved by Elizalde. We also show that most of thepatterns behave similarly to the least avoided one.</p><p> 2013 Elsevier Inc. All rights reserved.</p><p>Introduction</p><p>Let Sn be the symmetric group of permutations of length n. Consider a permutation =1, . . . ,n) Sn and a pattern = (1, . . . , m) Sm .For any sequence of different positive integers X = (x1, . . . , xk), we dene the standardizationX , st(x1, . . . , xk), as the permutation in Sk obtained by relabeling each element of X on the set</p><p>, . . . ,k} such that the relative order among the elements of X is preserved. For instance, the se-ence (3,8,4,1) standardizes to (2,4,3,1).A permutation Sn contains as a consecutive pattern if there exists 0 i n m such that</p><p>(i+1, . . . ,i+m) = , that is, there are m consecutive elements in that have the relative orderescribed by . For instance, if = (1,2, . . . ,m), then contains as a consecutive pattern if andly if it contains m consecutive increasing elements (a run of length m). A permutation Sn islled -avoiding if it does not contain as a consecutive pattern. Denote by n( ) the number ofrmutations in Sn that are -avoiding.The problem of determining n( ) is inspired by the problem of nding the number of permu-</p><p>tions of length n that avoid a pattern non-necessarily in consecutive positions. A permutation Sn contains if there exist 1 i1 &lt; &lt; im n such that st(i1 , . . . ,im ) = . Clearly, if </p><p>E-mail address: guillem.perarnau@ma4.upc.edu.</p><p>97-3165/$ see front matter 2013 Elsevier Inc. All rights reserved.</p><p>tp://dx.doi.org/10.1016/j.jcta.2013.02.004</p></li><li><p>G. Perarnau / Journal of Combinatorial Theory, Series A 120 (2013) 9981011 999</p><p>avanmw</p><p>beesevis</p><p>exth</p><p>upAnrethCo</p><p>Co</p><p>foov</p><p>fo</p><p>Co</p><p>geth</p><p>thfutu</p><p>oftoab</p><p>Thoids , then also avoids as a consecutive pattern. Knuth [11] introduced the latter problemd exactly determined the number of permutations avoiding some pattern of length 3. There areany interesting results in the area (see e.g. [2,1]) as well as the famous StanleyWilf conjecturehich was solved by Marcus and Tardos [12].To provide an exact formula for n( ) is a tough problem for large n. Asymptotic formulas canderived for some special patterns as shown by Elizalde and Noy [7]. The authors also provide an</p><p>timation of n( ) for every pattern of length 3 and also for some patterns of length 4. However,en for the case of length 4, there are still some patterns for which the asymptotic behavior of n( )not known.Elizalde [5] showed that for every Sm , the following limit</p><p> = limn</p><p>(n( )</p><p>n!)1/n</p><p>,</p><p>ists and that 0.7839 &lt; &lt; 1 if m 3. A stronger result is given in [4], where the authors showat n( ) cn n!, for some constant c that only depends on .Even if the exact computation of for every Sm is a hard problem, it is possible to provideper and lower bounds in terms of m. Besides, it is interesting to see which patterns are extremal.pattern of length m is called monotone if it is either (1,2, . . . ,m) or (m, . . . ,2,1). It is clear that(1,2, . . . ,m) = n(m, . . . ,2,1), since = (1, . . . ,n) is (1,2, . . . ,m)-avoiding if and only if itsversing (n, . . . ,1) is (m, . . . ,2,1)-avoiding. It was conjectured in [7] that monotone patterns aree most avoided ones among all patterns of length m, when n is large enough. This is known as thensecutive Monotone Pattern (CMP) Conjecture.</p><p>njecture 1 (CMP conjecture). (See [7].) For every Sm, (1,2,...,m).</p><p>The results in [7], determining for every S3, settle in the armative the CMP conjecturer patterns of length 3. Elizalde and Noy [8] show that the conjecture is true for the class of non-erlapping patterns.Regarding the least avoided pattern among all the patterns of length m, Nakamura [13] posed the</p><p>llowing conjecture:</p><p>njecture 2. (See [13].) For every Sm, (1,2,...,m2,m,m1).</p><p>Both conjectures have been recently proved by Elizalde [6]. The proofs are based on computing thenerating function for the number of -avoiding permutations, P (z) =n( ) znn! , combined withe cluster method of Goulden and Jackson [9].Here we will use a completely different approach to the consecutive pattern avoiding problem</p><p>rough the so-called probabilistic method. While this approach is not as precise as the generatingnction technique, it provides simpler alternative proofs of some known results, as the CMP conjec-re, and allows one to obtain more general results.Henceforth, m will be a xed integer while n will be considered to tend to innity. For the sakesimplicity, however, we will use asymptotic notation on m. If this is the case, we will consider mtend to innity while n will be an arbitrarily large function of m. Our rst result bounds fromove when the pattern is not monotone.</p><p>eorem 3. For every Sm \ {(1,2, . . . ,m), (m, . . . ,2,1)},</p><p> 1 1 + O(</p><p>12</p><p>).m! m m!</p></li><li><p>1000 G. Perarnau / Journal of Combinatorial Theory, Series A 120 (2013) 9981011</p><p>mth</p><p>N</p><p>Co(m</p><p>thTh</p><p>ev</p><p>ev</p><p>cola</p><p>Th</p><p>Th2,ca</p><p>deth</p><p>Th</p><p>w</p><p>corepoth</p><p>deThe proofs of this and all the following results, have a probabilistic avor. To prove this theorem weake use of Suens inequality [15], a powerful tool that provides an upper bound on the probabilityat none of the events of a certain collection happen simultaneously.By comparing the upper bound given by Theorem 3 with the result obtained by Elizalde and</p><p>oy [8] for (1,2,...,m) we get the following corollary:</p><p>rollary 4. There exists an integer m0 , such that for any mm0 and every pattern Sm \ {(1,2, . . . ,m),, . . . ,2,1)},</p><p>1 (1+ O</p><p>(1</p><p>m</p><p>))(1 (1,2,...,m)).</p><p>This provides an alternative probabilistic proof for the CMP conjecture. Analyzing more carefullye proof of Theorem 3 for small values of m, we can show that the CMP conjecture holds for m 5.e same approach, however, does not provide meaningful results for the case m = 4.This corollary also gives a lower estimation of the minimum gap between (1,2,...,m) and , forery Sm \ {(1,2, . . . ,m), (m, . . . ,2,1)}.Theorem 3 can be extended to the whole set of patterns, Sm , by weakening the upper bound: forery Sm ,</p><p> 1 1m! + O</p><p>(1</p><p>m m!).</p><p>One could think that the ideas that appear in this paper could be applied for the study of non-nsecutive pattern avoiding in permutations. Unfortunately, due to the strong dependency among arge part of the dened events, these techniques seem useless there.The second part of the paper is devoted to the proof of a general lower bound on for Sm .</p><p>eorem 5. For every Sm,</p><p> 1 1m! O</p><p>(m 1(m!)2</p><p>).</p><p>To prove this lower bound we use a one-sided version of the Lovsz Local Lemma (see [14]).is bound is asymptotically tight and an extremal example is provided by the pattern (1,2, . . . ,m m,m 1). Unlike in the case of the upper bound and the CMP conjecture, our proof of Theorem 5nnot be adapted to extract a proof of Conjecture 2.As Theorems 3 and 5 give bounds for the value of in terms of m, a natural question is totermine how most of the patterns behave. In this direction a much stronger upper bound, close toe general lower bound, is shown to hold for most of the patterns.</p><p>eorem 6. Let Sm be chosen uniformly at random. Then, for each 2 km/2,</p><p> 1 1m! + O</p><p>(4m</p><p>(m k)!m!),</p><p>ith probability at least 1 2(k+1)! m2m/2 .</p><p>This theorem shows that when m is large enough, for most of the patterns the value of isncentrated close to the lower bound provided by Theorem 5. The idea behind the proof of thissult is that the number of permutations avoiding a pattern depends on the maximum overlappingsition of this pattern. It can be shown that almost all patterns do not have a large overlap and thus,ey are far from the upper bound attained by monotone patterns, the ones with maximum overlap.The paper is organized as follows. Theorem 3 and Corollary 4 are proved in Section 2. Section 3 is</p><p>voted to the proof of Theorem 5. Finally, we provide in Section 4 the proof of Theorem 6.</p></li><li><p>G. Perarnau / Journal of Combinatorial Theory, Series A 120 (2013) 9981011 1001</p><p>2.</p><p>anfr</p><p>su</p><p>as</p><p>an</p><p>w</p><p>(s</p><p>Th</p><p>exifgi</p><p>ofnuX</p><p>win</p><p>Boprlesh</p><p>Le|i</p><p>PrthdeAn upper bound on </p><p>Consider the set of events A= {A1, . . . , AN } with associated indicator random variables X1, . . . , XNd let X = Ni=1 Xi . The events in A are considered to be bad and the aim is to bound, eitherom above or from below, the probability that none of these bad events occurs. We will denote by= E(X) =Ni=1 Pr(Ai), the expected number of bad events.A dependency graph of A is a graph H with vertex set V (H) = {1, . . . ,N} where if two disjointbsets S, T [N] share no edges then {Ai}iS and {A j} jT are independent.In order to control the dependencies among all events, the following two parameters are usually</p><p>sociated to the dependency graph H . To measure the global effect of the dependencies, we consider</p><p> =</p><p>i jE(H)Pr(Ai A j),</p><p>d for the local one,</p><p> = max1iN</p><p>j: i jE(H)</p><p>Pr(A j),</p><p>here E(H) denotes the edge set of H .We will use the following version of Suens inequality which holds for any dependency graph H</p><p>ee e.g. Theorem 2 in [10]):</p><p>eorem 7 (Suens inequality). With the above notation,</p><p>Pr(X = 0) = Pr(</p><p>Ai) exp</p><p>{(1 e</p><p>2</p><p>)</p><p>}. (1)</p><p>Suens inequality bounds from above the probability of having no bad events in terms of thepected number of events, but also takes into account the pairwise dependence of the events. Thus,there are no many dependencies among events or these dependencies are weak, we will be able tove a meaningful upper bound on such probability.Let Sn be chosen uniformly at random, and let Sm be a xed pattern. We consider the setevents A = {A0, . . . , Anm} where Ai := {st(i+1, . . . ,i+m) = }. As before, we let X denote thember of events in A which are realized. Then avoids as a consecutive pattern if and only if= 0, that is, no copy of the pattern appears. We have</p><p>n( ) = Pr(X = 0)n!,here the dependency of X on will be clear from the context. In particular we will be interested</p><p> = limnPr(X = 0)</p><p>1/n. (2)</p><p>unding from above the number of edges in a dependency graph H is crucial in order to give aoper upper bound on the probability that none of the events happen simultaneously. The followingmma shows that we can choose a dependency graph H which contains many pairs of sets of eventsaring no edges.</p><p>mma 8. Let S, T {0,1, . . . ,n m} be two disjoint subsets such that for each (i, j) S T , we have j|m. Then, the events {Ai}iS and {A j} jT are independent.</p><p>oof. The support of S , supp(S) = {k1 &lt; &lt; ks}, is composed by the elements k {1, . . . ,n} suchat there exists i S with i + 1 k i +m. Every event A in the sigma-algebra generated by {Ai}iS</p><p>pends only on the standardization of (k1 , . . . ,ks ), and is therefore independent of the entries of</p></li><li><p>1002 G. Perarnau / Journal of Combinatorial Theory, Series A 120 (2013) 9981011</p><p>si</p><p>A</p><p>ww</p><p>n/</p><p>By</p><p>Si</p><p>Ade</p><p>natistno</p><p>N(1{1</p><p>Le</p><p>Prpa</p><p>Siifin [n] \ supp(S). Let supp(T ) = {1 &lt; &lt; t} be the support of T . Similarly, every event A in thegma-algebra generated by {A j} jT depends only on the standardization of (1 , . . . ,t ).Observe that supp(S) supp(T ) = , by the assumptions on S and T . This implies that the eventsand A are independent. According to the previous lemma, the circulant graph H with vertex set V (H) = {0,1, . . . ,n m},</p><p>here i j E(H) if and only if 0 &lt; |i j| &lt; m, is a dependency graph for A. Throughout the paper,e will use this circulant graph H as a dependency graph of A.A simple upper bound follows directly from the previous observation. Consider I = {km: 0 k ki + mi 2( ki 1</p><p>),</p><p>d</p><p>mi 2m k (m ki +m mi </p><p>(k ki</p><p>)).</p><p>bserving that the rst inequality is strict, we get</p><p>mi = i+1 + mi ki . By using this last lemma, we can provide an upper bound on the probability that two given occur-</p><p>nces of a pattern appear.</p><p>mma 11. For every Sm and any k O ,</p><p>Pr(Ai Ai+mk)(2(mk)</p><p>mk)</p><p>(2m k)! .</p><p>oof. Set = st(i+1, . . . ,i+2mk). Recall that Sn has been chosen uniformly at random, whichplies that is also uniformly distributed in S2mk . Moreover, satises Ai and Ai+mk if and only satises A0 and Amk .There are (2m k)! possible candidates for . We will count how many of them are such that theents A0 and Amk hold. By Lemma 10, we know that the elements {mk+1, . . . , m} are uniquelytermined by and k. Thus, one must select a subset of mk elements among the 2m2k availablees in order to build (1, . . . , mk). Since satises A0, once these elements have been chosen,ere is just one order such that st(1, . . . , m) = , and only one way to set the last m k elements , in order to satisfy Amk .Since = st(i+1, . . . ,i+2mk), for every permutation chosen uniformly at random in Sn ,</p><p>Pr( satises Ai Ai+mk) = Pr( satises A0 Amk)(2(mk)</p><p>mk)</p><p>(2m k)! . Now we are able to prove Theorem 3.</p><p>oof of Theorem 3. First of all we compute , and , needed to apply Suens inequality. Thepected number of occurrences of the pattern does not depend on and can be computed as</p><p> =nm</p><p>Pr(Ai) = n m + 1 n .</p><p>i=0 m! m!</p></li><li><p>1004 G. Perarnau / Journal of Combinatorial Theory, Series A 120 (2013) 9981011</p><p>ev</p><p>w</p><p>fo</p><p>Siim</p><p>fo</p><p>anofAssume that i &lt; j and j i = m k. Recall that by the choice of our dependency graph H , twoents Ai and A j share no edge if i j m.By Lemmas 8, 11 and (3), can be expressed as</p><p> =</p><p>i jE(H)Pr(Ai A j) =</p><p>nmi=0</p><p>m1k=1</p><p>Pr(Ai Ai+mk) nkO</p><p>(2(mk)mk</p><p>)(2m k)! . (5)</p><p>Since is not monotone, by Lemma 9 we have that m 1 /O . Thus, by using that(2aa</p><p>) 4a</p><p>a,</p><p>e have</p><p> nm2k=1</p><p>(2(mk)mk</p><p>)(2m k)!</p><p> nm2k=1</p><p>4mk(m k) (2m k)!</p><p> nm2k=1</p><p>4mk2(2m k)!</p><p>=(1+ 4</p><p>m + 3 + O(m2</p><p>)) 16n2(m + 2)!</p><p> 17n2(m + 2)! , (6)</p><p>r any m large enough.Observe that the degree of a vertex in the dependency graph H is at most 2(m 1). Then,</p><p> = max0inm</p><p>j: i jE(H)</p><p>Pr(A j) = 2(m 1)Pr(A j) = 2(m 1)m! </p><p>2</p><p>(m 1)! .</p><p>nce e2 e4/(m1)! 2 if m 4 and by using that ea 1 a1+a , for any a 0, both (1) and (2)ply that</p><p> exp(1 34</p><p>2(m+2)(m+1)m!</p><p>)</p><p> 11m! 342(m+2)(m+1)m!</p><p>1+ 1m! 1</p><p>(1 O</p><p>(1</p><p>m!))(</p><p>1</p><p>m! 34</p><p>2(m + 2)(m + 1)m!)</p><p> 1 1m! +</p><p>14</p><p>m2 m!r any large enough m. This completes the proof. We next proceed to prove Corollary 4. This is achieved by obtaining a lower bound on (1,2,...,m)d by showing that this bound is larger than the upper bound given in Theorem 3. A recent result</p><p>El...</p></li></ul>

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