-free posets, ascent sequences and pattern avoiding permutations

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  • Journal of Combinatorial Theory, Series A 117 (2010) 884909

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    00doJournal of Combinatorial Theory,Series A


    2+ 2)-free posets, ascent sequences and pattern avoidingermutations

    ireille Bousquet-Mlou a,1, Anders Claesson b,2, Mark Dukes c,ergey Kitaev b,2

    NRS, LaBRI, Universit Bordeaux 1, 351 cours de la Libration, 33405 Talence, Francehe Mathematics Institute, Reykjavik University, 103 Reykjavik, Icelandcience Institute, University of Iceland, 107 Reykjavik, Iceland

    r t i c l e i n f o a b s t r a c t

    ticle history:ceived 22 January 2009ailable online 4 January 2010

    ywords:+ 2)-free posetcent sequencettern avoiding permutationord diagramvolutioncodeumeraten-D-nite series

    We present bijections between four classes of combinatorialobjects. Two of them, the class of unlabeled (2 + 2)-free posetsand a certain class of involutions (or chord diagrams), alreadyappeared in the literature, but were apparently not known tobe equinumerous. We present a direct bijection between them.The third class is a family of permutations dened in terms ofa new type of pattern. An attractive property of these patternsis that, like classical patterns, they are closed under the actionof the symmetry group of the square. The fourth class is formedby certain integer sequences, called ascent sequences, which havea simple recursive structure and are shown to encode (2 + 2)-free posets and permutations. Our bijections preserve numerousstatistics.We determine the generating function of these classes of objects,thus recovering a non-D-nite series obtained by Zagier forthe class of chord diagrams. Finally, we characterize the ascentsequences that correspond to permutations avoiding the barredpattern 31524 and use this to enumerate those permutations,thereby settling a conjecture of Pudwell.

    2009 Elsevier Inc. All rights reserved.

    E-mail address: anders@ru.is (A. Claesson).MBM was supported by the French Agence Nationale de la Recherche, project SADA ANR-05-BLAN-0372.AC and SK were supported by grant no. 060005012 from the Icelandic Research Fund.

    97-3165/$ see front matter 2009 Elsevier Inc. All rights reserved.


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    teEnFig. 1. The bijections of the paper.


    This paper presents correspondences (see Fig. 1) between three main structures, seemingly unre-ted: unlabeled (2+ 2)-free posets on n elements, certain xed point free involutions (or chord dia-ams) on 2n elements introduced by Stoimenow in connection with Vassiliev invariants of knots [20],d a new class of permutations on n letters. An auxiliary class of objects, consisting of certain se-ences of nonnegative integers that we call ascent sequences, plays a central role in some of theserrespondences. Indeed, we show that both our permutations and (2+2)-free posets can be encodedascent sequences.A poset is said to be (2+ 2)-free if it does not contain an induced subposet that is isomorphic to

    + 2, the union of two disjoint 2-element chains. Fishburn [11] showed that a poset is (2+ 2)-freeecisely when it is isomorphic to an interval order. Amongst other results concerning (2 + 2)-freesets [9,10,17,8], the following characterization plays an important role in this paper: a poset is+ 2)-free if and only if the collection of strict principal down-sets can be linearly ordered byclusion [4]. Precise denitions will be given in Sections 3 and 7.The class of permutations we consider will be dened in Section 2, together with ascent sequences.sentially, it is a class of permutations that avoid a particular pattern of length three. This type ofttern is new in the sense that it does not admit an expression in terms of the vincular3 patternstroduced by Babson and Steingrmsson [3]. An attractive property of these new patterns is that, likeassical patterns, they are closed under the action of the symmetry group of the square. Vinculartterns do not enjoy this property. We show how to construct (and deconstruct) these permutationsement by element, and how this gives a bijection with ascent sequences.In Section 3 we perform a similar task for (2+ 2)-free posets. We present a recursive constructionthese posets, more sophisticated than that of permutations, which gives a bijection with ascentquences.In Section 4 we present a simple algorithm that, given an ascent sequence x, computes what

    e call the modied ascent sequence, denoted x. Some of the properties of the permutation ande poset corresponding to x are more easily read from x than from x. We also explain how todirectly between a given poset and the corresponding permutation as opposed to via the ascent

    quence. As an additional application of our machinery we show that the xed points under x xe in one-to-one correspondence with permutations avoiding the barred pattern 31524. We use thisaracterization to count these permutations, thus proving a conjecture of Pudwell [16].In Section 5 we prove that the bijections and respect numerous natural statistics.In Section 6 we determine the generating function of ascent sequences, and thus, of (2 + 2)-freesets and pattern avoiding permutations. Several authors have tried to count these posets before2,8,13], but did not obtain a closed expression for the generating function, which turns out to berather complicated, non-D-nite series. That our approach succeeds probably relies on the simpleructure of ascent sequences.

    Babson and Steingrmsson call these patterns generalized rather than vincular, but we wish to promote a change ofrminology here, since vincular is more descriptive. The adjective vincular is derived from the Latin noun vinculum (bond in


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    pabiThe generating function we obtain for (2 + 2)-free posets has, however, already appeared in theerature: it was shown by Zagier [23] to count certain involutions (or chord diagrams) introducedStoimenow to give upper bounds on the dimension of the space of Vassilievs knot invariants of

    given degree [20]. In Section 7 we present an alternative proof of Zagiers result by giving a directjection between (2+ 2)-free posets and Stoimenows involutions.Finally, in Section 8 we state some natural questions.Let us conclude with a few words on the genesis of this paper: we started with an investigationpermutations avoiding our new type of pattern. Patterns of length 2 being trivial, we moved to

    ngth 3, and discovered that the numbers counting one of our permutation classes formed the ratherysterious sequence A022493 of the on-line Encyclopedia of Integer Sequences [15]. From this arosee curiosity to clarify the connections between this class of permutations and (2+2)-free posets, butso between these posets and Stoimenows involutions, as this had apparently not been done before.e hope that the study of these new pattern-avoiding permutations will lead to other connectionsith interesting objects.

    Ascent sequences and pattern avoiding permutations

    Let (x1, . . . , xi) be an integer sequence. The number of ascents of this sequence is

    asc(x1, . . . , xi) ={1 j < i: x j < x j+1}.

    t us call a sequence x = (x1, . . . , xn) Nn an ascent sequence of length n if it satises x1 = 0 and xi ,1 + asc(x1, . . . , xi1)] for all 2 i n. For instance, (0,1,0,2,3,1,0,0,2) is an ascent sequence.e length (number of entries) of a sequence x is denoted |x|.Let Sn be the symmetric group on n elements. Let V = {v1, v2, . . . , vn} with v1 < v2 < < vnany nite subset of N. The standardisation of a permutation on V is the permutation std()[n] := {1,2, . . . ,n} obtained from by replacing the letter vi with the letter i. As an example,

    d(19452) = 15342. Let Rn be the following set of permutations:Rn =

    {1 . . . n Sn: if std(i jk) = 231 then j = i + 1 or i = k + 1


    uivalently, if ii+1 forms an ascent, then i 1 is not found to the right of this ascent. Thisass of permutations could be more descriptively written as Rn = Sn

    ( ), the set of permutations

    oiding the pattern in the diagram. Dark lines indicate adjacent entries (horizontally or vertically),hereas lighter lines indicate an elastic distance between the entries. Conversely, contains thisttern if there exists i < k such that k +1= i

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    Prto] and X and Y are subsets of [0,k]. An occurrence of p in a permutation = 1 . . .n on [n] isbsequence o = i1 . . .ik such that std(o) = and

    x X, ix+1 = ix + 1 and y Y , j y+1 = j y + 1,here {i1 , . . . ,ik } = { j1, . . . , jk} and j1 < j2 < < jk; by convention, i0 = j0 = 0 and ik+1 =+1 = n + 1. With this denition we have Rn = Sn((231, {1}, {1})). Note also that the number ofvincular permutations of length n is 4n+1n!.The classical patterns are those of the form p = ( ,,). Vincular patterns are of the form

    = ( , X,). Let p = ( , Xp, Yp) and q = ( , Xq, Yq) be any two patterns. If and have the samength, we dene their composition, or product, by p q = ( , XpYq, YpXq), where AB = B) (B A) is the symmetric difference. This operation is not associative, but it admits a rightentity, (id,,), and every element p = ( , X, Y ) has an inverse p1 = (1, Y , X); this turns thet of bivincular permutations of length n into a quasigroup with right identity. Also, reverse is de-ed by pr = ( r,n+1 X, Y ) and complement is dened by pc = ( c, X,n+1 Y ), in which k Anotes the set {k a: a A}. Thus the set of bivincular patterns has the full symmetry of a square.One simple instance of bivincular pattern avoidance that has already appeared in the literature is

    e set of irreducible permutations [1], that is, permutations such that i+1 = i 1 for all i. Withr terminology, these are the permutations avoiding (21, {1}, {1}). Similarly, the strongly irreduciblermutations of [2] are the (21, {1}, {1})- and (12, {1}, {1})-avoiding permutations.Let us now return to the set R := n Rn of permutations avoiding (231, {1}, {1}). Let be armutation of Rn , with n > 0. Let be obtained by deleting the entry n from . Then Rn1.deed, if ii+1 j is an occurrence of the forbidden pattern in (but not in ), then this impliesat i+1 = n. But then ii+1 j+1 would form an occurrence of the forbidden pattern in .This property allows us to construct the permutations of Rn inductively, starting from the emptyrmutation and adding a new maximal value at each step. (This is the generating tree approach,stematized by West [21].) Given = 1 . . . n1 Rn1, the sites where n can be inserted in soto produce an element of Rn are called active. It is easily seen that the site before 1 and the siteter n1 are always active. The site between the entries i and i+1 is active if and only if i = 1i 1 is to the left of i . Label the active sites, from left to right, with labels 0, 1, 2 and so on.

    bserve that the site immediately to the left of the maximal entry of is always active.Our bijection between permutations of Rn and ascent sequences of length n is dened recur-

    vely on n as follows. For n = 1, we set (1) = (0). Now let n 2, and suppose that Rn istained by inserting n in the active site labeled i of a permutation Rn1. Then the sequencesociated with is () := (x1, . . . , xn1, i), where (x1, . . . , xn1) = ().

    ample 1. The permutation = 61832547 corresponds to the sequence x = (0,1,1,2,2,0,3,1),nce it is obtained by the following insertions (the subscripts indicate the labels of the active sites):

    011x2=1 01122x3=1 0113 22x4=1 0113 2243x5=1 0113 225 43x6=1 06 113 225 43x7=1 06 113 225 4374x8=1 6 1 8 3 2 5 4 7.

    eorem 1. The map is a bijection from Rn to the set of ascent sequences of length n.

    oof. Since the sequence () encodes the construction of , the map is injective. We want

    prove that the image of Rn is the set An of ascent sequences of length n. Let s() denote the

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    PrtwLeis(2mber of active sites of the permutation . Our recursive description of the map tells us that= (x1, . . . , xn) (Rn) if and only if

    x = (x1, . . . , xn1) (Rn1) and 0 xn s(1

    (x)) 1 (1)

    ecall that the leftmost active site is labeled 0, so that the rightmost one is s() 1).We will prove by induction on n that for all Rn , with associated sequence () = x =

    1, . . . , xn), one has

    s() = 2+ asc(x) and b() = xn, (2)here b() is the label of the site located just before the maximal entry of . Clearly, this willnvert the above description (1) of (Rn) into the denition of ascent sequences, thus concludinge proof.So let us focus on the properties (2). They obviously hold for n = 1. Now assume they hold forme n 1, with n 2, and let Rn be obtained by inserting n in the active site labeled i Rn1. Then () = x = (x1, . . . , xn1, i) where () = x = (x1, . . . , xn1). Every entry of aller than n is followed in by an active site if and only if it was followed in by an active

    te. The leftmost site also remains active. Consequently, the label of the active site preceding n inis i = xn , which proves the second property. Thus, in order to determine s(), the only questionwhether the site following n is active in . There are two cases to consider. Recall that, by theduction hypothesis, s( ) = 2+ asc(x) and b( ) = xn1.ase 1: If 0 i b( ) = xn1 then asc(x) = asc(x) and the entry n in is to the left of n 1. So thember of active sites remains unchanged: s() = s( ) = 2+ asc(x) = 2+ asc(x).

    ase 2: If i > b( ) = xn1 then asc(x) = 1+ asc(x) and the entry n in is to the right of n 1. Thete that follows n is thus active, and s() = 1 + s( ) = 3 + asc(x) = 2 + asc(x). This concludes theoof. Ascent sequences and unlabeled (2+ 2)-free posets

    Let Pn be the set of unlabeled (2 + 2)-free posets on n elements. In this section we shall give ajection between Pn and the set An of ascent sequences of length n. As in the previous section, thisjection encodes a recursive way of constructing (2 + 2)-free posets by adding one new (maximal)ement. There is of course a corresponding removal operation, but it is less elementary than in these of permutations. Before giving these operations we need to dene some terminology.Let D(x) be the set of predecessors of x (the strict down-set of x). Formally,

    D(x) = {y: y < x}.is well knownsee for example Bogart [4]that a poset is (2 + 2)-free if and only if its sets ofedecessors, {D(x): x P }, can be linearly ordered by inclusion. For completeness we prove thissult here.

    mma 2. A poset P is (2 + 2)-free if and only if the set of strict downsets of P can be linearly ordered byclusion.

    oof. If the set...