On bijections for pattern-avoiding permutations

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  • Journal of Combinatorial Theory, Series A 116 (2009) 12711284

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    Joa 6b C







    00doJournal of Combinatorial Theory,Series A


    n bijections for pattern-avoiding permutations

    nathan Blooma, Dan Saracino b

    5 Trevor Court, Rochester, NY 14610, United Statesolgate University, Department of Mathematics, 13 Oak Drive, Hamilton, NY, United States

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

    ticle history:ceived 6 February 2008ailable online 17 April 2009

    ywords:ectionttern-avoiding permutationed pointcedance

    By considering bijections from the set of Dyck paths of length 2nonto each of Sn(321) and Sn(132), Elizalde and Pak in [S. Elizalde,I. Pak, Bijections for rened restricted permutations, J. Combin.Theory Ser. A 105 (2004) 207219] gave a bijection : Sn(321) Sn(132) that preserves the number of xed points and the numberof excedances in each Sn(321). We show that a directbijection : Sn(321) Sn(132) introduced by Robertson in [A.Robertson, Restricted permutations from Catalan to Fine and back,Sm. Lothar. Combin. 50 (2004) B50g] also preserves the numberof xed points and the number of excedances in each . Wealso show that a bijection : Sn(213) Sn(321) studied in [J.Backelin, J. West, G. Xin, Wilf-equivalence for singleton classes,Adv. in Appl. Math. 38 (2007) 133148] and [M. Bousquet-Melou,E. Steingrimsson, Decreasing subsequences in permutations andWilf equivalence for involutions, J. Algebraic Combin. 22 (2005)383409] preserves these same statistics, and we show that ananalogous bijection from Sn(132) onto Sn(213) does the same.

    2009 Elsevier Inc. All rights reserved.


    We write the permutation Sn in one-line notation as 12 . . . n . If m n and Sm thene say that is -avoiding if there do not exist indices 1 i1 < < im n such that i j < ik iffj < k for 1 j,km. We denote by Sn() the set of all Sn that are -avoiding. For example,452 S5(321), but 31452 / S5(132).It was shown by Knuth in [4] that |Sn()| is independent of S3. Three decades later a rene-

    ent of this result was given by Robertson, Saracino, and Zeilberger [7], by taking into account the

    E-mail address: dsaracino@mail.colgate.edu (D. Saracino).

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


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    Exarmber of xed points of . If Skn() denotes the set of Sn() with exactly k xed points, it wasown in [7] that |Skn(321)| = |Skn(132)| = |Skn(213)| and |Skn(231)| = |Skn(312)| for all 0 k n.The proofs in [7] were nonbijective, and Elizalde and Pak soon improved the result |Skn(321)| =

    kn(132)| by giving in [3] a bijection : Sn(321) Sn(132) that preserves not only the number ofed points but also the number of excedances (is such that i > i). The bijection is obtainedcombining two previously known bijections: a bijection of Knuth [4] between Sn(321) and the

    t Dn of Dyck paths of length 2n, and (in modied form) a bijection of Krattenthaler [5] between(132) and Dn. (Knuths bijection involves the RobinsonSchenstedKnuth correspondence betweenrmutations and Young tableaux.)On the other hand, Robertson, in [6], introduced a direct bijection : Sn(321) Sn(132) thatoceeds by starting with Sn(321) and systematically converting occurrences of 132-patterns into1-patterns. (Elizalde and Pak compute the example (23514687) = 67435281. It turns out that(23514687) = 86452371, so and are distinct bijections.) Robertson conjectured at the 2005tegers conference that preserves the number of xed points. The main purpose of this paper isprove that preserves the number of xed points and the number of excedances. As an ingredientthe proof, we also show that (1) = ( ( ))1 for all Sn(321). We also give variations ofe algorithm for computing ( ).The rst version of this paper dealt exclusively with , and we thank one of the referees for

    aking us aware that [2] contains a proof (via a result analogous to our Lemma 2) that a bijection : Sn(213) Sn(321) similar to satises (1) = (( ))1. At the end of Section 1 we willscuss the connections between the results of that section and the results of [2], and in Section 2e will prove that preserves the number of xed points and the number of excedances in each Sn(213).The outline of our paper is as follows. In Section 1 we prove our results about inverses and aboutriations in the algorithm for computing , all modulo Lemma 2, whose proof is deferred to Sec-on 3. In Section 2 we prove that preserves the numbers of xed points and excedances, the mainea being an inductive argument that we sketch in an example before beginning the proof. We thene a variant of this argument to prove that the bijection preserves the numbers of xed pointsd excedances, and we use a different argument to show that an analogous bijection from Sn(132)to Sn(213) does the same.

    ( ) and (1)

    Although the denition of ( ) in [6] is stated for Sn(321), the denition applies to all Sn ,follows.We use to denote the lexicographic ordering of triples of positive integers. If i jk and

    xyz are 132-patterns in we say that i jk is p-smaller than xyz if (i, j,k) (x, y, z) (p forosition).For any permutation , we dene a permutation g as follows. If Sn(132) then g = . If/ Sn(132) we take the p-smallest 132-pattern i jk in and let g be the permutation obtainedom by replacing i, j, k by, respectively, j, k, i . Thus g converts the p-smallest 132-pattern , if there is any, to a 321-pattern.

    ct. (See [6].) If g / Sn(132) and i jk and (g)x(g)y(g)z are the p-smallest 132-patterns inand g , then (i, j,k) (x, y, z).

    It follows that for every Sn there is a smallest nonnegative integer r such that gr Sn(132).e call this smallest r the -rank of , and let ( ) = gr . Note that ( ) = gr+m for all non-gative integers m, and we have ( ) = (g).

    ample. Let = 14237586 S8(321). Then 1 = 13426857. The calculations of ( ) and (1)


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    i( ( ): (1):14237586 1342685742137586 3241685772135486 6241583775134286 8241573678134265 8431572678324165 8731562478624153 8751462378643152 8754261378643521 87546312

    Note that here and 1 have the same -rank. We will prove that this is the case for all Sn(321) (and for some other s).In this example we see that indeed (1) is ( ( ))1. But more is true: we can arrive at

    (1) by carrying out any number of steps we like in the calculation of ( ), then taking anverse, and then continuing with the calculation of . For instance, if we take the fourth line in thelculation of ( ), namely 75134286, its inverse is 36452817, and applying g ve more times to thisverse yields 87546312.We will explain this exibility in the calculation of (1) near the end of this section.

    Our main goal in this section is to prove by induction on the -rank of that (1) = ( ( ))1r all Sn(321). We want to take advantage of the fact that, unless g = , g has -rank oness than that of ; but we are blocked by the fact that g may not be in Sn(321). Accordingly, weed to work with a set of permutations larger than Sn(321) that is closed under the application of g ,d it will turn out that we also need this larger set to be closed under inverses.We use the set Sn(1432) of all permutations that avoid the pattern 1432. It is easy to see that this

    t includes Sn(321) and is closed under inverses.

    mma 1. If Sn(1432) then g Sn(1432).

    oof. If Sn(132) then g = so the result is clear.Now suppose contains 132-patterns, and let i jk be the p-smallest one. Suppose that)w(g)x(g)y(g)z is a 1432-pattern in g .If w < i then (g)w = w and w jk contradicts our choice of i jk in unless k < w . But

    en at most one of i, j, k can be greater than w , so at least two of x, y, z are not in {i, j,k}.is, with w , gives us a 132-pattern in that is p-smaller than i jk , a contradiction.If w = i then all of (g)x, (g)y, (g)z are greater than (g)i = j , so none of x, y, z is in {i, j,k}.us since i < j , ixyz is a 1432-pattern in , a contradiction.So w > i. Note that if w = k or if w = j and k / {x, y, z}, then ixyz is a 1432-pattern in .w = j and k {x, y, z} then (g)w = k and i {(g)x, (g)y, (g)z} so i > k , a contradiction.w / { j,k}. Since Sn(1432) at least one of x, y or z must be in { j,k}.Since w < min{(g)x, (g)y, (g)z} and (g)t t for t = i, w < min{x, y, z}. Since (1432) we must have a < b either for a = x, b = y or for a = y, b = z. Note that we cannot have= j and b = k (since j > k), while if a { j,k} and b / { j,k} then (g)a < a and (g)b = b, so)a < (g)b, contradicting the fact that (g)w(g)x(g)y(g)z is a 1432-pattern. So a / { j,k} andus b { j,k} since a < b but (g)a > (g)b. If b = j then (g)a > (g)b says a > k , so iakntradicts our choice of i jk. If b = k then (g)a > (g)b says a > i , so i < a < k. If j (g)y and y = a / { j,k}, x > y = a.) In proving that (1) = ( ( ))1 for all Sn(1432) we will use a variant of the function g . If

    jk and xyz are 132-patterns in we say that i jk is v-smaller than xyz if (i, j, k)

    x, y, z) in the lexicographic ordering (v for value).

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    ikSoi =




    PrFor any permutation , we dene h just as we did g , except that we use the v-smallest 132-ttern in , if any.The following lemma is the foundation for all of our results about . It will be proved in Section 3.

    mma 2. Let Sn(1432). Then:

    ) Either g = h or gh = hg .) If g = h then -rank( ) 2.

    Lemma 2(a) may fail if / Sn(1432). For example, let = 4152763. We compute g = 7152643,= 4521763, gh = 7521643, and hg = 7452631. Thus g = h and gh = hg .Both alternatives in Lemma 2(a) are needed, even for Sn(321). For = 13254 we have g == 32154, but gh = 52143 = 32541 = hg . For = 2143 we have g = 4132 = 2431 = h , but = 4321 = hg .

    mma 3. Let Sn(1432). Then ( ) = (h), and if -rank( ) > 0 we have -rank(h) =-rank( ) 1.

    oof. We use induction on -rank( ). If -rank( ) = 0 then Sn(132) so h = and the resultobvious.Now assume the result for members of Sn(1432) whose -rank is less than that of . By Lemma 2

    e have either g = h or gh = hg . If g = h then ( ) = (g) = (h), and -rank(h) =-rank(g) = -rank( ) 1. If g = h but gh = hg then, using Lemma 1 to enable us toply the induction hypothesis to g , we have ( ) = (g) = (hg) = (gh) = (h). Fore statement on -ranks, note that since g = h we have -rank( ) 2 by Lemma 2(b). Thusthe induction hypothesis applied to g , -rank(hg) = -rank(g) 1, i.e., -rank(gh) =

    -rank(g) 1, so -rank(h) = -rank(g) = -rank( ) 1. To proceed further we need the following description of h .

    mma 4. If Sn(1432) then h = (g1)1.

    oof. If Sn(132) then h = and g1 = 1, so h = = (g1)1.Suppose / Sn(132). We must show that if i jk is the v-smallest 132-pattern in then ikj is

    e p-smallest 132-pattern in 1. Suppose it is not, and xzy is p-smaller in 1. We know i jkv-smaller than xyz in .If i < x then i occurs to the left of x in 1, a contradiction.Suppose i = x , and j < y . Then j occurs to the left of y in 1, so since xzy is p-smaller than

    j in 1, z occurs to the left of k in 1. Thus z < k . Since z > x = i we have i < z < k.if z > j then the 132-pattern i jz contradicts our choice of i jk in . We conclude thatx< y < z < j < k and that (recall y > j) iy jk is a 1432-pattern in , a contradiction.Finally, suppose i = x , j = y , and k < z. Then k occurs to the left of z in 1, contradicting

    e assumption that xzy is p-smaller than ikj in 1. We can now prove that (1) = ( ( ))1 for all Sn(1432). Note that this result may fail / Sn(1432). For example, if = 31542 then 1 = 25143 and (1) = 54123, while ( ) =312 and ( ( ))1 = 45321. In this example we have -rank( ) = 2 and -rank(1) = 1, so thesertion about -ranks in the following theorem may also fail if / Sn(1432).

    eorem 1. If Sn(1432) then (1) = ( ( ))1 and -rank( ) = -rank(1).

    oof. We proceed by induction on -rank( ). If -rank( ) = 0 then and 1 are in Sn(132) so1 1 1 1( ) = = ( ( )) and -rank( ) = 0.

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    thIninvabiAssume the result for all elements of Sn(1432) of -rank smaller than that of . Then ( ( ))1 =(g))1 = ((g)1) by the induction hypothesis, and this is (h1) by Lemma 4. By Lemma 3,(h1) = (1).For the assertion about -ranks, we have by the induction hypothesis -rank((g)1) =

    -rank(g), i.e., by Lemma 4, -rank(h1) = -rank(g). By Lemma 3 this says -rank(1) = -rank( ) 1, so -rank(1) = -rank( ). If Sn(1432) then by Lemmas 1 and 4 we obtain h1 = (g)1, h21 = (g2)1, . . . ,1 = (gm)1 for all m. Letting m = -rank( ) = -rank(1), we obtain hm1 = ( ( ))1 =(1). Since 1 accounts for all elements of Sn(1432) this says that for all Sn(1432), ( )n be calculated via m successive applications of h, where m = -rank( ). But more is true: we cantually apply any combination of m g s and hs to , in any order, and obtain ( ).

    eorem 2. Suppose Sn(1432) and -rank( ) = m. Then if f i {g,h} for i = 1,2, . . . ,m, we have( ) = f1 f2 . . . fm . If k

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    ththcoFixed points, excedances, and antiexcedances

    If Sn, an antiexcedance in is an i such that i < i. We denote by a( ), f ( ), and e( ) thembers of antiexcedances, xed points, and excedances in . If i is an antiexcedance (respectively,xed point or an excedance) then we will also call i a low value (respectively, a xed value or a highlue).Our rst goal in this section is to prove

    eorem 3. If Sn(321) then a( ) = a( ( )), f ( ) = f ( ( )) and e( ) = e( ( )).

    The example from Section 1 will be useful as an illustration in giving the proof of Theorem 3. Thelues of the triple (a, f , e) at each of the nine steps in the calculation of (14237586) are:


    Note rst that for the fth step of the computation, 78134265 S8(1432), the values a = 6, f = 0,= 2 are all different from the values a = 4, f = 1, e = 3 for (78134265). So the results of Theo-m 3 may fail for Sn(1432).Notice, though, that the value a = 4 stays xed through the third step of the computation, where

    e have the permutation = g2 = 72135486. This is the rst step in the computation in whichere is no 132-pattern starting with the rst entry. If we let be the rst entry in ( = 7 in theample) and let be the permutation obtained from by deleting and lowering all values greateran by 1 (so = 2135476 in the example), then a( ) = 4 = 3+ 1 =...