We characterise and enumerate permutations that are sortable by n-4 passes
through a stack. We conjecture the number of permutations sortable by n-5
passes, and also the form of a formula for the general case n-k, which involves
a polynomial expression.Comment: 6 page

Claesson, Anders

Dukes, Mark

Steingrimsson, Einar

English

arXiv

The subject of pattern avoiding permutations has its roots in computer science, namely in the problem of sorting a permutation through a stack. A formula for the number of permutations of length n that can be sorted by passing it twice through a stack (where the letters on the stack have to be in increasing order) was conjectured by West, and later proved by Zeilberger. Goulden and West found a bijection from such permutations to nonseparable planar maps, and later, Jacquard and Schaeffer presented a bijection from these planar maps to certain labeled plane trees, called beta(1,0)-trees. Using generating trees, Dulucq, Gire and West showed that nonseparable planar maps are equinumerous with permutations avoiding the (classical) pattern 2413 and the barred pattern 41\bar{3}52; they called these permutations nonseparable. We give a new bijection between beta(1,0)-trees and permutations avoiding the dashed patterns 3-1-4-2 and 2-41-3. These permutations can be seen to be exactly the reverse of nonseparable permutations. Our bijection is built using decompositions of the permutations and the trees, and it translates seven statistics on the trees into statistics on the permutations. Among the statistics involved are ascents, left-to-right minima and right-to-left maxima for the permutations, and leaves and the rightmost and leftmost paths for the trees. In connection with this we give a nontrivial involution on the beta(1,0)-trees, which specializes to an involution on unlabeled rooted plane trees, where it yields interesting results. Lastly, we conjecture the existence of a bijection between nonseparable permutations and two-stack sortable permutations preserving at least four permutation statistics

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Permutations sortable by n-4 passes through a stack

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Strathprints Institutional RepositorySteingrimsson, Einar (2010) Permutations sortable by n-4 passes through a stack. Annals ofCombinatorics, 14. pp. 45-51.Strathprints is designed to allow users to access the research output of the University of Strathclyde.Copyright c© and Moral Rights for the papers on this site are retained by the individual authorsand/or other copyright owners. You may not engage in further distribution of the material for anyprofitmaking activities or any commercial gain. You may freely distribute both the url (http://strathprints.strath.ac.uk/) and the content of this paper for research or study, educational, ornot-for-profit purposes without prior permission or charge.Any correspondence concerning this service should be sent to Strathprints administrator:mailto:strathprints@strath.ac.ukhttp://strathprints.strath.ac.uk/arXiv:0812.0143v2  [math.CO]  3 Feb 2009PERMUTATIONS SORTABLE BYn− 4 PASSES THROUGH A STACKANDERS CLAESSON, MARK DUKES AND EINAR STEINGRI´MSSONAbstract. We characterise and enumerate permutations that are sortable byn − 4 passes through a stack. We conjecture the number of permutationssortable by n − 5 passes, and also the form of a formula for the general casen− k, which involves a polynomial expression.1. BackgroundWe view permutations as words without repeated letters; if pi is a permutationof an n element set and pi(i) = pii, then we write pi = pi1 . . . pin. The stack sortingoperator S can be defined recursively on permutations of finite subsets of {1, 2, . . .}as follows. If pi is empty then S(pi) = pi. If pi is nonempty write pi as the concatena-tion pi = LnR, where n is the greatest element of pi and L and R are the subwordsto the left and right of n respectively. ThenS(pi) = S(L)S(R)n.For example, S(42513) = 24135. We say that a permutation pi is k-stack sortable ifSk(pi) = id, where Sk = S ◦ Sk−1, S0 is the identity operator and id is the identitypermutation 12 . . . n. Let the (stack sorting) complexity of pi, denoted sscpi, be thesmallest k such that pi is k-stack sortable. Let Sn be the set of permutations of{1, . . . , n}. For the permutations in S3 we havepi = 123 132 213 231 312 321sscpi = 0 1 1 2 1 1Let Wn,k be the set of all k-stack sortable permutations in Sn; in other words,Wn,k is the set of permutations in Sn whose complexity is at most k. Let En,k bethe set of permutations in Sn whose complexity is exactly k. Note thatEn,k =Wn,k −Wn,k−1;Wn,k = Sn − (En,k+1 ∪ · · · ∪ En,n−1).It is easy to see that Wn,n−1 = Sn. Knuth [1, 2.2.1.5] leaves as an exercise tothe reader to show that Wn,1 = Sn(231). In his PhD thesis West [3] showed thatWn,2 = Sn(2341, 35241),where Sn(2341, 35241) is the set of permutations in Sn that avoid the pattern2341 and the “barred” pattern 35241; for information on these, see [2], especiallySection 7. West also showed that Wn,n−2 are precisely those permutations thatdo not have suffix n1. This last statement is easily shown by proving that thepermutations in En,n−1 are those with suffix n1.In addition, West characterized En,n−2. To state that result it is convenient tointroduce some notation for special sets of words over the alphabet {1, 2, . . .}. Letan asterisk (∗) stand for any word of zero or more characters, and let a questionThe first and third authors were supported by grant no. 060005013 from the Icelandic ResearchFund. The second author acknowledges funding by the EC’s Research Training Network ‘AlgebraicCombinatorics in Europe’, grant HPRN-CT-2001-00272 while at Universite´ Bordeaux 1, France.2 ANDERS CLAESSON, MARK DUKES AND EINAR STEINGRI´MSSONmark (?) stand for any single letter. These conventions are adopted from the socalled glob patterns in computer science. For a word w over {∗, ?} ∪ {1, 2, . . .}, let〈w 〉 denote the set of words of the form w. For instance, 〈 ∗n1 〉 consists of all wordswith suffix n1, and S5 ∩ 〈 ∗51? 〉 = {23514, 24513, 32514, 34512, 42513, 43512}. Let〈w1, . . . , wk 〉 = 〈w1 〉 ∪ · · · ∪ 〈wk 〉.West’s characterisation of En,n−1 and En,n−2 can then be stated as in the followingtwo lemmas, which follow from Theorems 4.2.4 and 4.2.17 in [3] and their proofs.Lemma 1.1. For all n ≥ 2,En,n−1 = Sn ∩ 〈 ∗n1 〉.Thus, the cardinality of En,n−1 is (n− 2)!.Lemma 1.2. For all n ≥ 4,En,n−2 = Sn ∩ 〈 ∗n2, ∗(n− 1)1n, ∗n1?, ∗n?1, ∗n∗(n− 2)∗(n− 1)1 〉.Thus, the cardinality of En,n−2 is (n− 3)!(7n− 12)/2.By subtracting the cardinalities in Lemmas 1.1 and 1.2 from n!, we get thefollowing result.Proposition 1.3. For all n ≥ 4 the cardinality of Wn,n−3 is(n− 3)!2(2n3 − 6n2 − 5n+ 16).2. Permutations requiring exactly (n− 3)-stack sortsTheorem 2.1. For all n ≥ 6, the set of permutations En,n−3 are those given incolumn labeled “Type” of Table 1. The number of such permutations is(n− 4)!3(47(n− 62)+ 194(n− 61)+ 297). (2.1)To prove this we require some terminology and results from West [3, §4.2]. Aforbidden pattern of order k in a permutation pi is a triple (B, c, a), where B is asubsequence of length k in pi and (c, a) is a pair of entries in pi such that for everyb ∈ B the subsequence bca is an occurrence of the pattern 231, that is, a < b < c.In such a situation we say that the pair (c, a) witnesses the forbidden pattern B.We call a forbidden pattern (B, c, a) uninterrupted if there is no subsequence bxb′in pi where b, b′ ∈ B and x > c.Lemma 2.2. [3, Theorems 4.2.10 and 4.2.14] Let pi be a permutation.(i) ssc(pi) ≤ k if pi does not contain any forbidden pattern of order k;(ii) ssc(pi) > k if pi contains an uninterrupted forbidden pattern of order k.Proof of Theorem 2.1. Let pi ∈ Sn. Using the contrapositive of (i) and (ii) inLemma 2.2, if ssc(pi) = n − 3, then pi contains a forbidden pattern of order n − 4and does not contain an uninterrupted forbidden pattern of order n − 3. Theforbidden pattern of order n− 4 must be witnessed by entries that appear in sometwo of the positions n − 3, n − 2, n − 1 or n of the permutation. We will firstcondition on the position of the largest entry in the permutations, then conditionon those permutations that contain a forbidden pattern of the required order, andfinally single out those permutations that are in En,n−3.Suppose pin = n. A permutation pi = pi′n is a member of En,n−3 precisely whenpi′ is a member of En−1,n−3. Thus entries 1(a)–1(e) of Table 1 immediately followfrom Lemma 1.2.PERMUTATIONS SORTABLE BY n − 4 PASSES THROUGH A STACK 3Case Type Numberpin = n1(a) ∗(n− 1)2n (n− 3)!1(b) ∗(n− 1)?1n (n− 3)!1(c) ∗(n− 1)1?n (n− 3)!1(d) ∗(n− 1)∗(n− 3)∗(n− 2)1n (n− 3)!/21(e) ∗(n− 2)1(n− 1)n (n− 4)!pin−1 = n2(a) ∗n3 (n− 2)!2(b) ∗(n− 1)1n? (n− 5)(n− 4)!2(c) ∗(n− 2)1n(n− 1) (n− 4)!pin−2 = n3(a) ∗n2? (n− 3)(n− 3)!3(b) ∗n?2 (n− 3)(n− 3)!pin−3 = n4(a) ∗n??1, but not ∗n(n− 2)(n− 1)1 (n− 2)!− (n− 4)!4(b) ∗n?1? (n− 2)!4(c) ∗n1?? (n− 2)!4(d) ∗n(n− 2)(n− 1)2 (n− 4)!pin−i = n and i > 35(a) ∗nA(n− 2)B(n− 1)2, where A ∪B 6= ∅ (n− 2)!/2− (n− 4)!5(b) ∗n∗(n− 3)∗(n− 1)(n− 2)1 (n− 3)!/25(c) ∗n∗(n− 1)∗(n− 3)∗(n− 2)1 (n− 2)!/65(d) ∗(n− 1)∗n∗{(n− 3)∗(n− 4)(n− 4)∗(n− 3)}∗(n− 2)1 (n− 2)!/125(e) ∗n∗(n− 2)∗(n− 1){1??1}(n− 4)(n− 3)!5(f) ∗n∗(n− 3)∗(n− 1)1(n− 2) (n− 3)!/25(g) ∗n∗(n− 3)∗(n− 2)1(n− 1) (n− 3)!/25(h) ∗(n− 2)∗n∗{(n− 3)∗(n− 4)(n− 4)∗(n− 3)}∗(n− 1)1 (n− 2)!/12Table 1. Permutations in En,n−34 ANDERS CLAESSON, MARK DUKES AND EINAR STEINGRI´MSSONSuppose pin−1 = n. We must have pin ≥ 3, for otherwise, by Lemmas 1.1 and 1.2,we would have pi ∈ En,n−1 or pi ∈ En,n−2. If pin = 3 then the permutation is inEn,n−3, and hence we get 2(a). If k = pin > 3 then it is impossible for (n, k)to witness a forbidden pattern F of order n − 4 since there could be at mostn− k − 1 < n− 4 elements in F . Thus the forbidden pattern F must be witnessedby (pin−3, pin−2). As there are now n − 2 values from which to form the forbiddenpattern F , we are forced to choose the extreme values from this set as the valuesfor (pin−3, pin−2). Consequently, if k = n − 1 then we must have (pin−3, pin−2) =(n− 2, 1), giving case 2(c). Otherwise 3 < k < n− 1 and (pin−3, pin−2) = (n− 1, 1)which gives case 2(b). One easily verifies that all such permutations are in En,n−3.Suppose pin−2 = n. By Lemma 1.2 the value 1 cannot be to the right of n. Let kbe the smaller of the two values pin−1 and pin. If k > 2 then the pair (n, k) witnessesa forbidden pattern of order at most n− k − 1− 1 < n− 4. For the same reason,(pin−1, pin) cannot witness a forbidden pattern of order n− 4. So exactly one of theentries to the right of n must be 2, giving 3(a) and (b). One easily verifies that allsuch permutations are in En,n−3.Suppose pin−3 = n. If 1 is to the right of n in pi then (n, 1) witnesses a forbiddenpattern of order n−4. However, if pi ∈ 〈 ∗n(n−2)(n−1)1 〉, it has one of the formsgiven in Lemma 1.2. In all other cases the permutation is in En,n−3. This gives4(a), (b) and (c).Alternatively, if 1 is to the left of n, let k be the value of the smallest entry tothe right of n. Since there are at most n− 5 entries to the left of n in pi that takevalues between k and n, the pair (n, k) cannot witness a forbidden pattern of ordern − 4. In order for pi to contain a forbidden pattern F of order n − 4 it must bewitnessed by the pair (pin−1, pin) and have as the block of n − 4 values the valuepin−2 along with the n − 5 elements to the left of n that are not 1. Thus pin = 2,pin−1 = n− 1 andpi ∈ 〈 ∗nk(n− 1)2 〉for some k 6= 1. One easily checks that the only value of k for which pi ∈ En,n−3 isk = n− 2. This gives 4(d).Suppose pin−i = n for some i > 3. Then (n, k) can never witness a forbiddenpattern of order n− 4. In the remainder of the proof, we condition on the relativepositions of (n−2), (n−1) and n. If pi = AnB(n−1)C(n−2)D where |B∪C∪D| ≥ 2,thenS(pi) = S(A)S(B)S(C)S(D)(n − 2)(n− 1)n.This givespi ∈ En,n−3 ⇐⇒ S(A)S(B)S(C)S(D) ∈ En−3,n−4⇐⇒ S(A)S(B)S(C)S(D) ∈ 〈 ∗(n− 3)1 〉.Since |B ∪ C ∪D| ≥ 2, the ways in which this can happen are restricted to(i) D = 1 and n− 3 ∈ C, giving case 5(c),(ii) D = 1, C = ∅ and n− 3 ∈ B, giving case 5(b),(iii) D = ∅, C = 1 and n− 3 ∈ B, giving case 5(f).If pi = AnB(n− 2)C(n− 1)D where |B ∪C ∪D| ≥ 2, thenS(pi) = S(A)S(B)S(C)(n − 2)S(D)(n− 1)n.This implies that pi belongs to En,n−3 if and only ifS(A)S(B)S(C)(n − 2)S(D) ∈ En−2,n−4.If D = ∅ then pi belongs to En,n−3 if and only if S(A)S(B)S(C) belongs toEn−3,n−4, which happens if and only if C = 1 and (n − 3) ∈ B, from whichPERMUTATIONS SORTABLE BY n − 4 PASSES THROUGH A STACK 5we get 5(g). Otherwise S(A)S(B)S(C)(n − 2)S(D) belongs to the types listed inLemma 1.2. There are 3 cases to consider:(i) S(D) = 2, so pi = AnB(n− 2)C(n− 1)2 where B ∪C 6= ∅, giving case 5(a);(ii) S(D) = 1k, so D = 1k or D = k1, giving case 5(e); and(iii) it is not possible that S(A)S(B)S(C)(n− 2)S(D) matches the last type inLemma 1.2 since this would mean that S(D) ends in 1, its smallest entry.If pi = A(n − 1)BnC(n − 2)D where |C ∪ D| ≥ 3, then a forbidden patternof order n − 4 can only be witnessed by the pair (n − 2, 1). Furthermore, allelements 2 through n − 3 must be to the left of n − 2, so pin−1 = n − 2 andpin = 1, that is pi = A(n − 1)BnC(n − 2)1 where |C| ≥ 2. From this we haveS2(pi) = S(S(A)S(B))S(S(C)1)(n − 2)(n− 1)n andpi ∈ En,n−3 ⇐⇒ pi′ = S(S(A)S(B))S(S(C)1) ∈ En−3,n−5.The conditions on pi′ (and therefore pi) are easily derived by comparing pi′ to thetypes in Lemma 1.2. Since |C| ≥ 2 one cannot have (n−3)2 as a suffix of S(S(C)1).Similarly, (n− 3)1k and (n− 3)k1 cannot be suffixes of S(S(C)1). Also, it is onlypossible thatpi′ ∈ 〈 ∗(n− 3)∗(n− 5)∗(n− 4)1 〉if C = ∅, which is not allowed. The only possibility for pi′ is that it has (n−4)1(n−3)as a suffix, and so n−4, n−3 ∈ C since |C| ≥ 2. Under these conditions, S2(pi) hassuffix (n−4)1(n−3)(n−2)(n−1)n and pi ∈ En,n−3. ThusD = 1 and n−4, n−3 ∈ C,from which we get case 5(d).If pi = A(n− 2)BnC(n− 1)D where |C ∪D| ≥ 3 thenS(pi) = S(A)S(B)(n− 2)S(C)S(D)(n− 1)n.Now, pi belongs to En,n−3 if and only if pi′ = S(A)S(B)(n − 2)S(C)S(D) belongsto En−2,n−4. Since |C ∪D| is at least 3, the entry n− 2 has at least 3 entries to itsright in pi′. By comparing this to the possible types that it may take in Lemma 1.2,we find that the only possibility ispi′ ∈ 〈 ∗(n− 2)∗(n− 4)∗(n− 3)1 〉.This happens if and only if D = 1 and n− 4, n− 3 ∈ C, which gives case 5(h).For the two final cases in which n − 2 and n − 1 are to the left of n, it is notpossible for pi to contain a forbidden pattern of order n − 4, since at least one ofn− 2, n− 1 and n is needed in order to witness such a pattern, and none of themare in the rightmost four positions of the permutation.The number of permutations of each type is shown in column three of Table 1.Adding these gives (47n2 − 223n+ 240)(n− 4)!/6 which may be be rewritten as informula (2.1). Corollary 2.3. For n ≥ 6, the collection of (n− 4)-stack sortable permutations inSn are those permutations that are not of the types listed in Lemma 1.1, Lemma 1.2or Table 1. The number of these is (n− 4)!(3n4 − 18n3 − 4n2 + 158n− 192)/3.It is now straightforward to write down (a lengthy expression for) the descentpolynomial of the (n−4)-stack sortable permutations, that is, the polynomial whosek-th coefficient is the number of (n − 4)-stack sortable permutations with exactlyk descents (a descent in pi = a1a2 . . . an is an i such that ai > ai+1).The following conjecture is based on computer generated data for n ≤ 13.Conjecture 2.4. For all n ≥ 8, the number of permutations in En,n−4 is(n− 5)!10(854(n− 83)+ 5099(n− 82)+ 12545(n− 81)+ 16130),6 ANDERS CLAESSON, MARK DUKES AND EINAR STEINGRI´MSSONor equivalently, the number of (n− 5)-stack sortable permutations is(n− 5)!60(60n5 − 600n4 + 506n3 + 11241n2 − 38369n+ 34236).We end with a conjecture about the form of an expression for the number ofpermutations needing exactly n − k stack sorts. This has been verified for alln ≤ 14 and all relevant k.Conjecture 2.5. For all n ≥ 2k, the number of permutations in En,n−k may bewritten as(k − 1)!(n− k − 1)!(2(k − 1))!k−1∑i=0ai(n− 2ki)where ai ∈ N.References[1] D. E. Knuth, The art of computer programming. Vol. 1: Fundamental algorithms, AddisonWesley Publishing Co., Reading, Mass.-London-Don Mills, Ont, 1969.[2] E. Steingr´ımsson: Generalized permutation patterns — a short survey, “Permutation Pat-terns, St Andrews 2007,” S.A. Linton, N. Ruskuc, V. Vatter (eds.), LMS Lecture Note Series,Cambridge University Press, to appear.[3] J. West, Permutations with forbidden subsequences; and, Stack sortable permutations, Ph.D.thesis, Massachusetts Institute of Technology, 1990.The Mathematics Institute, Reyjav´ık University, 103 Reykjav´ık, Iceland.Science Institute, University of Iceland, 107 Reykjav´ık, Iceland.

Generalized permutation patterns — a short survey, “Permutation Patterns, St Andrews

ımsson: Generalized permutation patterns — a short survey, “Permutation Patterns, St Andrews

Permutations with forbidden subsequences; and, Stack sortable permutations,

The art of computer programming.

Permutations sortable by n-4 passes through a stack

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