We establish an exact formula for the length of the shortest permutation containing all layered permutations of length n, proving a conjecture of Gray

Albert, Michael

Engen, Michael

Pantone, Jay

Vatter, Vincent

English

arXiv

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Marquette University e-Publications@Marquette Mathematical and Statistical Science Faculty Research and Publications Mathematical and Statistical Science, Department of 2018 Universal Layered Permutations Michael Albert Michael Engen Jay Pantone Vincent Vatter Follow this and additional works at: https://epublications.marquette.edu/math_fac Universal layered permutationsMichael Albert Michael EngenDepartment of Computer Science Department of MathematicsUniversity of Otago University of FloridaDunedin, New Zealand Gainesville, Florida USAmalbert@cs.otago.ac.nz engenmt@ufl.eduJay Pantone Vincent Vatter∗Department of Mathematics Department of MathematicsDartmouth College University of FloridaHanover, New Hampshire USA Gainesville, Florida USAjaypantone@dartmouth.edu vatter@ufl.eduSubmitted: Oct 12, 2017; Accepted: Jun 28, 2018; Published: Aug 10, 2018c©The authors. Released under the CC BY-ND license (International 4.0).AbstractWe establish an exact formula for the length of the shortest permutation con-taining all layered permutations of length n, proving a conjecture of Gray.Mathematics Subject Classifications: 05A05, 06A07We establish the following result, which gives an exact formula for the length of theshortest n-universal permutation for the class of layered permutations, verifying a conjec-ture of Gray [3]. Definitions follow the statement.Theorem 1. For all n, the length of the shortest permutation that is n-universal for thelayered permutations is given by the sequence defined by a(0) = 0 anda(n) = n+ min{a(k) + a(n− k − 1) : 0 6 k 6 n− 1} (†)for n > 1.∗Vatter’s research was partially supported by the National Security Agency under Grant NumberH98230-16-1-0324. The United States Government is authorized to reproduce and distribute reprintsnot-withstanding any copyright notation herein.the electronic journal of combinatorics 25(3) (2018), #P3.23 1Up to shifting indices by 1, the sequence a(n) in Theorem 1 is sequence A001855 in theOEIS [6]. It seems to have first appeared in Knuth’s The Art of Computer Programming,Volume 3 [4, Section 5.3.1, Eq. (3)], where it is related to sorting by binary insertion.Knuth shows there that (in our indexing conventions),a(n) = (n+ 1)dlog2(n+ 1)e − 2dlog2(n+1)e + 1.This formula also shows that the minimum in (†) is attained when k = bn/2c. We referthe reader to the OEIS for further information about this old and interesting sequence.The permutation pi of length n, thought of in one-line or list notation, is said to containthe permutation σ of length k if pi has a subsequence of length k in the same relativeorder as σ. Otherwise, pi avoids σ. For example, pi = 432679185 contains σ = 32514, aswitnessed by the subsequence 32918. We frequently associate the permutation pi with itsplot, which is the set {(i, pi(i))} of points in the plane.A set of permutations that is closed downward under the containment order is calleda permutation class. Thus, C is a class if for all pi ∈ C and all permutations σ containedin pi, σ also lies in C. If pi has length m and σ has length n, we define the sum of pi andσ, denoted pi ⊕ σ, as the permutation of length m+ n defined by(pi ⊕ σ)(i) ={pi(i) for 1 6 i 6 m,σ(i−m) +m for m+ 1 6 i 6 m+ n.The plot of pi ⊕ σ can therefore be represented as below.pi ⊕ σ =piσA permutation is layered if it can be expressed as a sum of decreasing permutations,which are themselves called layers. For example, the permutation 3214657 is layered, asit can be expressed as 321⊕ 1⊕ 21⊕ 1 (note that the sum operation is associative and sothere is no ambiguity in this expression).Given a permutation class C, the permutation pi is said to be n-universal for C ifpi contains all of the permutations of length n in C (the alternate term superpattern issometimes used in the literature). A permutation is said to be simply n-universal if itcontains all permutations of length n.To date, the best bounds on the length of the shortest n-universal permutation are thatit lies between n2/e2 (a consequence of Stirling’s Formula, because if such a permutationhas length m, the inequality(mn)> n! must hold) and(n+12), which was establishedby Miller [5]. Before Miller’s result was established, Eriksson, Eriksson, Linusson, andWa¨stlund [2] had conjectured that this length is asymptotic to n2/2.Theorem 1 is concerned with n-universal permutations for the class of layered permu-tations. Until this work, the best bounds on the length of the shortest such permutationswere obtained by Gray [3], who showed that their lengths lie between n lnn− n + 2 andnblog2 nc+n. Similar bounds were obtained independently by Bannister, Cheng, Devanny,and Eppstein [1]. By Theorem 1 we see that this length is asymptotic to n log2 n.the electronic journal of combinatorics 25(3) (2018), #P3.23 2One very useful property that layered permutations have, and which is used throughoutour proofs, is that when determining whether the layered permutation λ = λ1 ⊕ · · · ⊕ λ`(where each λi is a decreasing permutation) is contained in the layered permutation pi, itsuffices to take a greedy approach: embed the first layer λ1 as far to the left as possible,then embed the second layer λ2 as far left as possible, and so on.Our first result establishes that among the shortest n-universal permutations for theclass of layered permutations, there is one that is layered. This is a key component in ourproof of Theorem 1.Proposition 2. Given any permutation pi of length m, there is a layered permutation oflength m that contains every layered permutation contained in pi.Proof. We prove the claim by induction on m. Note that the base case is trivial. LetD denote a decreasing subsequence of pi of maximum possible length. Because D is amaximal decreasing subsequence, every entry of pi that is not in D must either lie to thesouthwest of an entry of D or to the northeast of such an entry, but not both. Let D−denote the set of entries that lie to the southwest of an entry of D and let D+ denote theset of entries that lie to the northeast of such an entry, so that D, D−, and D+ togetherconstitute a partition of the entries of pi. An example of this decomposition is shown onthe leftmost panel of Figure 1.Define pi− (resp., pi+) to be the permutation in the same relative order as the entriesof D− (resp., D+), let δ = |D| · · · 21, and set pi∗ = pi− ⊕ δ⊕ pi+. Thus in some sense pi∗ isa “straightened-out” version of pi; an example is shown in the central panel of Figure 1.•••••••••••D−D+••••••••••••••••••••••Figure 1: The steps in the proof of Proposition 2. From left to right, the drawings showan example of pi, of pi∗, and of the layered permutation τ− ⊕ δ ⊕ τ+.We claim that every layered permutation contained in pi is also contained in pi∗. Sup-pose λ = λ1⊕· · ·⊕λ` is a layered permutation contained in pi, where each λi is a decreasingpermutation, and fix an embedding of λ into pi. Choose j maximally so that in this em-bedding of λ into pi, the layers λ1⊕ · · · ⊕ λj−1 are embedded entirely using entries in D−.It follows that the entries of λj+1 ⊕ · · · ⊕ λ` are embedded entirely using entries of D+.Since λj certainly embeds into D and consequently into δ, we have that λ 6 pi∗.Finally, by induction we see that there are layered permutations µ− and µ+ whichcontain all of the layered permutations contained in pi− and pi+, respectively. It followsthat µ− ⊕ δ ⊕ µ+ is layered and contains all of the layered permutations contained inpi∗, which in turn contains all of the layered permutations contained in pi, proving theproposition. An example of this final construction is shown in the rightmost panel ofFigure 1.the electronic journal of combinatorics 25(3) (2018), #P3.23 3• We can now prove our main result.Proof of Theorem 1. We prove the theorem by induction on n. As the base case is trivial,suppose that the statement is true for all values less than n.Let a(n) denote the length of the shortest permutation which is n-universal for thelayered permutations. First we show thata(n) 6 n+ min{a(k) + a(n− k − 1) : 0 6 k 6 n− 1}.Choose k so that a(k) + a(n − k − 1) is minimized. Then choose a permutation σ oflength a(k) containing all layered permutations of length k and a permutation τ of lengtha(n− k − 1) containing all layered permutations of length n− k − 1. We claim that thepermutation pi = σ ⊕ (n · · · 21)⊕ τ is n-universal for the layered permutations. Considerany layered permutation λ = λ1 ⊕ · · · ⊕ λ`, where each λi is a decreasing permutation, oflength n = |λ1|+ · · ·+ |λ`|, and choose j so that|λ1| + · · · + |λj−1| 6 k,|λ1| + · · · + |λj−1| + |λj| > k.Note that this implies that |λj+1| + · · · + |λ`| 6 n − k − 1. Since σ is k-universal forthe layered permutations, it contains λ1 ⊕ · · · ⊕ λj−1. Clearly λj is contained in n · · · 21.Finally, as τ is (n−k−1)-universal for the layered permutations, it contains λj+1⊕· · ·⊕λ`.This verifies that pi contains λ, as desired.To establish the reverse inequality, suppose that the permutation pi is n-universal forthe class of layered permutations and is as short as possible subject to this constraint.Proposition 2 allows us to assume that pi is itself layered, so we may decompose pi aspi1 ⊕ · · · ⊕ pi` where each pii is a decreasing permutation.Because pi contains all layered permutations of length n, it contains n · · · 21. Moreover,as every embedding of n · · · 21 into pi may use entries from only one layer, there must besome index j such that |pij| > n. Now choose k so thata(k) 6 |pi1 ⊕ · · · ⊕ pij−1| < a(k + 1).Thus there is at least one layered permutation λ of length k + 1 that does not embedinto pi1 ⊕ · · · ⊕ pij−1. Therefore the earliest that λ may embed into pi is if it embeds intopi1 ⊕ · · · ⊕ pij. Now let µ denote an arbitrary layered permutation of length n − k − 1.Because λ⊕µ has length n, it must embed into pi, and by our observations about where λmay embed into pi we see that µ must embed into pij+1 ⊕ · · · ⊕ pi`. As µ was an arbitrarylayered permutation of length n− k− 1, pij+1 ⊕ · · · ⊕ pi` is therefore (n− k− 1)-universalfor the class of layered permutations, so we have|pi| = |pi1 ⊕ · · · ⊕ pij−1|+ |pij|+ |pij+1 ⊕ · · · ⊕ pi`|> a(k) + n+ a(n− k − 1)> a(n),completing the proof of the theorem.the electronic journal of combinatorics 25(3) (2018), #P3.23 4• Proposition 2 implies that among the shortest permutations which are n-universal forthe layered permutations, there is at least one that is itself layered. Computation showsthat this property is not shared by all permutation classes. For example, one of the5-universal permutations of minimum length for the class of 231-avoiding permutations is1 5 11 9 3 2 8 4 7 6 10,However, this permutation contains 231, and there is no 231-avoiding permutation oflength 11 which is 5-universal for the class of 231-avoiding permutations. The shortest 231-avoiding permutations which are 5-universal for the class of 231-avoiding permutationsinstead have length 12; one such permutation is1 11 3 2 10 7 5 4 6 9 8 12.On the other hand, computational evidence leads us to suspect that the class of 321-avoiding permutations does possess this property:Conjecture 3. For all n, among the shortest permutations that are n-universal for theclass of 321-avoiding permutations there is at least one which avoids 321 itself.References[1] M. J. Bannister, Z. Cheng, W. E. Devanny, and D. Eppstein. Superpatterns anduniversal point sets. J. Graph Algorithms Appl., 18(2):177–209, 2014.[2] H. Eriksson, K. Eriksson, S. Linusson, and J. Wa¨stlund. Dense packing of patterns ina permutation. Ann. Comb., 11(3-4):459–470, 2007.[3] D. Gray. Bounds on superpatterns containing all layered permutations. Graphs Com-bin., 31(4):941–952, 2015.[4] D. E. Knuth. The Art of Computer Programming, Volume 3. Addison-Wesley, Read-ing, Massachusetts, 1973.[5] A. Miller. Asymptotic bounds for permutations containing many different patterns.J. Combin. Theory Ser. A, 116(1):92–108, 2009.[6] The On-line Encyclopedia of Integer Sequences (OEIS). Published electronically athttp://oeis.org/.the electronic journal of combinatorics 25(3) (2018), #P3.23 5

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