Shortest sequence containing all permutations

Question

Given an integer $n$, define $s(n)$ to be the length of the shortest sequence $S = (a_1, \cdots a_{s(n)})$ such that every permutation of $\{1,\cdots,n\}$ is a subsequence of $S$.

If $n=1$, then $S = (1)$ is the shortest sequence containing all permutations of $\{1\}$, so s(1) = 1. If $n=2$, then $S = (1, 2, 1)$ contains all permutations of $\{1,2\}$ as a subsequence, so $s(2)=3$.

Is there a general formula for $s(n)$?

There are several different variations on this problem, depending on what alphabet you allow the sequence to use, and whether you insist the permutations be contained in contiguous sequences or simply in subsequences. Michael Engen and I collected these different versions of the problem in an article called Containing all permutations. — Vince Vatter, Mar 23 '21 at 21:36
These sequences are called superpermutations: see https://en.wikipedia.org/wiki/Superpermutation — a3nm, Jun 05 '23 at 06:22

score 11 · Accepted Answer · answered Jul 01 '11 at 03:09

11

Just so you know, after a quick Google search I found that your question is listed as an open problem on the Open Problem Garden.

A recent partial result gives a lower bound of $n^2-2n+3$ for $s(n)$.

answered Jul 01 '11 at 03:09

enfield

130

Thanks, I had no idea this was a known problem. It came up in an applied context. – Fixee Jul 01 '11 at 03:17
OEIS has the first four terms 1,3,9,33 and suggests it may be $\sum k!$ – Ross Millikan Jul 01 '11 at 03:21
@Ross: I don't think 9 is right for $s(3)$. I think $2312321$ (length 7) contains all length-3 permutations of ${1,2,3}$ as subsequences. – enfield Jul 01 '11 at 03:23
@enfield: As Yongyi Chen points out, 9 is for substrings, not subsequences. – Ross Millikan Jul 01 '11 at 03:31
@Ross and anybody interested: this question is about substrings. – A. De Luca Jul 01 '11 at 11:59
2

The OEIS sequence is for substrings, not subsequences. – Yongyi Chen Jul 01 '11 at 03:27
9 is correct for $s^\prime(3)$ where $s^\prime$ is the substring version of the problem. It's obvious any string of this type has to be at least of length 8, so that it has enough substrings. But if it's of length 8 then every substring has to be different, so you start constructing the string 12312ohcrap. 123121321, of length 9, works. – Michael Lugo Jul 02 '11 at 01:25
It seems that the problem was removed from the Open Problem Garden. Do you know why? Has it been solved? And can you please provide a reference for this lower bound? I need to cite it. – Mateus Araújo May 02 '14 at 17:09
As can be seen from http://oeis.org/A062714, it’s not a lower bound, but an upper bound, and one can do a bit better. – Emil Jeřábek May 06 '14 at 14:47
The first link in the answer is now broken – Andrew Savinykh Oct 26 '18 at 01:30

score 2 · Answer 2 · edited Apr 13 '17 at 12:58

An update: the page linked above doesn't exist anymore but the problem still seems to be open, see sequence A062714 on OEIS and references therein. The question was asked recently on mathoverflow and more information was given. I briefly report here the answer.

A better upper then the one shown above was proven by Radomirovic in 2012: $$s(n)≤⌈n^2−\frac{7}{3}n+\frac{19}{3}⌉.$$ The best lower bound seems to be the one found by Kleitman and Kwiatkowski in 1976: $$s(n)≥n^2−C_{\epsilon}n^{7/4+\epsilon},$$ where $C_{\epsilon}$ is a positive constant for any $ϵ>0$.

Shortest sequence containing all permutations

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