On a conjecture of H. Gupta

Abstract

Denote by r(n) the length of a shortest integer sequence on a circle containing all permutations of the set {1, 2, ..., n} as subsequences. Hansraj Gupta conjectured in 1981 that r(n) n 2 2 . In this paper we confirm the conjecture for the case where n is even, and show that r(n) < n 2 2 + n 4 − 1 if n is odd.

Containing All Permutations

“…Gupta conjectured that one could always construct a rosary of length at most n 2 /2. This conjecture was discussed by Guy [20,Problem E22] and proved in the case where n is even by Lecouturier and Zmiaikou [34]. Gupta also considered the variant where one is allowed to traverse the rosary both clockwise and counterclockwise (see the right of Figure 5); he conjectured that one can always construct a rosary of length at most 3n 2 /8 + 1/2 in this version of the problem.…”

An asymptotically tight lower bound for superpatterns with small alphabets

“…Hence, by our bounds on F (k, n) we get a contradiction for large k. In fact, essentially repeating the analysis from Section 5.1, we can show that if σ ∈ [k] n contains each τ ∈ S k as a BCP, then n ⩾ k 2 2 − k 7/4+o(1) . In 2012, Lecouturier and Zmiaikou proved that there exists σ ∈ [k] k 2 /2+O(k) which contain each τ ∈ S k as a circular pattern (and hence as a BCP), thus our bound is tight up to lower-order terms [LZ12].…”

An asymptotically tight lower bound for superpatterns with small alphabets

“…Hence, by our bounds on F (k, n) we get a contradiction for large k. In fact, essentially repeating the analysis from Section 5.1, we can show that if σ ∈ [k] n contains each τ ∈ S k as a BCP, then n ≥ k 2 2 − k 7/4+o (1) . In 2012, Lecouturier and Zmiaikou proved that there exists σ ∈ [k] k 2 /2+O(k) which contain each τ ∈ S k as a circular pattern (and hence as a BCP), thus our bound is tight up to lower-order terms [12].…”