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Partitioning permutations into monotone subsequences

David Wärn

Abstract

A permutation is $k$-coverable if it can be partitioned into $k$ monotone subsequences. Barber conjectured that, for any given permutation, if every subsequence of length $k+2 \choose 2$ is $k$-coverable then the permutation itself is $k$-coverable. This conjecture, if true, would be best possible. Our aim in this paper is to disprove this conjecture for all $k \ge 3$. In fact, we show that for any $k$ there are permutations such that every subsequence of length at most $(k/6)^{2.46}$ is $k$-coverable while the permutation itself is not.

Partitioning permutations into monotone subsequences

Abstract

A permutation is -coverable if it can be partitioned into monotone subsequences. Barber conjectured that, for any given permutation, if every subsequence of length is -coverable then the permutation itself is -coverable. This conjecture, if true, would be best possible. Our aim in this paper is to disprove this conjecture for all . In fact, we show that for any there are permutations such that every subsequence of length at most is -coverable while the permutation itself is not.

Paper Structure

This paper contains 7 sections, 15 theorems, 3 equations, 4 figures.

Key Result

Lemma 1

Let $\pi$ be a permutation, and suppose the length of the longest increasing subsequence of $\pi$ is $s$. Then $\pi$ can be partitioned into $s$ decreasing subsequences. Similarly, if the length of the longest decreasing subsequence of $\pi$ is $r$, then $\pi$ can be partitioned into $r$ increasing

Figures (4)

  • Figure 1: $\pi_{12}$
  • Figure 2: $\pi_9$
  • Figure 3: $\pi_{15}$
  • Figure 4: Dividing one triangle into four

Theorems & Definitions (29)

  • Lemma 1
  • proof
  • Lemma 2
  • proof
  • Lemma 3
  • proof
  • Lemma 4
  • proof
  • Lemma 5
  • proof
  • ...and 19 more