The Problem Clusters Shifts

Shift Equivalence in Consecutive PatternAvoidance

Brian K. Miceli

Trinity UniversityMathematics Department

AMS Fall SectionalUniversity of Wisconsin, Eau Claire

September 21, 2014

Miceli Shift Equivalence

The Problem Clusters Shifts

Outline

1 The Problem

2 The Cluster Method

3 Shift Equivalence

Miceli Shift Equivalence

The Problem Clusters Shifts

Outline

1 The Problem

2 The Cluster Method

3 Shift Equivalence

Miceli Shift Equivalence

The Problem Clusters Shifts

Outline

1 The Problem

2 The Cluster Method

3 Shift Equivalence

Miceli Shift Equivalence

The Problem Clusters Shifts

Background

Joint work with Jay Pantone.

Follows off the work in two articles:

“Rationality, irrationality, and Wilf equivalence in g.f.o.,” EJC(2009), Kitaev, Liese, Remmel, and Sagan.“On the rearrangement conjecture for generalized factor orderover P,” arXiv (2014), Pantone and Vatter.

Miceli Shift Equivalence

The Problem Clusters Shifts

Definitions for wordsThe basics

We let N = {1, 2, 3, . . .} and let N∗ denote the set of all wordsover N.

We say that u ∈ N∗ is a factor of v ∈ N∗ if there exist w1,w2 ∈ N∗such that v = w1uw2.

Miceli Shift Equivalence

The Problem Clusters Shifts

Definitions for WordsEmbeddings

Given two words u,w ∈ N∗ with |u| = k , we say that there is anembedding u into w if there exists a factor z = z1z2 · · · zk of wsuch that for every 1 ≤ i ≤ k , ui ≤ zi .

For example, let u = 132 and w = 27311231454.

Then there are three embeddings of u into w :

We would say that the last two embeddings are overlapping.

Miceli Shift Equivalence

The Problem Clusters Shifts

Definitions for WordsStatistics

Given w = w1w2 · · ·wk ∈ N∗, we define three statistics on w :

i. The length of w is |w | = k .

ii. The norm of w is ||w || = w1 + w2 + · · ·+ wk .

iii. The number of embeddings of u into w is eu,w .

We then define A(u; x , y , z) =∑w∈N∗

x |w |y ||w ||zeu,w .

Miceli Shift Equivalence

The Problem Clusters Shifts

Wilf-EquivalenceK.L.R.S. version

In the K.L.R.S. paper, they define that words u and v areWilf-equivalent, denoted by u ∼ v , if

A(u; x , y , 0) = A(v ; x , y , 0).

We say that u and v are strongly Wilf-equivalent, denoted byu ∼s v , if

A(u; x , y , z) = A(v ; x , y , z).

We can see that u ∼s v ⇒ u ∼ v .

Miceli Shift Equivalence

The Problem Clusters Shifts

Wilf-EquivalenceConjectures & questions from K.L.R.S.

Using theorems from K.L.R.S., we could show that

31425 ∼ 31524 ∼ 52413 ∼ 42513

and32415 ∼ 32514 ∼ 51423 ∼ 41523.

However, a computational result from that paper is that these 8words actually belong to the same Wilf-equivalence class.

Miceli Shift Equivalence

The Problem Clusters Shifts

Wilf-EquivalenceConjectures & questions from K.L.R.S.

There are also two relevant conjectures from that paper:

(Rearrangement Conjecture) If u ∼ v , then u and v arerearrangements of one another.

If u is a permutation of {1, 2, . . . , n}, then the size of theWilf-equivalence class of u is 2k for some k ∈ N.

Note that the the converse of the Rearrangement Conjecture is nottrue, as 123 ∼ 132 ∼ 321 ∼ 231 6∼ 213 ∼ 312.

In their paper, Pantone and Vatter show that the RearrangementConjecture does hold for strongly Wilf-equivalent words.

Miceli Shift Equivalence

The Problem Clusters Shifts

The Cluster MethodWhat does a cluster look like here?

In this situation, an m-cluster of u is a word w that contains mmarked, overlapping embeddings of u.

There are two facts to note here.

i. A word w can be both an i-cluster and an j-cluster of u, i 6= j ,although these are considered different clusters.

ii. Increasing any letter of an m-cluster gives a new word that isstill m-cluster.

Consequently, we define a minimal m-cluster to be one in whichreducing the size of any letter destroys one of our markedembeddings.

Miceli Shift Equivalence

The Problem Clusters Shifts

The Cluster MethodHow do we construct a minimal cluster?

Let u = 1522414. Then a minimal 3-cluster of u, where themarked embeddings occur with starting position 1, 2, and 4 wouldbe w as constructed below.

w =

Miceli Shift Equivalence

The Problem Clusters Shifts

The Cluster MethodA g.f. for strong Wilf-equivalence

The Cluster Method is a g.f. version of Inclusion/Exclusion.

In this instance, we get that

A(u; x , y , z) =1

1− xy1−y − C (u; x , y , z − 1)

,

where if MC(u) denotes the set of all minimal m-clusters of u,then

C (u; x , y , z) =∑m≥1

zm∑

w∈MC(u)

x |w |y ||w ||.

Miceli Shift Equivalence

The Problem Clusters Shifts

The Cluster MethodStrong Wilf-equivalence

Accordingly, u ∼s v if and only if C (u; x , y , z) = C (v ; x , y , z).

Pantone and Vatter then prove that if C (u; x , y , z) = C (v ; x , y , z),u and v must be rearrangements.

As we can see, it is enough to find a weight-preserving bijectionbetween all minimal m-clusters of u and v .

Miceli Shift Equivalence

The Problem Clusters Shifts

The Shift OperationMotivation

Going back to the earlier conjecture of K.L.R.S., when we see thatthe equivalence classes have size 2k , the combinatorial mind goesto, “There should be some sort of operation one can either do ornot do to a word to maintain Wilf-equivalence.”

Taking the reverse a word is such an operation.

We define another operation on words call a shift operation.

Miceli Shift Equivalence

The Problem Clusters Shifts

The Shift OperationBlock diagrams

To define a shift operation, we first replace a word with itscorresponding block diagram.

u  =  134224  

Miceli Shift Equivalence

The Problem Clusters Shifts

The Shift Operation“Definition” of acceptable shift

Miceli Shift Equivalence

The Problem Clusters Shifts

The Shift Operation“Definition” of shift-equivalence

We will say that two words, u and v , are shift-equivalent, denotedby u ≈ v , if the block diagram of one can be obtained by asequence of reverses and shifts of the other.

Which words are shift equivalent to u = 134224?

134224 ≈ 342241 ≈ 422431 ≈ 142243

134224 ∼ 342241 ∼ 422431 ∼ 142243

Miceli Shift Equivalence

The Problem Clusters Shifts

Shift vs. Wilf-EquivalenceWhat’s the relationship?

Theorem

If u ≈ v, then u ∼ v.

Proof: There is a weight-preserving bijection on the set of minimalclusters.

In fact, we can show something even stronger happens...

Miceli Shift Equivalence

The Problem Clusters Shifts

Shift vs. Wilf-EquivalenceWhat’s the relationship?

Let u = 134224 and v = 342241, and notice that v is a shift of u.

Consider the minimal 2-clusters of both with embeddings startingin positions 1 and 3.

1 3 4 2 2 41 3 4 2 2 4

1 3 4 3 4 4 2 4

3 4 2 2 4 13 4 2 2 4 1

3 4 3 4 4 2 4 1

Theorem

Suppose u and v are words such that v = Γ(u) for some shift Γ. Ifc is a minimal m-cluster of u with embeddings starting at positionss1, s2, . . . sm, then the minimal m-cluster of v with embeddingsstarting at positions s1, s2, . . . sm is Γ(c).

Miceli Shift Equivalence

The Problem Clusters Shifts

Shift vs. Wilf-EquivalenceA return to our questions and conjectures

From K.L.R.S. we know that we have these Wilf-equivalences:

31425 ∼ 31524 ∼ 52413 ∼ 42513

and32415 ∼ 32514 ∼ 51423 ∼ 41523.

However, 31425 ≈ 41523, giving the needed connection betweenthese two sets.

There are other theorems in this paper that follow immediatelyfrom the fact that u ≈ v ⇒ u ∼ v .

Miceli Shift Equivalence

The Problem Clusters Shifts

Shift vs. Wilf-EquivalenceA return to our questions and conjectures

Let’s make a conjecture of our own.

Conjecture

u ≈ v if and only if u ∼ v.

What would this allow us to do?

The two conjectures from K.L.R.S. follow immediately.

We could quickly and easily compute all Wilf-equivalenceclasses.

Miceli Shift Equivalence

The Problem Clusters Shifts

The End

Thanks for listening!Questions

Miceli Shift Equivalence