Shift Equivalence in Consecutive Pattern...
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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
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The Problem Clusters Shifts
Outline
1 The Problem
2 The Cluster Method
3 Shift Equivalence
Miceli Shift Equivalence
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The Problem Clusters Shifts
Outline
1 The Problem
2 The Cluster Method
3 Shift Equivalence
Miceli Shift Equivalence
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The Problem Clusters Shifts
Outline
1 The Problem
2 The Cluster Method
3 Shift Equivalence
Miceli Shift Equivalence
![Page 5: Shift Equivalence in Consecutive Pattern Avoidanceramanujan.math.trinity.edu/bmiceli/research/EauClaire.AMS.web.pdf · The ProblemClustersShifts Shift Equivalence in Consecutive Pattern](https://reader031.fdocuments.us/reader031/viewer/2022022716/5c15be2f09d3f28f1e8b77b5/html5/thumbnails/5.jpg)
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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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The Problem Clusters Shifts
The Shift Operation“Definition” of acceptable shift
Miceli Shift Equivalence
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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
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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
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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
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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
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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
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The Problem Clusters Shifts
The End
Thanks for listening!Questions
Miceli Shift Equivalence