deflatability of permutation classes · 2015-05-12 · 1.show that every permutation of av(ˇ) is...
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Deflatability of permutation classes
Michael Albert
Department of Computer Science, University of Otago.
Joint work withMike Atkinson (UO), Cheyne Homberger (UF) and Jay Pantone (UF).
Permutation Patterns 2014, Johnson City
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The definitions
A permutation is a set of points in the plane (no two on ahorizontal or vertical line)
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The definitions
I lied , it’s the equivalence class of such a set of points underarbitrary vertical and horizontal rescaling.
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The definitions
A (permutation) class is a set of permutations closed undererasing points.
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The definitions
A permutation is simple if no box contains at least two pointsand is not cut horizontally or vertically by an exterior pointunless it contains all the points.
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The definitions
An interval in a permutation is a box (or the elements itcontains) that is not cut horizontally or vertically by any pointnot in it.
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The definitions
So a permutation is simple if its only intervals contain 0, 1 or allthe points (these are trivial intervals).
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The definitions
A permutation class is deflatable if it has a proper subclasscontaining the same simple permutations.
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The definitions
Equivalently, a class C is not deflatable if every permutation in Chas a simple extension in C.
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What’s the point of deflatability?
I If a class C is deflatable, then the class D formed byclosing its simples under subpermutation is strictly smaller.
I Understand D.I Understand the allowed inflations to get C.
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The question (and answer)
Question: Which principal permutation classes Av(π) aredeflatable?
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The question (and answer)
Question: Which principal permutation classes Av(π) aredeflatable?
Answer : It’s complicated.
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The question (and answer)
Question: Which principal permutation classes Av(π) aredeflatable?
Answer : It’s complicated, and we don’t know the whole story.
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Intervals and extensions
α
2 ≤ |α| < |ω|
ω
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Intervals and extensions
α
x
x cuts α
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Intervals and extensions
α
x
x is separated from α
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An observation
Let ω be a permutation containing an interval α and supposethat ω′ = ω ∪ {x} where x cuts and is separated from α. If xbelongs to a proper interval of ω′ then at least one of thefollowing must hold:
I ω is decomposable,I α is not a maximal interval of ω.
If neither holds:I ω′ is indecomposable,I any proper interval of ω′ is a proper interval of ω, andI α is not an interval of ω′.
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An observation
Let ω be a permutation containing an interval α and supposethat ω′ = ω ∪ {x} where x cuts and is separated from α. If xbelongs to a proper interval of ω′ then at least one of thefollowing must hold:
I ω is decomposable,I α is not a maximal interval of ω.
If neither holds:I ω′ is indecomposable,I any proper interval of ω′ is a proper interval of ω, andI α is not an interval of ω′.
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A plan
Let |π| ≥ 4. To show Av(π) is not deflatable:1. Show that every permutation of Av(π) is contained in an
indecomposable one.2. Show that if α is a maximal interval of a non-simple
element, ω, of Av(π) then there is an extension ω′ = ω∪{x}also in Av(π) where α is cut by and separated from x.
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A plan
Let |π| ≥ 4. To show Av(π) is not deflatable:1. Show that every permutation of Av(π) is contained in an
indecomposable one.2. Show that if α is a maximal interval of a non-simple
element, ω, of Av(π), then there is an extensionω′ = ω ∪ {x} also in Av(π) where x cuts and is separatedfrom α.
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Av(2143): Embedding in an indecomposable
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Av(2143): Breaking an interval (I)
Suppose there is a 21 below and left of the interval α. Let b bethe 1 of the leftmost such 21. Let a = minα and add x justabove a and just to the left of b. Check that it can’t make a 2143.
a
b
xα
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Av(2143): Breaking an interval (II)
I We can assume the area below and left of α is increasing.
I To avoid decomposability, above left or below right must benon empty (say the former by symmetry).
I Now let b be the rightmost element of that area.
α
b
a
x
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Av(2143): Breaking an interval (II)
I We can assume the area below and left of α is increasing.I To avoid decomposability, above left or below right must be
non empty (say the former by symmetry).
I Now let b be the rightmost element of that area.
α
b
a
x
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Av(2143): Breaking an interval (II)
I We can assume the area below and left of α is increasing.I To avoid decomposability, above left or below right must be
non empty (say the former by symmetry).I Now let b be the rightmost element of that area.
α
b
a
x
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Av(2143): Breaking an interval (II)
I We can assume the area below and left of α is increasing.I To avoid decomposability, above left or below right must be
non empty (say the former by symmetry).I Now let b be the rightmost element of that area.
α
b
a
x
![Page 27: Deflatability of permutation classes · 2015-05-12 · 1.Show that every permutation of Av(ˇ) is contained in an indecomposable one. 2.Show that if is a maximal interval of a non-simple](https://reader034.vdocuments.us/reader034/viewer/2022042419/5f358742a12bdc5ca44409d1/html5/thumbnails/27.jpg)
Av(2143): Breaking an interval (II)
I We can assume the area below and left of α is increasing.I To avoid decomposability, above left or below right must be
non empty (say the former by symmetry).I Now let b be the rightmost element of that area.
α
b
a
x
![Page 28: Deflatability of permutation classes · 2015-05-12 · 1.Show that every permutation of Av(ˇ) is contained in an indecomposable one. 2.Show that if is a maximal interval of a non-simple](https://reader034.vdocuments.us/reader034/viewer/2022042419/5f358742a12bdc5ca44409d1/html5/thumbnails/28.jpg)
Towards part 2 of the argument
I As my Olympiad students would say “case bashing”.
I From experience in the length four case, concentrate ondecomposable π, say π = λ⊕ ρ.
I Try adding “cutting and separated” x in obvious places(near the edge of α, separated by at most one elementetc.)
TheoremIf π = λ⊕ ρ and |λ|, |ρ| ≥ 2 then Av(π) is not deflatable.
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Towards part 2 of the argument
I As my Olympiad students would say “case bashing”.I From experience in the length four case, concentrate on
decomposable π, say π = λ⊕ ρ.
I Try adding “cutting and separated” x in obvious places(near the edge of α, separated by at most one elementetc.)
TheoremIf π = λ⊕ ρ and |λ|, |ρ| ≥ 2 then Av(π) is not deflatable.
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Towards part 2 of the argument
I As my Olympiad students would say “case bashing”.I From experience in the length four case, concentrate on
decomposable π, say π = λ⊕ ρ.I Try adding “cutting and separated” x in obvious places
(near the edge of α, separated by at most one elementetc.)
TheoremIf π = λ⊕ ρ and |λ|, |ρ| ≥ 2 then Av(π) is not deflatable.
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Towards part 2 of the argument
I As my Olympiad students would say “case bashing”.I From experience in the length four case, concentrate on
decomposable π, say π = λ⊕ ρ.I Try adding “cutting and separated” x in obvious places
(near the edge of α, separated by at most one elementetc.)
TheoremIf π = λ⊕ ρ and |λ|, |ρ| ≥ 2 then Av(π) is not deflatable.
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More on the decomposable case
A bond is an interval of size 2. Bonds can be increasing ordecreasing.
TheoremIf π = 1⊕ ρ and ρ does not contain both an increasing and adecreasing bond, then Av(π) is not deflatable.
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More on the indecomposable case
Ummm . . .
Theorem (MA + VV)Av(2413) is not deflatable.
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More on the indecomposable case
Ummm . . .
Theorem (MA + VV)Av(2413) is not deflatable.
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Deflatable examples
To show that Av(π) is deflatable one needs to find someω ∈ Av(π) without a simple extension. This is greatly facilitatedby:
TheoremSuppose that ω ∈ Av(π) contains a bond α and there is no onepoint extension ω ∪ {x} ∈ Av(π) where x cuts and is separatedfrom α. Then ω has no simple extension in Av(π).
Which is most easily illustrated using PermLab.
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Deflatable examples
To show that Av(π) is deflatable one needs to find someω ∈ Av(π) without a simple extension. This is greatly facilitatedby:
TheoremSuppose that ω ∈ Av(π) contains a bond α and there is no onepoint extension ω ∪ {x} ∈ Av(π) where x cuts and is separatedfrom α. Then ω has no simple extension in Av(π).
Which is most easily illustrated using PermLab.
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Deflatable examples
To show that Av(π) is deflatable one needs to find someω ∈ Av(π) without a simple extension. This is greatly facilitatedby:
TheoremSuppose that ω ∈ Av(π) contains a bond α and there is no onepoint extension ω ∪ {x} ∈ Av(π) where x cuts and is separatedfrom α. Then ω has no simple extension in Av(π).
Which is most easily illustrated using PermLab.
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Remaining issues
I The indecomposable case, in particular is deflatability,non-deflatability, or “it depends” typical?
I Systematic construction of deflatable examples (we havean infinite set but all very ad hoc)
I In the principal case, do one point extensions suffice (toreduce the number of intervals)? They do not in general(see PermLab again).
Thank you!
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Remaining issues
I The indecomposable case, in particular is deflatability,non-deflatability, or “it depends” typical?
I Systematic construction of deflatable examples (we havean infinite set but all very ad hoc)
I In the principal case, do one point extensions suffice (toreduce the number of intervals)? They do not in general(see PermLab again).
Thank you!