Thermodynamic classification of three-dimensional Kitaev ...
Permutations, sequences, and partially ordered sets Sergey Kitaev Reykjavik University Joint work...
Transcript of Permutations, sequences, and partially ordered sets Sergey Kitaev Reykjavik University Joint work...
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Permutations, sequences, and partially ordered sets
Sergey KitaevReykjavik University
Joint work with
Mireille Bousquet-Mélou Anders Claesson Mark Dukes
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Overview of results
Bijections (respecting several statistics) between the following objects
unlabeled (2+2)-free posets on n elements
pattern-avoiding permutations of length n
ascent sequences of length n
linearized chord diagrams with n chords = certain involutions
Closed form for the generating function for these classes of objects
Pudwell’s conjecture (on permutations avoiding 31524) is settled using modified ascent sequences
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Ascent sequencesNumber of ascents in a word: asc(0, 0, 2, 1, 1, 0, 3, 1, 2, 3) = 4
(0,0,2,1,1,0,3,1,2,3) is not an ascent sequence, whereas (0,0,1,0,1,3,0) is.
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Unlabeled (2+2)-free posets
A partially ordered set is called (2+2)-free if it contains no induced sub-posets isomorphic to (2+2) =
Such posets arise as interval orders (Fishburn):
P. C. Fishburn, Intransitive indifference with unequal indifference intervals, J. Math.Psych. 7 (1970) 144–149.
bad guy good guy
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Unlabeled (2+2)-free posets
Theorem. (Not ours!) A poset is (2+2)-free iff the collection of strict down-sets may be linearly ordered by inclusion.
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Unlabeled (2+2)-free posets
How can one decompose a (2+2)-free poset?
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Unlabeled (2+2)-free posets
2
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Unlabeled (2+2)-free posets
1 1 3
1 0 1
Read labels backwards: (0, 1, 0, 1, 3, 1, 1, 2) – an ascent sequence!
Removing last point gives one extra 0.
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Theorem. There is a 1-1 correspondence between unlabeled (2+2)-free posets on n elements and ascent sequences of length n.
(0, 1, 0, 1, 3, 1, 1, 2)
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Permutations avoiding
31524 avoids 32541 contains
How can one decompose such permutations?
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Permutations avoiding
61832547 avoids . Remove the largest element, 8:
61 32 54 7
61 32 54 72 410 3
8 corresponds to the position labeled 1.
61 32 54 210 3
7 corresponds to the position labeled 3. Etc.
Read obtained labels starting from the recent one, to get (0, 1, 1, 2, 2, 0, 3, 1) – an ascent sequence!
Remove 7:
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Theorem. There is a 1-1 correspondence between permutations avoiding on n elements and ascent sequences of length n.
(0, 1, 1, 2, 2, 0, 3, 1)6 1 8 3 2 5 4 7
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Restricted permutations versus (2+2)-free posets
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Modified ascent sequences
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Some statistics preserved under the bijections
(0, 1, 0, 1, 3, 1, 1, 2)
(0, 1, 0, 1, 3, 1, 1, 2 )
(0, 1, 0, 1, 3, 1, 1, 2) 3 1 7 6 4 8 2 5
3 1 7 6 4 8 2 5
3 1 7 6 4 8 2 5
min zeros leftmost decr. run
min maxlevel
last element
0 1 2
label in fron ofmax element
(0, 3, 0, 1, 4, 1, 1, 2)
Level distri-bution
letter distributionin modif. sequence
element distributionbetween act. sites
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Some statistics preserved under the bijections
(0, 1, 0, 1, 3, 1, 1, 2)
(0, 1, 0, 1, 3, 1, 1, 2)
(0, 1, 0, 1, 3, 1, 1, 2) 3 1 7 6 4 8 2 5
3 1 7 6 4 8 2 5
3 1 7 6 4 8 2 5
highestlevel
number of ascentsnumber of ascentsin the inverse
2 7 1 5 8 4 3 6
(0, 3, 0, 1, 4, 1, 1, 2)
right-to-left maxright-to-left maxin mod. sequencemax
compo-nents
Components inmodif. sequence components
(0, 3, 0, 1, 4, 1, 1, 2)
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RLCD versus unlabeled (2+2)-free posets
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RLCD versus unlabeled (2+2)-free posets
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Sketch of a proof for surjectivity
(2+2)-free poset interval order
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Sketch of a proof for surjectivity
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Posets avoiding and
Ascent sequences are restricted as follows:
m-1, where m is the max element here
Catalan many
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An open problemFind a bijection between (2+2)-free posets and permutations
avoiding using a graphical way, like one suggested below.
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Thank you for your attention!Any questions?