restricted symmetric signed permutations · 2019. 4. 30. · september 28, 2011. signed...
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Restricted Symmetric Signed PermutationsEnumerations of Pattern-Avoiding Signed Permutations Invariant
Under Certain Symmetry Subgroups
Andy Hardt and Justin M. Troyka
Carleton College
September 28, 2011
Signed Permutations
Definition
A permutation of length n is an ordering of the numbers from 1 to n. Forexample, the permutations of length 3 are 123, 132, 213, 231, 312, and 321.
Definition
A signed permutation is a permutation where we can put bars over someof the entries. For example, the signed permutations of length 2 are12, 12̄, 1̄2, 1̄2̄, 21, 21̄, 2̄1, and 2̄1̄.
Hardt and Troyka (Carleton College) Signed Permutations September 28, 2011 2 / 12
Pattern Avoidance
Definition
A signed permutation π contains another signed permutation (called apattern) ρ if there is a substring of π with the same relative ordering andbar configuration as ρ. If π does not contain ρ, we say π avoids ρ.
yi
The pattern 2̄1
ssccc
sc s
627̄4̄8̄15̄3 contains the pattern 21̄
Hardt and Troyka (Carleton College) Signed Permutations September 28, 2011 3 / 12
Symmetric Invariance
scsc c
scs
16̄25̄4̄73̄8 is invariant under R180.
s sc
cc
cs s
358̄2̄7̄1̄46 is invariant under R90.
s cs s
c cs c
75̄136̄8̄42̄ is invariant under R180.
s s s cc s
c s
1324̄7̄85̄6 is invariant under D.
s cs
ss
cc c
64̄8271̄5̄3̄ is invariant under D.
s cc
cs
ss c
64̄8̄2̄7153̄ is invariant under D and D′.
Hardt and Troyka (Carleton College) Signed Permutations September 28, 2011 4 / 12
Simplifications
Some symmetry operations,such as reflection over a verticalline, fix no permutations.
If the permutation π avoids thepattern ρ and is invariant underthe symmetry g , then π alsoavoids g(ρ).
If H is a symmetry subgroup, Ris a set of patterns, and g is asymmetry in the normalizer ofH, then for all n ≥ 0,|BH
n (R)| = |BHn (g(R))|.
r rb b r
r rb
r r bb b b
r r
rb b
b rr r
b
←→
←→
←→
br r
r bbr r
r r bb b b
r r
br r
r bb b
r
Hardt and Troyka (Carleton College) Signed Permutations September 28, 2011 5 / 12
Permutations Invariant Under R180
Result
|B1802k (2̄1̄, 12)| =
(2kk
).
|B1802k+1(2̄1̄, 12)| =
2(2kk
).
Terms
1, 2, 2, 4, 6, 12, 20, 40, 70, 140
sc c
s sc c
s
Hardt and Troyka (Carleton College) Signed Permutations September 28, 2011 6 / 12
Permutations Invariant Under R180
Result
|B1802k (2̄1̄, 12)| =
(2kk
).
|B1802k+1(2̄1̄, 12)| = 2
(2kk
).
Terms
1, 2, 2, 4, 6, 12, 20, 40, 70, 140
sc c
s sc c
s
Hardt and Troyka (Carleton College) Signed Permutations September 28, 2011 6 / 12
Permutations Invariant Under R180
Result
|B1802k (2̄1̄, 2̄1, 21)| =
(2k+1k
).
Even terms
1, 3, 10, 35, 126, 462, 1716, 6435ssc c
s scc
Hardt and Troyka (Carleton College) Signed Permutations September 28, 2011 7 / 12
Permutations Invariant Under R180
Result
|B1802k (2̄1̄, 2̄1, 21)| =
(2k+1k
).
Even terms
1, 3, 10, 35, 126, 462, 1716, 6435ssc c
s scc
Hardt and Troyka (Carleton College) Signed Permutations September 28, 2011 7 / 12
Permutations Invariant Under D and D ′
Result
|BH2k(2̄1̄, 2̄1, 21̄, 21)| =
2k .
Even terms
1, 2, 4, 8, 16, 32, 64, 128
Result
|BH2k(2̄1̄)| =
3|BH2(k−1)(2̄1̄)|
+2(k − 1)|BH2(k−2)(2̄1̄)|.
Even terms
1, 3, 11, 45, 201, 963, 4899, 26253
s s c s s c s s
s c c s
s c c s
Hardt and Troyka (Carleton College) Signed Permutations September 28, 2011 8 / 12
Permutations Invariant Under D and D ′
Result
|BH2k(2̄1̄, 2̄1, 21̄, 21)| = 2k .
Even terms
1, 2, 4, 8, 16, 32, 64, 128
Result
|BH2k(2̄1̄)| =
3|BH2(k−1)(2̄1̄)|
+2(k − 1)|BH2(k−2)(2̄1̄)|.
Even terms
1, 3, 11, 45, 201, 963, 4899, 26253
s s c s s c s s
s c c s
s c c s
Hardt and Troyka (Carleton College) Signed Permutations September 28, 2011 8 / 12
Permutations Invariant Under D and D ′
Result
|BH2k(2̄1̄, 2̄1, 21̄, 21)| = 2k .
Even terms
1, 2, 4, 8, 16, 32, 64, 128
Result
|BH2k(2̄1̄)| =
3|BH2(k−1)(2̄1̄)|
+2(k − 1)|BH2(k−2)(2̄1̄)|.
Even terms
1, 3, 11, 45, 201, 963, 4899, 26253
s s c s s c s s
s c c s
s c c s
Hardt and Troyka (Carleton College) Signed Permutations September 28, 2011 8 / 12
Permutations Invariant Under D and D ′
Result
|BH2k(2̄1̄, 2̄1, 21̄, 21)| = 2k .
Even terms
1, 2, 4, 8, 16, 32, 64, 128
Result
|BH2k(2̄1̄)| = 3|BH
2(k−1)(2̄1̄)|+2(k − 1)|BH
2(k−2)(2̄1̄)|.
Even terms
1, 3, 11, 45, 201, 963, 4899, 26253
s s c s s c s s
s c c s
s c c s
Hardt and Troyka (Carleton College) Signed Permutations September 28, 2011 8 / 12
Permutations Invariant Under D and D ′
Result
|BW2k (1̄2̄, 12)| =
2k .
Even terms
1, 2, 4, 8, 16, 32, 64, 128
Result
|BW2k (1̄2̄, 1̄2, 12)| =
( kbk/2c
).
Even terms
1, 1, 2, 3, 6, 10, 20, 35
s cc c
s ss c
s sc
sc
sc c
Hardt and Troyka (Carleton College) Signed Permutations September 28, 2011 9 / 12
Permutations Invariant Under D and D ′
Result
|BW2k (1̄2̄, 12)| = 2k .
Even terms
1, 2, 4, 8, 16, 32, 64, 128
Result
|BW2k (1̄2̄, 1̄2, 12)| =
( kbk/2c
).
Even terms
1, 1, 2, 3, 6, 10, 20, 35
s cc c
s ss c
s sc
sc
sc c
Hardt and Troyka (Carleton College) Signed Permutations September 28, 2011 9 / 12
Permutations Invariant Under D and D ′
Result
|BW2k (1̄2̄, 12)| = 2k .
Even terms
1, 2, 4, 8, 16, 32, 64, 128
Result
|BW2k (1̄2̄, 1̄2, 12)| =
( kbk/2c
).
Even terms
1, 1, 2, 3, 6, 10, 20, 35
s cc c
s ss c
s sc
sc
sc c
Hardt and Troyka (Carleton College) Signed Permutations September 28, 2011 9 / 12
Permutations Invariant Under D and D ′
Result
|BW2k (1̄2̄, 12)| = 2k .
Even terms
1, 2, 4, 8, 16, 32, 64, 128
Result
|BW2k (1̄2̄, 1̄2, 12)| =
( kbk/2c
).
Even terms
1, 1, 2, 3, 6, 10, 20, 35
s cc c
s ss c
s sc
sc
sc c
Hardt and Troyka (Carleton College) Signed Permutations September 28, 2011 9 / 12
Permutations Invariant Under D
Result
|BD2k(2̄1̄, 2̄1, 21)| =
Ck .
Even terms
1, 1, 2, 5, 14, 42, 132, 429
s scsc c
s c
Hardt and Troyka (Carleton College) Signed Permutations September 28, 2011 10 / 12
Permutations Invariant Under D
Result
|BD2k(2̄1̄, 2̄1, 21)| = Ck .
Even terms
1, 1, 2, 5, 14, 42, 132, 429
s scsc c
s c
Hardt and Troyka (Carleton College) Signed Permutations September 28, 2011 10 / 12
Open Questions
Combinatorial proofs:
I |B1802k (2̄1̄, 2̄1, 21)| =
(2k+1k
)I |BW
2k (1̄2̄, 1̄2, 12)| =(
kbk/2c
)r -colored permutations for r > 2.
Avoidances of length 2 and length 3
Hardt and Troyka (Carleton College) Signed Permutations September 28, 2011 11 / 12
Acknowledgements
Carleton’s HHMI Grant
Carleton College Department of Mathematics
Our Adviser Eric Egge
Hardt and Troyka (Carleton College) Signed Permutations September 28, 2011 12 / 12