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

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Acknowledgements

Carleton’s HHMI Grant

Carleton College Department of Mathematics

Our Adviser Eric Egge

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