Pattern Avoiding and Quasisymetric function
Food for thought talk 2016
Kuang Sittipong Thamrongpairoj
Outline
What is pattern avoiding permutation?
What are quasisymmetric functions?
Outline
What is pattern avoiding permutation? What are quasisymmetric functions?
Outline
What is pattern avoiding permutation? What are quasisymmetric functions?
Acknowledgement
Pattern avoiding permutation
What is a pattern in permutation? Consider σ = 21534 ∈ S5
Then, we get rid of some numbers in σ σ0 = 254
Pattern avoiding permutation
What is a pattern in permutation? Consider σ = 21534 ∈ S5
Then, we get rid of some numbers in σ σ0 = 254
Pattern avoiding permutation
What is a pattern in permutation? Consider σ = 21534 ∈ S5
Then, we get rid of some numbers in σ σ0 = 254
Pattern avoiding permutation
If we can get τ from σ via the process, we say σ contains pattern τ .
We say that σ avoids τ , if σ does not contain τ . Let Sn(τ ) be the set of σ ∈ Sn avoiding τ
Pattern avoiding permutation
If we can get τ from σ via the process, we say σ contains pattern τ . We say that σ avoids τ , if σ does not contain τ .
Let Sn(τ ) be the set of σ ∈ Sn avoiding τ
Pattern avoiding permutation
If we can get τ from σ via the process, we say σ contains pattern τ . We say that σ avoids τ , if σ does not contain τ .
Let Sn(τ ) be the set of σ ∈ Sn avoiding τ
Pattern avoiding permutation
If we can get τ from σ via the process, we say σ contains pattern τ . We say that σ avoids τ , if σ does not contain τ .
Let Sn(τ ) be the set of σ ∈ Sn avoiding τ
Pattern avoiding permuation
.
Pop quiz
:what is |Sn(12)|?
Pattern avoiding permuation
.
Pop quiz
: what is |Sn(12)|?Pattern avoiding permuation
.
Pop quiz
: what is |Sn(12)|?Pattern avoiding permutation
Given two patterns τ1 and τ2, we say that τ1 and τ2 are Wilf-equivalent, if
|Sn(τ1)| = |Sn(τ2)| for all n.
Let’s look at equivalence classes of pattern of length 3. First, it is clear that 123 and 321 are equivalent
Proof: If σ avoids 123, the “reverse” of σ will avoid 321. Second, 132, 231, 312, 213 are equivalent.
Pattern avoiding permutation
Given two patterns τ1 and τ2, we say that τ1 and τ2 are Wilf-equivalent, if
|Sn(τ1)| = |Sn(τ2)| for all n.
Let’s look at equivalence classes of pattern of length 3.
First, it is clear that 123 and 321 are equivalent
Proof: If σ avoids 123, the “reverse” of σ will avoid 321. Second, 132, 231, 312, 213 are equivalent.
Pattern avoiding permutation
Given two patterns τ1 and τ2, we say that τ1 and τ2 are Wilf-equivalent, if
|Sn(τ1)| = |Sn(τ2)| for all n.
Let’s look at equivalence classes of pattern of length 3. First, it is clear that 123 and 321 are equivalent
Proof: If σ avoids 123, the “reverse” of σ will avoid 321.
Second, 132, 231, 312, 213 are equivalent.
Pattern avoiding permutation
Given two patterns τ1 and τ2, we say that τ1 and τ2 are Wilf-equivalent, if
|Sn(τ1)| = |Sn(τ2)| for all n.
Let’s look at equivalence classes of pattern of length 3. First, it is clear that 123 and 321 are equivalent
Proof: If σ avoids 123, the “reverse” of σ will avoid 321. Second, 132, 231, 312, 213 are equivalent.
Pattern avoiding permutation
Theorem (Knuth, 1973)
123 and 132 are equivalent. Therefore, there is only one equivlance class for patterns of length 3.
Pattern avoiding permutation
Theorem (Knuth, 1973)
123 and 132 are equivalent. Therefore, there is only one equivlance class for patterns of length 3.
Pattern avoiding permutation
Theorem (Knuth, 1973)
123 and 132 are equivalent. Therefore, there is only one equivlance class for patterns of length 3.
Pattern avoiding permutation
Before we move on...
For any σ ∈ Sn, let Des(σ) = {i | σi > σi +1}
For example, if σ = 51423, then Des(σ) = {1, 3}
Pattern avoiding permutation
Before we move on...
For any σ ∈ Sn, let Des(σ) = {i | σi > σi +1}
For example, if σ = 51423, then Des(σ) = {1, 3}
Pattern avoiding permutation
Before we move on...
For any σ ∈ Sn, let Des(σ) = {i | σi > σi +1}
For example, if σ = 51423, then Des(σ) = {1, 3}
Pattern avoiding permutation
Before we move on...
For any σ ∈ Sn, let Des(σ) = {i | σi > σi +1}
For example, if σ = 51423, then Des(σ) = {1, 3}