Identifier
- St000677: Permutations ⟶ ℤ
Values
[1,2] => 0
[2,1] => 1
[1,2,3] => 0
[1,3,2] => 1
[2,1,3] => 1
[2,3,1] => 0
[3,1,2] => 0
[3,2,1] => 1
[1,2,3,4] => 1
[1,2,4,3] => 2
[1,3,2,4] => 2
[1,3,4,2] => 1
[1,4,2,3] => 1
[1,4,3,2] => 2
[2,1,3,4] => 2
[2,1,4,3] => 3
[2,3,1,4] => 1
[2,3,4,1] => 2
[2,4,1,3] => 0
[2,4,3,1] => 3
[3,1,2,4] => 1
[3,1,4,2] => 4
[3,2,1,4] => 2
[3,2,4,1] => 3
[3,4,1,2] => 1
[3,4,2,1] => 2
[4,1,2,3] => 2
[4,1,3,2] => 3
[4,2,1,3] => 3
[4,2,3,1] => 2
[4,3,1,2] => 2
[4,3,2,1] => 3
[1,2,3,4,5] => 1
[1,2,3,5,4] => 2
[1,2,4,3,5] => 2
[1,2,4,5,3] => 1
[1,2,5,3,4] => 1
[1,2,5,4,3] => 2
[1,3,2,4,5] => 2
[1,3,2,5,4] => 3
[1,3,4,2,5] => 1
[1,3,4,5,2] => 2
[1,3,5,2,4] => 0
[1,3,5,4,2] => 3
[1,4,2,3,5] => 1
[1,4,2,5,3] => 4
[1,4,3,2,5] => 2
[1,4,3,5,2] => 3
[1,4,5,2,3] => 1
[1,4,5,3,2] => 2
[1,5,2,3,4] => 2
[1,5,2,4,3] => 3
[1,5,3,2,4] => 3
[1,5,3,4,2] => 2
[1,5,4,2,3] => 2
[1,5,4,3,2] => 3
[2,1,3,4,5] => 2
[2,1,3,5,4] => 3
[2,1,4,3,5] => 3
[2,1,4,5,3] => 2
[2,1,5,3,4] => 2
[2,1,5,4,3] => 3
[2,3,1,4,5] => 1
[2,3,1,5,4] => 2
[2,3,4,1,5] => 2
[2,3,4,5,1] => 1
[2,3,5,1,4] => 1
[2,3,5,4,1] => 2
[2,4,1,3,5] => 0
[2,4,1,5,3] => 3
[2,4,3,1,5] => 3
[2,4,3,5,1] => 2
[2,4,5,1,3] => 2
[2,4,5,3,1] => 1
[2,5,1,3,4] => 1
[2,5,1,4,3] => 2
[2,5,3,1,4] => 4
[2,5,3,4,1] => 1
[2,5,4,1,3] => 3
[2,5,4,3,1] => 2
[3,1,2,4,5] => 1
[3,1,2,5,4] => 2
[3,1,4,2,5] => 4
[3,1,4,5,2] => 1
[3,1,5,2,4] => 3
[3,1,5,4,2] => 2
[3,2,1,4,5] => 2
[3,2,1,5,4] => 3
[3,2,4,1,5] => 3
[3,2,4,5,1] => 2
[3,2,5,1,4] => 2
[3,2,5,4,1] => 3
[3,4,1,2,5] => 1
[3,4,1,5,2] => 2
[3,4,2,1,5] => 2
[3,4,2,5,1] => 1
[3,4,5,1,2] => 1
[3,4,5,2,1] => 2
[3,5,1,2,4] => 2
[3,5,1,4,2] => 1
[3,5,2,1,4] => 3
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Description
The standardized bi-alternating inversion number of a permutation.
The bi-alternating inversion number $i(\pi)$ is $\sum_{1\leq y < x\leq n} (-1)^{y+x} \mathrm{sgn}(\pi(x)-\pi(y))$, see Section 3.1 of [1].
It takes values in between $-\lfloor\frac n2\rfloor^2$ and $\lfloor\frac n2\rfloor^2$, and $i(\pi)+\lfloor\frac n2\rfloor^2$ is even, so we standardize it to $(i(\pi)+\lfloor\frac n2\rfloor^2)/2$.
For odd $n$ the bi-alternating inversion is invariant under rotation, that is, conjugation with the long cycle.
The bi-alternating inversion number $i(\pi)$ is $\sum_{1\leq y < x\leq n} (-1)^{y+x} \mathrm{sgn}(\pi(x)-\pi(y))$, see Section 3.1 of [1].
It takes values in between $-\lfloor\frac n2\rfloor^2$ and $\lfloor\frac n2\rfloor^2$, and $i(\pi)+\lfloor\frac n2\rfloor^2$ is even, so we standardize it to $(i(\pi)+\lfloor\frac n2\rfloor^2)/2$.
For odd $n$ the bi-alternating inversion is invariant under rotation, that is, conjugation with the long cycle.
References
[1] Even-Zohar, C. The Writhe of Permutations and Random Framed Knots arXiv:1511.09469
Code
def statistic(pi):
"""
sage: all(statistic(Permutation([e+1 if e < len(pi) else 1 for e in [pi[-1]] + pi[:-1]])) == statistic(pi) for r in range(1,8,2) for pi in Permutations(r))
True
"""
return 1/2*((len(pi)//2)^2+sum((-1)^(y+x) if pi(x) > pi(y) else -(-1)^(y+x) for x in range(1,len(pi)+1) for y in range(1,x)))
Created
Dec 30, 2016 at 14:38 by Martin Rubey
Updated
Dec 30, 2016 at 14:38 by Martin Rubey
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