Question

σ=$\left(\begin{array}{cccccccccccccccccccccccccccccccccccccccccccccccccccc}1&2&3&4&...&n&n+1&n+2&n+3&...&2n\\1&3&5&7&...&2n-1&2&4&6&...&2n\\\end{array}\right)$
m(σ)=?
Σ(σ)=sgn m(σ)=?
Va rog am mare nevoie sa stiu cum se rezolva!

Answer 1

Inversiunile sunt
$\sigma(1)\rightarrow cu\ nimeni \\ \sigma(2)\rightarrow cu\ \sigma(n+1) \\ \sigma(3)\rightarrow cu\ \sigma(n+1)si(\sigma(n+2) \\ . \\ . \\ . \\ \sigma(n)\rightarrow cu\ \sigma(n+1);\sigma(n+2);\sigma(n+3)...\sigma(2n-1) \\ Total:\ 0+1+2+3+...+2n-1= \frac{(2n-1)2n}{2}=n(2n-1)\ inversiuni$

Deci m=n(2n-1).
Daca n este numar par, atunci m este numar par. Daca n este numar impar, atunci m este numar impar.

$\Sigma(\sigma)=(-1)^m= \left \{ {{1\ pentru\ n\ numar\ par} \atop {-1\ pentru\ n\ numar\ impar}} \right.$