#Do not post homework
#William Wang, 10/9/2020, Assignment 10

#1.
#i.
P1 := randperm(9);
               P1 := [7, 4, 2, 9, 3, 8, 6, 5, 1]

P2 := randperm(9);
               P2 := [5, 2, 6, 8, 7, 1, 3, 9, 4]

#P1*P2 = [3,8,2,4,6,9,1,7,5]
#P2*P1 = [3,4,8,5,6,7,2,1,9]
MulPers(P1, P2);
                  [3, 8, 2, 4, 6, 9, 1, 7, 5]

MulPers(P2, P1);
                  [3, 4, 8, 5, 6, 7, 2, 1, 9]

#They are the same!

#ii.
P3 := randperm(9);
               P3 := [4, 9, 8, 6, 2, 1, 3, 7, 5]

#The inverse permutation of P3 can be obtained by listing the index of 1, followed by the index of 2,.... index of 9
#1/P3 = [6, 5, 7, 1, 9, 4, 8, 3, 2]
InvPer(P3);
                  [6, 5, 7, 1, 9, 4, 8, 3, 2]

#They are the same!

#iii.
P4 := randperm(9);
               P4 := [5, 7, 1, 4, 9, 8, 3, 6, 2]

#Cycle structure of P4 is (4)(8,6)(9,2,7,3,1,5)
PtoC(P4);
               [[4], [8, 6], [9, 2, 7, 3, 1, 5]]

#They are the same!

#iv.
P5 := randperm(9);
               P5 := [7, 1, 3, 5, 2, 8, 9, 6, 4]

#Number of inversions of P5 should be 6 + 0 + 1 + 2 + 0 + 2 + 2 + 1 + 0 = 14
#Major index should be 1 + 4 + 7 + 8 = 20

inv(P5);
                               14

maj(P5);
                               20

#They are the same!