# permutations

• December 7th 2007, 11:42 PM
anncar
permutations
Suppose we have a deck of 10 cards - say the ace to ten of spades
(denoted 1,. . . , 10 for definitiveness). Permutations of the deck may be
regarded as elements of S(10).

Define s to be the ‘shuffle’ which hides the top card:
s = 1 2 3 4 5 6 7 8 9 10
2 4 6 8 10 1 3 5 7 9 .

Let t be the ‘shuffe’ that leaves the first card unchanged:
t = 1 2 3 4 5 6 7 8 9 10
1 3 5 7 9 2 4 6 8 10.

Finally, let c be the ‘cut’:

c = 1 2 3 4 5 6 7 8 9 10
6 7 8 9 10 1 2 3 4 5 .

(a) Write s, t, c, cs and scs using cycle notation, as a product of disjoint
cycles, and find the sign of each.

(b) For each of these basic and combined shuffes s, t, c, cs and scs, how many times must it be repeated before the cards are returned to their original positions?
• December 8th 2007, 05:01 AM
kalagota
Quote:

Originally Posted by anncar
Suppose we have a deck of 10 cards - say the ace to ten of spades
(denoted 1,. . . , 10 for definitiveness). Permutations of the deck may be
regarded as elements of S(10).

Define s to be the ‘shuffle’ which hides the top card:
s = 1 2 3 4 5 6 7 8 9 10
2 4 6 8 10 1 3 5 7 9 .

Let t be the ‘shuffe’ that leaves the first card unchanged:
t = 1 2 3 4 5 6 7 8 9 10
1 3 5 7 9 2 4 6 8 10.

Finally, let c be the ‘cut’:

c = 1 2 3 4 5 6 7 8 9 10
6 7 8 9 10 1 2 3 4 5 .

(a) Write s, t, c, cs and scs using cycle notation, as a product of disjoint
cycles, and find the sign of each.

(b) For each of these basic and combined shuffes s, t, c, cs and scs, how many times must it be repeated before the cards are returned to their original positions?

a)

$s = (1,2,4,8,5,10,9,7,3,6)$
$t = (2,3,5,9,8,6)(4,7)$
$c = (1,6)(2,7)(3,8)(4,9)(5,10)$

$
cs = (1,6)(2,7)(3,8)(4,9)(5,10)(1,2,4,8,5,10,9,7,3,6)
$

$cs = \left( {\begin{array}{cccccccccc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 7 & 9 & 1 & 3 & 5 & 6 & 8 & 10 & 2 & 4 \end{array} } \right) = (1,7,8,10,4,3)(2,9)$

$scs = (1,2,4,8,5,10,9,7,3,6)(1,6)(2,7)(3,8)(4,9)(5,10)(1 ,2,4,8,5,10,9,7,3,6)$

$scs = \left( {\begin{array}{cccccccccc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 3 & 7 & 2 & 6 & 10 & 1 & 5 & 9 & 4 & 8 \end{array} } \right) = (1,3,2,7,5,10,8,9,4,6)$

b) n times where n is the length of the longest cycle..
• December 12th 2007, 10:27 AM
Wilmer
Nice!