Understanding Permutations through the 10-card Shuffle, Study notes of Algebra

The concept of permutations through the example of the 10-card shuffle. It covers topics such as cycle notation, order of a permutation, and composing permutations. The document also includes exercises for the reader to practice.

Typology: Study notes

Pre 2010

Uploaded on 08/19/2009

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Shuffles
The “10-card shuffle” (Note that I am not playing with a full deck. Ha, ha.) is performed as
follows:
The cards 1, 2, 3, . . . , 9, 10 are in order, with 1 being the top card and 10 on the bottom.
Cut the cards in the middle, making two piles: Pile A contains the cards 1 5 and Pile B
contains the cards 6 10.
Now interleave these two decks, alternating between Pile A and Pile B, first laying down the
5 at the bottom of Pile A, followed by the 10 at the bottom of Pile B, etcetera, ending with the
6 from Pile B on top.
5
4
3
2
1
last
8th
6th
4th
2nd
Pile A
10
9
8
7
6
9th
7th
5th
3rd
top
Pile B
The new deck, from top to bottom is
12345678 9 10
6172839410 5
If we shuffle, following this method, a second time, we get
123456 7 8910
3691471025 8
A third shuffle gives
12 3 45678910
7310629518 4
1
pf3

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Shuffles

The “10-card shuffle” (Note that I am not playing with a full deck. Ha, ha.) is performed as

follows:

The cards 1, 2, 3,... , 9, 10 are in order, with 1 being the top card and 10 on the bottom.

Cut the cards in the middle, making two piles: Pile A contains the cards 1 – 5 and Pile B

contains the cards 6 – 10.

Now interleave these two decks, alternating between Pile A and Pile B, first laying down the

5 at the bottom of Pile A, followed by the 10 at the bottom of Pile B, etcetera, ending with the

6 from Pile B on top.

last

8th

6th

4th

2nd

Pile A

9th

7th

5th

3rd

top

Pile B

The new deck, from top to bottom is

1 2 3 4 5 6 7 8 9 10 6 1 7 2 8 3 9 4 10 5

If we shuffle, following this method, a second time, we get

1 2 3 4 5 6 7 8 9 10 3 6 9 1 4 7 10 2 5 8

A third shuffle gives

1 2 3 4 5 6 7 8 9 10 7 3 10 6 2 9 5 1 8 4 1

2

A fourth shuffle gives 1 2 3 4 5 6 7 8 9 10 9 7 5 3 1 10 8 6 4 2

A fifth shuffle gives 1 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1

A sixth shuffle gives 1 2 3 4 5 6 7 8 9 10 5 10 4 9 3 8 2 7 1 6

A seventh shuffle gives 1 2 3 4 5 6 7 8 9 10 8 5 2 10 7 4 1 9 6 3

An eighth shuffle gives 1 2 3 4 5 6 7 8 9 10 4 8 1 5 9 2 6 10 3 7

An ninth shuffle gives 1 2 3 4 5 6 7 8 9 10 2 4 6 8 10 1 3 5 7 9

An tenth shuffle gives 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10

Construct the multiplication table for Z× 11. Do you recognize these permutations?

A simpler way to write permutations is to track what each number maps to, starting with 1,

until you eventually get back to 1.

1 → 6 → 3 → 7 → 9 → 10 → 5 → 8 → 4 → 2 → 1

This cycle is usually written with commas instead of the arrows:

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

Notice that we don’t write 1 at the end, as it is understood from the diagram that the last

element, 2, points to the first element of the cycle, 1.

Compute the order of 6 in Z× 11. What numbers in the ten shuffles do you see?

The permutation σ^2 is computed by applying the permutation σ to itself ( 1 2 3 4 5 6 7 8 9 10 σ(σ(1)) σ(σ(2)) σ(σ(3)) · · · · · σ(σ(9)) σ(σ(10))

or

σ^2 =

the “second” shuffle above.