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 Faulty Card Shuffler (Posted on 2013-07-24)
This card shuffler, instead of randomizing the positions of cards in the deck, always rearranges any given set of cards into the same new order.

If for example one puts all the hearts into the shuffler, in the order A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, and runs that through the shuffler first once, and then, without rearranging the cards, runs it through again, the resulting sequence after the second shuffle is always:

10, 9, Q, 8, K, 3, 4, A, 5, J, 6, 2, 7

The question is:

What was the order of the cards after the first run through this non-randomizing shuffler?

From the Skeptics' Guide to the Universe podcast episode of June 22, 2013 (episode 414).

 See The Solution Submitted by Charlie Rating: 4.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: ? Solution for any set of subcycles? | Comment 8 of 10 |
(In reply to ? Solution for any set of subcycles? by Larry)

If

Applying a single shuffle to
A 2 3 4 5 6 7 8 9 10 J Q K  (0 shuffles)
Results in
2 3 4 A 6 7 8 9 10 J Q K 5 (1 shuffle)
After two shuffles it is at
3 4 A 2 7 8 9 10 J Q K 5 6 (2 shuffles)

The first shuffle (the solution and therefore not given), represented in position cycles, is:

(2,1,4,3)(6,5,13,12,11,10,9,8,7)

When this is iterated, to give the position after the 2nd shuffle (the puzzle position, given), it is the permutation:

(3,1)(2,4)(6,13,11,9,7,5,12,10,8)

If the results of the second shuffle were presented as the puzzle in this case, and you kept iterating this latter permutation, positions 1 and 3 would just keep going back and forth, as would positions 2 and 4. You'd never get to the condition of the 2 getting to the Ace position and the Ace getting into the 4th position.

That's why that method wouldn't work in producing the solution to this type puzzle.

 Posted by Charlie on 2013-07-29 10:05:25

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