All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 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).

 Submitted by Charlie Rating: 4.0000 (2 votes) Solution: (Hide) 9, A, 4, Q, J, 7, 3, 2, 10, 5, K, 8, 6 After two "shuffles", the first postion has gone from holding an Ace to holding a 10. The 10 position has gone from holding a 10 to holding a Jack, the Jack (11th) position has gone from holding a Jack to holding a 6, etc. That sequence of "position is replaced by" can be represented as follows: A -> 10 -> J -> 6 -> 3 -> Q -> 2 -> 9 -> 5 -> K -> 7 -> 4 -> 8 -> A The "square root" of this transformation can be obtained by going forward 7 or backward 6 in the cycle, so that the first position would go from Ace to 9, and the 9 to the 10, and the 10 to the 5, etc.: A -> 9 -> 10 -> 5 -> J -> K -> 6 -> 7 -> 3 -> 4 -> Q -> 8 -> 2 -> A So after one iteration instead of two, the Ace has been replaced by a 9, the deuce by an ace, the trey by a 4, etc: 9, A, 4, Q, J, 7, 3, 2, 10, 5, K, 8, 6 which is the solution. To verify: The 9, in the A position after one "shuffle" is moved to the 2 position as required. The A, in the 2 position after one "shuffle" is moved to the 8 position as required. The 4, in the 3 position after one "shuffle" is moved to the 7 position as required. The Q, in the 4 position after one "shuffle" is moved to the 3 position as required. The J, in the 5 position after one "shuffle" is moved to the 10 position as required. The 7, in the 6 position after one "shuffle" is moved to the K position as required. The 3, in the 7 position after one "shuffle" is moved to the 6 position as required. The 2, in the 8 position after one "shuffle" is moved to the Q position as required. The 10, in the 9 position after one "shuffle" is moved to the A position as required. The 5, in the 10 position after one "shuffle" is moved to the 9 position as required. The K, in the J position after one "shuffle" is moved to the 5 position as required. The 8, in the Q position after one "shuffle" is moved to the 4 position as required. The 6, in the K position after one "shuffle" is moved to the J position as required.

 Subject Author Date re(3): ? Solution for any set of subcycles? Steve Herman 2013-07-29 23:37:59 re(2): ? Solution for any set of subcycles? Larry 2013-07-29 21:42:52 re: ? Solution for any set of subcycles? Charlie 2013-07-29 10:05:25 ? Solution for any set of subcycles? Larry 2013-07-28 15:03:18 Cycle length determined Steve Herman 2013-07-25 09:13:57 Brians solution broll 2013-07-25 03:09:11 Cycle length is not obvious Steve Herman 2013-07-25 00:38:57 re: Solution Ady TZIDON 2013-07-24 17:00:42 Solution Brian Smith 2013-07-24 16:28:59 The Hard Way? (spoiler) Steve Herman 2013-07-24 14:37:37

 Search: Search body:
Forums (0)
Random Problem
Site Statistics
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox: