In Klondike solitaire, what is the fewest number of cards that can be placed in a beginning setup to...
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Problem
In Klondike solitaire, what is the fewest number of cards that can be placed in a beginning setup to guarantee the player will not be able to win? The rules: 7 piles Deal in 'standard' solitaire way From deck show every 3rd card, then repeat (if the last portion of cards is not a multiple of 3 - kill remaining 1 or 2 cards and restart) Aces placed in 4 piles (by suit) - cards may be moved freely from Aces stack to main stack (as long as normal rules are followed). So, for example, I could easily place 50 of the cards where I wished, and have the remaining two placed randomly in the two remaining spots, and guarantee a loss. I feel I could also do this with 49 cards, 48 cards, etc... But what is the SMALLEST number of cards that you can place, with the rest placed randomly (or in the most opportune way for the player), that will guarantee no solution? The best we could come up with (without delving very deeply into it) is this: Place each of the original top 7 cards (eg: 2 2 2 2 4 4 4). Then place every third card in the deck so that it is not an Ace or 3. This would need the placing of 14 cards (7 top ones and every third of the remaining 23). Is it possible to do it better? -- Stephen Dillinger <driver@mcs.com> (April 1994)
Show solution
I think you could do it placing the following four cards. Put an ace on the bottom of one pile. Now put both threes colored opposite to the ace on top of it. Now put the two that is the same suit as the ace on top of that. So, in other words: put the ace of hearts on the bottom, the three of spades and clubs on top of that, and the two of hearts on top of them. As far as I can figure, that should work. You'll never be able to play the two of hearts. -- eliel kelly suzanne <keliel@uxa.cso.uiuc.edu> (April 1994)