Unbeatable

The idea to beat FreeCell sprang from a message posted by sci.math regular Russell Turpin on July 24, 1994, with the definitive title "FreeCell is NOT always winnable." In his message, Turpin provided an example of an unwinnable FreeCell deal and asked fellow group members to help prove or disprove his gripe...Turpin’s proof-of-concept hand was indeed unwinnable, but it also wasn’t one of the 32,000 included in FreeCell—possibly for that reason...Ring had made up his mind. They would try to solve FreeCell. All 32,000 hands of it.

  1. In the late 90s I was interested enough in FreeCell to start playing through all the hands in order — if I recall, the version included with Windows 95 kept win/loss statistics for you, and that was incentive enough — but I don’t remember how many I played before eventually needing to buy another computer. (Apparently it was fewer than 11,982 of them.) And I never noticed until now that the version of XP I have installed since then doesn’t seem to include the game at all.

    But at the time, at least, FreeCell seemed like a revelation, a solitaire game that was a true puzzle game, with no hidden information and thus no luck in play beyond the initial deal.

  2. This has always amazed be about FreeCell. I assume the game number is just the seed for the pseudo-random number generator that produces the starting deal. I’d expect that either all starting configurations are winnable, or else some significant percentage aren’t. But instead, there’s just one out of tens of thousands that isn’t. That seems so contrary to how the universe in general tends to work.

    There’s another way of looking at the problem, too, which is how many free cells do you need to win each hand? The game gives you 4, and many hands are winnable with fewer. I think the single unwinnable hand becomes winnable with 5 free cells. If you look at all possible starting hands (not just the 32,000 the game can generate), how many need 5 free cells to win? Do any need 6 or more?

    And what is it about the arrangement of the cards that makes this hand more difficult than all the others? Is it possible to then compute the most difficult FreeCell hand that there could ever be?

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