Calculation is a solitaire variant known for its large skill ceiling; supposedly, the large majority of games can be won by skilled human players, whereas the games to be won by uninformed play are in a slim minority (of these two facts, I can only attest to the latter!).

I imagine that, like pretty much every other solitaire variant, there should be starting configurations of Calculation that are unwinnable even when you know all 48 cards in the deck in advance. Can you provide an example of such a configuration, with a human-verifiable proof that it cannot be won? Or can you prove that every configuration is winnable?

This is to assume no waste pile, and four tableaux.

Rules of Calculation: Suit never matters in this game, only rank. Take out an ace, two, three, and a four, and use them to start four foundation piles, and make the other 48 cards into a shuffled deck. Four tableaux piles start empty. The goal is to build the four sequential foundations:

  • A,2,3,4,5,6,7,8,9,10,J,Q,K ("one foundation");
  • 2,4,6,8,10,Q,A,3,5,7,9,J,K ("two foundation");
  • 3,6,9,Q,2,5,8,J,A,4,7,10,K ("three foundation");
  • 4,8,Q,3,7,J,26,10,A,5,9,K ("four foundation").

A legal move consists of either (1) drawing a card from the deck, and placing it on one of the four tableaux, or (2) taking a card from the top of a tableau and moving it to a foundation (but not to another tableau) which is ready to receive it. Play ends when you run out of legal moves, at which point you win if and only if all four foundations have been completed.

  • $\begingroup$ What do you do with a card that does not fit? Put on top of what is called tableau? below? $\endgroup$
    – Moti
    Jul 25 '21 at 15:29
  • $\begingroup$ A card that does not fit on any foundation, drawn from the deck? It must go to a tableau first, at any rate. A card on top of a tableau that does not fit on any foundation? There it must stay until it can, and you decide to place it (it is often better to wait until necessary). $\endgroup$
    – Feryll
    Jul 25 '21 at 18:17
  • $\begingroup$ Looks like Solvitaire doesn't support Calculation at the moment, the foundations incrementing by one is hardcoded in. $\endgroup$ Jul 27 '21 at 3:56

Partial answer (without a proof, only some assumptions):

The configuration (all 48 cards placed in order) like
Q Q Q Q 9 9 9 9 10 10 10 10 3 3 3 5 5 5 5 7 7 7 7 J J J J A A A K K K K 2 2 2 4 4 4 6 6 6 6 8 8 8 8 (or a variation) should be likely unwinnable at all (and if it isn't, I would like to see a solution for it!).


Since all the deuces, fours, sixed and eights are placed in the very end of the deck, you cannot move any cards to the foundation until you draw all the other cards (so you have to arrange all of them on the tableaux), and particularly all the kings and aces. So, at least 2 of the 4 tableaux will be "blocked" with an ace and a king respectively (you cannot place all of them in a single tableau, because aces come before the kings, thus they should be placed first, but all the aces cannot be buried under kings since it would be impossible to place the aces onto the foundations). But the aces are "tricky" since they cannot be used early in the game (an ace can be at least 7th card on a foundation).
The same can be said about the jacks, and (at lesser extent) fives and sevens. Since they come after the "early" cards like threes and queens in the deck, so it would be difficult (if possible at all) to place them on the tableaux without blocking the "early" cards which are more useful.


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