![]() ![]() If the reader knows of any way to beat the trivial 92.6% then please let me know □Īnother interesting problem, not mentioned in Bart’s link is the Spider and Fly problem. think of two diagonally adjacent squares on a chessboard). I suspect some kind of trickery is needed, such as diagonal cuts or tiles that are connected at a vertex only (e.g. Note that all of them assume tiles are the union of unit squares. ![]() The following diagram shows many ways to achieve 50/54 = 92.6%. It is trivial to cover 50 squares out of 54 with any number of colourings. If I had to invent a similar puzzle with a Spider Solitaire theme, then I would probably replace the Y-shaped pentomino with the following: They are asking for >86% but I can’t even beat the trivial 80%. An example of the latter is the following:įor some reason, this type of puzzle ain’t my thing. This includes a chess problem where White wins even though neither side wants the Black king to be checkmated, the hackneyed group of logicians trying to deduce the colour of their own hat or tessellation problems involving dividing a geometric shape into two or more congruent pieces. My good friend Bart Wright has emailed be a link about “wacky and offbeat problems” If you have any other ideas for Spider Solitaire variants, please let me know in the comments □ ![]() I’m not sure if I’m keen for a game of Five-Suit at this stage, but I would like to thank Bart for the suggestion anyways. Obviously this is tedious – and the reason why software versions were invented in the first place – but at least it can be done in theory.Ĭlearly, the N-suit version can be implemented with physical playing cards in the same manner, where N = 5,6,7. Basically, if you expose a card then (1) you know if it is vertical or horizontal if another card of same rank and suit is already exposed, otherwise (2) flip a coin. If the cards are not distinguishable then we can use a random number generator. Provided you choose one or the other consistently then you are quids in. Whenever a card is exposed (either by dealing from the stock or turning over in the tableau), we can lay plain cards horizontally and magic-eye-happy-stars cards vertically (or the other way round). Without loss of generality assume the backs of the cards are plain or magic-eye-happy-stars. First, we assume the two decks are distinguishable. Interestingly, the eight-suit version can be easily realised with two decks of physical playing cards. If you prefer Tarot Cards to happy stars, then your eight suits would be Clubs, Diamonds, Hearts, Spades, Cups, Pentacles, Wands and Swords. Obviously, having more than four suits would imply having to invent new suits such as Happy Stars in addition to Clubs, Diamonds, Hearts and Spades. I am aware that three suit versions exist. Anyways, you get the idea.īart pointed out the holy grail of eight-suit spider has not been implemented in actual software. One can even use vectors like to represent a partially complete standard four-suited game with one suit moved to the foundations. Continuing our examples, the standard four-suit difficulty can be represented as and the standard two-suit and one-suit difficulties are and respectively. Assuming we have 104 cards, then a game is possible if we have a vector of positive numbers whose sum is 8. In mathematical terms, Bart is proposing a game which means there are five suits A,B,C,D,E where there are two of A,B,C plus one of D,E. But we still have 104 cards which is not a multiple of 5 so something’s gotta give. As the name implies, this variant makes the game challenging by adding a fifth suit. Obviously not now with an important game of Bingo Solitaire in progress, but food for thought if you pardon the cliché. Bart suggested I try a variant of Five-Suit Spider some time. ![]()
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