Rules for Sudoku
Fit the 10 pentomino pieces in the two 5x5 grids without overlapping. Pieces can be mirrored and rotated. Then solve the resulting Jigsaw Sudokus.
Tuesday, 20 November 2012
Puzzle #140: Broken Sudoku
At first I had tried to find a subset of 10 different pentominoes to create two 5x5 squares, but I wasn't able to do it. I'm not sure if it is even possible. It wouldn't be the first time I've tried to do something which wasn't possible at all. I think though that repeated shapes make it a bit more interesting as that gives more options for the 2 puzzles. The puzzle isn't too hard, but you have to remember that shapes can be rotated and mirrored, so you can't be sure of the placement of some numbers as certain pieces have symmetry. For 5x5 Sudokus I think they're pretty nice this way.
Rules for Sudoku
Fit the 10 pentomino pieces in the two 5x5 grids without overlapping. Pieces can be mirrored and rotated. Then solve the resulting Jigsaw Sudokus.
Rules for Sudoku
Fit the 10 pentomino pieces in the two 5x5 grids without overlapping. Pieces can be mirrored and rotated. Then solve the resulting Jigsaw Sudokus.
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Much as I hate to be annoying, but after a little doodling I have an example of two 5x5 squares tiled by 10 different pentominoes. Up to various symmetries, I'm fairly sure it's a unique solution for this set of 10. Sadly 12 choose 10 is quite a big number so I'm not going to make any sweeping statements.
ReplyDeleteMaybe I should refrain from giving any more details immediately in case you want to find this (or maybe another!) solution yourself!
That's okay. Pentomino packing has always been a really weak thing for me. I think that's why I like pentomino genres where you pack them through other clues (like sum or A-E) as I'm then able to do it much easier.
DeleteI'm going to go ahead and make a sweeping statement, and say that, ignoring rotation, reflection and square order,there are 12 solutions. Assuming I reduced symmetries properly that is.
DeleteAll solvable subsets require the use of FILPTUVY, though there doesn't appear to be one which resolves uniquely, unless you're willing to additionally ignore the I pentomino, and only consider the remaining 5*4 rectangle.
Oh, and 66 possible piece combinations isn't particularly terrifying (though certainly far more than I'd like to manually check for the sake of a comment).