Episode #125 of the Stack Overflow podcast is here. We talk Tilde Club and mechanical keyboards. Listen now

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Proof with brute-force procedure. Here I used fact that we for sure can cover with 16 pentominoes, so I tried to cover half with 8 or less and then see if two such half-covers cover the whole board. It takes about 15 seconds on my PC to get the answer. #include <iostream> #include <vector> const int kHalfUpperBound = 8; const int kSide = 8; ...


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I can prove that the answer is exactly Several people, including Jaap Scherphuis, have shown that the square can be covered with this many pentominoes, so it only remains to show that at least this many pentominoes are needed. (A matching lower bound). Let us start with the magic board given by A. Rex: As a first lower bound, However, Therefore, we can ...


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I was going to post a different answer, only to realize it was the same as Magma's. So I had to find a new one:


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Note: Answer:


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Partial answer, Tasks 2-4 Finished Task 2 Finished Task 3 (what a doozy) Finished Task 4 (easier than 3)


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Partial Answer Finished Task 1


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Here's another proof of the lower bound in Sriotchilism O'Zaic's answer.


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For completeness, all the solutions, excluding rotations and reflections.


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People have given some good upper bounds, how about a lower bound. However we can improve this ... However we can improve this ...


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Answer: Reasoning:


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This is a solution with 15 pentominos. Here is a proof that this is optimal. Consider the top and bottom rows, left and right columns, which form the border of the area to cover. Most of the time a W-pentomino covers one or two cells of the border. There are only two ways for a piece to cover more than two border cells. These occur only at the corner of ...


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Here is a solution with 16 pentominos:


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The X-pentomino tiles the plane, so that tiling is a good way to start. There are two ways to cut an 8x8 region out of that tiling. If one of the 4 central squares of the 8x8 region has an X centred on it, you get this or else you get this The latter can be easily improved by replacing the ones at the edges to give this A different way to get the same ...


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I’m thinking: The solution:


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And here are a few 2x14 solutions:


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Simlar to Dr Xorile, I think there are many solutions. Here are several:


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Here's 2 fundamentally different solutions: I suspect there are many because these are literally the first two things I tried as I was playing around and in both cases I was able to just shove the pieces in and get a solution. I did it in the numerical order shown.


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Sooo much NSFW! But thanks anyway :) Task 1 Task 2 Task 3 Task 4 - not solved yet.


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Like @Omega Krypton I have solved one part, Task 2:


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Partial Answer - finished Task 1


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