12 lines
1.1 KiB
Markdown
12 lines
1.1 KiB
Markdown
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Pretty cool problem, I have to say.
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It's one of these problem that you can easily make dynamic. In the middle-left of this program you see the number `6` surrounded with `#`.
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This is the "amount of resulting numbers" parameter of our program. For this Euler-problem you need to set this to `6`.
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But for debugging purposes I mostly tested it with `3`. And if you want you can try it out with bigger numbers `7` or `8`.
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*(But then you need to move the code down a bit - otherwise the cache-part will intersect with the code-part)*.
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With this number we first generate all the [polygon-numbers](https://en.wikipedia.org/wiki/Polygonal_number) from `3` to `SIDES_COUNT + 2` with the formula `(n-2) * (n^2 - n)/2 + n`.
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Then we go recursively through all these numbers. First we test triangle[1] with square[1], then with square[2] and so on. *(The recursion is - as always - just a stack and a stack pointer)*.
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A last note: This was *(until now)* probably the hardest program to fit in the befunge 80x25 size restriction.
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I *barely* managed to get it (and now its exactly 80x25) and I had to use quite some trickery.
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