1.1 KiB
Pretty cool problem, I have to say.
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 #.
This is the "amount of resulting numbers" parameter of our program. For this Euler-problem you need to set this to 6.
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.
(But then you need to move the code down a bit - otherwise the cache-part will intersect with the code-part).
With this number we first generate all the polygon-numbers from 3 to SIDES_COUNT + 2 with the formula (n-2) * (n^2 - n)/2 + n.
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).
A last note: This was (until now) probably the hardest program to fit in the befunge 80x25 size restriction. I barely managed to get it (and now its exactly 80x25) and I had to use quite some trickery.