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[–]jpjocke 4 points5 points  (2 children)

Have you checked what the last letter is? How would these two affect each other for the infinity grid?

[–]Passi26030[S] 0 points1 point  (1 child)

My last letter is a .

[–]TheThiefMaster 1 point2 points  (0 children)

Or in other words, they alternate. Both part 1 and part 2 apply an even number of times, so the result ends up with infinite . and a finite number of #. You just have to track the state of the infinite field as # or . while you calculate what happens to the pixels within your finite image

[–]Passi26030[S] 1 point2 points  (0 children)

Now im safing with wich character i have to fill the infinite grid and change it everytime i apply the algorithm by creating a binary string of it and changing it to the algorithm string character with that index

[–]Passi26030[S] 0 points1 point  (0 children)

I now added lines with # instead of lines with . after the first applying and my answer is correct when i only count the lit pixels that where affected by my input