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Optimizing the O(n^3) Gaussian Elimination e.g. by grouping booleans to integers took me to level ~400. Then I redesigned my algorithm. Now it has less than O(n^2) complexity and solves even the last levels in less than 20 seconds using Java without any optimization at all. Algorithm beats implement...

What!?! After weeks of hard thinking and headache I finally came up with the c-$ idea today and passed the challenge with a 94 instructions solution and now I'm supposed to do it in 81?!? How can one ever reach this

Of course the angle is different. Only comparing the pixels won't do it.

I didn't bother reading the words but only compared the pictures to each other.

I had the same problem. Took me some time until I finally tried the english google.

I used principal component analysis to calculate the rotation and didn't reduce the size of the images but sorted them according to the sum of all pixel's color values and only correlated each picture with 100 neighbors. I expected that too many similar images correlate well and already planned what...

The linked machine seems to have a strange carry over behavior. When all three rotors are rotated, this takes place in two steps and the second one is rotated twice. Is this a bug in the swf or did I understand something wrong and it is supposed to be like that?

I pushed x zeros and y ones on the stack, then took the x-th value from the stack and returned the corresponding number x or y. This worked without arithmetic operations or extensive memory but only for numbers between 0 and a certain limit and the program size was about four times that limit.

The first breakout could be solved with the level codes that are printed out when BreakoutGame.class is run. But for Breakout Extreme I also edited the bytecode with a HexEditor cause I didn't manage to save my changes with a bytecode editor (was the code too ugly?)

BWT is the first step in the compression algorithm of bzip2. It sorts the letters such that those appearing in a similar surrounding come together. E.g. a 'q' is always followed by a 'u', so the q's will appear only within a certain region of the BWT-sequence. Therefore this sequence is easier to co...

I suppose, 34 defines the beginning of the original text. Since the BWT ist rotation invariant (abcde has the same bwt-result as e.g. bcdea or deabc), you cannot otherwise know where the text starts in the backtransformed sequence. But I thought that this should be the position of the last (or the f...

This site was recommended in the most popular German computer magazine c't about one month ago. Therefore many Germans (me too) joined this forum within the last weeks and started solving challenges. And the magazine said that everyone who solved more than 50 challenges should apply for a job there :)

I have to admit that I cheated at this challenge... only checked memory cells 0 to 9 for the maximum and submitted until I passed the five tests :-*) The deterministic version of my program has size 28. Does anybody have a 'real' solution?