hacker.org

Hi,

did you really have access to a system/370?
I considered using an emulator (hercules), but finally managed to convert the floats to their binary representation and that to EBCDIC. Very nice and highly informative challenge!

My solution: 8c8c67*C1c67*1+C528**C8c9c67*C4c67*1+C528**C8c9c67*C4c67*1+C528**C8c4c67*C528**C 9c9c67*C8c67*1+C528**C9c8c67*C7c67*1+C528**C9c9c67*C4c67*1+C528**C9c9c67*C1c67*1+C528**C 8c4c67*C528**C8c8c67*C2c67*1+C528**C8c8c67*C1c67*1+C528**C9c9c67*C2c67*1+C528**C8c8c67*C 5c67*1+C528**C Far from ligh...

Finding the CIL code rather menacing, I opted for a C# decompiler instead of a mere disassembler. "Exemplar" from the Anakrino9 package did the trick for me, after having delighted an otherwise quite useful vmware image with the dotnet framework... shudder. My formula (Unix "bc" ...

Chapeau!
Creating this challenge must have been at least as hard as solving it!

No idea about the solution, but I have to say: Nice image!

Ah! The memories... I feel young again... thanks for that challenge, man!
(And no way those youngsters will figure this out :twisted::twisted:)

Hm. I really appreciate the effort made to guide us towards solving homophone ciphers. However, I feel that the warm up is actually harder than the original challenge: - real English does have much more properties than the proposed xyzw-"plaintext"; not only groups of two letters do have t...

Actually, I deciphered the opcodes manually. Never thought about a disassembler, which would indeed have been a nice aid. Anyway, the numbers representing the BREAKs are next to the BNEs with no instruction between...

What do you mean with unusable labels?

Well, I solved the challenge some days ago (and answered the question correctly). Now the challenge is gone, at least from my perspective. Accessing the challenge url directly tells my that I am not ready for this challenge. What happened?

Hi gfoot, I guessed from the challenge name that this was a homophone cipher. I found some way of clustering the homophones, i.e. finding the numbers that denote the same letter. After that, it was a matter of simply feeding the now "normal" substitutiion cipher into a substitution cipher ...

Should not the program end at the first break? From what I read about the MIPS branch opcodes, the instruction immediately following the branch instruction (in the so called delay slot) is always executed first, no matter if the branch takes place or not; the branching takes place only thereafter. (...

Suggest you stop it.
Read the challenge again. It does not ask for the fibonacci number itself.

@deutronomium: Your approach did work just fine for me. (Well, I had perl feed expressions to fresh instances of a php interpreter. In your case, could previous eval()s have influenced the result of the current one?)

@rmplpmpl: Normally, I'd agree. In this particular case, not...

MerickOWA wrote:This was a pretty fun challenge. Its method of "encryption" is not complicated. Theres no complicated dictionary/key/ or other external information needed.

Sigh. That is what I feared most.

Anyone who would like to talk about the Blizzard challenge? I sincerely hope it is ok to mention things that do not work. For me, that includes: - interpreting the black and white pixels as a bit stream of 7-bit or 8-bit ascii characters, be it lsb or msb first, be black "1" and white &quo...