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This is a code from an old home computer magazine. The cipher used for it is still unknown, although they gave the solution:

145 211 145 185 255 186 112 88
183 174 224 34 145 126 226 178
51 207 191 129 188 234 4 191
199 175 178 243 197 16 118 43
210 198 166 241 237 194 211 94
213 171 252 246 233 178 12 218
210 203 172 129 133 219 23 186
206 170 203 141 126 246 117 203
190 250 212 206 22 160 197 161
182 183 246 20 53 141

The solution is:

"Behind every good programmer there stands an Atari computer. Congratulations!"

What cipher was used?

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    $\begingroup$ Hmm.. weird that the cipher text is one number longer than the number of characters in the solution (or one shorter if you count the "s)... $\endgroup$
    – Alconja
    Commented Dec 13, 2017 at 2:18
  • $\begingroup$ yes, there is big chance that had a typo, but maybe still a cipher could be identified $\endgroup$
    – devwebcl
    Commented Dec 13, 2017 at 2:20
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    $\begingroup$ A quick search turned up this image of the original contest, which has a little more information (found it on an atariage.com thread asking a similar question) if it's useful for anyone trying this $\endgroup$
    – user40308
    Commented Dec 13, 2017 at 8:55
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    $\begingroup$ 211-145 or 211^145 is B. $\endgroup$
    – Carl Löndahl
    Commented Mar 8, 2018 at 22:09
  • $\begingroup$ Can I confirm - is the solution that you printed in the exact case that it was given in the magazine? Obviously upper case / lower case characters have different ASCII values. $\endgroup$
    – Astralbee
    Commented Mar 27, 2018 at 13:29

1 Answer 1

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Some thoughts, hope this helps:

The cipher is 1 character longer than the plaintext. This could suggest one of these (also other possibilities I didn't think of):

  • The cipher relies on some state, that you initialize and then update each time a character is produced, for example start with $st=145$ and then at each step you read $c$, output $st-c$ (works for 1st letter "B"), and update $st$ (tried a few things but didn't find something coherent for the second letter.
  • The cipher is a block cipher, and they needed to pad the message to reach the block size. $78=2*3*13$ so the block size can be one of $2,3,6,13,26,39,78$.
    I tried with size $2$, hence trying to decode "145 211" into "Be", it looks like "B" is $211-145$ and "e" is $145+211$ (everything done mod 255), but this doesn't give anything for second couple of chars.

After a bit more analysis, the frequencies of couples in the cipher and plaintext do not match. They could match for blocksize 3, so i tried to find a matrix that would map the first 9 plaintext chars to first 9 ciphertext chars, but can't conclude because the system is not invertible

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