## Security Corner

Apr 28 2010   1:38AM GMT

# Security Fun: Homophonic Cipher Security Challenge

Profile: Ken Harthun

Time to lighten up a bit. Well, some people may consider this anything but light; however, real Geeks enjoy this kind of thing. Consider this homophonic ciphertext:

32 25 00 75 67 94 63 57 96 43 73 90 91 97 90 45 92 52 00 34 24 42 78 17 92 19 04 97 65 16 06 57 64 04 92 81 05 63 69 65 99 27 05 38 65 07 91 83 62 41 83 95 23 55 29 96 96 54 83 43 39 07 63 06 65 17 83 89 90 63 26 79 51 46 30 52 07 63 88 59 07 66 17 65 57 27

Here’s the challenge: Decipher the message above and post the resulting plaintext in the comments. Everyone who gets it right wins free lifetime access to my Geek Toolkit (\$37 value). Hint: There are tools on the web that will decipher this if you know where to look.

If you don’t know what a homophonic cipher is, better do some homework. First, it’s a simple substitution cipher in which plaintext letters map to more than one ciphertext symbol. Typically, the highest-frequency plaintext symbols–such as the letter e in the English language–are given more equivalents than letters that appear less frequently. This makes it much more difficult to use frequency analysis to break the cipher. But this isn’t always the case. Take Poe‘s “The Gold-Bug” for a literary example where a whole set of symbols was invented to describe the location of buried treasure. See this Wikipedia entry for more information.

This won’t help you much, but here is the message “The quick brown fox jumps over the lazy dogz.” I put the extra z there because it’s the least-used letter in the English language and has only one substitution in a homophonic cipher, so that’s a clue. But e has 12 substitutions, so you won’t find that one. Here’s the ciphertext: 17 68 82 94 63 70 13 04 48 29 54 60 59 31 72 28 15 63 27 95 96 90 34 14 77 30 50 24 26 33 02 52 03 54 06 02.

Have fun!