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| jdege | May 25 2008, 02:27 PM |
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Another one that turned out to be not very difficult - monome/dinome. In this, you need a permutation of the alphabet and a permutation of the digits. Use a keyword to shuffle the alphabet. Write out the keyword, followed by the rest of the alphabet, leaving out J and Z, in three rows of eight columns each. Use a keyword to shuffle the digits. Write the first two digits to the left of the second and third rows the text, write the other eight over the top of the columns. Plaintext letters in the first row are replaced by the digit above, letters in the second and third row are replaced by the digit to the left followed by the digit above. This is clearly in the same family as Donald's Spread Cipher. The trick is to determine which are the two row digits. Once these are identified, you can convert this into a Pat. In the above case, replace 70 with A, 71 with B, 72 with C, 74 with D, etc. Skip 73, because 3 is your other row digit. Then replace 30 with J, 31 with K, etc. And when you're done with that, replace each remaining digit with a letter. Now you have a simple substitution cipher. The question is how to identify the row digits. Two things we know - they should be fairly frequent, and they should never appear in pairs or together. Frequent because they are each involved in a third of the letters in the alphabet, never in pairs because they don't appear as column digits. So start by identifying the paired digits. 11, 33, 66. Eliminate these. They cannot be the row digits. Then do a frequency count of the digits. Look at the high-frequency digits that never appear paired. Check each possible pair to see if they ever appear adjacent to each other. If they do, they can't be the row digits. The highest frequency digits that never appear doubled and never appear adjacent to each other are very likely the row digits. Use them to convert the ciphertext to a Pat, and see if the result looks like a monoalphabetic substitution. Worst case there are only a hundred possibilities. In the problem I was trying to solve, there were four digits that appeared doubled, which cuts it down to 36 possibles. Look at them in order of descending frequency. In my case, the highest-frequency pair of remaining letters never appeared adjacent to each other, so I did a test conversion to letters. The result contained a match for the crib, so I attacked it, and found the solution. (If you don't have a crib, you should be able to judge whether a test conversion is correct using the IC.) What's fascinating to me is how many of these ciphers can be solved by converting them to Pats. |
| When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl. | |
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| Sometimes Things Are Easier Than They Look · General | |
3:40 PM Nov 26
