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November 20, 2003Challenge Five SolutionWell done to everyone who tried Challenge Five. Our German friends were devious in their choice of cipher, but not devious enough to foil Aunt Aggie, nor to fool many of you. It was a transposition cipher, based on the keyword STONE. So the text was arranged in five columns headed STONE and the columns rearranged to give ENOST. To decipher it you needed to rearrange them back again. The first clue was that the number of characters was 645=5x3x43. Hence the keyword must have length 3, 5, 15, 43, 129 or 215. While a 43, 129 or 215 character keyphrase is possible it is not likely, and 3 characters is not enough. In this case a five letter keyword has been used. Fortunately in this case the text was already grouped in 5 letter blocks making it easy to experiment with the rearrangements. There are precisely 5!=5x4x3x2x1 possible rearrangements of the columns, so if you know German a brute force attack would work, trying each of the 120 possibilities in turn. A better approach is to look for a crib, a word or part of a word you expect to see. Aunt Aggie was looking for SUEVIC, and it wasn't too hard to see the block CUVIE suggesting we should rearrange it as UEVIC. This cracks the code. Posted by harry at November 20, 2003 11:45 AM |
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