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| jdege | May 9 2008, 07:54 PM |
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Have you read: http://z13.invisionfree.com/Crypto/index.php?showtopic=456? You split the ciphertext into n slices, where n is your provisional keylength. So if you're testing to see if the keylength is 3, you split the ciphertext into three slices, where slice 0 has letters 0,3,6,9,12,...slice 1 has letters 1,4,7,10,..., and slice 2 has letters 2,5,8,11,... The for each slice we do a frequency count, and then calculate the Incidence of Coincidence of that slice, and then we average the IC's for the n slices. You need two frequency counts, one of slice you're testing, and one of a piece of standard English text that you're using as a reference. Suppose you have F1: {A=3, B=1, C=1, D=1, E=12, F=3,...} and F2: {A=8, B=1, C=3, D=0, E=21, F=8,...}. Where the sum of the counts for F1 is N1: 105, and the sum of the counts for F2 is N2: 306. You multiple the count for A in F1 by the count for A in F2, and add to it the product of the count for B in F1 and B in F2, etc. Then divide by the product of the sums.
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| When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl. | |
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| Cracking A Vig With Ic · General | |
11:47 AM Nov 25
