WORLD WARII CIPHERS

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Table 2.5: Matched Plaintext and Ciphertext

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Plaintext Ciphertext Plaintext Ciphertext Plaintext Ciphertext

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0 1 2 3 4 5 6 7 8 P E A R L H A R B J K F H N V P G P T 0 R A T 0 R A T K P L T H D W V J M I D W A Y I S L X H T E S A G N K

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91011 0 R X G P Y 0 R A L 0 P A N D 0 L I

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Furthermore, suppose the messages in Table 2.5 form a 3-legged depth. Then the permutation at each position of the three messages is the same. Let Pi be the ith permutation generated by this cipher for this particular key. Then we have some information on the first several permutations. For example, we know that Po maps plaintext P to ciphertext J , plaintext T to ciphertext K and plaintext M to ciphertext X. Also, PI maps E to K and 0 to P, and I to H, and so on. In this way, we can partially reconstruct the permutations, and the more legs of depth that are available, the more information on each permutation we obtain. While it was clear to Rowlett and his team that Purple was a substitution cipher, it was unclear how the permutations were generated. The cryptanalysts had knowledge of previous Japanese cipher machines, as well as knowledge of other cryptographic devices of the time, including rotor machines. The knowledge of an earlier Japanese cipher known as Red proved most valuable. The Red cipher employed an unusual split of the alphabet into sixes and twenties, which was carried over into Purple. Initially, the Red cipher split the alphabet into the six vowels, AEIOUY, and the remaining twenty consonant. Substituting vowels for vowels and consonants for consonants was also used in some other ciphers of the time. This split reduced the cost of cabling the ciphertext messages, since the resulting messages were considered pronounceable ---even though the ciphertext was gibberish--and therefore were charged a lower rate than messages consisting of random letters [52]. But encrypting vowels to vowels is a serious weakness, since some niessages can be inferred simply based on the placement of vowels within the ciphertext. The Japa,nese apparently realized this was a weakness and usage of the Red cipher was modified so that any six letters could act as the sixes. This is precisely the same situation as with Purple. The ciphertext sixes would be expected to each occur with the average probability of the corresponding plaintext sixes, and similarly, each ciphertext twenties let,ter would occur with t,he average probability of the plaintext

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2.3 PURPLE

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twenties. For most random selections of six letters from the alphabet, the selected letters will occur at either a higher or lower average frequency than the remaining letters (see Problems 14 and 15). For example, if the plaintext letter E is among the sixes, and the other sixes are all letters of average frequency, then the expected frequency of each of the ciphertext sixes will be higher than the expected frequency of each of the ciphertext twenties. Therefore, the ciphertext sixes can usually be determined simply from a frequency count of individual ciphertext letters. That is, the six highest-frequency or six lowest-frequency letters are most likely the sixes. Once the sixes have been isolated, it is relatively easy to determine the sixes permutations from a small amount of known plaintext, since in Purple there are only 25 sixes permutations. Using their knowledge of the Red cipher, Rowlett s team was quick to realize that Purple also employed a 6-20 split. They were then able to reconstruct the sixes permutations. But the twenties proved far more difficult to crack. The output permutations generated by a switch-based cipher, such as Purple, have certain identifiable characteristics. For example, consider the permutations in Table 2.6, which were generated by a process analogous to that used by Purple to generate its twenties permutation. In this example, there are three banks of permutations, as with the twenties permutations in Purple, but here each of the switches contains just three permutations (as opposed to 25 for Purple), giving a cycle length of 33 = 27. Table 2.6: Successive Permutations

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Consider permutations P and Ps in Table 2.6. The element in position 0 4 is 4, while the 4 is in position 2 of P5. That is, the first element

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4 3 3 3 4 4 4 2 2 5

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