ENIGMA in .NET

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2.2 ENIGMA
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let Pi be the permutation at step i, that is, Pi is the permutation determined by the composition of the three rotors, followed by the reflector, followed by the three rotors-in the opposite direction--at step i. Using the notation in (2.1), we obtain
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s-~R,~R,~R~~TR~R,R,s,
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where, to simplify the notation, we ignore the dependence of Re, R,, and R,r on i. Since Pi is a permutation, its inverse, exists. Due to the rotation of the rotors, the permutation varies with each letter typed. Consequently, Pi does indeed depend on i. The Enigma attack we present here exploit,s cycles that occur in the known plaintext and corresponding ciphertext. Consider, for example, the column labeled 8 in Table 2.2. The plaintext letter A passes through the stecker, then through P and, finally, through S- to yield the ciphertext M, g that is, S - l P g s ( A ) = M which we can rewrite as P g S ( A ) = S(M). Then from the known plaintext in Table 2.2, we have
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which can be combined to yield the cycle
Suppose that we select one of the possible initial settings for the machine, neglecting the stecker. Then all P and PtF1 that correspond to this setting i are known. Now suppose that we guess, say, S(E) = G , that is, we guess that E and G are connected by a cable in the stecker plugboard. If it is actually the case that the stecker has a wire connecting E and G, and if our guess for the initial settings of the machine is correct, then from (2.2) we must have
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If we try all 26 choices for S ( E ) and (2.2) is never satisfied, then we know that our guess for the rotor settings is incorrect and we can eliminate this choice. We would like to use this observation to reduce the number or rotor settings, ideally, to just one. However, if we find any guess for S(E) for which (2.2) holds, then we cannot rule out the current rotor settings. Unfortunately, there are 26 possible guesses for S ( E ) and for each, there is a 1/26 chance that (2.2) holds at random. Consequently, we obtain no reduction in the number of possible keys from this one cycle.
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Fortunately, all is not lost. We can easily find an additional cycle involving S ( E ) , which can then be used in combination with (2.2) to reduce the number of possible rotor settings. For example, we can combine the four equations
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S ( E ) = P3S(R)
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s(w) P 1 4 S ( R ) = s ( W ) = P7S(M)
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S ( E ) = P6;S(M)
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S ( E ) = P&'P7PC1S(E).
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Now if we guess, say, S ( E ) = G, we have two equations that must hold if this guess is correct. There are still 26 choices for S ( E ) , but with two cycles, there is only a (1/26)2 charice that they both hold at random. Therefore, with two cycles in S ( E ) , we can reduce the number of viable machine settings (that is, keys) by a factor of 26. We can easily develop an attack based on this observation. Using only two cycles, the attack is outlined in Table 2.3. However, several additional cycles would be required to uniquely determine the key. Table 2.3: Enigma Attack Given: Cycles Co and C1 for S ( E ) (Lo,L I , . . . , L25)= (A, B, . . . , Z) f o r each rotor setting Conipute required permutations to test Co and C1 f o r j = 0 to 25 S ( E ) = L, i f Co and CI hold then save putative rotor settings arid S ( E ) value L,
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