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for all j = 2, , J - 1 and k = 2, , K - 1.

Secondary Phase Refinement To reduce the secondary work factor requires some amount of known plaintext. Here, we only outline the plan of attack, leaving the details as a challenge problem. Thv irit,eraction of the control rotors and the index rotors is illustrated in Figure 2.22. where we have collapscd the five control rotors into a single permutation (denoted as control ) and, similarly, the five index rotors are corisidcred as a single permutation (denoted as index ). The four inputs

to the control permutation, F, G, H, and I, are activated at each step. This results in four active outputs, which are combined as indicated before being fed into the index permutation. At least one-and at most four-inputs to the index permutation will be active. The outputs from the index permutation are combined in pairs, as indicated, and these determine which of the cipher rotors, COthrough C4, step. At least one cipher rotor will step, and at niost four will step.

8.24 As you move your mouse over your Web Journal, group boxes for each image or text group appear with extra controls.

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Figure 2.22: Control and index permutations. In Figure 2.22, the control permutation changes with each letter typed, but the index permutation is fixed for the entire message. Since the control permutations are changing, we model their output as uniformly random, that is, we assume that each of the combinations of output letters is equally likely at each step. Then, due to the way that the control rotor outputs are ORed together, the inputs to the index permutation are not uniform. For example, input 8 will be active much more often than inputs 1, 2, or 9 and input 0 is never active. The outputs of the index rotors are ORed in pairs, and the results determine the cipher rotor stepping. Therefore, if we have sufficient information on the frequency of the stepping of individual cipher rotors-which is available from known plaintext using the putative cipher rotor settings obtained in the primary phase of the attack-we can assign probabilities to the index permutations. For example, suppose that the index permutation is 5479381026, that is, input 0 is mapped to output 5, input 1 is mapped to output 4, and so on. Then by considering the pairs of outputs that determine the cipher rotor stepping, we can see that some cipher rotors will step more often than others. The data in Table 2.9 makes this more precise. For example, cipher rotor C d steps if either output 1 or 2 (or both) of the index permutation is active. For the index permutation in Table 2.9, inputs 6 and 8 are mapped to outputs 1 and 2, respectively. Inputs 6 and 8 of the index permutation correspond to outputs 6 and 8 of the current control permutation, and at

The standard training algorithm for SOMs is stochastic, where codebook vectors are updated after each pattern is presented to the network. For each neuron, the associated codebook vector is updated as Wkj (t + l)= wkj (t) + hmn,kj (t) [zp - wkj (t)] (4.19)

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least one of these outputs is active if one or more of the 10 letters L, M, N, 0, U, V, W, X, Y, or Z, which are connected t o these outputs, are active. On the other hand, CZ steps only when output A of the control permutation is active. As a result, C will step much more often than Cz. 4 Table 2.9: Index Permutation 5479381026 Cipher Rotor

(1,2) (6,8) 10

where mn is the row and column index of the winning neuron. The winning neuron is found by computing the Euclidean distance from each codebook vector to the input vector, and selecting the neuron closest to the input vector. That is,

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