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the key from the information accumulated to this point. For the remainder of this section, we assume that in the primary phase we uniquely determined T ( 0 ) ;if not, we simply repeat this secondary phase for each candidate T ( 0 ) . In the chosen plaintext attack, discussed above, we recovered T , without finding the key. In contrast, in the secondary phase of the known plaintext attack presented here, we recover the key, using the A table obtained in the primary phase. From (4.7) we have
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T ( z )= c[(s(z)K6) @
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is an element of the Cave Table. Consequently,
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for some z1 in the Cave Table. We select a putative ( K GK7), and a particu, lar z. Then we test each z1 that is in the Cave Table, by computing
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and looking up the value of Ax,u.If for any z we find that every z1 in the Cave Table yields Ax,v = 0, then the choice of ( K G , K must be incorrect. This ~) will reduce the number of possible partial keys ( K 6 ,K7), with the number of survivors depending on the amount of information available in A, which, in turn, depends on the number of known plaintexts. Typical results for this secondary phase of the SCMEA attack are given in Table 4.9. Table 4.9: Number of Surviving Known plaintext blocks Partial keys (K6,K7)
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For each putative ( K G , K ~ ) , can determine putative ( K 4 , K s )values we from the pair of equations
S(z)= C [ ( R ( Z@ K4) ) T ( z ) c[(s(Z) = @ K6)
+ K5] + + K7] + z.
To accomplish this, for each candidate (K4,Ks), compute we
K7) and we use the putative pair ( K G , to find
We are now in precisely the same position as discussed above, except that here we rule out ( K 4 ,K s ) key pairs instead of ( K GK7) pairs. , Suppose that we find a total of n pairs (Kfi,K7).Then for each of these we would expect to find about the same number of ( K 4 , K s )pairs, since we are relying on the same A table in both cases. The attack can be extended to find putative (K2,K s ) and putative ( K OK l ) ,which enables us to obtain pu, tative keys ( K OK I ,. . . , K7). The expected number of such keys is about n4, , where n is the number of ( K 6 , K 7 )pairs. Of course, n depends on the number of known plaintext blocks available. From Table 4.9 we see that if we have 150 known plaintext blocks available, then we can expect to recover 16 putative keys. This clearly shows that the SCMEA cipher is extremely weak, and in the next section we show that these results are easily extended to CMEA-although the known plaintext requirement increases significantly. However, 150 known plaintext bytes niay be unrealistic in practice. With just 75 known plaintext blocks available, the number of putative keys would be almost 244. While this is a significant improvement over an exhaustive search, where there are 264 possible keys, it is worth considering whether we can do better, particularly since the equivalent CMEA attack will require more known plaintext. We now discuss an alternative approach to the secondary phase of this attack. This alternative is slightly more complex, but it results in a lower known plaintext requirement.
Secondary Phase: Meet-in-the-Middle
An alternative way to complete the secondary phase is a meet-in-the-middle attack, as discussed in 11521. For this attack to be practical, we must have determined T ( j )for at least four distinct values of j during the primary phase. This would be indicated by four rows of the A table that each contain a single one. The expected number of such rows depends on the number of known plaintext blocks available- typical numbers appear in Table 4.10. From these tabulated results, we see that this attack will be possible provided somewhat more than 50 known plaintext blocks are available. However, the attack can also be used if wf: do not have four uniquely determined values, provided that we have at least four rows of A , each of which has a small number of possible T ( j ) .