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This is consistent with shifting the previous value of H ( X ) one position and filling in the new bit with 1. Therefore, Trudy retains this fill and tries to extend it further at the next iteration. For any initial guess ( H ( A ) H ( B ) )Trudy can solve for a consistent value , of H ( X ) . Consequently, Trudy can only discover whether any guess was correct or not when she tries to extend the fills beyond the first byte. And it is possible that that some false positives will occur, that is, some fills will be consistent with the keystream for a few steps before failing. We carefully analyze these probabilities below. Note that, in effect, the attack we have described performs a breadth-first search. However, a depth-first search works equally well. The attack algorithm is outlined in Table 3.7. This algorithm must be repeated for each of the 2 guesses for the initial 16 bits of ( H ( A ) H ( B ) ) . , Recall that e ( A , j ) is our notation from Table 3.5 for the j t h extension of register A. Once an iteration of the attack in Table 3.7 returns a solution, there is no need to continue searching, provided that a sufficient number of keystream bytes are provided to uniquely determine the key. Below, we show that with just six keystream bytes we only expect one surviving set of initial fills, and

Note that the width of the con dence interval for Y 0 x0 is a function of the value speci ed for x0. The interval width is a minimum for x0 x and widens as 0 x0 x 0 increases. EXAMPLE 11-5 We will construct a 95% con dence interval about the mean response for the data in Example 11-1. The tted model is Y 0 x0 74.283 14.947x0, and the 95% con dence interval on Y 0 x0 is found from Equation 11-31 as

1.18 c

// //

1 20

74.283 B

1.18 c

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Given: keystream bytes ko, k l , k 2 , . . . , kiv, table L , and a guess for initial ( H ( A ) , ( B ) ) H H ( X )= (ko - L [ H ( A )- L [ H ( B ) ](mod 256) ] ) for i = 1 to N // for each keystream byte for each (X, , B ) // putative fill A TO extend X with 0 = TI = extend X with 1 for j = 0 to 11 // for each possible extension T x = (ki - L[H(e(A,j))l L [ H ( e ( B , j ) )(mod 256) l) if Tx == H(T0)then save (To,e(A,j),( B , j ) ) next iteration e for end if if TX== H(T1)then , , for save ( T Ie ( A , j )e ( B , j ) ) next iteration end if next j next putative fill next i

1 20

99 Oxygen purity y (%)

after 25 bytes we expect to have determined all of the bits of the (shifted) initial fills. Finally, we analyze the performance of this attack. For any initial choice of H ( A )and H ( B ) , we can use ko to solve for a consistent value of H ( X ) . This implies that with just a single keystream byte available, we would obtain 65,536 consistent fills. In other words, the first keystream byte yields no reduction in the number of potential fills. However, if we have ko and k l , then for each of the 65,536 fills (X, A,B ) obtained in the first step, the pair ( A , ) B can be extended in 12 different ways, and for each of these, the implied extension of X is computed. Each valid extension must match in the seven rightmost bits of the shifted H ( X ) . Consequently, on average, only one in 128 of the extensions will survive, assuming we can model the byte comparisons as random. Since L is a permutation it is reasonable t o model the computed value as a random selection from {0,1,2,. . . ,255). The bottom line is that using only ko and k l , the expected number of valid fills remaining is 12.65,536 = 6144. 128

Figure 11-7 Scatter diagram of oxygen purity data from Example 11-1 with tted regression line and 95 percent con dence limits on Y 0 x0 .

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