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Figure 3.8: Keystream generator. Lct Ix he the length of the cycle grrierated by the shift register X , and define l y and t z similarly. Then the keystream generated by stream cipher in Figure 3.8 has a cycle length of lcm(i x, t y , i z ) . For the registers in Figure 3.8, the cycle length are i ,y = 7, l y = 15, and l , = 31, provided that none of the

0 0 (11-5)

1 0, 1

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initial fills are all zero and, consequently, the cycle length of this keystream generator is 3255. To use the keystream generator in Figure 3.8 as a stream cipher, we require a 12-bit key, which is used to determine the initial fills of the registers, with the restriction that no initial fill can be all zero. If Trudy can recover the initial fills, then she has broken the stream cipher. Of course, in this example, Trudy can simply use an exhaustive search to recover the initial fills, but we use this simple example to illustrate a shortcut attack. Suppose that the initial fill of register X is 011, register Y is 0101 and register 2 is 11100. Then the key that Trudy wants to recover consists of these 12-bits of initial fill. For these initial fill bits, the first 31 output bits for registers X , Y , and 2 in Figure 3.8 are given in the first three rows of Table 3.3, with the keystream given in the fourth row. Table 3.3: Register Bits and Keystream Bits

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+03:00

n 0

(11-6)

Equations 11-6 are called the least squares normal equations. The solution to the normal equations results in the least squares estimators 0 and 1.

0 , 1 , 2 , . . . ,29,30

De nition The least squares estimates of the intercept and slope in the simple linear regression model are

x 1

0101100100011110101100100011110 1111001001100101100010110101011 By Kerckhoffs Principle, we assume that Trudy knows the LFSR feedback functions and the nonlinear boolean function f . Trudy s attack will take advantage of certain properties of the function f . The truth table for the boolean function f(z,y, z ) = zy @ yz @ z is given in Table 3.4. Note that f ( z , y , z ) = z and f ( z , y , z ) = z both occur with probability 314. Trudy can take advantage of this fact to efficiently recover the initial fills (that is, the key) for the keystream generator in Figure 3.8. Table 3.4: Truth Table for f ( z ,y, 2 ) = zy @ yz @ z

(11-7)

(11-8)

x 0 0 0 0 1 1 1 1

y

x 1

(11-9)

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g 0 0 1 1 0 0 1 1

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yi is called the residual. The residual describes the error in the t of the where ei yi model to the ith observation yi. Later in this chapter we will use the residuals to provide information about the adequacy of the tted model. Notationally, it is occasionally convenient to give special symbols to the numerator and denominator of Equation 11-8. Given data (x1, y1), (x2, y2), p , (xn, yn), let

x2 2

0 0 0 1 0 0 0 1

(11-10)

x2 2

0 1 0 0 0

(11-11)

EXAMPLE 11-1

We will t a simple linear regression model to the oxygen purity data in Table 11-1. The following quantities may be computed:

1 1 -

20 20

i 1 20

The attack we consider here requires known plaintext. Since we are dealing with a stream cipher, known plaintext immediately gives Trudy the corresponding bits of the keystream. Suppose that Trudy, via known plaintext, recovers the 31 keystream bits in the last row of Table 3.3. Trudy can then simply try all possible initial fills for the X register, and for each of these, she can generate the first 31 bits of the corresponding X sequence. For the correct initial fill of X she expects to find Ici = xi with probability 3/4, and for an incorrect initial fill she expects that the keystream will match the X register sequence with a probability of about l/2. Therefore, Trudy can recover the X register initial fill by simply trying all 23 possibilities and computing the correlation with the known keystream bits for each putative fill. For example, suppose that Trudy guesses the initial fill of X is 111. Then the first 31 bits of X would be 1110010111001011100101110010111.

1.1960 y

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