4.5 CMEA

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and, according to (4.13), (4.14), and the solution to Problem 11, the corresponding ciphertext is

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((1CB ( T ( 0 ) 1))- T ( 0 ) v c1 = 1 - T ( T ( 0 ) 1) v

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c2 = T ( 0 ) T(((1 CE

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( T ( 0 ) 1)) 1) @ 2). v

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We therefore encrypt chosen plaintexts of the form

and any of these that satisfy

cg = (1 @

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(20 V

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1))- 2 0 =

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0 255

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and 2 0 is even and xo is odd

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are consistent with xo = T ( 0 ) .To further reduce the false alarm rate, we use the fact that if zo= T ( 0 )then 1 - ~1 and

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v 1)

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(4.15) (4.16)

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(((1@ ( 2 0 v 1)) 1) CB 2)

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must both be in the Cave Table. Any false alarms that survives these tests will be discovered quickly. After having found a putative 2 0 = T(O), choose plaintext using (4.12), we with = 0 for j = 1 , 2 , 3 , .. . ,255. For each j , we recover a putative value for xj = T ( j ) 1 from (4.13). If neither xj - j nor (xj@ 1)- j is in the Cave V Table, then we have detected a false alarm, and we discard zo and continue searching for T ( 0 ) . Assuming that 2 0 = T(O),then if only one of xj - j or (xj@ 1)- j is in the Cave Table, we have unambiguously determined T ( j ) . On the other hand, for each case where both xj - .j and (zj CB 1) - j are in the Cave Table, the low-order bit of T ( j )is ambiguous. To resolve the ambiguous low-order bit of a recovered T ( j ) , can make we use of (4.14). First, we recover all T ( j ) , j = 1 , 2 , . . . ,255, using T ( 0 )and for the method described in the previous paragraph. We also maintain an auxiliary array, A , where Aj = 0 if the low-order bit of T ( j )is known, and Aj = 1 if the low-order bit of T ( j )is ambiguous. We can use A to resolve the ambiguous cases as follows. Suppose that the low-order bit of T ( k ) is ambiguous, that is, T ( k )- k and ( T ( k )@ 1) - k are both in the Cave Table. Then we set A k = 1. Now we find a t and j such that

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k = t @ ( T ( j ) 1) and At = 0 V

(4.17)

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(that is, T ( t )is not ambiguous). If such a t and j are found, we then set po = ( t CE 1) - T ( 0 )

p1 =

P2 =

( j 6 2) 9 0

(t @ 1) - T ( t )

and encrypt this plaintext block using CMEA, to obtain the the corresponding ciphertext block (co,e l , c2). From (4.14), we have

T ( tQ ( T ( j ) 1)) = ( j @ a ) - (t CB 1) - C I , V

which. by our choice of t and j gives

T ( k ) = ( j (22)

( t CE 1) - c1.

There is no ambiguity in this equation, and consequently we have resolved the low-order bit of T ( k ) . Problem 14 explores the probability that this part of the attack will fail. Note that this part of the attack fails if for an ambiguous T ( k ) ,we cannot find t and j satisfying (4.17). Now we provide a careful analysis of the expected number of chosen plaintexts required in this attack. Each time we test an element in the Cave Table to see whether it is a possible T(O),there is a chance of a false alarm. As noted above, letting ! j = 0 in (4.12), a plaintext of the form =

(PO,Pl,lnZ)

yields

= (1 - T(O), - T(O),O) 1

0 255

if T ( 0 )is even if T ( 0 )is odd.

Another interesting and related property of CMEA encryption is considered in Problem 12. Since, on average, we require 82 iterations before we can determine T(O), the probability of false alarms can be approximated by a binomial distribution with n = 81 and p = 1/128. Therefore, the expected number of false alarms is about n p = 81/128 M 0.63. If we include a check that both (4.15) and (4.16) are in the Cave Table, then the expected number of' false alarms drops to 0.63( 164/256)' M 0.258. Recall that 164 of the 256 elenients in the Cave Table are distinct. Also, we have that T ( i )- i is in the Cave Table, for i = 0 , 1 , 2 , . . . ,255. Since T ( 0 ) is in the Cave Table, about 82 chosen plaintexts are required before we expect to find T ( 0 ) .Once T ( 0 )has been recovered, one chosen plaintext is required to determine each of the remaining values T ( i ) , i = 1 , 2 , 3 , .. . ,255. This for gives a total of 337 chosen plaintexts. However, some of the recovered T ( i )will be ambiguous in the low order bit. The low order bit of T ( i )is known if either T ( i )- i or ( T ( i ) 1) - i @