Anonymous
timer Asked: May 6th, 2020

Question Description

linear congruential pseudo-random number generator (LCG). An LCG generates successive random

numbers r1,r2,r3,... by picking three secret numbers A,B,N, and using the formula

rk+1 ≡(A·rk +B)modN. (1)

The LCG must be given an initial value r0, called a seed. The sender and the receiver share the values A, B, N, r0. Encryption and decryption work as follows:

  • Given plaintext blocks t1,t2,t3,..., the sender generates the LCG keystream r1,r2,r3,..., and obtains the corresponding ciphertexts c1, c2, c3, . . . by xor’ing them, as in Table 1.
  • Similarly, the receiver uses A,B,N,r0 to generate the same keystream r1,r2,r3,... and re- covers the plaintexts t1,t2,t3,... from c1,c2,c3,....The security of the Vernam cipher depends on keeping A,B,N secret. Otherwise, the following attack is possible:
    • If the adversary knows any of the plaintexts ti, he can immediately recover the keystream value ri from the ciphertext ci.
    • Hecangeneratealllatervaluesri+1,ri+2,ri+3,...,andrecovertheplaintextsti+1,ti+2,ti+3,... from the ciphertexts ci+1, ci+2, ci+3, . . .There are three unknowns A,B,N in the linear congruence represented in Equation 1. In the domain of linear equations, we could solve for three unknowns, given a set of three independent linear equations. We can break the congruence using a similar strategy, given a sufficient number of the ri. We can get these ri from the ciphertexts, if we know the corresponding plaintexts.Your ChallengeYour task is to break the given Vernam cipher using this strategy. You are given a ciphertext, which was an e-mail message intercepted at about 1:15pm on April 19, 2000. You know the following:
    1. The keystream comes from a LCG of the form ri+1 =(ri∗A+B)modN.
    2. The ciphertext values are generated as (ti + ri) mod N. This is not an xor, but regularaddition.
    Break the cipher. Show in detail the steps that you followed.Hints: The blanks in the encrypted version below have been inserted for clarity, and are not part of the original message. Each character is printed as a decimal integer after encryption. For your convenience, the ASCII character codes are given.2

The ciphertext is as follows:

140 91 200 218 25 59 219 201 172 20 42 223 32 13 135 110 12 113 184 191 36 115 41 187 127 216 141 61 168 230 40 163 195 46 17 13 51 45 208 137 113 103 162 75 237 48 77 160 44 104 79 179 88 108 149 130 120 151 1 147 99 94 238 129 215 175 166 103 160 174 155 16 6 1 215 122 41 170 240 128 101 113 246 133 233 146 17 234 63 164 120 207 248 41 89 249 45 200 7 188 153 65 58 8 87 42 81 197 164 4 223 127 237 67 210 22 99 242 106 15 198 10 0 6 88 146 186 37 32 124 29 210 18 140 246 232 222 112 162 254 125 226 62 37 117 86 35 32 112 77 158 24 13 203 86 86 181 190 115 244 216 70 141 152 48 131 197 30 222 49 243 91 130 229 75 55 171 104 81 209 116 52 240 255 14 14 173 152 233 86 244 14 195 244 236 62 246 240 250 159 149 162 210 45 63 151 94 10 66 53 98 18 165 223 151 221 76 202 58 9 220 101 83 157 124 20 215 28 80 143 7 127 57 87

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