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Lecture 6 STREAM CIPHER SYSTEM

This lecture discusses the statistical tests applied to evaluate the security and randomness of stream cipher systems. Key tests covered include the frequency test, which checks the distribution of 0's and 1's in a bit sequence, and the serial test that examines overlapping pairs within sequences. An illustrative example demonstrates the application of the poker test to assess how well a bit sequence meets randomness criteria. Understanding these tests is fundamental for cryptanalysis and ensuring the robustness of encryption methods.

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Lecture 6 STREAM CIPHER SYSTEM

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  1. Lecture 6STREAM CIPHER SYSTEM By: NOOR DHIA AL- SHAKARCHY 2012-2013

  2. STREAM CIPHER SYSTEMS X2 = Statistical Tests: 1-frequency test: 0 < x2 < 3.84 n = length of sequence. n0= number of 0's in sequence. n1= number of 1's in sequence. X2 = 2- serial test: x2 ≤ 5.99 n = length of sequence. n0= number of 0's in sequence. n1= number of 1's in sequence. nij = numbers of ij in sequence.

  3. STREAM CIPHER SYSTEMS Example:- Test the sequence : 0101011101100011111001101001000 1- frequency test: (-1)2 1 = = 31 31 (15 - 16 )2 X2 = n = 31 n0= 15 n1= 16 31 0 < x2 < 3.84 pass n00 = 6 n01= 8 n10= 8 n11 = 8 2- serial test: 4 2 n X2 = (n002+n012+n102+n112) - (n02+n12) + 1 n-1 4 2 31 = (62+82+82+82) - (152+162) + 1 30 = 30 – 4 -31.032 +1 = 0.368 ≤ 5.99 pass

  4. STREAM CIPHER SYSTEMS Statistical Tests: 3- poker test: 2m F (xi)2 (mi)) X2 = ∑im=0 - F n m F = m i m! (m-i) !i ! F = the number of segment in sequence. m = the length oF the segment. Xi = the number of segments contain i of 1's. m-i = the number of segments contain i of 0's, (default with xi).

  5. STREAM CIPHER SYSTEMS m= 5 01010 11101 10001 11110 01101 00100 0 Example poker test: ignore 31 5 F = = 6 X0 = 0 X1 = 1 X2 = 2 X3 = 1 X4 = 2 X5 = 0 5 0 5! (5-0) ! 0! 5! (5-0) ! 0 ! = = = 1 5 1 5! (5-1) ! 1 ! 5*4! 4 ! 1 ! = = = 5 5 2 5! (5-2) ! 2 ! 5*4*3! 3 ! 2 ! = = = 10 5 3 5! (5-3) ! 3 ! 5*4*3! 2 ! 3 ! = = = 10 5 4 5! (5-4) ! 4 ! 5*4! 1 ! 4 ! = = = 5 5 5 5! (5-5) ! 5! 5! (0) ! 5 ! = = = 1 25 6 (xi)2 )5i) X2 = ∑i5=0 - 6 32 6 0 1 1 5 4 10 1 10 4 5 0 1 = + + + + + pass 2+4+1+8 10 = 5.33 - 6 = 1.99

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