1 / 9

Lecture 5 STREAM CIPHER SYSTEM

Lecture 5 STREAM CIPHER SYSTEM. By: NOOR DHIA AL- SHAKARCHY 2012-2013. Shift register. I 0. r n. r n-1. r 1. t n. n. t n-1. t 1. +. Linear Feedback Shift Register (LFSR). STREAM CIPHER SYSTEMS. 1- Linear Feedback Shift Register (LFSR):

varden
Download Presentation

Lecture 5 STREAM CIPHER SYSTEM

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lecture 5STREAM CIPHER SYSTEM By: NOOR DHIA AL- SHAKARCHY 2012-2013

  2. Shift register I0 rn rn-1 ...... r1 tn n tn-1 ...... t1 + Linear Feedback Shift Register (LFSR) STREAM CIPHER SYSTEMS 1- Linear Feedback Shift Register (LFSR): Is a one method used to generate a stream key . this method dependent on n-stage feedback shift register with linear operation that represent the relationship between the bits in the stage.

  3. STREAM CIPHER SYSTEMS Ri' = ri+1 i = 1,2,....., n-1 Rn' = TR = ∑in=1 tiri mod 2 = t1r1 mod 2 Ө t2r2 mod 2 Ө ..... Ө tnrn mod 2 An n-stage LFSR can generated pseudo random bits with a period of 2n-1 (maximum period) if the polynomial : T(x) = tnxn + tn-1xn-1 + ….. + t1x +1 is primitive. A primitive polynomial of degree n is an irreducible polynomial that divides X2n-1 , but not Xd+1 for any d that divides 2n-1 Some of primitive polynomial are: T(x) = x4 + x + 1 T(x) = x4 + x3 + 1 T(x) = x3 + x + 1 T(x) = x3 + x2 + 1

  4. Shift register I0 r4 r3 r2 r1 n + 4-stage Linear Feedback Shift Register (LFSR) STREAM CIPHER SYSTEMS Example:- Use the 4-stage LFSR and the polynomial T(x) = x4 + x + 1 and the I0=0001 to generate key stream. 24 -1 = 15 I0 = 0001 Key = 100011110101100

  5. STREAM CIPHER SYSTEMS • H.W:-Use the 4-stage LFSR and the polynomial T(x) = x4 + x3 +x2 + 1 and the I0=1001 to generate key stream. • Randomness: If a cryptanalyst intercept part of the sequence, and have no information on how to predict what comes next, the sequence is called pseudo-random sequence. St : any binary sequence. RUN: is the string of consecutive identical sequence elements which is neither preceded nor succeeded by sum symbol. Gap: a RUN of zeros (0). Block: a RUN of ones (1). • Example:- St = O 1 1 1 0 0 1 RUN RUNRUNRUN =1 =3 =2 =1 gap block gap block

  6. STREAM CIPHER SYSTEMS • Auto Correlation Function:- Suppose; st : is a binary sequence of period P. ‪ Ţ: any fixed. We compare the first P terms of st and it's translated (st + Ţ). If; A: is the number of positions which these two sequences agree. D: (P –A) is the number of positions in which theydisagree. Then autocorrelation function C(Ţ) = A – D P Clearly C (Ţ + P) = C(Ţ) for all Ţ, so 0 ≤ Ţ ≤P. When Ţ =0 ( A=P , D =0, C( 0) = p - 0 =1) We have In- Phase Autocorrelation. p Ţ ≠ 0 we have out-of-autocorrelation .

  7. STREAM CIPHER SYSTEMS Example: P =7, Ţ =3; st = 1001010 St = 1 0 0 1 0 1 0 S1 S2 S3 S4 S5 S6 S7 First P term 1 0 0 1 0 1 0 1 0 0 1 0 1 0 St = 1 0 0 1 0 1 0 s(t+T ) = 1 0 1 0 1 0 0 A =3 ; D = 7-3=4 C(T) = C(3) = = = - A – D P 3 – 4 7 1 7 H.W:- C(3+7) == C(3)

  8. STREAM CIPHER SYSTEMS • Golomb Randomness Postulates: R1: if P is even the cycle of length P shall cantainan equal nmber of zeros and ones. If P is odd number then the number of zeros shall be one more or less then the number of ones. R2: in the sycle of length P, half length 1, a guarter have length 2, an aither have length 3. R3: the out- of- phase autocorrelation is a constant. Example: 100010 111 0110001110010 R1: p = n = 22 (even) n0 = 11 n1 = 11 then R1 true R2: no. of RUN = 12 (1/2) no. of RUN = 6 { must be that length 1} no. of RUN that length (1 ) = 6 {true} (1/4) no. of RUN = 3 { must be that length 2} no. of RUN that length (2 ) = 2 {false} (1/3) no. of RUN = 4 { must be that length 3} no. of RUN that length (3 ) = 4 {true}

  9. STREAM CIPHER SYSTEMS R3: T = 1 A = 10 D = P – A = 22-10 = 12 C(T = 1) = (A- D ) / P = (10- 12 ) / 22 = -2/22 = -1/11 T = 2 A = 10 D = P – A = 22-10 = 12 C(T = 2) = (A- D ) / P = (10- 12 ) / 22 = -2/22 = -1/11 T = 3 A = 8 D = P – A = 22-8 = 14 C(T = 1) = (A- D ) / P = (8- 14 ) / 22 = -6/22 = -3/11 That's mean the function not constant and R3 is false The sequence not good randomness

More Related