2.7 The Friedman and Kasiski Tests

Transcription

1 Math 135, Summer The Friedman and Kasiski Tests 1. In each of the following suppose you have a ciphertext with the given number of letters n and the given index of coincidence I. Assuming that the Vigenère encipherment was used on English, estimate the length of the keyword. (1) n 473, I.0422 (2) n 1284, I.0518 (3) n 583, I.0611 (4) n 20849, I.0648 Solution: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Calculate the index of coincidence for the following ciphertext and use it to estimate the keyword length. NGTSA IPNGE PBSFW NCPBN RSAGF ASGEW JSVSF NRCNI WKGPW KIGEW PSWWM SYEWN NFUNG EPBSF WNCPB NRSWS PWSWR GMNRS USVGH VSBRG MNRSU KCPTS SDNSP BSBMR CNNRS IFHIY INWCF SNGMR IKRGP SWMSR CVSCN NCKRS BNCTG GWCPB MRCNN RSUKC PNSCK REWCT GENNR SFCVI WRIPA MGFHB MSRCV SNRSX FIVIH SASNG IPRCT IN Solution: The ciphertext has a total of 237 characters and the following table gives the letter frequencies: Letter # of times Letter # of times Letter # of times A 5 J 1 S 34 B K 7 T 5 C 18 L 0 U 4 D 1 M 9 V 7 E 7 N 29 W 17 F O 0 X 1 G 17 P 15 Y 2 H 4 Q 0 Z 0 I 13 R 21 Using the above information we can calculate I: Z ia I n i(n i 1) n(n 1)

2 Using the Friedman test we can use this to estimate the keyword length:.0265 n (.065 I) + n (I.0385) ( ) + n ( ) We estimate the keyword length to be one (that is, we think this is a monoalphabetic substitution cipher). 8. Use the Kasiski test to estimate the keyword length used to produce the following ciphertext. ANIIX GFXVH XCGZD CWYTU KLZKW HFXSG FARXX GLBRH PTWLI IXGFX VHXCG ZDCWY TUKLZ KWCJY OBWHC CCCSF ZDPBO BPTCG FWVXZ LOUJK PRLCE KCCKO CXDRZ YPAPP WQFVV GFANM EFLBV MYQLL LQPLL NYIJZ MCCXE FWCEW EPMGV RERZY POYCO QYICC WPVVJ RZCEK HYKNY IJIFY NLZUW PVVJR ZCEKH MJEVH EFWLV ALQFI VRRGF YVVTL NICZP BWRTI AREUP FPFWQ RWHMJ EZRRR ZYIIE MG Solution: The ciphertext string IIXGF XVHXC GZDCW YTUKL ZKW occurs at positions 3 and 45. We estimate that the keyword length divides 42. The ciphertext string WPVVJ RZCEK H occurs at positions 181 and 205. We estimate that the keyword length divides 24. The greatest common divisor of these two number is 6. So we estimate that the keyword length must be a divisor of 6. The choices are 1, 2, 3, and Hill Ciphers; Matrices 1. Calculate the matrix products modulo the given modulus [ 0 1 0] (mod 7) (mod 12) (mod 2) (mod 2) 6. 5 [15 ] 20

3 (mod 17) Solution: (mod 7) 5 3 [ 12 ] (mod 12) (mod 2) [ [ ] 0 1] (mod 2) [15 ] 5 [ 11 ] [ 1 4 ] 2 6 [ ] [ 8 3 ] (mod 17) 3. Calculate the determinants of the following matrices modulo the given modulus. 1 0 (a) A, m 2 Solution: det(a) 1 (mod 2) [ ] 4 3 (b) B, m 5 Solution: det(b) (mod 5) [ 1 3 ] 6 (c) C, m 11 Solution: det(c) (mod 11) [ 5 3 ] 11 (d) D, m 26 Solution: det(d) [ 7 ] (e) E, m 26 Solution: det(e) [ 6 8] 0 1 (f) F, m 26 Solution: det(f )

4 5. Encipher the message WASHINGTON using the Hill Cipher with key matrix A Solution: We break the plaintext into pairs of letters and position encode them: [ [ [ W 22 S 18 I 8 G 6 O 14 A 0 H] 7 N] 13 T 19] N 13 and then multiply them by the matrix A: [ ] Which gives us the encrypted text OQZQLUVYDW. [ O Q] Z Q [ L U] V Y D W 6. Decipher the message VPNNNHTPQF, which was enciphered by the Hill cipher with key matrix 3 5 A. 7 8 Solution: The first step is to find A 1 this will allow us to decrypt the ciphertext. det(a) So det(a) 1 7. A We [ now ] [ break the ciphertext [ up [ into ] [ pairs ] and position [ encode: V 21 N N 13 T 19 Q 16 P 15] N 13] H 7 P 15] F 5 In order to find the plaintext that corresponds to the pairs we multiply by A 1 on the left: [ B O] [ N A] [ P E] T I [ [ T 3 5] 5 23 X]

5 and we get the plaintext BONAPETITX. Or with spaces and minus the padding letter at the end Bon apetit. 9. Suppose you know that ABLE enciphers as ZXTQ by using the Hill cipher with a key matrix A. Determine A. [ 0 Solution: We have that A 1] Multiply both sides of this equation by on the right to get: A and A 23 4 A [ We concatenate these two systems and get: 16 [ 25 ] ] 1 [ 5 ] Calculate the determinants of the following matrices modulo the given modulus, m. For those with determinant relatively prime to m, find the inverse matrix (a) A, m (b) A 0 2 0, m (c) A 1 0 0, m Solution: (a) () () det(a) 1 det + 2 det (mod 4) A (mod 4) (b) det(a) (mod 13) A (mod 13) 0 0 (c) det(a) 1 A 1 A 2 The last one can be seen as the matrix A just switches the rows of anything it multiplies: x 1 x 2 x 3 x for all possible vectors x. We call any such matrix a permutation matrix. x 3 x Encipher AUTUMN LEAVES by the matrix A

6 2. Decipher OASUDC using the above encryption scheme. Solution: We break the plaintext up into blocks of three letter and then position encode: A U 0 20 U M L E 11 4 V E 21 4 T 19 N 13 A 0 S 18 Then we multiply the resulting vectors by the matrix A to get the corresponding ciphertext: M U 9 13 N U S 3 19 T L T P F H 8 17 R Which gives the encrypted text MUNUS TLTPF HR. To decrypt we first must find A 1. We know from a previous exercise that det(a) 1 1. So the inverse is: A O 14 U 20 We break the ciphertext into blocks of three letters and position encode: A 0 D 3 S 18 C 2 And then multiply them by A 1 on the left K G C O L D This gives us KGCOLD as the plaintext.