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Vigenère Cipher

Vigenère Cipher. a polyalphabetic version of the shift cipher Key: (k 1 ,k 2 ,…,k m ) 2 Z 26 m , where m is the key length Encryption in blocks: (x 1 , x 2 , …, x m )  (x 1 +k 1 , x 2 +k 2 , …, x m +k m ) 2 Z 26 m Example: encrypt cryptography using key duh How to decrypt ?.

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Vigenère Cipher

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  1. Vigenère Cipher • a polyalphabetic version of the shift cipher • Key: (k1,k2,…,km) 2 Z26m, where m is the key length • Encryption in blocks: • (x1, x2, …, xm)  (x1+k1, x2+k2, …, xm+km) 2 Z26m • Example: • encrypt cryptography using key duh • How to decrypt ?

  2. Cryptanalysis of the Vigenère Cipher • - if m small: • exhaustive search might be ok • how many keys ? • known plaintext attack: easy / hard ? • ciphertext-only attack • long thought safe against this attack • what do we know now ? • first find the key length m: counting coincidences (see Section 2.3.1), Kasiski test, index of coincidence

  3. Kasiski test • two identical segments of ciphertext of length ¸ 3 likely correspond to the same segments of plaintext • if two such segments are ± positions apart, what is the likely relationship between m and ± ? • Example (Stinson, Exercise 1.21 1(b) using Vigenère): • KCCPKBGUFDPHQTYAVINRRTMVGRKDNBVFDETDGILTXRGUD • DKOTFMBPVGEGLTGCKQRACQCWDNAWCRXIZAKFTLEWRPTYC • QKYVXCHKFTPONCQQRHJVAJUWETMCMSPKQDYHJVDAHCTRL • SVSKCGCZQQDZXGSFRLSWCWSJTBHAFSIASPRJAHKJRJUMV • GKMITZHFPDISPZLVLGWTFPLKKEBDPGCEBSHCTJRWXBAFS • PEZQNRWXCVYCGAONWDDKACKAWBBIKFTIOVKCGGHJVLNHI • FFSQESVYCLACNVRWBBIREPBBVFEXOSCDYGZWPFDTKFQIY • CWHJVLNHIQIBTKHJVNPIST • Trigram HJV: positions 108, 126, 264, 318, and 330. Conclusions ?

  4. Index of Coincidence • Reasoning: • Let y = y1y2…yn be the ciphertext. • For a given m, we define these substrings of y: • y1 = y1ym+1y2m+1… • y2 = y2ym+2y2m+2… • : : : : • ym = ymy2my3m… • If m is divisible by the key length, what property do we expect from these strings ? • If m not divisible by the key length, then what ?

  5. Index of Coincidence • Example: • For m=6 and the text from slide 3: • y1 = KGQNGVGGTGCQWAWQHNJEPJTKQFWAP… • y2 = CUTRRFIUFEKCCKRKKCVTKVRCDRSFR… • y3 = CFYRKDLDMGQWRFPYFQAMQDLGZLJSJ… • y4 = PDATDETDBLRDXTTVTQJCDASCXSTIA… • Y5 = KPVMNTXKPTANILYXPRUMYHVZGWBAH… • Y6 = BHIVBDROVGCAZECCOHWSHCSQSCHSK…

  6. Index of Coincidence Def: Suppose that x = x1x2…xn is a string of n alphabetic characters. The index of coincidence of x, denoted Ic(x), is defined to be the probability that two random elements of x are identical. Thus, Ic(x) = where fi, i2 Z26 is the number of occurrences of the symbol i in the string x.

  7. Index of Coincidence For English text, the index of coincidence is approximately i=025 pi2 = 0.066 where pi is the probability of occurrence of symbol i (see Table 2.1). What is the index of coincidence of shift ciphertext, if plaintext is English ? For a completely random text, the index of coincidence is approximately Idea: find m so that the index of coincidence close to 0.066

  8. Index of Coincidence • Example from slide 3: • m = 1: 0.041 • m = 2: 0.038, 0.047 • m = 3: 0.056, 0.048, 0.048 • m = 4: 0.037, 0.042, 0.037, 0.050 • m = 5: 0.043, 0.043, 0.031, 0.035, 0.043 • m = 6: 0.063, 0.084, 0.049, 0.065, 0.042, 0.071 • m = 7: 0.031, 0.044, 0.043, 0.038, 0.044, 0.044, 0.041 • What m do we choose ?

  9. Computing the Key • need to find (k1,k2,…,km) • let fi be the number of occurrences of symbol i, i2 Z26 in yi; let n’ be the length of yi • Idea 1: use frequencies to determine ki • Idea 2: • more precise but needs more computation • for a shift g, we expect the normalized frequencies (divide by n’) • fg/n’, fg+1/n’, fg+2/n’, …, fg+25/n’ • to be “close to” p0,p1,…,p25 iff g = ki

  10. Computing the Key • Idea 2, continued: • let • Mg = i=025 pifg+i/n’ • if g=ki, what do we expect Mg to be ? • if g ki, what do we expect Mg to be ? • algorithm to determine the key ?

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