Cryptography - Sage

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Examples of CryptanalysisLet’s begin with the substitution ciphertext constructed previously:QWMMPQDVKUVFDTXJQVDBOPIDUHDQQUGDLAMWJGXBGURRBPBURMKULDVXOOKUJUOVDJQDGBWHLDJQQMUODQUBIMWBOVWUVXPBUBIOKUBGXBGURROKUJUOVDJQVPWMMDJOUQDVKDBVKDCDAQXEDFKXOKLPWBIQVKDQDOWJXVAPTVKDQAQVDHXQURMKULDVXOOKUJUOVDJQVKDJDTPJDVKDVPVURBWHLDJPTCDAQXQPTDBPJHPWQQXEDBDNDJVKDRDQQFDFXRRQDDVKUVQXHMRDQWLQVXVWVXPBXQNDJAQWQODMVXLRDVPOJAMVUBURAVXOUVVUOCQWe find the following character counts, scaled to that of a 1000 character input text.A 24.0964B 54.2169C 9.0361D 129.5181E 6.0241F 12.0482G 18.0723H 18.0723I 12.0482J 54.2169K 51.2048L 24.0964M 36.1446N 6.0241O 57.2289P 45.1807Q 96.3855R 39.1566S 0T 15.0602U 81.3253V 99.3976W 39.1566X 57.2289Y 0Z 0The distributions look like a frequency preserving substitution cipher. We guess thatthe enciphering takes E ↦→ D and T ↦→ V or T ↦→ Q. The most frequent characters are D, V,Q, V, U, O, J, K, B and E, N, S, Y, Z are of lowest frequency.andEquating high frequency and low frequency characters, we might first guess{E, I, O, A, T, N, R, S} ↦→ {D, V, Q, U, O, J, K, B}{J, K, Q, X, Z} ↦→ {E, N, S, Y, Z}How would you go about reconstructing the entire text?Index of CoincidenceIn the 1920’s William Friedman introduced the index of coincidence as a measure of thevariation of character frequencies in text from a uniform distribution. The index of coincidenceof a text space (e.g. that of all plaintext or ciphertext) is defined to be theprobability that two randomly chosen characters are equal. In a language over an alphabet3.2. Cryptanalysis by Frequency Analysis 21
of size n, suppose that p i is the probability of a random character is the i-th character ina string of length N. Then, in the limit as the string length N goes to infinity, the indexof coincidence in that language is:n∑p 2 i .Over an alphabet of 26 characters, the coincidence index of random text is∑26i=1i=1(1/26) 2 = 1/26 ∼ = 0.0385.For English text, the coincidence index is around 0.0661. For a finite string of length N,we the index of coincidence is defined to be:∑ ni=1 n i(n i − 1),N(N − 1)where n i is the number of occurrences of the i-th character in the string.Theorem 3.1 The expected index of coincidence of a ciphertext of length N, output froma period m cipher, which is defined by m independent substitutions ciphers at each positionin arithmetic progression the i + jm, is( )( )1 N − m (m − 1) Nim N − 1 X + i n ,m N − 1where i X is the index of coincidence of the space X, the size of the alphabet is n, and andi n = 1/n is the index of coincidence of random text in that space.Example 3.2 Consider the ciphertext enciphered with a Vigenère cipher:OOEXQGHXINMFRTRIFSSMZRWLYOWTTWTJIWMOBEDAXHVHSFTRIQKMENXZPNQWMCVEJTWJTOHTJXWYIFPSVIWEMNUVWHMCXZTCLFSCVNDLWTENUHSYKVCTGMGYXSYELVAVLTZRVHRUHAGICKIVAHORLWSUNLGZECLSSSWJLSKOGWVDXHDECLBBMYWXHFAOVUVHLWCSYEVVWCJGGQFFVEOAZTQHLONXGAHOGDTERUEQDIDLLWCMLGZJLOEJTVLZKZAWRIFISUEWWLIXKWNISKLQZHKHWHLIEIKZORSOLSUCHAZAIQACIEPIKIELPWHWEUQSKELCDDSKZRYVNDLWTMNKLWSIFMFVHAPAZLNSRVTEDEMYOTDLQUEIIMEWEBJWRXSYEVLTRVGJKHYISCYCPWTTOEWANHDPWHWEPIKKODLKIEYRPDKAIWSGINZKZASDSKTITZPDPSOILWIERRVUIQLLHFRZKZADKCKLLEEHJLAWWVDWHFALOEOQWThe coincidence index of this text is 0.0439. This would suggest a period of approximately5. We will see that this is a bad estimate. Note that this doesn’t disprove the theorem, itjust shows that the statistical errors are too great and that we would need a much largersample size to converge to this theoretical expectation, or that the substitutions employedwere not independent.22 Chapter 3. Elementary Cryptanalysis
Page 1 and 2: Author (David R. Kohel) /Title (Cry
Page 4 and 5: CONTENTS1 Introduction to Cryptogra
Page 6: PrefaceWhen embarking on a project
Page 10 and 11: information. We introduce here some
Page 12 and 13: ut strings in A ∗ map injectively
Page 14 and 15: CHAPTERTWOClassical Cryptography2.1
Page 16 and 17: LV MJ CW XP QO IG EZ NB YH UA DS RK
Page 18 and 19: As a special case, consider 2-chara
Page 20 and 21: Note that if d k = 1, then we omit
Page 22: ExercisesSubstitution ciphersExerci
Page 25: Ciphertext-only AttackThe cryptanal
Page 29 and 30: Note that ZKZ and KZA are substring
Page 31: Checking possible keys, the partial
Page 34 and 35: sage: X = pt.frequency_distribution
Page 36 and 37: CHAPTERFOURInformation TheoryInform
Page 38 and 39: For each of these we can extend our
Page 40 and 41: in terms of the cryptosystem), then
Page 42 and 43: CHAPTERFIVEBlock CiphersData Encryp
Page 44 and 45: Deciphering. Suppose we begin with
Page 46 and 47: The Advanced Encryption Standard al
Page 48 and 49: 1. Malicious substitution of a ciph
Page 50 and 51: locks M j−1 , . . . , M 1 as well
Page 52: where X = K ⊕ M = (X 1 , X 2 , X
Page 55 and 56: 6.2 Properties of Stream CiphersSyn
Page 57 and 58: Exercise. Verify that the equality
Page 59 and 60: n 2 n − 11 12 33 74 155 316 637 1
Page 61 and 62: Exercise 6.6 In the previous exerci
Page 63 and 64: Exercise 6.9 Compute the first 8 te
Page 65 and 66: which holds since −4 = 17 + (−1
Page 67 and 68: must therefore have a divisor of de
Page 69 and 70: Shrinking Generator cryptosystemLet
Page 72 and 73: CHAPTEREIGHTPublic Key Cryptography
Page 74 and 75: Initial setup:1. Alice and Bob publ
Page 76 and 77:
We apply this rule in the RSA algor
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the discrete logarithm problem (DLP
Page 80 and 81:
Man in the Middle AttackThe man-in-
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Exercise 8.6 Fermat’s little theo
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k < p − 1 with GCD(k, p − 1) =
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CHAPTERTENSecret SharingA secret sh
Page 90:
using any t shares (x 1 , y 1 ), .
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sage-------------------------------
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sage: x.is_unit?Type:builtin_functi
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Python (hence SAGE) has useful data
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sage: n = 12sage: for i in range(n)
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sage: I = [55+i for i in range(3)]
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sage: I = [7, 4, 11, 11, 14, 22, 14
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ExercisesRead over the above SAGE t
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102
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Solution. The block length is the n
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Solution.below.The coincidence inde
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analysis of the each of the decimat
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arbitrary permutation of the alphab
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In order to understand naturally oc
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We do this by first verifying the e
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Solution.None provided.Linear feedb
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Multiplying each through by the con
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Solution. The linear complexity of
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If a, b, and c are as above, then f
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Exercise 8.5 Use SAGE to find a lar
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Solution. Now we can verify that e
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which has no common factors with p
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sage: p = 2^32+61sage: m = (p-1).qu
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sage: a5 := a^n5sage: c5 := c^n5sag
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The application of this function E
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5. (∗) How many elements a of G h
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1. The value f(0) of the polynomial
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Cryptography - Sage

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