Cryptography - Sage

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# 1 2 3 4 5 6 7 8 9 10 113 Z A Z K2 N D L W T2 P W H W E2 V S Y E2 L W CNow let’s guess what the translations are at each of the periods. The following is a tableof common characters in each of the 11 decimations of period 11, organized by the numbersof their appearances.i 9 8 7 6 51 S,W,Z V2 L A3 F T4 D O R5 L I,W6 W L7 T H W8 I,S,X E9 E H10 Z E,Y11 IThese characters are not themselves the characters in the key, but if we assume thatone of these frequently occuring characters is the image of E, then we can make a guessat the key. The table below gives the enciphering characters which take the correspondingcharacter in the previous table to E.i 9 8 7 6 51 M,I,F J2 T E3 Z L4 B Q N5 T W,I6 I T7 L X I8 W,M,H A9 A X10 F A,G11 W3.3. Breaking the Vigenère Cipher 25
Checking possible keys, the partial key I****IL*A*W gives the following text which issuggestive of English:1 2 3 41 W****OS*I*I N****ND*M*N E****BE*T*E E****LL*H*D2 A****YV*E*T H****UN*E*P E****BU*W*E N****EP*N*R3 E****HE*L*O K****EE*N*D A****BR*G*T A****IG*T*N4 D****IR*C*E D****TH*U*H O****AD*W*H A****DO*H*A5 K****GH*H*W W****TH*S*A D****OR*F*R M****YS*O*T6 O****LE*R*A Y****TH*M*C H****RE*L*G H****NT*U*S7 E****EY*S*H Y****ES*I*E S****WW*U*D I****IN*E*E8 S****ES*E*M A****LO*K*N G****EE*N*H E****NG*A*W9 H****DE*D*I G****YF*I*I M****EC*S*A D****OU*H*E10 A****EE*O*S I****ES*E*E S****ST*Y*L L****AR*N*G11 H****SE*T*L L****TH*E*N D****TS*R*G H****SW*E*D12 R****DO*H*W T****E3.4 Cryptanalysis of Transposition CiphersA transposition cipher can easily be recognized by an analysis of character frequencies. Iteratingtransposition ciphers can greatly increase security, but as with substitution ciphers,almost all such ciphers can be broken. Although many modern cryptosystems incorporatetransposition ciphers, the operation on large blocks has the disadvantage of requiring a lotof memory.3.5 Statistical MeasuresSo far we have focused on Vigenère ciphers, and their reduction to monoalphabetic substitutions.Here we show how to use SAGE to complete the final step of breaking theseciphers. Recall that the reduction to monoalphabetic substitution is done by the process ofdecimation, by which we lose all 2-character frequency structure of the language. A moresophisticated approach will be necessary for breaking more complex ciphers.Correlation. We first introduce the concept of correlation of two functions. Let X and Ybe discrete random variables on a space Ω of n symbols, with values (x 1 , x 2 , . . . , x n ) and(y 1 , y 2 , . . . , y n ), respectively. For simplicity we assume that all n symbols of Ω occur withequal likihood. We define the correlation of the two sequences to beCorr(X, Y) =∑ ni=1 (x i − µ(X))(y i − µ(Y))σ(X)σ(Y)26 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 and 26: Ciphertext-only AttackThe cryptanal
Page 27 and 28: of size n, suppose that p i is the
Page 29: Note that ZKZ and KZA are substring
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
Page 78 and 79: the discrete logarithm problem (DLP
Page 80 and 81:
Man in the Middle AttackThe man-in-
Page 82:
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 ), .
Page 93 and 94:
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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