Cryptography - Sage

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With a value of 128, the modulus is of size 2 128 , or about 39 decimal digits. Each of theprimes is of size approximately 20 decimal digits. This particular example can be easilybroken by the factorization:sage: x = K[0]; x86398677368792768067556452456311743331sage: x.factor()6046864213681032211 * 14288178850339607921Exercise 8.7 Use the above factorization to reproduce the private key L (generated but notprinted above) for this K.Solution.Given the factorization86398677368792768067556452456311743331= 6046864213681032211 · 14288178850339607921,we can find the inverse to the exponente = 49338921862830381807760291818994034053.sage: e = 49338921862830381807760291818994034053sage: p = 6046864213681032211sage: q = 14288178850339607921sage: d = inverse_mod(e,lcm(p-1,q-1))sage: d285484457605725559400259141876035917It is now possible to verify as above that (e, n) and (d, n) server as inverse RSA keys.Exercise 8.8 Why is the choice for which key is the public key and which key is theprivate key arbitrary? Practice encoding, decoding, enciphering, and deciphering with theRSA cryptosystem. Why do the member functions enciphering and deciphering returnthe same values?Solution.Provided that e is chosen as a random number in the range1 ≤ e ≤ lcm(p − 1, q − 1),127
which has no common factors with p − 1 or q − 1, then its inverse is a similarly randomvalue in this range. Therefore after creation, the decision of which key to publish as thepublic key, and which key to guard as the private key is arbitrary.N.B. Sometimes a special value, such as 3, 5, 17, 257, or 65537, is chosen as the publicexponent. These are each of the form 2 r + 1, so that the enciphering can be done rapidlyusing only r squarings and one multiplication. In such a case it is clear that no such“obvious” value is suitable for the private key, and the symmetry of the choice betweenpublic and private keys is broken.An ElGamal cryptosystem is based on the difficulty of the Diffie–Hellman problem:Given a prime p, a primitive element a of (Z/pZ) ∗ = {c ∈ Z/pZ : c ≠ 0}, and elementsc 1 = a x and c 2 = a y , find the element a xy in (Z/pZ) ∗ .Exercise 8.9 Recall the discrete logarithm problem: Given a prime p, a primitive elementa of (Z/pZ) ∗ , and an element c of (Z/pZ) ∗ , find an integer x such that c = a x . Explainhow a general solution to the discrete logarithm problem for p and a implies a solution tothe Diffie–Hellman problem.Solution. Suppose that the discrete logarithm problem has an efficient solution. Then,given a primitive element a of F p , for every a x and a y we could solve for x = log a (a x ) andfor y = log a (a y ). It follows that we could then produce the value a xy , which solves theDiffie-Hellman problem.Exercise 8.10 Fermat’s little theorem tells us that a p−1 = 1 for all a in (Z/pZ) ∗ . Recallthat a primitive element a has the property that Z/(p − 1)Z → (Z/pZ) ∗ given by x ↦→ a xis a bijection.1. Show that a is primitive if and only if a x = 1 only when p − 1 divides x.2. Let p be prime 2 32 + 15. Show that a = 3 is a primitive element of (Z/pZ) ∗ . Use theSAGE function log to compute discrete logarithms of elements of FiniteField(p)with respect to a.3. Let p be the prime 2 32 + 61. Show that the element a = 2 is a primitive element for(Z/pZ) ∗ . Use the SAGE function log to compute discrete logarithms of elements ofFiniteField(p) with respect to a.Solution. The statement of the definition of primitive is a formal statement equivalent tothat which follows. An element a of Z/pZ is primitive if and only if1, a, a 2 , . . . , a p−2are all distinct, and therefore enumerate all nonzero elements of Z/pZ.128 Appendix C. Solutions to Exercises
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Author (David R. Kohel) /Title (Cry
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CONTENTS1 Introduction to Cryptogra
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PrefaceWhen embarking on a project
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information. We introduce here some
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ut strings in A ∗ map injectively
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CHAPTERTWOClassical Cryptography2.1
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LV MJ CW XP QO IG EZ NB YH UA DS RK
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As a special case, consider 2-chara
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Note that if d k = 1, then we omit
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ExercisesSubstitution ciphersExerci
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Ciphertext-only AttackThe cryptanal
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of size n, suppose that p i is the
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Note that ZKZ and KZA are substring
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Checking possible keys, the partial
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sage: X = pt.frequency_distribution
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CHAPTERFOURInformation TheoryInform
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For each of these we can extend our
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in terms of the cryptosystem), then
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CHAPTERFIVEBlock CiphersData Encryp
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Deciphering. Suppose we begin with
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The Advanced Encryption Standard al
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1. Malicious substitution of a ciph
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locks M j−1 , . . . , M 1 as well
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where X = K ⊕ M = (X 1 , X 2 , X
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6.2 Properties of Stream CiphersSyn
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Exercise. Verify that the equality
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n 2 n − 11 12 33 74 155 316 637 1
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Exercise 6.6 In the previous exerci
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Exercise 6.9 Compute the first 8 te
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which holds since −4 = 17 + (−1
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must therefore have a divisor of de
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Shrinking Generator cryptosystemLet
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CHAPTEREIGHTPublic Key Cryptography
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Initial setup:1. Alice and Bob publ
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We apply this rule in the RSA algor
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the discrete logarithm problem (DLP
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Man in the Middle AttackThe man-in-
Page 82: Exercise 8.6 Fermat’s little theo
Page 85 and 86: k < p − 1 with GCD(k, p − 1) =
Page 88 and 89: CHAPTERTENSecret SharingA secret sh
Page 90: using any t shares (x 1 , y 1 ), .
Page 93 and 94: sage-------------------------------
Page 95 and 96: sage: x.is_unit?Type:builtin_functi
Page 97 and 98: Python (hence SAGE) has useful data
Page 99 and 100: sage: n = 12sage: for i in range(n)
Page 101 and 102: sage: I = [55+i for i in range(3)]
Page 103 and 104: sage: I = [7, 4, 11, 11, 14, 22, 14
Page 105 and 106: ExercisesRead over the above SAGE t
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Page 109 and 110: Solution. The block length is the n
Page 111 and 112: Solution.below.The coincidence inde
Page 113 and 114: analysis of the each of the decimat
Page 115 and 116: arbitrary permutation of the alphab
Page 117 and 118: In order to understand naturally oc
Page 119 and 120: We do this by first verifying the e
Page 121 and 122: Solution.None provided.Linear feedb
Page 123 and 124: Multiplying each through by the con
Page 125 and 126: Solution. The linear complexity of
Page 127 and 128: If a, b, and c are as above, then f
Page 129 and 130: Exercise 8.5 Use SAGE to find a lar
Page 131: Solution. Now we can verify that e
Page 135 and 136: sage: p = 2^32+61sage: m = (p-1).qu
Page 137 and 138: sage: a5 := a^n5sage: c5 := c^n5sag
Page 139 and 140: The application of this function E
Page 141 and 142: 5. (∗) How many elements a of G h
Page 143: 1. The value f(0) of the polynomial
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Cryptography - Sage

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