Cryptography - Sage

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CryptosystemsSpecific cryptosystem can be created in SAGE with the following commands:sage: S = AlphabeticStrings()sage: SFree alphabetic string monoid on A-Zsage: E = SubstitutionCryptosystem(S)sage: T = TranspositionCryptosystem(S,15)sage: TTransposition cryptosystem on Free alphabetic string monoid on A-Zof block length 15The latter constructor (of the transposition cryptosystem) specifies a key length of 15characters. The SAGE cryptosystem represents a ...sage: K = E.random_key()sage: KHYTIKQRWXUPZBJSCANEFODLVMG # randomsage: K = S(’HYTIKQRWXUPZBJSCANEFODLVMG’)sage: e = E(K)sage: eHYTIKQRWXUPZBJSCANEFODLVMGsage: e(S(’THECATINTHEHAT’))FWKTHFXJFWKWHFsage: m = E.encoding("This is sample message text to be encoded.")sage: mTHISISSAMPLEMESSAGETEXTTOBEENCODEDsage: c = e(m)sage: cFWXEXEEHBCZKBKEEHRKFKVFFSYKKJTSIKIsage: L = E.inverse_key(K)sage: E.enciphering(L,c)THISISSAMPLEMESSAGETEXTTOBEENCODEDsage: E.deciphering(K,c)THISISSAMPLEMESSAGETEXTTOBEENCODED99
ExercisesRead over the above SAGE tutorial and become familiar with the concepts of type, parent,assignment, basic constructions of integers, rationals, and strings, and with simple loopingand boolean operations.Exercise B.1 For the function strip encoding, type the function name followed by a ?at the SAGE prompt to display the docstring for the function. Do the same forSubstitutionCryptosystem and TranspositionCryptosystem.What are the components of this information, and what does it tell you?Create a cryptosystem E and typeto read the corresponding docstrings.E.encoding?, E.random key? and E.enciphering?Solution.The docstring details of the path for the function and a description of its use.Exercise B.2 Create the string in SAGE“I am standing up at the water’s edge in my dream”and assign it to a variable W . Next create the string monoid S of AlphabeticStrings.Apply the member function encoding to W , while reassign W to be the output. What isthe encoded plaintext that you obtain?Solution.The command encoding gives IAMSTANDINGUPATTHEWATERSEDGEINMYDREAM.Exercise B.3 Define K to be the output of E.random key() for a substitution cryptosystem.What is your key? Using the cipher E(K), to find the enciphering of W with respectto the key K.Solution. The substitution key UVLOIDTGKYZCRHBPMJQWXNFSAE enciphers the above plaintextas KURQWUHOKHTXPUWWGIFUWIJQIOTIKHRAOJIUR.Exercise B.4 Show that the deciphering map with respect to K is also a simple substitution.What is the inverse substitution key with respect to your particular K? Verify thisby creating the inverse key and enciphering the ciphertext with respect to it.100 Appendix B. SAGE Cryptosystems
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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
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
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: sage: I = [7, 4, 11, 11, 14, 22, 14
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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 and 132: Solution. Now we can verify that e
Page 133 and 134: which has no common factors with p
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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