Cryptography - Sage

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Note that if d k = 1, then we omit this term, and the identity permutation can be writtenjust as 1. This notation gives more information about the permutation π and is morecompact for simple permutations such as transpositions.Simple columnar transpositionA classical example of a transposition cipher is an (r, s)-simple columnar transposition.In this cryptosystem the plaintext is written in blocks as r rows of fixed length s. Theciphertext is read off as the columns of this array. Suppose we begin with the plaintext:I was riding on the MayflowerWhen I thought I spied some landI yelled for Captain ArabI have yuh understandWho came running to the deckSaid, "Boys, forget the whaleLook on over yonderCut the enginesChange the sailHaul on the bowline"We sang that melodyLike all tough sailors doWhen they are far away at seaStripped to our plaintext alphbet and written in lines of 36 characters each, we have theplaintext:IWASRIDINGONTHEMAYFLOWERWHENITHOUGHTISPIEDSOMELANDIYELLEDFORCAPTAINARABIHAVEYUHUNDERSTANDWHOCAMERUNNINGTOTHEDECKSAIDBOYSFORGETTHEWHALELOOKONOVERYONDERCUTTHEENGINESCHANGETHESAILHAULONTHEBOWLINEWESANGTHATMELODYLIKEALLTOUGHSAILORSDOWHENTHEYAREFARAWAYATSEAReading off the columns, we obtain the following ciphertext under the columnar transpositioncipher:2.2. Transposition Ciphers 15
IIHDYOOWSAEONUAPVCNTGSIEKDHHREYSEESIDUARBADSHICOIIOUDUWLNMNBTLOGEDOTIROLEYHNSNARSEEDTNSFEWOHDTONEWEIARGSHMYNGIAEAEDENNNYLWTEGTFLHTSTHLEOHCHEODCEHAYWFAWATAEOMHNMRRREAGEEWCRLELFHAUETOAEPNLHDRNTNOEYAIAIOSLWTINKAIAHNGOIKYOATNLEAUROOHATGATVALSHBHEULETIERLTASimple columnar transpositions impose unnecessarily restrictive conditions on the form ofthe transposition, but were widely used as a component of ciphers used until the 1950’s.More general columnar transpositions allow for permuations of the columns before readingthem off.General transposition ciphersA general transposition cipher of block length m allows m! different permuations.m = 7, this is a mere 5040 permutations, but for block length 36, this givesFor371993326789901217467999448150835200000000possibilities. One such permutation, given in cycle notation, is:(1, 12, 5, 36, 30, 31, 4, 28, 33, 22, 26, 17, 10, 16, 14, 23, 18, 35, 32)(2, 9, 3, 25, 15, 7, 21, 6, 29, 34, 11, 27, 19, 24)(8, 13, 20).Applied to the above ciphertext this gives a the ciphertextNNWNTIOTAMERLEDHGHRIIHYWEAFUGHSIWOOTAMCTIADNPYPEEOSDEBRODALSIELRANIIFLAIRNRNEICSVNNYOMHTDHEUUUWAADHOTGEHAETNSBLOROEFCGLSHHIOOEADAETERETOVOKDWYNKETEELSHENIHECNCNTUGURTEOGNSHAIDYAHLAELLYTLAWTADEHMOEILEWBOGNSNTALKHOTNEIDOFAAWYOGERSERIWREELAATUHNHTSYHOASAADespite the large number of possible permutations, the unmasked stucture of the plaintextpermits an adversary to decipher transposition ciphertext.16 Chapter 2. Classical <strong>Cryptography</strong>
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 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 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
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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-
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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
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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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