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MASON ECE 646 - Lecture 6 Historical Ciphers

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1Historical CiphersECE 646 - Lecture 6 2Review of Lecture 5•GCD and a mod p = c •The Ring m•Properties of Rings•Euclid's Algorithm•Extended Euclidean Algorithm3Why (not) to study historical ciphers?AGAINSTFORNot similar to modern ciphersLong abandonedBasic components becamea part of modern ciphersUnder special circumstancesmodern ciphers reduceto historical ciphersInfluence on world eventsThe only ciphers you can break! 4Secret WritingSteganography(hidden messages)Cryptography(encrypted messages)SubstitutionTransformationsTranspositionCiphers(change the orderof letters)Codes SubstitutionCiphers(replace words)(replaceletters)5Selected world events affected by cryptology1 - trial of Mary Queen of Scots - substitution cipher1 - Zimmermann telegram, America enters World War I1939-1945 Battle of England, Battle of Atlantic, D-day - ENIGMA machine cipher1944 – world’s first computer, Colossus - German Lorentz machine cipher1950s – operation Venona – breaking ciphers of soviet spies stealing secrets of the U.S. atomic bomb – one-time pad 6Ciphers used predominantly in the given period(1)Electromechanical machine ciphers(Complex polyalphabetic substitution ciphers)1919Vigenère cipher(Simple polyalphabetic substitution ciphers)CryptographyCryptanalysis1586 Invention of the Vigenère CipherMonoalphabetic substitution cipherHomophonic ciphersInvention of rotor machinesXVIII c.Black chambers1863Kasiski’s method1918Index of coincidenceWilliam FriedmanShift ciphers100 B.C.IX c.Frequency analysisal-Kindi, Baghdad1926 Vernam cipher (one-time pad)7Ciphers used predominantly in the given period(2)CryptographyCryptanalysisDES19772001AESTriple DES193219772001Rejewski, PolandReconstructing ENIGMA19391949Shennon’s theoryof secret systemsPolish cryptological bombs,and perforated sheetsPublication of DES1945British cryptological bombs, Bletchley Park, UKBreaking Japanese “Purple” cipher1990DES crackersone-time padStream CiphersS-P networks 8Simple Monalphabetic substitution ciphersA. Caesar Cipherci = f(mi) = mi + 3 mod 26No keyB. Shift Cipherci = f(mi) = mi + k mod 26Key = kNumber of keys = 26mi = f-1(ci) = ci - 3 mod 26mi = f-1(ci) = ci - k mod 269Coding characters into numbersA ⇔ 0B ⇔ 1C ⇔ 2D ⇔ 3E ⇔ 4F ⇔ 5G ⇔ 6H ⇔ 7I ⇔ 8J ⇔ 9K ⇔ 10L ⇔ 11M ⇔ 12N ⇔ 13O ⇔ 14P ⇔ 15Q ⇔ 16R ⇔ 17S ⇔ 18T ⇔ 19U ⇔ 20V ⇔ 21W ⇔ 22X ⇔ 23Y ⇔ 24Z ⇔ 25 10Caesar Cipher: ExamplePlaintext:Ciphertext:I C A M E I S A W I C O N Q U E R E D8 2 0 12 4 8 18 0 22 8 2 14 13 16 20 4 17 4 3 11 5 3 15 7 11 21 3 25 11 5 17 16 19 23 7 20 7 6 L F D P H L V D Z L F R Q T X H U H G11Simple Monalphabetic substitution ciphers (2)C. Affine Cipherci = f(mi) = k1 ⋅ mi + k2 mod 26Key = (k1, k2)Number of keys = 12⋅26 = 312gcd (k1, 26) = 1mi = f-1(ci) = k1-1 ⋅ (ci - k2) mod 26 12General Substitution Cipher1. Monalphabetic (simple) substitution cipherM = m1 m2 m3 m4 . . . . mNC = f(m1) f(m2) f(m3) f(m4) . . . . f(mN)Generally f is a random permutation, e.g.,f = a b c d e f g h i j k l m n o p q r s t u v w x y zs l t a v m c e r u b q p d f k h w y g x z j n i oKey = fNumber of keys = 26! ≈ 4 ⋅ 1026 ≈ 28913General Substitution Cipher (2)•#keys = 288•Q: Is a brute-force attack possible?i.e., trying of all possible keys•Q: Which other attack is possible? 14Most frequent single letters Average frequency in a long English text:E — 13%T, N, R, I, O, A, S — 6%-9%D, H, L — 3.5%-4.5%C, F, P, U, M, Y, G, W, V — 1.5%-3%B, X, K, Q, J, Z — < 1%= 0.038 = 3.8%Average frequency in a random string of letters:12615Digrams:TH, HE, IN, ER, RE, AN, ON, EN, ATTrigrams:THE, ING, AND, HER, ERE, ENT, THA, NTH, WAS, ETH, FOR, DTHMost frequent digrams, and trigrams 1602468101214A B C D E F G H I J K L M N O P Q R S T U V W X Y ZRelative frequency of letters in a long English textby Stallings7.251.253.54.2512.75323.57.750.250.53.752.757.757.52.750.58.569.2531.51.50.52.250.251702468101214a b c d e f g h i j k l m n o p q r s t u v w x y z02468101214a b c d e f g h i j k l m n o p q r s t u v w x y zCharacter frequencyin a long English plaintextCharacter frequencyin the corresponding ciphertextfor a shift cipher 1802468101214a b c d e f g h i j k l m n o p q r s t u v w x y zCharacter frequencyin a long English plaintextCharacter frequencyin the corresponding ciphertextfor a general monoalphabetic substitution cipher 02468101214a b c d e f g h i j k l m n o p q r s t u v w x y z1902468101214a b c d e f g h i j k lm n o p q r s t u vw x y z02468101214a b c d e f g h i jk lm n o p q r s tu v w x y z02468101214a b c d e f g h I j k lm n o p q r s t u v w x y z02468101214a b c d e f g h I j k l m n o p q r s t u vw x y zLong English text TCiphertext of the long English text TShort English message MCiphertext of the short English message MFrequency analysis attack: relevant frequencies 20Ciphertext:FMXVE DKAPH FERBN DKRXR SREFM ORUDSDKDVS HVUFE DKAPR KDLYE VLRHH RH A B C D E F G H I J K L M N O P Q R S T U V W X Y ZR - 8D - 7E, H, K - 5Frequency analysis attack (1)Step 1: Establishing the relative frequency of letters in the ciphertext21f(E) = Rf(T) = Df(4) = 17f(19) = 3Frequency analysis attack (2)Step 2: Assuming the relative frequency of letters in the corresponding message, and deriving the corresponding equationsAssumption: Most frequent letters in the message: E and TCorresponding equations:E → RT → D4 → 1719 → 3 22f(4) = 17f(19) = 3Frequency analysis attack (3)Step 3: Verifying the assumption for the case of affine cipher 4⋅k1 + k2 ≡ 17 (mod 26)19⋅k1 + k2 ≡ 3 (mod 26) 15⋅k1 ≡ -14 (mod 26) 15⋅k1 ≡ 12 (mod 26)23Substitution Ciphers (2)2. Polyalphabetic substitution cipherM = m1 m2 … mdKey = d, f1, f2, …, fdNumber of keys for a given period d = (26!)d ≈ (4 ⋅ 1026)dmd+1 md+2 … m2dm2d+1 m2d+2 … m3d…..C = f1(m1) f2(m2) … fd(md)f1(md+1) f2(md+2) … fd(m2d )f1(m2d+1 ) f2( m2d+2) … fd(m3d )…..d is a period of the cipher 2402468101214a b c d e f g h i j k l m n o p q r s t u v w x y zCharacter frequencyin a long English


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