Cryptography Lecture 2: Classic Ciphers Piotr Faliszewski

Previous Class Highlights Last class  Historical perspective  Kerckhoff’s principle  Requirements for information security Confidentiality Data integrity Authentication Non-repuditation  Attacks on confidentiality Ciphertext only attack Plaintext only attack Chosen ciphertext attack Chosen plaintext attack Key-only attack

Convention Plaintext alphabet and ciphertext alphabet ABCDEFGHIJKLM NOPQRSTUVWXYZ

Modular Arithmetic Modulo arithmetic  Modulus m  “Clock”-like arithmetic – numbers “wrap around” after reaching m  Calculate in the world of remainders modulo m 2 = 14 mod (12) 26 = 0 (mod 26) 7*11 ≡ 77 (mod 5) ≡ 2 (mod 5) ≡ 2 * 1 (mod 5) -5 (mod 11) ≡ 6 (mod 11)

Monoalphabetic Substitution Each letter is replaced by another letter from the alphabet Key:  1-1 mapping of letters  How many keys are there? Various flavors  Shift cipher  Affine cipher  Atbash  Kama-sutra cipher

Shift Cipher Substitution cipher  Key k  N  Formula: y = (x + k) mod 26  26  size of the alphabet Cryptanalisis?  Brute force! Popular culture  2001: Space Odyssey  HAL 9000

Shift Cipher Substitution cipher  Key k  N  Formula: y = (x + k) mod 26  26  size of the alphabe Cryptanalisis?  Brute force! Popular culture  2001: Space Odyssey  HAL 9000 ABBCCDDEEFFGGHHIIJJKKLLMMNABBCCDDEEFFGGHHIIJJKKLLMMN

Shift Cipher Substitution cipher  Key k  N  Formula: y = (x + k) mod 26  26  size of the alphabe Cryptanalisis?  Brute force! Popular culture  2001: Space Odyssey  HAL 9000 ABBCCDDEEFFGGHHIIJJKKLLMMNABBCCDDEEFFGGHHIIJJKKLLMMN

Shift Cipher Substitution cipher  Key k  N  Formula: y = (x + k) mod 26  26  size of the alphabe Cryptanalisis?  Brute force! Popular culture  2001: Space Odyssey  IBM 9000 ABBCCDDEEFFGGHHIIJJKKLLMMNABBCCDDEEFFGGHHIIJJKKLLMMN

Affine Cipher Substitution cipher  Key ( ,  )  Formula: y = (  x +  ) mod 26 How to select keys? Cryptanalysis? A  0 B  1 C  2 D  3 E  4 F  5 G  6 H  7 I  8 J  9 K  10 L  11 M  12 N  13 O  14 P  15 Q  16 R  17 S  18 T  19 U  20 V  21 W  22 X  23 Y  24 Z  25

General Substitution Cipher Key  Permutation of the alphabet  Large key space! … but not used efficiently! Does not hide various properties of the underlying text!

General Substitution Cipher Key  Permutation of the alphabet  Large key space! … but not used efficiently! Does not hide various properties of the underlying text! How to attack a substitution cipher?  Frequency attack Digrams, trigrams  “fingerprint” of a language

Frequency Attack Frequencies of letters  Letters appear with different frequencies in natural texts  Exceptions: Gadsby by E.V. Wright Disparation by George Perec (A Void, translation G. Adair)  Breaking substitution ciphers A  0.082N  B  0.015O  C  0.028P  D  0.043Q  E  0.127R  F  0.022S  G  0.020T  H  0.061U  I  0.070V  J  0.002W  K  0.008X  L  0.040Y  M  0.024Z  0.001

Frequencies vs Subst’n Cipher Shift and affine cipher  usually enough to locate ‘E’  gives the key  for affine cipher: One more letter can be helpful Full substitution cipher  Might need digram and trigram frequencies A  0.082N  B  0.015O  C  0.028P  D  0.043Q  E  0.127R  F  0.022S  G  0.020T  H  0.061U  I  0.070V  J  0.002W  K  0.008X  L  0.040Y  M  0.024Z  0.001

Example: Shift Cipher The following frequencies were found:  Count the letters  Normalize the frequencies  Compare to the natural one A : 27 B : 6 C : 5 D : 12 E : 14 F : 1 G : 2 H : 8 I : 5 J : 14 K : 16 L : 5 M : 2 N : 13 O : 14 P : 19 Q : 6 R : 3 S : 5 T : 0 U : 4 V : 0 W : 17 X : 4 Y : 6 Z : 8

Example: Shift Cipher There are 216 letters  Divide counts by 216 A : B : C : D : E : F : G : H : I : J : K : L : M : 0.009

Example: Shift Cipher There are 216 letters  Divide counts by 216 Compare to natural frequencies  Natural guess: shift 4  Others plausible shift A : B : C : D : E : F : G : H : I : J : K : L : M : A : B : C : D : E : F : G : 0.02 H : I : 0.07 J : K : L : 0.04 M : 0.024

Example: Shift Cipher There are 216 letters  Divide counts by 216 Compare to natural frequencies  Natural guess: shift 4  Others plausible shift A : B : C : D : E : F : G : H : I : J : K : L : M : A : B : C : D : E : F : G : 0.02 H : I : 0.07 J : K : L : 0.04 M : 0.024

Example: Shift Cipher There are 216 letters  Divide counts by 216 Compare to natural frequencies  Natural guess: shift 4  Others plausible shift -2 but the rest of frequencies mismatch A : B : C : D : E : F : G : H : I : J : K : L : M : A : B : C : D : E : F : G : 0.02 H : I : 0.07 J : K : L : 0.04 M : 0.024

Example: Shift Cipher There are 216 letters  Divide counts by 216 Compare to natural frequencies  Natural guess: shift 4  Others plausible shift -2 but the rest of frequencies mismatch A : B : C : D : E : F : G : H : I : J : K : L : M : A : B : C : D : E : F : G : 0.02 H : I : 0.07 J : K : L : 0.04 M : 0.024

Dot-Product Method Treat the list of frequencies as vector  A 0 – list of frequencies  A i – list of frequencies shifted by i Example  A 0 = ( 0.82, 0.015, 0.028, 0.043,... )  A 2 = ( 0.20, 0.001, 0.082, 0.015,... ) Dot-product  A i · A j  multiply elements position wise and add

Dot-Product Method Property  A i · A j  largest when i = j Method  Compute frequency vector W for our text W approximates some A j  Compute W · A i for each i  Shift is the i that maximizes the value

Vigenere Cipher How to make the shift cipher more difficult to break?  Problem: each letter shifted by the same ammount  Solution: pick different shifts for different letters! Vigenere cipher  A sequence of shift ciphers

Vigenere Cipher Key  a vector (k 1,..., k n )  each k i is a letter (or equivalently, a small integer) Cipher:  plaintext: x 1 x 2... x m  ciphertext: y j = x j + k j mod n (mod 26)

Vigenere Cipher: Cryptanalisis Known ciphertext attack  dljhswbesidtyjfcqpjhrxfmdxipdabhordg itutnzfgiibhspzjcmovdlfkskfcovfrstit beosdlbigixxvpugixpqbibzsruwogmpcwsd yqkjcxudcifwyafpccuwswjh Finding the key:  Find the key length  displacement method  Break a sequence of shift ciphers

Displacement Method dljhswbesidtyjfcqpjhrxfmdxipdabhordgitutnzfgiibhsp 3 matches dljhswbesidtyjfcqpjhrxfmdxipdabhordgitutnzfgiibhsp 6 matches dljhswbesidtyjfcqpjhrxfmdxipdabhordgitutnzfgiibhsp 0 matches dljhswbesidtyjfcqpjhrxfmdxipdabhordgitutnzfgiibhsp 7 matches

Vigenere Cipher: Cryptanalisis After we get the key length  Break a series of shift ciphers... ... but we just have a sample of English letters for each cipher. Example: with key length 6, for each cipher we get every 6th letter of the message  Can still match the frequencies dot-product method

Unbreakable cipher Is it possible to create an unbreakable cipher?

Unbreakable cipher Is it possible to create an unbreakable cipher? One-time pad  Plaintext: x 1 x 2 x 3... x n  Random string: b 1 b 2 b 3... b n  Ciphertext: y i = x i  b i Cryptanalisis? Applications? This message is completely unreadable. I have encrypted it with the toughest cipher ever, one-time pad. TWICE! -- found in a cryptography discussion on the internet

One-Time Pad Keys Generate random sequence  Hardware generators Thermal noise from a semiconductor device Random fluctuations in disk sector latency times Etc.  Software generators Deterministic Initiated „randomly”  System clock  Elapsed time between keystrokes  Etc.

Pseudorandom Numbers Linear congruential generator  x i = ax i-1 + b (mod m)  Dangerous for cryptography! Blum-Blum-Shub generator  x i = x i-1 2 (mod n)  u i = x i (mod 2) Many others...