1. Introduction CS 303 Algorithmic Number Theory and Cryptography
September 4, 1997
Introduction CS 303 Algorithmic Number Theory and Cryptography
Jeremy R. Johnson

2. KHOOR

3. HELLO

4. Introduction to Classical Cryptography
September 4, 1997
Introduction to Classical Cryptography
Objective: Introduction to Cryptography and Cryptanalysis
Shift Cipher
Substitution Cipher
Attacks and adversaries
Frequency analysis

5. Classical Cryptography
Basic problem: Secure communication over an insecure channel
Solution: private key encryption
m  E(k,m) = c  D(k,c) = m
Shannon provided a rigorous theory of perfect secrecy based on information theory
Adversary has unlimited computational resources, but key must be as long as message

6. Communication Scenario
Encryption key
Decryption key
ciphertext
Alice
Encrypt
Decrypt
Bob
plaintext
Eve

7. Shift Cipher hello all hail ceasar
September 4, 1997
Shift Cipher
hello
all hail ceasar
If he had anything confidential to say, he wrote it in cipher, that is, by so changing the order of the letters of the alphabet, that not a word could be made out. If anyone wishes to decipher these, and get at their meaning, he must substitute the fourth letter of the alphabet, namely D, for A, and so with the others.
— Suetonius, Life of Julius Caesar 56

8. Shift Cipher m  E(k,m) = c = m+k mod 26 c  D(k,c) = c-k mod 26 KHOOR
September 4, 1997
Shift Cipher
KHOOR
DOO KDLO FHDVDU
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 z
D E F G H I J K L M N O P Q R S T U V W X Y Z A B C
Key = 3
m  E(k,m) = c = m+k mod 26
c  D(k,c) = c-k mod 26

9. Possible Attacks Kerchoff’s Principle Ciphertext only Known plaintext
Chosen plaintext
Chosen ciphertext

10. Cryptanalysis of Shift Ciphers
Number of keys = size of alphabet
Ciphertext only
Brute force
Look for most frequently occurring symbol
Known plaintext
m + k = c (mod 26)
Chosen plaintext
1 + k = c (mod 26)
Chosen ciphertext
m + k = 1 (mod 26)

11. September 4, 1997
Frequency Analysis
en.wikipedia.org/wiki/Frequency_analysis_(cryptanalysis)
scottbryce.com/cryptograms

12. War and Peace 3,223,402 Characters

13. freq.pl
infile = file(filename,'r') freq = {} count = 0 for c in string.ascii_lowercase: freq[c] = 0 for line in infile: for c in line.lower(): if c.isalpha(): freq[c] = freq[c] + 1 count = count + 1 keys = freq.keys() keys.sort() for c in keys: print "freq[" + c + "] = ",float(freq[c])/count

14. Substitution Cipher m  E(k,m) = c = (m) c  D(k,c) = -1(c) NYVVZ
September 4, 1997
Substitution Cipher
NYVVZ
JVV NJSV RYJDJC
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 z
J E R M Y O H N S T U V W X Z A B C D F G I K L P Q
Key – bijection from {0..25}  {0..25}: i  (i)
m  E(k,m) = c = (m)
c  D(k,c) = -1(c)

15. Cryptanalysis of Substitution Ciphers
Number of keys = n! where n is the size of the alphabet
Ciphertext only
Brute force (too expensive)
Frequency analysis
Known plaintext
m  (m) = c, for known (m,c). Partial permutation
Chosen plaintext
m  (m) = c, m = 0..n-1
Chosen ciphertext
-1(c)  m, c = 0..n-1

16. One Time Pad Pad = b1  bn  {0,1}* chosen randomly m = m1  mn
E(Pad,m) = c = m  Pad
D(Pad,c) = c  Pad = (m  Pad)  Pad = m
m,c PrPad[E(Pad,m) = c] = 1/2n
No information gained from seeing c
However, E(Pad,m)  E(Pad,m’) = m  m’