Rolling Dev's blog

Information Security: 3. Classical Encryption Technique

yoonmin's photo
yoonmin
·

3 min read

Table of contents

3.1 Symmetric Cipher Model

  • Steps:
    • Plaintext input
    • Encryption algorithm: Y = E(K, X)
    • Ciphertext
    • Decryption algorithm: X = D(K, Y)
    • Plaintext output
  • uses same key (symmetric)
  • classic model

Attacking Crypto System

  1. Cryptanalysis
  2. Brute-force attack

An Encryption Algorithm is said...

  1. Unconditionally secure
    • unsolvable
  2. Competitionally secure
    • solvable, but high cost & time consuming

3.2 Substitution Techniques

3.2.1 Caesar Cipher

  • c = E(k, p) = (p + k) mod 26
  • p = D(k, c) = (c - k) mod 26

3.2.2 Monoalphabetic Cipher

  • key space: 26!
  • brute force -> impossible
  • use relative frequency of letters in english text
  • Digram (substitute double)

3.2.3 Playfair Cipher

  • Key and table
    • if key = monarchy, then table is...
      m  o  n  a  r
c  h  y  b  d
e  f  g  ij k
l  p  q  s  t
u  v  w  x  z
  • Plaintext
    • split by every 2 letters (if odd, insert 'x' anywhere)
    • ballon -> ba lx lo on
  • Encryption: if two letters on the...
    • same row: right (ex: ux -> vz)
    • same column: below (ex: is -> sx)
    • else: make rectangle, go horizontal (ex: hs -> bp)
  • Not much effective

3.2.4 Hill Cipher

  • Matrix math
    • Inverse element: A + I = E
      A = 5 8
17 3
B = 9 2
1 15
AB = 53 130
 156 79
AB mod 26 = 1 0
        0 1
  • 0~25 on alphabets (if bigger, mod 26)
  • Encryption: C = PK mod 26
    • Key: m*m matrix => K
    • Plaintext: 1*m matrix
    • Cipher: 1*m matrix
    • Example
      K = (5 8)
   (17 3)
P = "HI" = (7 8)
C = PK mod 26 = (171 80) mod 26 = (15 2) = "PC"
  • Decryption: P = C (K^-1 mod 26)
    (15 2) (9 2)
       (1 15)
= (137 60) mod 26
= (7 8)
= "HI"
  • Is it good?
    • still has relative frequency problem
    • known plaintext attack

Known plaintext attack

  • Y = XK
  • If X and Y are known, the key is K = X^-1 Y.

3.2.5 Vigenere Cipher

  • From Caesar cipher, when len(K) != len(P) ...
    K = k[0] k[1] ... k[m-1]
P = p[0] p[1] ... p[n-1]
C = c[0] c[1] ... c[n-1]
C = (p[0]+k[0])mod26 ... (p[m-1]+k[m-1])mod26 (p[m]+k[0])mod26 ...
  • Example
    K = "deceptive"
P = "wearediscoveredsaveyourself"
C =
  wearedisc overedsav eyourself
  deceptive deceptive deceptive +
  _______________________________
  zicv...

  • Is it good?

    • still has relative frequency problem
    • all we need to know to decode cipher is the length of K.

  • Variant

    C =
  wearedisc overedsav eyourself
  deceptive wearedesc overedsav +
  _______________________________
  ...

Relative Frequencies

image.png

3.2.7 Vernam Cipher

c[i] = p[i] ⊕ k[i] p[i] = c[i] ⊕ k[i] 

p[i] = c[i] ⊕ k[i]
 = (p[i] ⊕ k[i]) ⊕ k[i]
 = p[i] ⊕ (k[i] ⊕ k[i])
 = p[i] ⊕ 0
 = p[i]
  • if c and p are known, k can be also found.

3.2.8 One-Time Pad

cA = E(A, kA) cB = E(B, kB) ...

  • In practice,
    • generate random key each time
    • or share keys

3.3. Transposition Techniques

3.3.1 Rail Fence

  • example
    meetmeaftertheparty
->
m e n a t r h p r y
e t e f e t r e a t

3.3.2

  • example
    P = meetmeaftertheparty
K = 4312567, xy
table: 4 3 1 2 5 6 7
     m e e t m e a
     f t e r t h e
     p a r t y x y
C = eer/trt/eta/mfp/mty/ehx/aey
  • If this is not enough, encode the first cipher in the same way again.
information security