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Cryptanalysis. Four kinds of attacks (recall) The objective: determine the key ( Herckhoff principle ) Assumption: English plaintext text Basic techniques: frequency analysis based on: Probabilities of occurrences of 26 letters Common digrams and trigrams.

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cryptanalysis
Cryptanalysis
  • Four kinds of attacks (recall)
  • The objective: determine the key (Herckhoff principle)
  • Assumption: English plaintext text
  • Basic techniques: frequency analysis based on:
    • Probabilities of occurrences of 26 letters
    • Common digrams and trigrams.
cryptanalysis statistical analysis
Cryptanalysis -- statistical analysis
  • Probabilities of occurrences of 26 letters
    • E, having probability about 0.120 (12%)
    • T,A,O,I,N,S,H,R, each between 0.06 and 0.09
    • D,L, each around 0.04
    • C,U,M,W,F,G,Y,P,B, each between 0.015 and 0.028
    • V,K,J,X,Q,Z, each less than 0.01
    • See table 1.1, page 26
  • 30 common digrams (in decreasing order):
    • TH, HE, IN, ER, AN, RE,…
  • 12 common trigrams (in decreasing order):
    • THE, ING,AND,HER,ERE,…
cryptanalysis of affine cipher
Cryptanalysis of Affine Cipher
  • Suppose a attacker got the following Affine cipher
    • FMXVEDKAPHFERBNDKRXRSREFNORUDSDKDVSHVUFEDKAPRKDLYEVLRHHRH
  • Cryptanalysis steps:
    • Compute the frequency of occurrences of letters
      • R: 8, D:7, E,H,K:5, F,S,V: 4 (see table 1.2, page 27)
    • Guess the letters, solve the equations, decrypt the cipher, judge correct or not.
      • First guess: Re, Dt, i.e., eK(4)=17, eK(19)=3
        • Thus, 4a+b=17  a=6, b=19, since gcd (6,26)=2, so incorrect.

19a+b=3

      • Next guess: Re, Et, the result will be a=13, not correct.
      • Guess again: Re, Ht, the result will be a=8, not correct again.
      • Guess again: Re, Kt, the result will be a=3, b=5.
        • K=(3,5), eK(x)=3x+5 mod 26, and dK(y)=9y-19 mod 26.
        • Decrypt the cipher: algorithmsarequitegeneraldefinitionsofarithmeticprocesses
      • If the decrypted text is not meaningful, try another guess.
  • Need programming: compute frequency and solve equations
  • Since Affine cipher has 12*26=312 keys, can write a program to try all keys.
cryptanalysis of substitution cipher
Cryptanalysis of substitution cipher
  • Final goal is to find the corresponding plaintext letter for each ciphertext letter.
  • Ciphertext: example 1.11, page 28
  • Steps:
    • Frequency computation, see table 1.3, page 29
      • Guess Ze, quite sure
      • C,D,F,J,M,R,Y are t,a,o,i,n,s,h,r, but not exact
    • Look at digrams, especially –Z or Z-.
      • Since ZW occurs 4 times, but no WZ, so guess Wd (because ed is a common digram, but not de)
      • Continue to guess
    • Look at the trigrams, especially THE, ING, AND,…
cryptanalysis of vigenere cipher
Cryptanalysis of Vigenere cipher
  • In some sense, the cryptanalysis of Vigenere cipher is a systematic method and can be totally programmed.
  • Step 1: determine the length m of the keyword
    • Kasiski test and index of coincidence
  • Step 2: determine K=(k1,k2,…,km)
    • Determine each ki separately.
kasisk i test determine keyword length m
Kasiski test—determine keyword length m
  • Observation: two identical plaintext segments will be encrypted to the same ciphertext whenever they appear  positions apart in plaintext, where 0 mod m. Vice Versa.
  • So search ciphertext for pairs of identical segments, record the distance between their starting positions, such as 1, 2,…, then m should divide all of i’s. i.e., m divides gcd of all i’s.
index of coincidence

25

Index of coincidence
  • Can be used to determine m as well as to confirm m, determined by Kasiskitest
  • Definition: suppose x=x1x2,…,xn is a string of length n. The index of coincidence of x, denoted by Ic(x), is defined to be the probability that two random elements of x are identical.
    • Denoted the frequencies of A,B,…,Z in x by f0,f1,…,f25

25

fi

2

( )

fi(fi-1)

i=0

i=0

=

( Formula IC)

--Ic(x)=

n

2

n(n-1)

()

slide8

Index of coincidence (cont.)

Suppose x is a string of English text, denote the expected

probability of occurrences of A,B,…,Z by p0,p1,…,p25 with

values from table 1.1, then Ic(x)  pi2 =0.0822+0.0152+…+0.0012=0.065

(since the probability that two random elements both are A is p02, both are B is p12,…)

Question: if y is a ciphertext obtained by shift cipher, what is the Ic(y)?

Answer: should be 0.065, because the individual probabilities will be

permuted, but the pi2 will be unchanged.

Therefore, supposey=y1y2…yn is the ciphertext from Vigenere cipher.

For any given m, divide yinto m substrings:

y1=y1ym+1y2m+1… if m is indeed the keyword length, then

y2=y2ym+2y2m+2… each yi is a shift cipher, Ic(yi) is about 0.065.

ym=ymy2my3m… otherwise, Ic(yi)  26(1/26)2 = 0.038.

slide9

Index of coincidence (cont.)

For purpose of verify keyword length m, divide the ciphertext into m substrings,

compute the index of coincidence by formula IC for each substring. If all IC values

of the substrings are around 0.065, then m is the correct keyword length. Otherwise

m is not the correct keyword length.

If want to use Ic to determine correct keyword length m, what to do?

Beginning from m=2,3, … until an m, for which all substrings have

IC value around 0.065.

Now, how to determine keyword K=(k1,k2,…,km)? Assume m is given.

slide10

f0f25

n′, , n′

f0+ki

n\'

p0

Determine keyword K=(k1,k2,…,km)

  • Determine each ki (from yi) independently.
  • Observation:
    • 2.1 let f0,f1,…,f25 denote the frequencies of A,B,…,Z in yi and n′=n/m
    • 2.2 then probability distribution of 26 letters in yi is:

2.3 if the shift key is ki, then f0+ki (i.e., A+ki) is the frequency of

a in the corresponding plaintext xi , …, f25+ki (note the subscript

25+ki should be computed by modulo 26) is the frequency

of z in xi. Since xi is normal English text, probability distribution

of

f0+kif25+ki

n′, , n′

should be “close to” ideal probability

distribution p0,p1,…,p25.

f25+ki

n\'

So:

p25

+…+

p0, …, p25

p02+…+p252 =0.065

slide11

Mg=i=025pi

n′

f0+g

n\'

p0

Determine keyword K=(k1,k2,…,km) (cont.)

On the other hand, for any g !=ki,

f25+g

n\'

will not be close to 0.065.

p25

+…+

Therefore, define:

fi+g

When g=ki, Mg will generally be around 0.065 (i.e., i=025pi2).

Otherwise Mg will be quite smaller than 0.065.

So let g from 0, until 25, compute Mg, and for some g, if Mg

is around 0.065, then ki=g.

Note: the subscript i+g should be seen as modulo 26.

cryptanalysis of vigenere cipher example

Mg=i=025pi

n′

Cryptanalysis of Vigenere cipher--example
  • Example 1.12, page 33.
    • Using Kasiski test to determine the keyword length
      • CHR appears five times at 1,166,236,276,286
      • the distance is 165, 235,275,285, the gcd is 5, so m=5.
    • Using index of coincidence to verify m=5.
      • Divide ciphertext into y1, y2, y3, y4, y5
      • Compute f0,f1,…,f25 for each yi and then Ic(yi), get 0.063,

0.068,0.069,0.061,0.072, so m=5 is correct.

    • Determine ki for i=1,…,5.
      • Compute Mg for g=0,1,…,25 and if Mg 0.065, then let ki=g. where

fi+g

As a result, k1=9,k2=0,k3=13,k4=4,k5=19, i.e., JANET

cryptanalysis of hill cipher
Cryptanalysis of Hill cipher
  • Difficult to break based on ciphertext only
  • Easily to break based on both ciphertext and plaintext.
  • Suppose given at least m distinct plaintext-ciphertext pairs: xj=(x1,j,x2,j,…,xm,j)

yj=(y1,j,y2,j,…,ym,j)

then define two matrices X=(xi,j) and Y=(yi,j)

Let Y=XK, if X is invertible, then K=X-1Y.

cryptanalysis of hill cipher example
Cryptanalysis of Hill cipher--example
  • Suppose plaintext is: friday

and ciphertext is: PQCFKU

and the m=2. Then eK(f,r)=(P,Q), eK(i,d)=(C,F).

That is:

15 16

2 5

5 17

8 3

( ) = ( )K

5 17

8 3

15 16

2 5

9 1

2 15

15 16

2 5

7 19

8 3

K=( )-1( )=( )( )=( )

Then using the third pair, i.e., (a,y) and (K,U) to verify K.

In case m is unknown, try m=2,3, …

cryptanalysis of lfsr stream cipher
Cryptanalysis of LFSR stream cipher
  • Vulnerable to known-plaintext attack.
  • Suppose m, plaintext binary string x1,x2,…,xn and ciphertext binary string y1,y2,…,yn are known, as long as n>2m, the key can be broken:
    • Keystream is: zi=(xi+yi) mod 2. (i=1,2,…,n)
    • Thenthe initialization vector of K is z1,…, zm.
    • Next is to determine coefficients (c0,c1,…,cm-1) of K

(recall that zi+m = m-1j=0cjzi+jmod 2 for all i1)

i.e,

z1 z2 … zm

z2z3 … zm+1

……………

zmzm+1 … z2m-1

  • (zm+1,zm+2,…,z2m)=(c0,c1,…,cm-1)
cryptanalysis of lfsr stream cipher cont

z1 z2 … zm

z2z3 … zm+1

……………

zmzm+1 … z2m-1

  • (c0,c1,…,cm-1)=(zm+1,zm+2,…,z2m)
Cryptanalysis of LFSR stream cipher (cont.)

Therefore:

-1

slide17

Cryptanalysis of LFSR stream cipher --example

Example 1.14, page 37. Suppose LFSR 5 with the following:

Ciphertext string: 101101011110010

Plaintext string: 011001111111000

Then keystream: 110100100001010

Therefore initialization vector is: 11010.

For next five key elements: 01000, set up equation

for coefficients (c0,c1,c2,c3,c4) and solve it.

The result is: (c0,c1,c2,c3,c4) =(1,0,0,1,0)

i.e., zi+5=(zi+zi+3) mod 2.

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