Information theory

Cryptography and Authentication

A.J. Han VinckEssen, 2003

Cryptographic model

sender receiver attacker

secrecy encrypt M decrypt M read M

find key

authentication sign test validity modify

generate

General (classical) Communication Model

source M encrypter C decrypter M destination

analyst M‘

Secure key channel

K K

no information providing ciphers

ShannonShannon (1949): Perfect secrecy condition

Prob. distribution (M) = Prob. distribution (M|C)

and thus: H(M|C) = H(M)

(no gain if we want to guess the message given the cipher)

Perfect secrecy condition

Furthermore: for perfect secrecy H(M) H(K)

H(M|C) H(MK|C) = H(K|C) + H(M|CK)

C and K M

= H(K|C) H(K)

H(M) = H(M|C) H(K) perfect secrecy!

Imperfect secrecy

How much ciphertext do we need to find

unique key-message pair

given the cipher?

Minimum is called unicity distance.

Imperfect secrecy

Suppose we observe a piece of ciphertext

Key K, H(K)

source M, H(M)

Cipher C

CL

K and ML determineCL

Key equivocation: H(K| CL) = H(K,CL) – H(CL)

H(CL ) Llog2|C|

Question: When is H(K| CL) = 0 ?

H(K| CL) = H(K) + H(CL|K) – H(CL)

= H(K) + H(ML) – H(CL)

Let HS(M) be the normalized entropy per output symbol

U = the least value of L such that H(K| CL) = 0

Using: H(CL ) Llog2|C|; H(ML) LHS(M)

U H(K) / [ log2|C|- HS (M)]

( K ML CL )

conclusion

Make HS (M) as large as possible:

USE DATA REDUCTION !!

H(K| CL)H(K)

L

U is called the unicity point

H(K) / [ log2|C|- HS (M)]

examples: U H(K) / [ log2|C|- HS(M) ]

Substitution cipher: H(K) = log2 26!

English: HS(M) 2; |M|= |C|=26; U 32 symbols

DES: U 56 / [ 8 – 2 ] 9 ASCII symbols

examples: U H(K) / [ log2|C|- HS(M) ]

Permutation cipher: period 26; H(K) = log226!

English: H(M) 2; |M|= |C|=26; U 32 symbols

Vigenere: key length 80, U 140 !

Plaintext-ciphertext attack

H(K,ML, CL) = H(K|ML, CL) + H( CL|ML ) + H(ML)

= H(CL|K,ML) + H( K|ML ) + H(ML)

H(K|ML, CL) = H(K) - H( CL|ML )

H( CL|ML ) Llog2|C|

thus: U H(K) / log2|C|

CL ← K,ML K independent from ML

Wiretapping model

sender noiselesss channel receiver

noise

wiretapper

Send: n binary digits

0: Xn of even weight

1: Xn of odd weight

Wiretapper:

Pe = P(Zn has odd # of errors)

= 1- ½(1+(1-2p)n)

S

Zn

Xn Xn

Wiretapping

Result:

for p ½ Pe ½

and H(S|Zn) = 1

for p 0, Pe np

and H(S|Zn) h(np)

Pe = 1- ½(1+(1-2p)n)

Wiretapping general strategy

Encoder: use R = k/n error correcting code C

carrier c { 2k codewords }

message m { 2nh(p) vectors as correctable noise }

select c at random and transmit c m

Note: 2k 2nh(p) 2n k/n 1 – h(p)

Communication sender-receiver

transmit = receive : c m

first decode: c

calculate m = c m c = m

m c

c m

Wiretapper:

receive z = c m n‘

- first decode: c

possible when m n‘ is decodable noise

- calculate: c m n‘ c = m n‘

m‘ = m n‘ is one of 2nh(p‘) messages

the # of noise sequences n‘ is |n‘ | ~ 2nh(p‘)

c m

n‘

z = c m n‘

Wiretapping general strategy

Result: information rate R = h(p)

p‘ small: c decodable and H(Sk|Zn) = nh(p‘)

p‘ p: H(Sk|Zn) = nh(p)H(Sk|Zn) nh(p)

P P‘

Wiretapping general strategy picture

Volume 2nh(p‘)

Volume 2nh(p)

codeword

2k codewords

2n vectors

authentication

Encryption table:

message X

Key K ( X, K ) Y

unique cipher: Y

Authentication: impersonation

message: 0 1 select y at random

00 00 10 Pi (y = correct) = ½

key 01 01 00 P(key = i ) = 1/4

10 11 01 P(message = i ) = 1/2

11 10 11 cipherPi is probability that an injected cipher is valid

Authentication: bound

Let : |X| # messages

|K| # keys

|Y| # ciphers

Pi prob (random cipher = valid) |X|/|Y|

= probability that we choose one of the colors in a specific row

´ specified by the key

|X|

|K|

Cont‘d

Since: ( Y, K) X H(X) = H(Y|K)

An improved (Simmons) bound gives:

Pi 2H(X)/2H(Y) = 2H(Y|K)-H(Y) = 2-I(Y;K)

Cont‘d

Pi 2-I(Y;K) = 2+ H(K |Y) - H(K)

For low probability of success: H(K|Y) = 0

For perfect secrecy: H(K|Y) = H(K)

Contradiction!

Cont‘d

0 1

0 00 01

1 10 11

0 1

0 00 01

1 01 00

prob success = ½ prob success = 1

H(K|Y) = 0 H(K|Y) = 1

no secrecy perfect secrecy

Prob ( key = 0 ) = Prob ( key = 0 ) = ½; Prob (X = 0) = Prob(X =

1) = 1/2

Authentication: impersonation

X= 0 1 P(X=0) = P(X=1)= ½

K= 0 0 1 P(K=0) = P(K=1)= ½

1 1 2 H(K) =1; H(K|Y) = ½

P(0,1,2) = (¼, ½, ¼) Pi = ½+2x¼x½= 0.75

random choise: Pi = 2/3

Pi 2H(Y|K)-H(Y) = 21-1.5 = 2 –0.5 = 0.7

Authentication: substitution

message

0 1

0 0 2

Key 1 1 3

2 0 3

3 1 2 cipher

Active wiretapping:

replace an observed cipher by another cipher

Example: observe 0 replace by 3

probability of success = ½ (accepted only if key = 2)

Authentication: substitution examples

0 1

0 0 2

1 1 3

2 0 3

3 1 2

0 1

0 3

1 2

2 1

3 0

0 1

0 2

1 0

3 1

2 3

Ps = ½

H(X|Y) = 0

Ps = 1

H(X|Y) = 1

Ps = ½

H(X|Y) = 1

Ps = probability( substitution is successful)

H(K) = 2; H(K|Y) = 1;Pi ½