Information Theory and Coding - Codes: The Guide to Secrecy from Ancient to Modern Times - page 435

then the probabilities for the ciphertexts, p C ( c j ) , for j =1 , 2 , 3 , 4 , is derived as
follows:

p c 1 = p k 1 · p s 1 + p k 2 · p s 4 + p k 3 · p s 3 =

0 . 3 · 0 . 1+0 . 3 · 0 . 4+0 . 4 · 0 . 3=0 . 27 ,

and similarly

p c 2 =0 . 25 , p c 3 =0 . 19 , and p c 4 =0 . 29 .
Now we may calculate some conditional probabilities to determine key equiv-
alence. First, we look at individual entropies.
Using the fact cited in Footnote 11.4 on page 432, we calculate the following:

p k 1 | c 1 = p k 1 ·

p s 1
p c 1

= 0 . 3

0 . 1
0 . 27 ≈

·

0 . 1111 ,

and similarly,

p k 1 | c 2 ≈ 0 . 2400; p k 1 | c 3 ≈ 0 . 4737; p k 1 | c 4 ≈ 0 . 4138;

p k 2 | c 1 ≈ 0 . 444; p k 2 | c 2 ≈ 0 . 1200; p k 2 | c 3 ≈ 0 . 3158; p k 2 | c 4 ≈ 0 . 3103;

p k 3 | c 1 ≈ 0 . 444; p k 3 | c 2 ≈ 0 . 6400; p k 3 | c 3 ≈ 0 . 2105; p k 3 | c 4 ≈ 0 . 2759 .

Thus, using Equation (11.3) as a formula, we get

H ( K | C ) ≈ 1 . 4342 .

Now we calculate, using Equation (11.1) ,

H ( K )= − 0 . 3log 2 (0 . 3) − 0 . 3log 2 (0 . 3) − 0 . 4log 2 (0 . 4) ≈ 1 . 5709 ,

and similarly, H ( M ) ≈ 1 . 8464 , H ( C ) ≈ 1 . 9831 . Therefore,

H ( K )+ H ( M ) − H ( C ) ≈ 1 . 4342 ,

which agrees with Equation (11.8) , as an illustration.

This tells us that when Eve intercepts encrypted conversation between Bob
and Alice, she obtains H ( ) information about the key, and this is deter-

mined bythe right-hand side of Equation (11.8). Moreover, this is, of course, a
ciphertext-onlyattack.
Now define

K | C

C

n to denote all n -grams (ciphertexts of length n ), and similarly

M

n will denote all n -grams of plaintext, with associated probabilitydistribu-
tions, then as with Equation (11.8), we have

n )= H ( n ) n ) ,

H ( K | C K )+ H ( M − H ( C (11.9)