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160 CHAPTER 8. POLYALPHABETIC CIPHERS
18. (a) Finish the computation of Φ for a text consisting of English with the
normal distribution started in section 8.3. (Use Figure 1.3. So there
are 8.2 A’s, 1.5 B’s, and N = 100.)
(b) Compute the value of Φ, if the plaintext consisted of English with the
normal distribution that was enciphered with a Caesar cipher with
key B.
(c) Compute the value of Φ, if the plaintext consisted of English with
the normal distribution that was enciphered with a monoalphabetic
cipher.
(d) Compute Φ for the ciphertext that has all 26 letters occurring an
equal number of times.
19. In this exercise we work out the formula for Φ given in Equation 8.3.
Assume we have a ciphertext of N letters enciphered with a Vigenère
cipher with keylength k. As we did in Section 7.7, lets assume we’ve
written the ciphertext in an array, so there are k rows, each containing all
the ciphtertext letters that were enciphered with the same keyletter.
Now Φ is the probability that two letters, chosen at random from the text,
are the same. Either these letters were chosen from the same row, or from
different rows.
Same row: There are N total letters and k rows, so about N/k letters per
row. Hence there are N k ( N k
− 1)/2 ways to choose two from any particular
row. There are k rows and so k N k ( N k
− 1)/2 ways to choose two letters a
row. Finally, the likelihood that two letters are the same in any of these
rows is 0.0656. So the contribution here is
.0656 ∗ k
N
k ( N k − 1) N(N − k)
= .0656
2
2k
Different rows: There are k(k − 1)/2 ways to pick to two rows. There are
N/k letters in each row, so there are N/k ∗ N/k ways to pick one letter
from each. Finally, we have no way of relating the keyletters from the
different rows, so we can only approximate the likelihood that these two
letters are the same with 1/26. So the contribution here is
1
26 ∗ N N k(k − 1)
∗ = 1 N 2 (k − 1)
k k 2 26 k
Adding the two probabilities, and simplifying,