Cryptography Reference
In-Depth Information
occurs approximately equally often). We achieve this by increasing the ciphertext
alphabet.
A possible homophonic code for use with English plaintexts can be devised
from Table 2.1 of English letter frequencies, as follows. Suppose that we choose a
ciphertext alphabet of 1000 symbols. This means that instead of 26 letters, each
ciphertext character is one of these 1000 symbols. We then secretly divide the
1000 ciphertext symbols into 26 groups. Each group is then associated with one
specific plaintext letter. From Table 2.1 we see that:
• Plaintext letter A occurs approximately 8
.
2% of the time, so we assign
82 ciphertext symbols for encrypting A.
• Plaintext letter B occurs approximately 1
.
5% of the time, so we assign
15 ciphertext symbols for encrypting B.
• Plaintext letter C occurs approximately 2
.
8% of the time, so we assign
28 ciphertext symbols for encrypting C.
We continue this process for the whole plaintext alphabet. Thus, for example, the
letter E is assigned 127 ciphertext symbols, the letter T is assigned 90 ciphertext
symbols, and the letters J, Q, Y and Z are all assigned only one ciphertext symbol.
To encrypt a plaintext letter, one of the ciphertext symbols in the group
associated with that letter is chosen at random by the sender Alice. It does not
matter which symbol is chosen, but it does matter that the process used to choose
it is random. The receiver Bob, who also knows how the 1000 ciphertext symbols
have been divided and assigned to plaintext letters, is able to decrypt this ciphertext
letter since there is only one plaintext letter assigned to any given ciphertext
symbol.
SINGLE LETTER FREQUENCY ANALYSIS OF HOMOPHONIC ENCODING
To see the advantage of homophonic encoding, think now about trying to conduct
single letter frequency analysis. We assume once again that the plaintext is in
English.
Suppose that we have a ciphertext consisting of 1000 symbols, which encrypts
a plaintext that consists of 1000 English letters. We know from Table 2.1 that
amongst these 1000 letters of the plaintext we can expect:
• One occurrence of the plaintext letter Z. Hence the ciphertext symbol
associated with Z should occur just once amongst the 1000 ciphertext
symbols.
• 127 occurrences of the plaintext letter E. However, every time that E is
encrypted using our homophonic code, it is replaced by a randomly selected
symbol from the group of 127 ciphertext symbols associated with E.
If we are
extremely lucky
then amongst these 1000 ciphertext symbols each
ciphertext symbol could occur just once. In practice (with very high probability)
this will not happen, since some of the ciphertext symbols will occur more than
once and a few will not occur at all. However, it should be clear that for very long
