Cryptography Reference
In-Depth Information
addition modulo 26; that is, all results are within the 0-25 (A-Z) range. If
the plaintext is “philosophy,” the process is as follows:
p
(16)
h
(8)
i
(9)
l
(12)
o
(15)
s
(19)
o
(15)
p
(16)
h
(8)
y
(25)
+
(5)
(5)
(5)
(5)
(5)
(5)
(5)
(5)
(5)
(5)
=
u
(21)
m
(13)
n
(14)
q
(17)
t
(20)
x
(24)
t
(20)
u
(21)
m
(13)
d
(4)
The resulting ciphertext would read “umnqtxtumd.”
Such an
algorithm
provides little security, because the key space is too
small, as there are only twenty-five possible keys, corresponding to the
twenty-five possible substitutions. To cryptanalyze the resulting ciphertext,
the adversary can merely use an
exhaustive key search
(or
brute-force
attack
), that is, try each key until a meaningful message emerges, assuming
of course he is aware that the message has been enciphered using this
technique.
The ciphertext would also easily fall to
statistical analysis
, because mono-
alphabetic substitution preserves all of the statistical properties of the
plaintext's language. These properties include the frequency distribution
of individual letters in the English language (
e
occurs about 13% of the
time;
v
,
k
,
j
,
x
,
q
, and
z
, less than 1%) as well as that of pairs and triplets
(the most frequent pairs in English are
th
and
er
). Cryptanalysis can thus
begin by simply replacing the most frequently occurring symbols of the
ciphertext with the most frequent letters of the plaintext language and
proceed on the basis of this partial solution.
Polyalphabetic Substitution
Here the substitution is more complex, as identical plaintext letters can
correspond to different symbols of the ciphertext. The classic Vigenère
cipher is based on the following algorithm: the plaintext is chopped into
blocks of
m
letters (say, 5) and a key of length
m
is chosen (say, “plato”).
The letters of the plaintext are then added, modulo 26, to the letters of
the key:
p
(16)
h
(8)
i
(9)
l
(12)
o
(15)
s
(19)
o
(15)
p
(16)
h
(8)
y
(25)
+
p
(16)
l
(12)
a
(1)
t
(20)
o
(15)
p
(16)
l
(12)
a
(1)
t
(20)
o
(15)
=
f
(6)
t
(20)
j
(10)
f
(6)
d
(4)
g
(7)
a
(1)
q
(17)
b
(2)
n
(14)
