Cryptography Reference
In-Depth Information
We will use this inverse to regain the plaintext of our ciphertext example. We get for the
first block
00001
00010
10000
00100
01000
8
8
18
7
19
19
7
8
18
8
=
for the second we get
00001
00010
10000
00100
01000
8
21
17
13
18
18
13
8
17
21
=
and for the third we get
00001
00010
10000
00100
01000
0
23
23
13
0
0
13
0
23
23
=
.
This produces the plaintext blocks
19 7 8 18 8
18 13 8 17 21
0 13 0 23 23
or
THISI
SNIRV
ANAXX.
7.3
COMBINATION SUBSTITUTION/TRANSPOSITION CIPHERS
When substitution ciphers are combined with transposition ciphers, the resulting cipher can
be very hard to crack, especially when the block sizes are different. Now that we have a con-
venient vehicle (matrices) for representing these ciphers, we will discuss how this is done.
Suppose we use a matrix cipher to map blocks of
n
characters from the plaintext
P
to the
ciphertext
C
. Then we regroup this ciphertext into blocks of size
m
and apply a transposi-
tion cipher to these blocks to produce another ciphertext
C
. This, of course, permutes the
characters in each
charac-
ter blocks, some characters find themselves in different blocks. Finally, to put another nail
in the coffin, we can encrypt
m
character block, and since the first encryption was done for
n
C
again using the first matrix cipher to produce the final
ciphertext
. For many ciphers, multiple encryption does not strengthen the cipher; hence
it is often just a waste of time, but in this case, it strengthens the cipher considerably. This
type of cryptosystem confounds any attempt at frequency analysis, and even makes a known
plaintext attack more difficult.
C
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