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and the second and third mappings follow:
19
a
+ 0
b
+
s
6 (mod 26)
19
c
+ 0
d
+
t
19 (mod 26)
2
a
+ 10
b
+
s
7 (mod 26)
2
c
+ 10
d
+
t
12 (mod 26)
She then rearranges the congruences to get two systems
0
a
+ 19
b
+
s
1 (mod 26)
19
a
+ 0
b
+
s
6 (mod 26)
2
a
+ 10
b
+
s
7 (mod 26)
and
0
c
+ 19
d
+
t
19 (mod 26)
19
c
+ 0
d
+
t
19 (mod 26)
2
c
+ 10
d
+
t
12 (mod 26).
She solves the first system to obtain values for
a
,
b
, and
s
, and the second system to get
the values for
c
,
d
, and
t
. Since the coefficients of the two systems are the same, she can solve
them simultaneously.
0191119
19 01619
2101712
She then proceeds to reduce the matrix, say first by multiplying row 2 by 11, an inverse
of 19 modulo 26, then by swapping row 2 with row 1, then by subtracting row 1 from row
3. This yields
.
1 0 11 14 1
019 1 119
010 5 510
She then multiplies row 2 by 11, then subtracts 10 times row 2 from row 3
.
1011141
0111111
0025250
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