Cryptography Reference
In-Depth Information
6.3
Example of Reduction of HS Scheme
Let
K
=
GF
(13) and let us consider a case of HS(
Q
13
(2)
,
3) as a toy example of
Proposition 3. (See
§
4.1 for the definition of
Q
13
(2)) A linear isomorphism
φ
is
defined by
φ
:
K
4
(
c
1
,c
2
,c
3
,c
4
)
→
c
1
+
c
2
i
+
c
3
j
+
c
4
ij
∈
Q
13
(2)
,
1. A central map
G
=(
g
2
, g
3
) is defined by, for
X
=
where
i
2
=2
,j
2
=
−
(
X
1
,X
2
,X
3
)
T
,
⎛
⎞
1+12
i
+11
j
1+8
j
0
0
⎝
⎠
X
g
2
(
X
)=
X
T
0
0
0
0
0
+(1+3
i
+9
j
+8
ij,
5+9
i
+1
j
+10
ij,
0)
X
+9+1
i
+2
j
+6
ij,
⎛
⎞
5+1
i
+2
j
+11
ij
3+10
i
+5
j
+6
ij
11+1
i
+11
j
+5
ij
9+9
i
+10
j
+6
ij
12+8
i
+4
j
+11
ij
12+8
i
+9
j
+12
ij
0
g
3
(
X
)=
X
T
⎝
⎠
X
0
0
+(4+11
i
+3
j
+5
ij,
7+3
i
+6
j
+8
ij,
12+1
i
+8
ij
)
X
+8+11
i
+9
j
Two ane transformation
A
1
:
K
8
K
8
and
A
2
:
K
12
K
12
, which are the
→
→
rest of secret key, are defined by
A
1
(
x
)=
L
1
y
+
v
1
,A
2
(
x
)=
L
2
x
+
v
2
,
⎛
⎝
⎞
⎠
11272 51212891134
91199 2 412552 06
1213971994464
127737381179125
7 063111 31227118
5410612750111
698840651054
10681001010761115
48916781151124
71000 9117 822 36
7 593 8 312567 28
5 099 4 811251189
⎛
⎝
⎞
⎠
7937111125
627115705
04852254
109 51110301
2571232212
101104 7 5711
0 011710241
012113 7 7011
L
1
=
,L
2
=
,
v
1
=(12
,
10
,
1
,
11
,
3
,
10
,
9
,
6)
T
,
2
=(5
,
2
,
11
,
9
,
1
,
11
,
12
,
4
,
8
,
7
,
4
,
9)
T
.
F
=(
f
5
, f
6
,..., f
12
) is described as
Then the public key
f
k
(
x
)=
x
T
A
k
x
+
b
k
x
+
c
k
