Cryptography Reference
In-Depth Information
2 1 t , we can transform a prover which succeeds the
Schnorr Identification Protocol with a honest verifier with probability P and time com-
plexity T into an algorithm which finds the secret key x in time complexity O ( T 2 t
Theorem 10.1. Fo r any P
>
/
P ) .
This is the standard way to prove that A must know x , otherwise fail the identification
protocol with probability at least 1
2 1 t . (Obviously, this statement makes sense
when t is small because of the 2 t
factor in the complexity reduction.)
The proof works as follows: let us assume that B follows the protocol correctly
and that A succeeds with probability P to pass the final check. Clearly A must output
a random r which only depends on the initial (random) state
ρ
of A (so let us denote
it r (
ρ
)), and a value s which depends on some e and on
ρ
(so let us denote it s (
ρ,
e )).
The probability is over the distribution of
ρ
and the distribution of e . Let Pr[
ρ
]bethe
probability of having
ρ
as an initial state. Let 1 S ( ρ, e ) be 1 if the final check succeeds
with r (
ρ
), e , and s (
ρ,
e ), and 0 if it fails. We have
2 t 1 S ( ρ, e ) Pr[
P
=
ρ
]
.
ρ, e
Let N (
ρ
) denote the number of possible e 's for which the protocol succeeds. We have
2 t N (
P
=
ρ
) Pr[
ρ
]
.
ρ
1, and 2 t
Clearly, by splitting the sum for N (
ρ
)
N (
ρ
)
2, we have
2 t Pr[ N (
P
ρ
)
1]
+
Pr[ N (
ρ
)
2]
.
Therefore we have
+ 1
2 t Pr[ N (
2 t
P
ρ
)
2]
P 2 t
1
P
hence Pr[ N (
ρ
)
2]
2 . We transform A as follows.
2 t
1. We generate r from a random
ρ
.
2. For all e 's we compute s (
e ) and we check if this succeeds.
3. If we have less than two succeeding e 's, we try again. Otherwise, we have ( e 1 ,
ρ,
s 1 )
and ( e 2 ,
s 2 ) such that
g s 1 y e 1
g s 2 y e 2
r
(mod p )
.
We thus compute
s 1
s 2
x
=
mod q
.
e 1
e 2
Since with probability Pr[ N (
2] we obtain two succeeding e 's, and that this proba-
bility is greater than 2 , this works within a complexity of
ρ
)
(2 t
O
/
P ) times the complexity
of A and B .
 
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