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4. In case of dispute,
can disclose the signer since it knows part of the private
key of each member of G .
In fact, as
T
knows the values of A i and C i of the signer U i ,byusingthe
equations in (6) defining the group signature, it can compute
T
C i
α λ i
F
A i ·
A i ·
·
= α λ i .
(mod n )
C i
C i
A i ·
Then,
T
can prove, without the collaboration of U i ,that
(mod n ) =
F
A i ·
( α λ i )= H i .
h
h
C i
4 Security Analysis
Moreover, the scheme is secure as no member of G ,say U i , knowing only his
own private key, ( b i ,d i ,A i ,C i ), and the shared public key, ( P = α a 0 + s·b 0 ,Q =
α s·c 0 + d 0 ), can determine neither the secret value s of
T
,noritsprivatekey
( a 0 ,b 0 ,c 0 ,d 0 ).
In fact, determining s from α and β
α s (mod n ), see formula (1), means
solving the discrete logarithm problem in the subgroup S r ,oforder r generated
by α , which is impossible as the size of r was chosen such that the SDLP was
unfeasible to solve, and moreover, the factorization of n is infeasible as well.
Moreover, the private key of
was generated at random and it is only known
that it verifies the equation (2), but computing any of the values of this key
implies solving the DLP in
T
Z n .
It is also impossible for any U i to determine the values of h = a i + s
·
b i ,and
c i + d i , as he only knows b i ,d i a i c i . In all cases, it is necessary to
solve a discrete logarithm problem.
Furthermore, two members of G ,say U i
k = s
·
and U j , could conspire and try to
compute any of the secret values of
: s, h, k, a 0 ,b 0 ,c 0 ,d 0 , or generate a false
signature for the group. To carry out any of these attacks, both could generate
their signatures for a message, say ( F i ,G i ,H i )and( F j ,G j ,H j ), respectively.
Then, from the verification identity (7), they have
T
Q m .
F i
·
G i
(mod n )
F j
·
G j = P
·
Hence, they obtain
C i
β b i
α m·d i
C j
β b j
α m·d j
A i ·
·
·
A j
·
·
·
(mod n ) ,
or equivalently,
α a i
β m·c i
β b i
α m·d i
α a j
β m·c j
β b j
α m·d j
·
·
·
·
·
·
(mod n ) ,
and as α has order r modulo n ,itresults
( a i + m
·
d i )+ s ( b i + m
·
c i )
( a j + m
·
d j )+ s ( b j + m
·
c j )(mod r ) .
 
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