Cryptography Reference
In-Depth Information
Alice computes
m
j
=
f
−
1
(
r
j
), and sends the values
x
j
=
m
j
⊕
2.
s
j
(where
⊕
is the mod 2 addition of the bitstrings), for
j
=1
,
2
,...,
,
to Bob.
3. Bob knows
m
k
=
f
−
1
(
f
(
r
k
)) =
r
k
, so Bob can compute
m
k
⊕
x
k
=
s
k
.
Show that Bob cannot compute
s
j
for
j
=
k
. Explain how Bob can cheat
Alice by altering his computation of
c
j
in step 1.
5.2. Explain how Kerberos (see pages 196 and 197), is vulnerable with respect
to (1) host security; (2) Carol's password (encryption keys); and (3) o^ine
attacks on Carol's ticket.
5.3. Show how the three-way authentication protocol described on page 198
would be vulnerable to the man-in-the middle attack (see Footnote 3.7 on
page 134), if the DSS is not employed. In other words, if signatures are
not employed, show how Mallory can impersonate Alice and successfully
convince Bob that he is talking to her.
In Exercises 5.4-5.7, use the Fiege-Fiat Shamir identification protocol pre-
sented on pages 202 and 203 to show that Bob should accept Alice's proof,
given the parameters in each case.
5.4. Let
p
= 523
1637 = 856151,
s
A
=5,
a
= 2, and assume that in round
1, Alice selects
m
= 651, and Bob chooses
c
= 0, while in round 2, Alice
picks
m
= 1516 and Bob selects
c
=0.
·
5.5. Let
p
= 613
2281 = 1398253,
s
A
=7,
a
= 2, and assume that in round
1, Alice selects
m
= 3291, and Bob chooses
c
= 1, while in round 2, Alice
picks
m
= 1923 and Bob selects
c
=1.
·
5.6.
2557 = 1889623,
s
A
= 25,
a
= 3, and assume that in
round 1, Alice selects
m
= 3681, and Bob chooses
c
= 1; in round 2, Alice
picks
m
= 111 and Bob selects
c
= 0; and in the third round Alice picks
m
= 38888 and Bob chooses
c
=1.
Let
p
= 739
·
5.7.
3323 = 2847811,
s
A
= 49,
a
= 3, and assume that in
round 1, Alice selects
m
= 333, and Bob chooses
c
= 1; in round 2, Alice
picks
m
= 723 and Bob selects
c
= 1; and in the third round Alice picks
m
= 111111 and Bob chooses
c
=1.
Let
p
= 857
·
In Exercises 5.8-5.11, employ the Schnorr identification protocol delineated
on page 205 to show that Bob should accept Alice's identity. As usual,
the parameters are artificially small to make computation reasonable. In
other words, we are not choosing
p
40
, for
instance. We will assume that all certificates and signature verifications
have taken place. All that is required is the calculation of the commitment,
the response, and the verification as outlined in steps 1-4 of the protocol
given on the aforementioned page.
≥
2
1024
,or
q>
2
t
with
t
≥
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