Cryptography Reference
In-Depth Information
11.11.
p
s
1
with
p
k
1
=0
.
3,
p
k
2
=0
.
3,
p
k
3
=0
.
4,
E
k
1
(
s
1
)=
c
1
,
E
k
1
(
s
2
)=
c
2
,
E
k
1
(
s
3
)=
c
3
,
E
k
1
(
s
4
)=
c
4
,
E
k
2
(
s
1
)=
c
4
,
E
k
2
(
s
2
)=
c
3
,
E
k
2
(
s
3
)=
c
2
,
E
k
2
(
s
4
)=
c
1
,
E
k
3
(
s
1
)=
c
2
,
E
k
3
(
s
2
)=
c
3
,
E
k
3
(
s
3
)=
c
4
, and
E
k
3
(
s
4
)=
c
1
.
11.12.
p
s
1
=0
.
3,
p
s
2
=0
.
3,
p
s
3
=0
.
3,
p
s
4
=0
.
1,
=0
.
2,
p
s
2
=0
.
2,
p
s
3
=0
.
2,
p
s
4
=0
.
4,
K
=
{
k
1
,k
2
,k
3
}
with
p
k
1
=0
.
1,
p
k
2
=0
.
1,
p
k
3
=0
.
8,
E
k
1
(
s
1
)=
c
1
,
E
k
1
(
s
2
)=
c
2
,
E
k
1
(
s
3
)=
c
3
,
E
k
1
(
s
4
)=
c
4
,
E
k
2
(
s
1
)=
c
2
,
E
k
2
(
s
2
)=
c
3
,
E
k
2
(
s
3
)=
c
4
,
E
k
2
(
s
4
)=
c
1
,
E
k
3
(
s
1
)=
c
1
,
E
k
3
(
s
2
)=
c
3
,
E
k
3
(
s
3
)=
c
2
, and
E
k
3
(
s
4
)=
c
4
.
K
=
{
k
1
,k
2
,k
3
}
11.13. Prove that if
M
is the message space and
C
is the ciphertext space for
a one-time pad, then
H
(
M
|
C
)=
H
(
M
). (See pages 439 and 440.)
11.14. Calculate the unicity distance of a block cipher with 56-bit keys and
64-bit blocks of plaintext, assuming use of the English language. (See page
439.)
11.15. In Section 11.5, we looked at error-correcting codes. This topic has an
ISBN number,
G.3
meaning an
International Standard Book Number
, given
by, 1
3. The first digit, 1 refers to the country, area, or
language group, in this case the fact that the topic is published in the
English language. The second group 58488 identifies the publisher. The
third group 470 identifies this particular topic, and the last digit 3 is a
checksum digit. In the case of an ISBN, what this means is that the last
digit is chosen such that the following occurs. Suppose that the ten digits
in the ISBN are
d
1
,d
2
,...,d
10
. Then we must have that the weighted sum
satisfies
−
58488
−
470
−
10
jd
j
≡
0 (mod 11)
.
j
=1
Note that the first nine digits
d
j
for
j
=1
,
2
,...,
9 are in
{
0
,
1
,
2
,...,
9
}
,
but
d
10
∈{
, but in the ISBN number a 10 will be represented
as the Roman numeral
X
. In the case of this topic,
0
,
1
,
2
,...,
10
}
1
·
1+2
·
5+3
·
8+4
·
4+5
·
8+6
·
8+7
·
4+8
·
7+9
·
0+10
·
3
≡
0 (mod 11)
,
as required. Are the following valid ISBN numbers?
1. 0
−
4523
−
2345
−
4.
2. 0
−
13
−
061817
−
9.
3. 1
−
23
−
098733
−
X
.
6.
G.3
The current ISBN system is reaching the end of its viability. On January 1, 2007,
it will be replaced by a 13-digit ISBN. Although the 10-digit ISBN system, designed for
printed topics in the late 1960s, has the capacity to assign a billion numbers, the internal
structure of the ISBN restricts the capacity of the system. The new 13-digit ISBN system
will be better suited to integrate with current bar-code technology. See
http://www.isbn-
international.org/en/revision.html
.
4. 2
−
432
−
23459
−
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