Information Technology Reference
In-Depth Information
1
Introduction
Let us consider a particular case when the main channel between legal parties is bi-
nary symmetric channel without memory (BSC) with the error probability
p
and
the wire-tap channel to eavesdropper is BSC too with the error probability
p
. Fol-
lowing to common tradition of cryptographic publications we will call two legal par-
ties by Alice and Bob, while eavesdropper by Eve.
One of legal parties (say Alice) transmits to another legal party (say Bob) some bit
string of the length
k
through the main BSCm whereas eavesdropper Eve receives
this string over the wire-tap channel BSCw. Thereafter both Alice and Bob exchange
messages with one to another through noiseless and public channel. Following to
some protocol agreed in advance Alice and Bob form the final bit strings
K
and
K
, respectively, of the same length
l
. If we denote by
U
the total Eve's knowl-
edge including the information received by her on BSCw, noiseless channels of pub-
lic discussion between Alice and Bob, and the full knowledge of protocol and key
computing algorithm, Eve receives some Shannon information
A
I
(
K
,
K
;
U
)
about
A
B
l
/
and key-
capacity
C
can be defined as the maximum possible key-rate taken over all possible
protocols for every
R
of any key sharing protocol is
k
the final key. Then the key-rate
P
=
P
(
K
≠
K
)
>
0
I
>
0
and sufficiently large
k
,
,
e
A
B
o
given
p
and
p
.
It has been proved in [1] that
(1)
=
h
(
p
)
−
h
(
p
)
,
k
m
p
=
p
+
p
−
2
p
p
,
h
(
p
)
=
−
p
log
p
−
(
−
p
)
log(
1
−
p
)
where
- the
m
w
m
w
entropy function.
It is worth to noting that
k
even in the case
p
>
p
, that is, when the
main channel is a
inferior
to the wire-tap channel. (So if we take
>
0
p
=0.1
and
m
p
=0.01
, the
C
by (1) is about 0.0249).
In a particular case when the main channel is a superior to the wire-tap channel
(
p
<
p
) it is possible to proceed without a public discussion to achieve the key-
capacity
.
(2)
But if we proceed using public discussion in the case
p
<
p
we get larger key-
capacity than (2), namely given by (1). On the other hand in the case
′
=
h
(
p
)
−
h
(
p
)
k
w
m
p
p
public
>
m
k
<
0
discussion is necessary because (2) does not longer work (
).
