Re: Probability - Cryptography
- From: rossum <rossum48@xxxxxxxxxxxx>
- Date: Sat, 01 Sep 2007 14:21:36 +0100
On Sat, 01 Sep 2007 10:23:40 -0000, Vivek B <vivekb1985@xxxxxxxxx>
wrote:
The problem is...Does K() always give the same output for the same input, or does the
I have a 3 bit number- say x
This is mapped to a 8 bit number using a function K().
We can assume that the function K() randomly maps the 3 bit number to
8 bit number and no two 3 bit number gets mapped to the same 8 bit
number.
random mapping change?
Anything from zero to 1.0 depending on the details of K().
Now lets name y as the first four bits of the 8 bit number.
What is the probability that two 3 bit numbers (x1 and x2), generates
the same 4 bit key (y) ??
If the bits of the 3-bit number are ttt then:
K(ttt) -> 0ttt1ttt -> y = 0ttt
(here K(T) = 17 * T + 8)
will always give different y's for different inputs.
However if we have:
K(ttt) -> 0101ttt0 -> y = 0101
(here K(T) = 2 * T + 80)
then all ttt are mapped to the 0101 four bit key y.
rossum
--------------
A solution:
z = K(x); z is 8 bit in size.
We can divide 16 bit numbers into 16 bins of 4 bit numbers.
bin of 0000
bin of 0001
.
.
.
bin of 1111
P(no 2 keys are same) = P( 8 keys generated by the function K() falls
in 8 different bins ).
= C(16,8)/16 power 8.
But i still doubt the solution is wrong!!!
HELP HELP :) ...
.
- References:
- Probability - Cryptography
- From: Vivek B
- Probability - Cryptography
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