Re: Probability - Cryptography
- From: A N Niel <anniel@xxxxxxxxxxxxxxxxxxxxx>
- Date: Sat, 01 Sep 2007 06:32:33 -0400
In article <1188642220.236279.75160@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Vivek B <vivekb1985@xxxxxxxxx> wrote:
The problem is...
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.
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) ??
--------------
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 :) ...
Could K map two different 3-bit numbers to the same 8-bit number? (No.)
In your solution, yes.
.
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