You are given a black box f : Z10 → Z10 that contains either a random permutation or a random function. Your distinguisher is allowed to invoke f twice. What is the best advantage you can achieve? Express your answer as a reduced fraction without any spaces (eg, 1/3 and not 12/36), or as 0 or 1 if appropriate.

Solution :

The function :       be a random permutation.

f is a permutation on ,  i.e. f is permutation 10.

Now we know that the total number of distinct permutation on to symbolize 10!.

Each of these 10! permutation to a permutation function    

Therefore, total number of permutation functions     are 10!.

Now we want the total number of permutation functioning :

  such that  f(0) = 0 and f(1)= 1

Now we notice that when f(0)=0 and f(1)=1, then two symbol '0' and '1 are fixed under permutation f.

So essentially when f(0) = 0 and f(1) = 1, f becomes permutation on 8 symbol.

Total number of permutation functioning    , f(0)=0 and f(1)=1 are 8!

Now we want the probability that a random permutation     satisfies f(0) = 0 and f(1) = 1.

The number of permutation function    , i.e.

The probability that a random permutation     satisfies f(0) = 0 and f(1) = 1 is

Therefore, the probability that a random permutation     satisfies f(0)= 0 and f(1)=1 is  

use the power rule  

( a b ) n = a n b n

to distribute the exponent.

apply the product rule to  

3 x 2 y 3 . ( 3 x 2 ) 2 ( y 3 ) 2

apply the product rule to  

3 x 2 . 3 2 ( x 2 ) 2 ( y 3 ) 2

raise  

3

to the power of  

2 . 9 ( x 2 ) 2 ( y 3 ) 2 multiply the exponents in   ( x 2 ) 2

.

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 2 ⋅ 2 ( y 3 ) 2

multiply  

2   by   2 . 9 x 4 ( y 3 ) 2

multiply the exponents in  

( y 3 ) 2 .

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 4 y 3

⋅

2

multiply  

3   by   2 . 9 x 4 y 6

step-by-step explanation: