precept4sol - Computer Science 340 Reasoning about...

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Computer Science 340 Reasoning about Computation Precept 4 Problem 1 Consider the family of hash functions discussed in class h a,b ( x ) = ( ax + b mod p )( mod n ) where a, b ∈ { 0 , . . . p - 1 } and a = 0. Consider a hash function h a,b drawn uni- formly and at random from this family. For any x 1 , x 2 ∈ { 0 , . . . , p - 1 } , x 1 = x 2 , show that Pr[ h a,b ( x 1 ) = h a,b ( x 2 )] 1 /n. Solution: Let f a,b ( x ) = ( ax + b mod p ). In class, we showed that for any x 1 , x 2 ∈ { 0 , . . . , p - 1 } , x 1 = x 2 , as a, b range over all values in { 0 , . . . p - 1 } , a = 0, the pair ( f a,b ( x 1 ) , f a,b ( x 2 )) ranges over all p ( p - 1) pairs ( y 1 , y 2 ) such that y 1 , y 2 ∈ { 0 , . . . , p - 1 } , y 1 = y 2 . Note that h a,b ( x ) = f a,b ( x ) mod n . Thus the number of h a,b for which h a,b ( x 1 ) = h a,b ( x 2 ) is exactly the number of pairs ( y 1 , y 2 ), y 1 , y 2 ∈ { 0 , . . . , p - 1 } , y 1 = y 2 such that y 1 = y 2 mod n . Let us count the number of such pairs as follows: There are p choices for y 1 . For each choice of y 1 , there are p/n - 1 choices of y 2 such that y 1 = y 2 mod n . Now p/n - 1 ( p + n - 1) /n - 1 = ( p - 1) /n . Hence the number of pairs ( y 1 , y 2 ) such that y 1 = y 2 mod p is at most p ( p - 1) /n . By the above discussion, the number of choices of h a,b such that h a,b ( x 1 ) = h a,b ( x 2 ) is at most p ( p - 1) /n , which is a 1 /n
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  • Fall '07
  • CharikarandChazelle
  • probability density function, 1 K, dt, k-1 dt

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