Question

You are given a hash function f1(x) = x % 5, where x is the value to be hashed and f(x) is the hash address. Double hashing is used to resolve collisions, f2(x)-1 + x%3. An ith probe collision is resolved by ( f(x) + i*f2(x) )965 The hash function receives the input (7, 6, 4, 12,10 in that order. Place each number in the hash table at its correct address. Address 0 1 234

Answer 1

7   ->  7 % 5 is 2. so, insert 7 at position 2
6   ->  6 % 5 is 1. so, insert 6 at position 1
4   ->  4 % 5 is 4. so, insert 4 at position 4
12  ->  12 % 5 is 2.
        there is already a value at position 2
        so, use second hash function.
        f2(x) = 1 + 12%3 = 1 + 0 = 1
        (f1(x) + 1*f2(x)) % 5 = (2 + 1) % 5 = 3
        so, insert 12 at index 3
10  ->  10 % 5 is 0. so, insert 10 at position 0

Address     value
-------------------
0           10
1           6
2           7
3           12
4           4