Double Hashing

Posted 2021-02-02 heifengli

Data Structure and Algorithm Analysis in C++

5.4.3 Double Hashing

In the beginning of chapter 5.4, the formula  h_i(x) = (hash(x) + f(i))  is mentioned and  f(i) is defined as f(i) = i ;

Here f(i) is defined as f(i) = i * hash₂(x) and hash₂(x) = R - (x mod R) where R is a prime smaller than TableSize. If R = 7 then hash₂(x) = 7 - (x mod 7).

The case it uses is as followed.  Figure 5.18 shows the result of inserting keys {89, 18, 49, 58, 69} into a hash table.

 

Hash(x), or hash₁(x) to be precise, is defined as hash₁(x) = x % 10  //  TableSize = 10. 

So if there‘s no collision, h(x) = hash₁(x);  otherwise hi(x) = { hash₁(x) + i * hash₂(x) } % TableSize .

The first collision occurs when 49 is inserted. hash2(49) = 7 ? 0 = 7, so 49 is inserted in position 6    // { 49%10 + (7 - 1* 49%7)  } % 10 .

hash2(58) = 7 ? 2 = 5, so 58 is inserted at location 3     // ( 8 + 1*5 )%10 .

Finally, 69 collides and is inserted at a distance hash2(69) = 7?6 = 1 away.

If we tried to insert 60 in position 0, we would have a collision. Since hash2(60) = 7 ? 4 = 3, we would then try positions 3, 6, 9, and then 2 until an empty spot is found.