Moving from Linear Probing to Quadratic Probing (hash collisons)

Question

My current implementation of an Hash Table is using Linear Probing and now I want to move to Quadratic Probing (and later to chaining and maybe double hashing too). I\'ve re

Answer 1

There is a particularly simple and elegant way to implement quadratic probing if your table size is a power of 2:

step = 1;

do {
    if(/* CHECK IF IT'S THE ELEMENT WE WANT */) {
        // FOUND ELEMENT

        return;
    } else {
        index = (index + step) % table_size;
        step++;
    }
} while(/* LOOP UNTIL IT'S NECESSARY */);

Instead of looking at offsets 0, 1, 2, 3, 4... from the original index, this will look at offsets 0, 1, 3, 6, 10... (the i^th probe is at offset (i*(i+1))/2, i.e. it's quadratic).

This is guaranteed to hit every position in the hash table (so you are guaranteed to find an empty bucket if there is one) provided the table size is a power of 2.

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Here is a sketch of a proof:

1. Given a table size of n, we want to show that we will get n distinct values of (i*(i+1))/2 (mod n) with i = 0 ... n-1.
2. We can prove this by contradiction. Assume that there are fewer than n distinct values: if so, there must be at least two distinct integer values for i in the range [0, n-1] such that (i*(i+1))/2 (mod n) is the same. Call these p and q, where p < q.
3. i.e. (p * (p+1)) / 2 = (q * (q+1)) / 2 (mod n)
4. => (p² + p) / 2 = (q² + q) / 2 (mod n)
5. => p² + p = q² + q (mod 2n)
6. => q² - p² + q - p = 0 (mod 2n)
7. Factorise => (q - p) (p + q + 1) = 0 (mod 2n)
8. (q - p) = 0 is the trivial case p = q.
9. (p + q + 1) = 0 (mod 2n) is impossible: our values of p and q are in the range [0, n-1], and q > p, so (p + q + 1) must be in the range [2, 2n-2].
10. As we are working modulo 2n, we must also deal with the tricky case where both factors are non-zero, but multiply to give 0 (mod 2n):
    • Observe that the difference between the two factors (q - p) and (p + q + 1) is (2p + 1), which is an odd number - so one of the factors must be even, and the other must be odd.
    • (q - p) (p + q + 1) = 0 (mod 2n) => (q - p) (p + q + 1) is divisible by 2n. If n (and hence 2n) is a power of 2, this requires the even factor to be a multiple of 2n (because all of the prime factors of 2n are 2, whereas none of the prime factors of our odd factor are).
    • But (q - p) has a maximum value of n-1, and (p + q + 1) has a maximum value of 2n-2 (as seen in step 9), so neither can be a multiple of 2n.
    • So this case is impossible as well.
11. Therefore the assumption that there are fewer than n distinct values (in step 2) must be false.

(If the table size is not a power of 2, this falls apart at step 10.)