When hashing a (key, value) pair where the key is a string, I have seen the following hash function in use:

E.g. $c_n + 256c_{n-1}+ 256^2c_{n-2}+...256^{n-1}c_1$, where this represents the string $c_1c_2..c_n$. I understand the equation above can be written as: $c_n + 256(c_{n-1} + 256(c_{n-2}+...256c_1))$, could someone explain how the following is valid, when I have to compute mod over the entire sum, how can we take it term by term?

r= 0;
for i = 1 to n do
r := (c[i] + 256*r) mod TableSize

1 Answer 1


Let's prove by induction that after $i$ iterations, $$ r = \sum_{j=1}^i 256^{i-j} c[j] \bmod m, $$ where $m$ is the size of the table.

The base case is $i = 0$, where $r = 0 = 0 \bmod m$. Now suppose that the formula holds after $i-1$ iterations. After $i$ iterations, we have \begin{align*} r &= \left[ c[i] + 256 \left(\sum_{j=1}^{i-1} 256^{i-1-j} c[j] \bmod m\right) \right] \bmod m \\ \\ &= \left[ c[i] + \sum_{j=1}^{i-1} 256^{i-j} c[j] \right] \bmod m \\ &= \sum_{j=1}^i 256^{i-j} c[j] \bmod m, \end{align*} using several times the formulas $$ (a + b) \bmod m = \left[(a \bmod m) + (b \bmod m)\right] \bmod m, \\ (ab) \bmod m = \left[(a \bmod m)(b \bmod m)\right] \bmod m. $$

For the proofs of these two formulas check: this and this.


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