A fast alternative to the modulo reduction

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A common operation in software is to take a machine word and map it to an integer value in a range [0,p) as fairly as possible. That is, you want that if all values of the machine word are equally likely then, as much as possible, all integer values in [0,p) are (almost) equally likely. This is common in hashing and probabilistic algorithms.

In computing, the modulo operation finds the remainder after division of one number by another (called the modulus of the operation). Given two positive numbers, a and n, a modulo n (abbreviated as a mod n) is the remainder of the Euclidean division of a by n, where a is the dividend and n is the divisor.

The standard approach to this problem is the modulo reduction (

x % p). Though a modulo reduction works fine and has several nice properties, it can be slow even on recent processors because it relies on an integer division.

Thankfully, there is a faster way: we can replace the modulo by a multiplication followed by a shift.

It has accelerated some operations in Google's Tensorflow by 10% to 20% (https://github.com/tensorflow/tensorflow/commit/a47a300185026fe7829990def9113bf3a5109fed).

Further reading : http://lemire.me/blog/2016/06/27/a-fast-alternative-to-the-modulo-reduction/

See also:

- Daniel Lemire, Fast Random Integer Generation in an Interval, ACM Transactions on Modeling and Computer Simulation (to appear)

This library provides a single portable header file that you should be able to just drop in your C/C++ projects. The API is simple:

#include "fastrange"// given a value word, produces an integer in [0,p) without division uint32_t fastrange32(uint32_t word, uint32_t p); uint64_t fastrange64(uint64_t word, uint64_t p); size_t fastrangesize(size_t word, size_t p); int fastrangeint(int word, int p);

For this code to give the desired result, the provided words should span the whole range (e.g., all 32-bit integers). The C

randfunction does not meet this requirement. If you must use the

randfunction, wrap it like so:

uint32_t get32rand() { return (rand() ^ (rand() << 15) ^ (rand() << 30)); }uint64_t get64rand() { return (((uint64_t)get32rand()) << 32) | get32rand(); }

However, the underlying idea is general: we only require that the word values span an interval of the form

[0,1<. It suffices then to do( x * range ) >> Lto get a fair map from[0,1< to[0,range). For example, if your word values span the interval [0,65536), then you could simply do( x * range ) >> 16.## Unbiased range functions

To generate an unbiased random number in a range, you can use an extension of this approach:

uint32_t random_bounded(uint32_t range) { uint64_t random32bit = random32(); //32-bit random number multiresult = random32bit * range; leftover = (uint32_t) multiresult; if(leftover < range ) { threshold = -range % range ; while (leftover < threshold) { random32bit = random32(); multiresult = random32bit * range; leftover = (uint32_t) multiresult; } } return multiresult >> 32; }See http://lemire.me/blog/2016/06/30/fast-random-shuffling/