Suppose you want to pick an integer at random in a set of *N* elements. Your computer has functions to generate random 32-bit integers, how do you transform such numbers into indexes no larger than *N*? Suppose you have a hash table with a capacity *N*. Again, you need to transform your hash values (typically 32-bit or 64-bit integers) down to an index no larger than *N*. Programmers often get around this problem by making sure that *N* is a power of two, but that is not always ideal.

We want a map that as fair as possible for an arbitrary integer *N*. That is, ideally, we would want that there are exactly 2^{32}/*N* values mapped to each value in the range {0, 1 ,…, *N* – 1} when starting from all 2^{32} 32-bit integers.

Sadly, we cannot have a perfectly fair map if 2^{32} is not divisible by *N*. But we can have the next best thing: we can require that there be either floor(2^{32}/*N*) or ceil(2^{32}/*N*) values mapped to each value in the range.

If *N* is small compared to 2^{32}, then this map could be considered as good as perfect.

The common solution is to do a modulo reduction: *x* mod *N*. (Since we are computer scientists, we define the modulo reduction to be the remainder of the division, unless otherwise stated.)

uint32_t reduce(uint32_t x, uint32_t N) { return x % N; }

How can I tell that it is fair? Well. Let us just run through the values of *x* starting with 0. You should be able to see that the modulo reduction takes on the values 0, 1, …, *N* – 1, 0, 1, … as you increment *x*. Eventually, *x* arrives at its last value (2^{32} – 1), at which point the cycle stops, leaving the values 0, 1, …, (2^{32} – 1) mod *N* with ceil(2^{32}/*N*) occurrences, and the remaining values with floor(2^{32}/*N*) occurrences. It is a fair map with a bias for smaller values.

It works, but a modulo reduction involves a division, and divisions are expensive. Much more expensive than multiplications. A single 32-bit division on a recent x64 processor has a throughput of one instruction every six cycles with a latency of 26 cycles. In contrast, a multiplication has a throughput of one instruction every cycle and a latency of 3 cycles.

There are fancy tricks to “precompute” a modulo reduction so that it can be transformed into a couple of multiplications as well as a few other operations, as long as *N* is known ahead of time. Your compiler will make use of them if *N* is known at compile time. Otherwise, you can use a software library or work out your own formula.

But it turns out that you can do even better! That is, there is an approach that is easy to implement, and provides just as good a map, without the same performance concerns.

Assume that *x* and *N* are 32-bit integers, consider the 64-bit product *x* * *N*. You have that (*x* * *N*) div 2^{32} is in the range, and it is a fair map.

uint32_t reduce(uint32_t x, uint32_t N) { return ((uint64_t) x * (uint64_t) N) >> 32 ; }

Computing (*x* * *N*) div 2^{32} is very fast on a 64-bit processor. It is a multiplication followed by a shift. On a recent Intel processor, I expect that it has a latency of about 4 cycles and a throughput of at least on call every 2 cycles.

So how fast is our map compared to a 32-bit modulo reduction?

To test it out, I have implemented a benchmark where you repeatedly access random indexes in an array of size *N*. The indexes are obtained either with a modulo reduction or our approach. On a recent Intel processor (Skylake), I get the following number of CPU cycles per accesses:

modulo reduction | fast range |

8.1 | 2.2 |

So it is four times faster! No bad.

As usual, my code is freely available.

What can this be good for? Well… if you have been forcing your arrays and hash tables to have power-of-two capacities to avoid expensive divisions, you may be able to use the fast range map to support arbitrary capacities without too much of a performance penalty. You can also generate random numbers in a range faster, which matters if you have a very fast random number generator.

So how can I tell that the map is fair?

By multiplying by *N*, we take integer values in the range [0, 2^{32}) and map them to multiples of *N* in [0, *N* * 2^{32}). By dividing by 2^{32}, we map all multiples of *N* in [0, 2^{32}) to 0, all multiples of *N* in [2^{32}, 2 * 2^{32}) to one, and so forth. To check that this is fair, we just need to count the number of multiples of *N* in intervals of length 2^{32}. This count must be either ceil(2^{32}/*N*) or floor(2^{32}/*N*).

Suppose that the first value in the interval is a multiple of *N*: that is clearly the scenario that maximizes the number of multiples in the interval. How many will we find? Exactly ceil(2^{32}/*N*). Indeed, if you draw sub-intervals of length *N*, then every complete interval begins with a multiple of *N* and if there is any remainder, then there will be one extra multiple of *N*. In the worst case scenario, the first multiple of *N* appears at position *N* – 1 in the interval. In that case, we get floor(2^{32}/*N*) multiples. To see why, again, draw sub-intervals of length *N*. Every complete sub-interval ends with a multiple of *N*.

This completes the proof that the map is fair.

For fun, we can be slightly more precise. We have argued that the number of multiples was maximized when a multiple of *N* appears at the very beginning of the interval of length 2^{32}. At the end, we get an incomplete interval of length 2^{32} mod *N*. If instead of having the first multiple of *N* appear at the very beginning of the interval, it appeared at index 2^{32} mod *N*, then there would not be room for the incomplete subinterval at the end. This means that whenever a multiple of *N* occurs before 2^{32} mod *N*, then we shall have ceil(2^{32}/*N*) multiples, and otherwise we shall have floor(2^{32}/*N*) multiples.

Can we tell which outcomes occur with frequency floor(2^{32}/*N*) and which occurs with frequency ceil(2^{32}/*N*)? Yes. Suppose we have an output value *k*. We need to find the location of the first multiple of *N* no smaller than *k* 2^{32}. This location is ceil(*k* 2^{32} / *N*) *N* – *k* 2^{32} which we just need to compare with 2^{32} mod *N*. If it is smaller, then we have a count of ceil(2^{32}/*N*), otherwise we have a count of floor(2^{32}/*N*).

You can correct the bias with a rejection, see my post on fast shuffle functions.

**Useful code**: I published a C/C++ header on GitHub that you can use in your projects.

**Further reading**:

- Daniel Lemire, Fast Random Integer Generation in an Interval, ACM Transactions on Modeling and Computer Simulation (to appear)
- Google Tensorflow adopted this approach through a contribution by David Andersen (see the commit Switching the presized cuckoo map from using strict mod and Google+ post).
- What is arguably the best Open Source Chess engine, Stockfish, also adopted this approach.
- The technique described in this blog post is in used within Microsoft Arriba.
- math/rand: speed up Int31n with multiply/shift instead of modulo (golang issue 16213), runtime: speed up fastrand() % n (golang commit)
- Agner Fog, Pseudo-Random Number Generators for Vector Processors and Multicore Processors, Journal of Modern Applied Statistical Methods, 2015.
- Kenneth A. Ross, Efficient Hash Probes on Modern Processors, IBM Research Report RC24100 (W0611-039) November 8, 2006

(Update: I have made the proof more intuitive following a comment by Kendall Willets.)

I’m missing something… are you saying that x % N and (x * N) div 2^32 are equivalent? This is trivially false for, e.g., 1 % 7 vs (1 * 7) div 2^32.

I am not saying that x % N is equal to (x * N) div 2^32 . This is obviously false, as you indicate. I am saying that they are both fair maps from the set of all 32-bit integers down to integers in [0,N).

It is not, unless you have numbers uniformly distributed from 0 to numeric_limits::max()

If you have, say, 31-bit integers, instead of 32-bit integers, as might happen with a call to

rand, you can adapt the map by shifting by 31 instead of 32.PS: Especially, if you use it for hashing and you numbers tend to be small then all your hash values will be horribly biased down. Perhaps, reducing div 2^32 to div 2^(some smaller degree of tw) may help.

A good hash function should be regular. That is, it should be so that all integers are “equally likely” (given a random input). If your hash values do not cover the whole 32-bit range, you need to adapt the map, probably as you describe.

What is the cost of computing a good hash value? 🙂 Perhaps, it should be included in the overall computation time. This may (drastically?) reduce the difference between two approaches.

I also suspect that in many cases, given a list of IDs you can get reasonable results by doing just ID % . In this case, you can get away without applying a hashing transformation. With your trick, you do need this.

For example, if you numbers are smaller than 2^20 (which is quite likely) and N < 2^16, then it looks like all your hash values will be zero.

by doing just ID * some perhaps prime number. Angle brackets were removed by HTML 🙂

What is the cost of computing a good hash value?Of course, you are correct that there are many cases where the modulo reduction will not be a bottleneck. In such cases, this fast map is useless.

However, there are real-world examples where people set their capacity to a power of two to improve performance. A latency of a couple of dozens of cycles (for a division) is not great.

I also suspect that in many cases, given a list of IDs you can get reasonable results by doing just ID % N.Yes. Moreover, as alluded to in my post, if N is known ahead of time, you can avoid the modulo reduction entirely and replace it with a faster function. See Hacker’s Delight.

Agree. Good point about N known in advance.

Would byte-swapping x help to preserve lower-order bits, e.g.:

uint32_t reduce2(uint32_t x, uint32_t N) {

return static_cast<uint64_t>(bswap_32(x)) * N >> 32;

}

On a second thought, if modulo reduction slows you down, you may want to try SIMD and bulk conversion. You can something like:

__m128 recip = __mm_set1_ps(1/float(N));

__m128 mult = __mm_set1_ps(N);

__m128 to_convert = __mm_loadu_ps(…);

__m128 tmp = _mm_mul_ps(mult, _mm_floor_ps(_mm_mul_ps(to_convert_,recip)));

__m128i res = _mm_cvtps_epi32(_mm_floor_ps(_mm_sub_ps(to_convert, tmp)));

Peraphs, an additional SIMD min is required to ensure the value is < N. Anyways, seems like it is worth trying such as solution.

Yes, Leonid, I suspect that you are quite right. There are cases where using floating-point numbers, especially in conjunction with SIMD instructions, could lead to great results.

To re-phase Daniel:

We want to generate an index (I) chosen uniformly from the range 0..N-1.

We have a 32-bit random value (R), and assume a uniform distribution.

When N is a power-of-two, then the good/easy/fair answer: mask off the higher bits of R. (I have used this more than a few times, as a programmer.)

When N is not a power-of-two, the easy answer is:

I = R mod N

Two problems: the modulus operation is somewhat expensive, and the distribution is not *quite* even (that last wrap-around) – though likely good enough.

Daniel’s solution is (in two steps):

R = R * N

We now have a value in the range to N * 2^32.

I = R >> 32

We now have a value in the range to N. The two operations (multiply and shift) are almost certainly cheaper than the modulus operation.

I suspect the values are not *quite* uniform, but likely good enough. (How uniformly distributed are the multiplication products in each of the N buckets?)

Might write a test program to measure. 🙂

If N is small compared to 2^32, then I submit to you that you won’t be able to measure any bias.

It’s a fairly simple proof. You stretch the [0, 2^32) random variable range out to multiples of N in [0, N*2^32] and then map [0,2^32) to 0, [2^32, 2*2^32) to 1, etc. by integer division (>>32). The number of multiples of N in each range is floor or ceil(2^32/N).

Thanks. I had a relatively short proof that used elementary number theory (a few lines), but it slightly got out of hand when I tried to make it accessible to people who may not master these concepts. A more direct approach like you suggest would have been better! Kudos!

I have updated my blog post with a more direct proof.

Thanks — I actually thought it through hoping to find a different method, and ended up with a proof of the same one. Oh well.

Note that expression `((uint64_t) x * (uint64_t) N) >> 32` on x86-64 compiles into a 64-bit multiplication that yields a 128-bit result, followed by the shift.

With a bit of inline assembly it can use a shorter 32-bit mul instruction that yields a 64-bit result in EDX:EAX. The required result is in EDX, no shift instruction is required.

That’s a very good point. I am somewhat disappointed that the compiler fails to exploit this optimization.

For the record, something like this could do the job…

uint32_t asm_highmult32to32(uint32_t u, uint32_t v) {

uint32_t answer;

__asm__ ("imull %[v]\n"

"movl %%edx,%[answer]\n"

: [answer] "+r" (answer) : [u] "a" (u), [v] "r" (v) :"eax","edx" );

return answer;

}

But it is unlikely to help performance as is. You’d probably need to rewrite a larger chunk of code using assembly.

For some reason, gcc does the correct thing with `uint128_t` operands.

See https://godbolt.org/g/kZ91Yu .

mulx instruction would be ideal here: takes 2 assembly instructions to produce the value in eax, no flags affected:

uint32_t reduce3(uint32_t x, uint32_t N) {

uint32_t r;

asm(“movl %[N], %%edx\n\t”

“mulxl %[x], %[r], %[r]”

: [r]”=r”(r)

: [x]”r”(x), [N]”r”(N)

: “edx”);

return r;

}

Doing this for floats (random within [0,x)) is also fairly easy since it just requires subtracting 32 from the exponent, which ldexp can do, eg ldexp(rand(), -32) gives a float in [0,1) (assuming rand() is 32-bit here).

There’s some loss of precision as floats discard lower-order bits automatically, unless we cast to a longer mantissa first.

Since I don’t read math proofs daily, this took me a bit of thinking to parse. At the end I realized all this was a variation of:

rand(1) * N

In order:

( rand(2^32) * N ) / 2^32

is equivalent to

( rand(2^32) / 2^32 ) * N

which is equivalent to

floor( (float)[0,1) * N )

Yes, for some fuzzy notion of “equivalent”. For example, rand(2^32) / 2^32 is not equivalent to picking a number in [0,1) fairly.

This is a godsend for FPGAs. Modulus by an arbitrary number is expensive in both time and space in hardware. Shifts are free and nearly all modern FPGAs have hard multiplier blocks. This is a perfect solution. Thanks !

I created a similar approach to fast range named “fast mod and scale”. I didn’t realize fast range until I started to survey alternatives for writing my blog post. http://www.idryman.org/blog/2017/05/03/writing-a-damn-fast-hash-table-with-tiny-memory-footprints/

I’m happy to see other people also interested in this problem. Even though solution is just one or two lines are code, it is still a important problem to solve!

It first mask the hash value to next power of two, then do the fast range (I named it scaling) described in your post. It cost a bit more cycles, but are small enough because of modern cpu pipelining. The major usage of fast mod and scale (my method) is to make hash table probing as easy to implement as using strait mod. For hash tables that doesn’t use probing, like cuckoo hashing or standard chaining, fast range is sufficient.

More analysis and implementation details are in the blog I posted above.

This also made it in the world’s strongest open source chess engine (stockfish) :

https://github.com/official-stockfish/Stockfish/commit/2198cd0524574f0d9df8c0ec9aaf14ad8c94402b

That’s amazing.

I’m using x64 laptop-windows 10 and try fast modulo function but not success. Result is always zero. Why??? Did I miss something?

Can you share the code that gives you always zero?

include

using namespace std;

uint32_t reduce(uint32_t x, uint32_t N)

{

return (((uint64_t)x * (uint64_t)N) >> 32);

}

int main()

{

cout<< reduce(12, 7);

return 0;

}

I tried in my laptop (x64) and “www.onlinegdb.com” but i received the same result is zero. Did i miss something?

What do you expect reduce(12,7) to do?

I tried many different values as 20, 33, 5 (modulo for 7) and I received the same result is zero

Did you try random 32-bit numbers? Please do. Here is a sample program…

https://gist.github.com/lemire/b9596313593dcb6aa311f5e5aa60f517

Wow! Thank you very much!

Isn’t this basically fixed point arithmetic where the bottom 32 bits are the fractional part?

Basically, yes. It is really not hard conceptually.

Unless

`x * N > (1<<shiftAmount)`

the result is 0. I don’t know how you are computing fairness but your post is misleading.In fact, you’re returning the quotient of the product by 2 raised to power of shiftAmount => reduce = floor[(x*N) / 2^shiftAmount]

Since, in a hash table, you are unlikely to use a power of 2 for the bucket’s count, this is not going to work well, and you’ll get a lot of collision if you’re using a power of 2 for the divisor. Since N can not be in [0 ; powerOfTwoRange] this is useless.

If you know that N is in [0; 2^32] range, nice. But that’s condition that rarely happens in real life.

Let’s say you have a hash table to have a more compact storage of ID => Value, then this function will produce a huge amount of collision on the low bucket’s indexes and almost no collision on the last buckets.

Obviously, in benchmark where N are as probable to happen in [0 2^32] range, this method is as fair as a modulo. Yet, these rarely happens in reality and the cost to handle collisions in a hash table is many times more important than the cost of the modulo.

You’d probably get a better result by SWARing the product before dividing (that is,

`p = x * N`

`p = p ^ (p>>32)`

and so on) then masking by`0xFFFFFFFF`

. But at some point, I doubt you’ll beat a modulo operation.Note that the trick described in this blog post is used in many real-life systems, some of which you are maybe relying upon. It definitively works.

If you know that N is in [0; 2^32] range, nice. But that’s condition that rarely happens in real life.You are expected to use this trick, like the modulo reduction, after hashing. You can hash to the [0, 2^32) range.

But at some point, I doubt you’ll beat a modulo operationYou need to hash your objects in all cases. You should never “just” use the modulo reduction. What you typically do is hash and then reduce. You can either reduce using the modulo reduction or using this trick.

Most hash functions, just like most random number generators, have output sizes that are powers of two.