I have a function ProdBigMod which computes the product of two double precise numbers \$x_1\$, \$x_2\$ (both of which are less than \$2^{53} - 1\$) and subsequently finds the remainder \$\mod p\$ (where \$p\$ is prime). I am not at liberty to use external libraries and must use double data type.

The main challenge comes into play when the product of \$x_1\$ and \$x_2\$ exceeds \$2^{53} - 1\$.

As @TobySpeights points out, 53 is important because it is the number of mantissa bits (also called significand hence the name of the constant Significand53) for double data types (see Double-precision). We can eliminate some of these issues by first ensuring that both \$x\$s are less than \$p\$ (this is achieved by immediately applying std::fmod). In fact, when p < sqrt(2^53 - 1), we know the product prodX = x1 * x2 < 2^53 - 1. To deal with the case when p > 2^53 - 1 and prodX > 2^53 - 1, we can take advantage of some properties of modular arithmetic.


$$(x_1 \cdot x_2) \pmod p = x_1 \cdot (x_{12} + x_{22} + ... + x_{n2}) \pmod p$$

Where \$(x_{12} + x_{22} + ... + x_{n2}) = x_2\$.

This gives:

$$\big( x_1 \cdot x_{12} \pmod p \big) \space\space + \space\space \big(x_1 \cdot x_{22} \pmod p\big) \space\space + ... + \space\space \big(x_1 \cdot x_{n2} \pmod p\big)$$

Now, given that each of the \$x_{i2}\$ (chunkSize below) are the same except for possibly the final one (i.e. \$x_{n2}\$), we can compute the following once:

$$x_1 \cdot x_{12} \pmod p \space\space = \space\space x_1 \cdot chunkSize \pmod p \space\space = \space\space chunkMod$$

To find the final \$x_{n2}\$ we first have to determine how many \$x_{i2}\$s will go into \$x_2\$:

numChunkMods = floor(x2 / chunkSize)

Now we can easily obtain xn2:

xn2 = (x2 - chunkSize * numChunkMods)  ==>>  part2 = x1 * xn2 (mod p)

Putting this altogether, Eq. (1) can be reduced to:

x1 * x2 (mod p) = (chunkMod * numChunkMods) + part2 (mod p)

This is a good start, but doesn't get us completely out of the woods, since we don't know for sure that the product numChunkMods * chunkMod < 2^53 - 1. To get around this, we continue the process above by setting x1 = chunkMod and x2 = numChunkMods, until the product is less than 2^53 - 1.

#include <iostream>
#include <cmath>

const double Significand53 = 9007199254740991.0;
const double SqrtSig53 = std::floor(std::sqrt(Significand53));

double PositiveMod(double x, double m) {
    if (x < 0)
        x = x + ceil(std::abs(x) / m) * m;
    else if (x > m)
        x = std::fmod(x, m);
    return x;

double ProdBigMod(double x1, double x2, double p) {
    double result = 0, prodX;

    x1 = PositiveMod(x1, p);
    x2 = PositiveMod(x2, p);
    prodX = x1 * x2;

    if (prodX < p) {
        result = prodX;
    } else if (p < SqrtSig53 || prodX < Significand53) {
        result = std::fmod(prodX, p);
    } else {
        double numChunkMods, part1 = Significand53;
        double chunkSize, chunkMod, part2;

        while (part1 >= Significand53) {
            chunkSize = std::floor(Significand53 / x1);  // Ensures chunkMod < 2^53 - 1
            chunkMod = std::fmod(x1 * chunkSize, p);
            numChunkMods = std::floor(x2 / chunkSize);
            part2 = std::fmod((x2 - chunkSize * numChunkMods) * x1, p);
            part1 = numChunkMods * chunkMod;
            x1 = chunkMod;
            x2 = numChunkMods;
            result = std::fmod(result + part2, p);

        result = std::fmod(part1 + result, p);

    return result;

Here is an example of how to call the function:

int main() {
    double test = ProdBigMod(914806066069, 497967734853, 732164213243);
    std::cout  << std::fixed;
    std::cout << test << "\n";
    return 0;

Output: 85635829849.000000

I have compared the above on random samples of numbers in the range 10^12 with a gmp analog and it seems to give correct results. However, I'm not exactly sure if my logic or implementation is bullet-proof. I'm also wondering if it could be more efficient.

Here is an online "Big Number Calculator" that you can test the results against: https://defuse.ca/big-number-calculator.htm

For example, input (914806066069 * 497967734853) % 732164213243 into the expression field and click "Calculate".

  • \$\begingroup\$ Do you have constraints about only using double? If you could convert to uint128_t... This reminds me a bit of double-double libraries, where before FMA (are you allowed to use fma?) multiplication would split each operand into a sum of short-mantissa numbers. \$\endgroup\$ – Marc Glisse Feb 4 '18 at 21:34
  • \$\begingroup\$ @MarcGlisse, for this particular case, I'm constrained to double. Besides, even without this constraint, I'm interested in improvements, as insights to this problem can be applied to other situations dealing precision. \$\endgroup\$ – Joseph Wood Feb 4 '18 at 21:58
  • 1
    \$\begingroup\$ Well, using one multiplication and one fma, you can rewrite a*b to c+d, so if you have already written AddBigMod... \$\endgroup\$ – Marc Glisse Feb 4 '18 at 22:08
  • \$\begingroup\$ @MarcGlisse Sadly, some fma() are not that precise: Is my fma() broken?. IMO, that case is non-compliant per the intent of the function. \$\endgroup\$ – chux Feb 10 '18 at 20:59


It should be made more clear that ProdBigMod() is to only work with whole number values and not the full range of double including values with fractional parts, NaN and infinites.

I would expect code to detect non-whole number values and exit/complain as needed - perhaps return NaN.


"... when p < sqrt(2^53 - 1) ..." should be "... when p <= sqrt(2^53) ...".
This allows for a slightly larger p.

Recall that something mod p will be at most, p - 1.

This is reflected in the incorrect code: x > m --> x >= m

double PositiveMod(double x, double m) {
    if (x < 0)
    // else if (x > m)
    else if (x >= m)
        x = std::fmod(x, m);
    return x;


// } else if (p < SqrtSig53 || ...
} else if (p <= SqrtSig53 || ...


Use of Significand53 and other code relies on double as a binary64. A simple test would prevent a number of errant compilations, even if not increase portability.

#if DBL_MANT_DIG != 53
  #error TBD code

Alternative precision test.

Code does prodX = x1 * x2; and various tests when code could test FE_INEXACT after the multiplication.

prodX = x1 * x2
if (fetestexcept(FE_INEXACT)) Handle_inexact_product();

Questionable code

I am not confident x = x + ceil(std::abs(x) / m) * m; work as expected for all x, m, especially when the math quotient is just over a whole number, but rounds down before ceil(). Clearer alternative:

double PositiveMod(double x, double m) {
  double y = std::fmod(x, m);
  if (y < 0.0) {
    y = (m < 0) ? y - m : y + m;
  return y;

Code alternate:

Use 64-bit or better integer math. See mulmodmax() in Modular exponentiation without range restriction. Good for at least 19 decimal digit integers vs. 12 here.

fmod() correctness.

Note: The specification of std::fmod() does not require the result to be the best possible answer, yet with IEEE Standard 754 adherence, it is. A good library would implement an exact result. Ref: Is fmod() exact when y is an integer? .


In all, using FP math to solve an integer problem has various unexpected corner concerns and so is hard to insure code is right for all x1,x2,p.

  • \$\begingroup\$ Great answer. How about 'at most, 1 less than p' -> 'at most \$ p - 1 \$'? it caught me off guard for a second. \$\endgroup\$ – Daniel Feb 10 '18 at 21:01
  • \$\begingroup\$ @Coal_ OK - answer amended. \$\endgroup\$ – chux Feb 10 '18 at 21:03
  • \$\begingroup\$ mulmodmax is a really nice function, however it is a bit esoteric. I have spent a good bit of time figuring out how it works and understand it, however I don't think it is immediately obvious to the reader. An explanation would be nice if you have time. Thanks for the thorough schooling! \$\endgroup\$ – Joseph Wood Feb 12 '18 at 2:48
  • \$\begingroup\$ @JosephWood Perhsp Avoiding overflow working modulo p would help. \$\endgroup\$ – chux Feb 12 '18 at 3:05
  • \$\begingroup\$ I think your off-by-one is itself off by one: p <= sqrt(2^53 - 1) or p < sqrt(2^53) (if I'm right that 2^53 is the lowest value that can't be exactly represented). That said, we know sqrt(2^53) is not an integer, so I'm splitting hairs here. \$\endgroup\$ – Toby Speight Feb 12 '18 at 11:05

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