Faster way of finding multiple of double

If have the following C function, used to determine if one number is a multiple of another to an arbirary tolerance

#include <math.h>

#define TOLERANCE 0.0001

int IsMultipleOf(double x,double mod)
{
    return(fabs(fmod(x, mod)) < TOLERANCE);
}

It works fine, but profiling shows it to be very slow, to the extent that it has become a candidate for optimization. About 75% of the time is spent in modulo and the remaining in fabs. I'm trying to figure a way of speeding things up, using something like a look-up table. The parameter x changes regularly, whereas mod changes infrequently. The number of possible values of x is small enough that the space for a look-up would not be an issue, typically it will be one of a f开发者_如何转开发ew hundred possible values. I can get rid of the fabs easily enough, but can't figure out a reasonable alternative to the modulo. Any ideas on how to optimize the above?

Edit The code will be running on a wide range of Windows desktop and mobile devices, hence processors could include Intel, AMD on desktop, and ARM or SH4 on mobile devices. VisualStudio 2008 is the compiler.

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Do you really have to use modulo for this?

Wouldn't it be possible to just result = x / mod and then check if the decimal part of result is close to 0. For instance:

11 / 5.4999 = 2.000003  ==> 0.000003 < TOLERANCE 

Or something like that.

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Division (floating point or not, fmod in your case) is often an operation where the execution time varies a lot depending on the cpu and compiler:

• gcc has a builtin replacement for
  that if you give it the right compile flags or if you use __builtin_fmod explicitly. This then might map the operation on a small number of assembler instructions.
• there may be special units like SSE on intel processors where this operation is implemented more efficiently

By such tricks, depending on your environment (you didn't tell which) the time may vary from some clock cycles to some hundred. I think best is to look into the documentation of your compiler and cpu for that particular operation.

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The following is probably overkill, and sub-optimal. But for what it is worth here is one way on how to do it.

We know the format of the double ...

• 1 bit for the sign
• 11 bits for the biased exponent
• 52 fraction bits

Let ...

• value = x / mod;
• exp = exponent bits of value - BIAS;
• lsb = least sig bit of value's fraction bits;

Once you have that ...

/*
 * If applying the exponent would eliminate the fraction bits
 * then for double precision resolution it is a multiple.
 * Note: lsb may require some massaging.
 */
if (exp > lsb)
    return (true);

if (exp < 0)
    return (false);

The only case remaining is the tolerance case. Build your double so that you are getting rid of all the digits to the left of the decimal.

• sign bit is zero (positive)
• exponent is the BIAS (1023 I think ... look it up to be sure)
• shift the fraction bits as appropriate

Now compare it against your tolerance.

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I think you need to inspect the bowels of your C RTL fmod() function: X86 FPU's have 'FPREM/FPREM1' instructions which computes remainders by repeated subtraction.

While floating point division is a single instruction, it seems you may need to call FPREM repeatedly to get the right answer for modulus, so your RTL may not use it.

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I have not tested this at all, but from the way I understand fmod this should be equivalent inlined, which might let the compiler optimize it better, though I would have thought that the compiler's math library (or builtins) would work just as well. (also, I don't even know for sure if this is correct).

#include <math.h>

int IsMultipleOf(double x, double mod) {
    long n = x / mod;  // You should probably test for /0 or NAN result here
    double new_x = mod * n;
    double delta = x - new_x;
    return fabs(delta) < TOLERANCE;  // and for NAN result from fabs
}

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Maybe you can get away with long long instead of double if you have comparable scale of data. For example long long would be enough for over 60 astronomical units in micrometer resolution.

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Does it need to be double precision ? Depending on how good your math library is, this ought to be faster:

#include <math.h>

#define TOLERANCE 0.0001f

bool IsMultipleOf(float x, float mod)
{
    return(fabsf(fmodf(x, mod)) < TOLERANCE);
}

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I presume modulo looks a little like this on the inside:

mod(x,m) {
  while (x > m) {
    x = x - m
  }
  return x
}

I think that through some sort of search i could be optimised: eg:

fastmod(x,m) {
  q = 1

  while (m * q < x) {
    q = q * 2
  }

  return mod((x - (q / 2) * m), m)
}

You might even choose to replace the finall call to mod with annother call to fastmod, adding the condition that if x < m then to return x.