A sign preserving modular distance

Question

I am looking for a modulus operator Mod(a,b,m) such that:

• Mod(2,6,7)=-3
• Mod(6,2,7)=3

That is, the operator avoids the 4-hop path between 2 and 6 and, instead, circles around via a path of length 3. The answer preserves the path direction if you leave from a and go to b.

The following code calculates this, but there is almost surely a better way.

#include <iostream>
#include <cmath>

double absmin(double a, double b){
  if(std::abs(a)<std::abs(b))
    return a;
  else
    return b;
}

double ModDist(double src, double dest, double m){
  if(dest<src)
    return absmin(dest+m-src, dest-src);
  else
    return absmin(dest-src, dest-m-src);
}

int main(){
  std::cout<<ModDist(2,6,7)<<std::endl;
  std::cout<<ModDist(6,2,7)<<std::endl;
}

Answer 1

If your direct distance is less than half the distance you wont do better by going the other way. Conversely if it is greater than half the distance you do better by changing direction.

This assumes 0<=src,dest<=m

double ModDist(double src, double dest, double m){
  double directDistance = dest - src;
  if(abs(directDistance)<m/2)
    return directDistance;
  return -(m - abs(directDistance))*sgn(directDistance);
}

Answer 2

The math library function remainder, as its name suggests, returns the remainder on dividing its first argument by it's second. Unlike fmod it returns a signed result, and indeed remainder(a,d) is the smallest number r, by absolute value, such that a-r is an integral multiple of d. What you want is this applied to b-a, ie:

double  ModDist( double a, double b, double m)
{ return remainder( b-a, m);
}

Answer 3

You can use the way in which the standard modulo function treats negative numbers to make it do the work of deciding which direction to go in:

#include <cmath>
double ModDist(double s, double d, double m) {
    return std::fmod((d - s) * 2, m) + s - d;
}