Question

How do I calculate the aspect ratio (formatted as integer:integer) by a given factor?

For example, aspect ratio 16:9 has a factor of 1.778, because 16 / 9 = 1.778. But how can I find the ratio by that factor? So

Dimension getAspectRatio(double factor) {
    ...
}

public static void main(String[] arguments) {
    Dimension d = getAspectRatio(16d / 9d);
    System.out.println(d.width + ":" + d.height);
}

should return

16:9

Answer 1

Disclaimer: These algorithms are silly and inefficient. I’m sure there’s a better one…

A silly, straightforward (not very efficient) algorithm to find an approximation is this:

double ratio = 1.778;
double bestDelta = Double.MAX_VALUE;
int bestI = 0;
int bestJ = 0;

for (int i = 1; i < 100; i++) {
  for (int j = 1; j < 100; j++) {
    double newDelta = Math.abs((double) i / (double) j - ratio);
    if (newDelta < bestDelta) {
      bestDelta = newDelta;
      bestI = i;
      bestJ = j;
    }
  }
}

System.out.println("Closest ratio: " + bestI + "/" + bestJ);
System.out.println("Ratio        : " + ((double) bestI / (double) bestJ));
System.out.println("Inaccurate by: " + bestDelta); 

Output.

Closest ratio: 16/9
Ratio        : 1.7777777777777777
Inaccurate by: 2.2222222222234578E-4

Update: Alternative algorithm

I’ve just thought of an alternative algorithm, which tries to close in on the approximation. Of course, it’s still not very efficient…

double bestDelta = Double.MAX_VALUE;
int i = 1;
int j = 1;
int bestI = 0;
int bestJ = 0;

for (int iterations = 0; iterations < 100; iterations++) {
  double delta = (double) i / (double) j - ratio;

  // Optionally, quit here if delta is "close enough" to zero
  if (delta < 0) i++;
  else j++;

  double newDelta = Math.abs((double) i / (double) j - ratio);
  if (newDelta < bestDelta) {
    bestDelta = newDelta;
    bestI = i;
    bestJ = j;
  }
}

System.out.println("Closest ratio: " + bestI + "/" + bestJ);
System.out.println("Ratio        : " + ((double) bestI / (double) bestJ));
System.out.println("Inaccurate by: " + bestDelta);

The output is the same

Should I stumble upon an efficient algorithm, I’ll post it here 🙂