Minimum Aspect Ratio

Question

I need to find the minimum size that has an aspect ratio of exactly (or within 0.001) of a certain value. Are there any quick math tricks or frameworks for this?

Here's the pseudo code for the current bad idea I was working with in O(n^2) :

 epsilon = 0.001; from x = 1 to MAX_X { from y = 1 to MAX_Y { if(Abs(x / y - aspectRatio) <= epsilon) { return new Size(x, y); } } } return Size.Empty; 

Answer 1

Unusual. You need to find the largest common divisor and divide its width and height. The algorithm is Euclid and is two thousand three hundred years old. More details here .

Answer 2

You can write aspectRatio as fraction (if you want it to be up to 0.001, you can use round (aspectRatio, 3) / 1000)

Then simplify this share . The resulting fraction is the x / y you are looking for.

Answer 3

A faster way, but still not stereotyped, would be to look only at the possible values ​​of y, and not at the iteration up to MAX_Y. eg:.

  static Size FindMinSize(double requiredRatio, double epsilon) { int x = 1; do { int y = (int)(x * requiredRatio); if (Test(x, y, requiredRatio, epsilon)) { return new Size(x, y); } y = (int)((x + 1) * requiredRatio); if (Test(x, y, requiredRatio, epsilon)) { return new Size(x, y); } x++; } while (x != int.MaxValue); return new Size(0, 0); } static bool Test(int x, int y, double requiredRatio, double epsilon) { double aspectRatio = ((double)y)/x; return Math.Abs(aspectRatio - requiredRatio) < epsilon; } 

Answer 4

Instead of checking all the possible combinations, simply increase the side that brings you closer to the aspect ratio:

 public static Size GetSizeFromAspectRatio(double aspectRatio) { double epsilon = 0.001; int x = 1; int y = 1; while (true) { double a = (double)x / (double)y; if (Math.Abs(aspectRatio - a) < epsilon) break; if (a < aspectRatio) { x++; } else { y++; } } return new Size(x, y); } 

Answer 5

Aspect ratio is the ratio between x and y. You can define the aspect ratio as x / y or y / x.

The minimum aspect ratio is 0/0.

It is necessary to determine some other minimum, or minimum x, or minimum y.

min x = (min y * x) / y

min y = (min x * y) / x