comparing doubles with adaptive approximately equal

approximate, c#, equals

Solution

`float` and `double` do not have precision in the way you are thinking. Programs fake precision by truncating trailing zeroes...but you can't make use of this trick for your purposes, since rounding errors will prevent such a trick from being reliable.

`decimal` does keep track of how many digits to place to the right of the decimal, but this is still worthless for implementing the algorithm you propose, as any an operation which introduces a representational error (e.g., dividing by 3) will tend to max out the the number of digits to the right of the decimal.

If you want to actually have fuzzy equality based on the known precision of your data, one way to do it would be to create your own number class. Something like this:

public class DoublePrecise
{
    public readonly double Error;
    public readonly double Value;
    public DoublePrecise(double error, double value) {
        Error = error;
        Value = value;
    }
    public static DoublePrecise operator -(DoublePrecise d1, DoublePrecise d2) {
        return new DoublePrecise(d1.Value - d2.Value, d1.Error + d2.Error);
    }
    //More stuff.
}

Basically, this is letting your represent numbers like 10.0±0.1. In that situation, you would treat two numbers as being approximately equal if their ranges overlap (though in reality, this would make your equality operation mean, "could be equal" and your inequality operation mean, "definitely not equal."

See also, Interval Arithmetic

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