Format Double as Fraction [closed]

Answer 1

I don't necessarily agree, base on the fact that Milhous wants to cover inches up to 1/64" Suppose that the program demands 1/64" precision at all times, that should take up 6 bits of the mantissa. In a float, there's 24-6 = 18, which (if my math is right), should mean that he's got a range of +/- 262144 + 63/64"

That might be enough precision in the float to convert properly into the faction without loss.

And since most people working on inches uses denominator of powers of 2, it should be fine.

But back to the original question, I don't know any libraries that would do that.

Answer 2

One problem you might run into is that not all fractional values can be represented by doubles. Even some values that look simple, like 0.1. Now on with the pseudocode algorithm. You would probably be best off determining the number of 64ths of an inch, but dividing the decimal portion by 0.015625. After that, you can reduce your fraction to the lowest common denominator. However, since you state inches, you may not want to use the smallest common denominator, but rather only values for which inches are usually represented, 2,4,8,16,32,64.

One thing to point out however, is that since you are using inches, if the values are all proper fractions of an inch, with a denominator of 2,4,8,16,32,64 then the value should never contain floating point errors, because the denominator is always a power of 2. However if your dataset had a value of .1 inch in there, then you would start to run into problems.

Answer 3

Function for this in a C-variant called LPC follows. Some notes:

1. Addition to input value at beginning is to try to cope with precision issues that otherwise love to wind up telling you that 5 is 4 999999/1000000.
2. The to_int() function truncates to integer.
3. Language has a to_string() that will turn some floats into exponential notation.

---

string strfrac(float frac) {
    int main = to_int(frac + frac / 1000000.0);
    string out = to_string(main);
    float rem = frac - to_float(main);
    string rep;
    if(rem > 0 && (to_int(rep = to_string(rem)) || member(rep, 'e') == Null)) {
        int array primes = ({ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 });
        string base;
        int exp;
        int num;
        int div;
        if(sscanf(rep, "%se%d", base, exp) == 2) {
            num = to_int(replace(base, ".", ""));
            div = to_int(pow(10, abs(exp)));
        } else {
            rep = rep[2..];
            num = to_int(rep);
            div = to_int(pow(10, strlen(rep)));
        }
        foreach(int prime : primes) {
            if(prime > num)
                break;
            while((num / prime) * prime == num && (div / prime) * prime == div) {
                num /= prime;
                div /= prime;
            }
        }
        out += " " + num + "/" + div;
    }
    return out;
}

Answer 4

To solve this problem (in one of my projects), I took the following steps:

• Built a dictionary of decimal/fraction strings.
• Wrote a function to search the dictionary for the closest matching fraction depending on the "decimal" part of the number and the matching criteria.

Answer 5

My code looks like this.

public static int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        else
            return gcd(b, a % b);
    }

public static String doubleToStringFraction(Double d)
    {
        StringBuffer result = new StringBuffer(" " + ((int) Math.floor(d)));
        int whole = (int) ((d - Math.floor(d)) * 10000);
        int gcd = gcd(whole, 10000);
        result.append(" " + (whole / gcd) + "/" + 10000 / gcd + " ");
        return result.toString();
    }

Answer 6

How about org.apache.commons.math ? They have a Fraction class that takes a double.

http://commons.apache.org/math/api-1.2/org/apache/commons/math/fraction/Fraction.html

You should be able to extend it and give it functionality for the 64th. And you can also add a toString that will easily print out the whole number part of the fraction for you.

Fraction(double value, int maxDenominator) Create a fraction given the double value and maximum denominator.