If you want to display the results of calculations as rational numbers
then the only 100% correct solution is to use rational arithmetic throughout all calculations, i.e. all intermediate values are stored as a pair of integers (numerator, denominator), and all additions, multiplications, divisions, etc are done using the rules for rational
numbers.
As soon as a result is assigned to a binary floating point number
such as Double, information is lost. For example,
let x : Double = 7/10
stores in x an approximation of 0.7, because that number cannot
be represented exactly as a Double. From
print(String(format:"%a", x)) // 0x1.6666666666666p-1
one can see that x holds the value
0x16666666666666 * 2^(-53) = 6305039478318694 / 9007199254740992
≈ 0.69999999999999995559107901499373838305
So a correct representation of x as a rational number would be
6305039478318694 / 9007199254740992, but that is of course not what
you expect. What you expect is 7/10, but there is another problem:
let x : Double = 69999999999999996/100000000000000000
assigns exactly the same value to x, it is indistinguishable from
0.7 within the precision of a Double.
So should x be displayed as 7/10 or as 69999999999999996/100000000000000000 ?
As said above, using rational arithmetic would be the perfect solution.
If that is not viable, then you can convert the Double back to
a rational number with a given precision.
(The following is taken from Algorithm for LCM of doubles in Swift.)
Continued Fractions
are an efficient method to create a (finite or infinite) sequence of fractions hn/kn that are arbitrary good approximations to a given real number x,
and here is a possible implementation in Swift:
typealias Rational = (num : Int, den : Int)
func rationalApproximationOf(x0 : Double, withPrecision eps : Double = 1.0E-6) -> Rational {
var x = x0
var a = floor(x)
var (h1, k1, h, k) = (1, 0, Int(a), 1)
while x - a > eps * Double(k) * Double(k) {
x = 1.0/(x - a)
a = floor(x)
(h1, k1, h, k) = (h, k, h1 + Int(a) * h, k1 + Int(a) * k)
}
return (h, k)
}
Examples:
rationalApproximationOf(0.333333) // (1, 3)
rationalApproximationOf(0.25) // (1, 4)
rationalApproximationOf(0.1764705882) // (3, 17)
The default precision is 1.0E-6, but you can adjust that to your needs:
rationalApproximationOf(0.142857) // (1, 7)
rationalApproximationOf(0.142857, withPrecision: 1.0E-10) // (142857, 1000000)
rationalApproximationOf(M_PI) // (355, 113)
rationalApproximationOf(M_PI, withPrecision: 1.0E-7) // (103993, 33102)
rationalApproximationOf(M_PI, withPrecision: 1.0E-10) // (312689, 99532)
Swift 3 version:
typealias Rational = (num : Int, den : Int)
func rationalApproximation(of x0 : Double, withPrecision eps : Double = 1.0E-6) -> Rational {
var x = x0
var a = x.rounded(.down)
var (h1, k1, h, k) = (1, 0, Int(a), 1)
while x - a > eps * Double(k) * Double(k) {
x = 1.0/(x - a)
a = x.rounded(.down)
(h1, k1, h, k) = (h, k, h1 + Int(a) * h, k1 + Int(a) * k)
}
return (h, k)
}
Examples:
rationalApproximation(of: 0.333333) // (1, 3)
rationalApproximation(of: 0.142857, withPrecision: 1.0E-10) // (142857, 1000000)
Or – as suggested by @brandonscript – with a struct Rational and an initializer:
struct Rational {
let numerator : Int
let denominator: Int
init(numerator: Int, denominator: Int) {
self.numerator = numerator
self.denominator = denominator
}
init(approximating x0: Double, withPrecision eps: Double = 1.0E-6) {
var x = x0
var a = x.rounded(.down)
var (h1, k1, h, k) = (1, 0, Int(a), 1)
while x - a > eps * Double(k) * Double(k) {
x = 1.0/(x - a)
a = x.rounded(.down)
(h1, k1, h, k) = (h, k, h1 + Int(a) * h, k1 + Int(a) * k)
}
self.init(numerator: h, denominator: k)
}
}
Example usage:
print(Rational(approximating: 0.333333))
// Rational(numerator: 1, denominator: 3)
print(Rational(approximating: .pi, withPrecision: 1.0E-7))
// Rational(numerator: 103993, denominator: 33102)