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# kieranb662/SwiftEquationSolvers.md

Last active Dec 2, 2021
[Polynomial Solvers] A set of polynomial equation solvers written in Swift. #Math

# Swift Equation Solvers

Most equations need to be solved numerically since no close-form expression representing their solutions can be obtained. For polynomial equations of order 1, 2, 3, 4 exact solutions can be obtained. I have created a series of solvers up to a cubic solver, that can be used to obtain most exact solutions. Of course with floating point errors, not everything is going to come out looking clean. To be able to handle complex numbers I made a simplified version of a complex number without all the mathematical operations.

If you call the `cubicSolve` function with `a = 0` then the solver falls back on the quadratic solver, the quadratic solver will fallback on the linear solver and linear solver will return an empty array(Thanks for catching that u\korbonix).

Example Usage

```let solutions = cubicSolve(a: 1, b: 1, c: 1, d: 1)

let realSolutions = solutions.filter({ \$0.isReal }) // filters for only real values
let complexSolutions = solutions.filter({ !\$0.isReal }) // filters for only complex values```

## Complex Number

Here I made a struct to house the real and imaginary portions of a complex number, I added an isReal property that returns a Bool depending on whether the imaginary portion is 0 or not.

```/// Complex Number
struct ComplexNumber {
var real: Double
var imaginary: Double

public init(_ real: Double, _ imaginary: Double) {
self.real = real
self.imaginary = imaginary
}

public init(_ real: Double) {
self.real = real
self.imaginary = 0
}

var isReal: Bool {
if imaginary == 0 { return true }
return false
}
}

extension ComplexNumber {
static func zero() -> Self {
return ComplexNumber(0, 0)
}
}

extension ComplexNumber: CustomStringConvertible {
var description: String {
return "(Real: \(real), Imaginary: \(imaginary)"
}
}```

## Linear Solver

```/// # Linear Equation Solver
///  Returns the solution to the equation ax + b = 0
func linearSolve(a: Double, b: Double) -> [ComplexNumber] {
if a == 0 {
return []
}

return [ComplexNumber(-b/a)]
}```

```/// # Quadratic Equation Solver
/// Returns the roots of the equation ax^2 + bx + c = 0
///
///
/// Procedure:
///
///     1. calculate discriminant `d = b^2 -4*a*c`
///         i. if d > 0 , equation has two distinct real roots exist.
///         ii. if d = 0, equation has two repeated real roots.
///         iii. if d < 0 equation has two complex roots.
///
///
func quadraticSolve(a: Double, b: Double, c: Double, threshold: Double = 0.0001) -> [ComplexNumber] {
if a == 0 { return linearSolve(a: b, b: c) }

var roots = [ComplexNumber]()

var d = pow(b, 2) - 4*a*c // discriminant

// Check if discriminate is within the 0 threshold
if -threshold < d && d < threshold { d = 0 }

if d > 0 {

let x_1 = ComplexNumber((-b + sqrt(d))/(2*a))
let x_2 = ComplexNumber((-b - sqrt(d))/(2*a))
roots = [x_1, x_2]

} else if d == 0 {

let x = ComplexNumber(-b/(2*a))
roots = [x, x]

} else if d < 0 {

let x_1 = ComplexNumber(-b/(2*a), sqrt(-d)/(2*a))
let x_2 = ComplexNumber(-b/(2*a), -sqrt(-d)/(2*a))
roots = [x_1, x_2]

}

return roots
}```

## Cubic Solver

```/// # Cubic Equation Solver
///
/// Solves the cubic equation of the form `ax^3 + bx^2 + cx + d = 0` - **eq: 1**
/// Should follow some variation of the following algorithm
/// - important:  a, b, c, d should all be real numbers.
/// - reference: [Cubic Equation](https://en.wikipedia.org/wiki/Cubic_equation)
///
///     Procedure:
///
///
///     1.  Divide both sides of equation by a.
///
///         i.  In the case that a == 0 fallback onto the `quadraticSolve`.
///         ii.  other wise move on to step 2. The equation should now look as follows x^3 + (b/a)x^2 + (c/a)x + d/a = 0.
///
///     2.  Calculate the discriminant D
///
///         i.  Let `a_1 = b/a `,  `a_2 = c/a` , and `a_3 = d/a`.
///         ii.  Let `q = (3*a_2 - a_1^2)/9` and `r = (9*a_1*a_2 - 27*a_3 - 2*a_1^3)/54`
///         iii.  Let ` s = cbrt(r + sqrt(q^3+r^2))` and `t = cbrt(r - sqrt(q^3+r^2))`
///         iiii.  Then `D = q^3 + f^2`
///
///         a.   If `D > 0` , one real root with 2 complex conjugate roots.
///         b.   if `D = 0` , all roots are real and atleast 2 are repeated.
///         c.   If `D < 0`, all roots are real and unequal.
///
///
///     2A.  D > 0`,  where `cbrt` and `sqrt`  are the cuberoot and squareroot respectively.
///
///         1.  The only real solution: `x_1 = ComplexNumber(s + t - (1/3)*a_1)`
///         2.  First complex  solution :  `x_2 = ComplexNumber(-(1/2)*(s+t) - (1/3)*a_1,  (1/2)*sqrt(3)*(s - t))`
///         3.  Second complex  solution:  `x_3 = ComplexNumber(-(1/2)*(s+t) - (1/3)*a_1, -(1/2)*sqrt(3)*(s - t))`
///
///     2B.  For `D = 0`  and `D < 0 ` Trigonometry can be used Let `theta  = acos(r/sqrt(-pow(q, 3)))`
///
///         1.`x_1 = ComplexNumber(2*sqrt(-q)*cos((1/3)*theta) - (1/3)*a_1)`
///         2.`x_2 = ComplexNumber(2*sqrt(-q)*cos((1/3)*theta + 2*Double.pi/3) - (1/3)*a_1)`
///         3.`x_3 = ComplexNumber(2*sqrt(-q)*cos((1/3)*theta + 4*Double.pi/3) - (1/3)*a_1)`
///
///
///
///
///
func cubicSolve(a: Double, b: Double, c: Double, d: Double, threshold: Double = 0.0001) -> [ComplexNumber] {
// if not a cubic fall back to quadratic
if a == 0 { return quadraticSolve(a: b, b: c, c: d) }

var roots = [ComplexNumber]()

let a_1 = b/a
let a_2 = c/a
let a_3 = d/a

let q = (3*a_2 - pow(a_1, 2))/9
let r = (9*a_1*a_2 - 27*a_3 - 2*pow(a_1, 3))/54

let s = cbrt(r + sqrt(pow(q, 3)+pow(r, 2)))
let t = cbrt(r - sqrt(pow(q, 3)+pow(r, 2)))

var d = pow(q, 3) + pow(r, 2) // discriminant

// Check if d is within the zero threshold
if -threshold < d && d < threshold { d = 0 }

if d > 0 {

let x_1 = ComplexNumber(s + t - (1/3)*a_1)
let x_2 = ComplexNumber(-(1/2)*(s+t) - (1/3)*a_1,  (1/2)*sqrt(3)*(s - t))
let x_3 = ComplexNumber(-(1/2)*(s+t) - (1/3)*a_1,  -(1/2)*sqrt(3)*(s - t))
roots = [x_1, x_2, x_3]

} else if d <= 0 {

let theta = acos(r/sqrt(-pow(q, 3)))
let x_1 = ComplexNumber(2*sqrt(-q)*cos((1/3)*theta) - (1/3)*a_1)
let x_2 = ComplexNumber(2*sqrt(-q)*cos((1/3)*theta + 2*Double.pi/3) - (1/3)*a_1)
let x_3 = ComplexNumber(2*sqrt(-q)*cos((1/3)*theta + 4*Double.pi/3) - (1/3)*a_1)
roots = [x_1, x_2, x_3]

}

return roots
}```

## Important Notes

The `threshold` value is used to clamp the discriminant to 0 so that the cubic and quadratic solvers will follow the correct path, without it real solutions end up having a non zero imaginary part.

These solvers aren't perfect and Im sure someone can find more features or optimizations. I just wanted to make a good base to work from that is not to complicated to deconstruct or understand.