So i have code in xcode playground and got the error error in code
and this is the excercise i do
Return the Factorial
Create a function that takes an integer and returns the factorial of that integer. That is, the integer multiplied by all positive lower integers.
and this is my code
func factorial(_ num: Int) -> Int {
var result :Int = 0
for _ in num...1 {
result = num * (num - 1)
}
return result
}
print(factorial(12)) //error in this line
but in terminal got this output:
The answer is in the error message - the upper bound of the range has to be greater than or equal to the lower.
Therefore you'll need to rewrite the method differently. You also have a fundamental flaw in your code - your maths logic is wrong - result will be overwritten every time, rather than accumulating a total. Adopting a similar approach to the question you can do this:
func factorial(_ num: Int) -> Int {
var result :Int = 1
guard num > 0 else {return result}. //0! is 1 not 0
var num = num
while num > 0 {
result *= num
num -= 1
}
return result
}
There is a more elegant way of performing this using recursion, which avoids the ugly var num=num
func fac2(_ num: Int, total: Int = 1) -> Int {
return num > 0 ? fac2(num-1, total: total * num) : total
}
I am trying to write a for loop, where I have to increment exponentially. I am using stride function but it won't work.
Here's c++ code, I am trying to write a swift version.
for (int m = 1; m <= high - low; m = 2*m){}
can you help me, to write this code in swift version?
A while-loop is probably the simplest solution, but here is an alternative:
for m in sequence(first: 1, next: { 2 * $0 }).prefix(while: { $0 <= high - low }) {
print(m)
}
sequence() (lazily) generates the sequence 1, 2, 4, ..., and prefix(while:) limits that sequence to the given range.
A slight advantage of this approach is that m is only declared inside the loop (so that it cannot be used later accidentally), and that is is a constant so that it cannot be inadvertently modified inside the loop.
There is no for loop in Swift, but you can achieve the same result with basic while loop
var m = 1 // initializer
while m <= high - low { // condition
...
m *= 2 // iterator
}
Based on #MartinR answer, only improvement is readability of the call:
// Helper function declaration
func forgen<T>(
_ initial: T, // What we start with
_ condition: #escaping (T) throws -> Bool, // What to check on each iteration
_ change: #escaping (T) -> T?, // Change after each iteration
_ iteration: (T) throws -> Void // Where actual work happens
) rethrows
{
return try sequence(first: initial, next: change).prefix(while: condition).forEach(iteration)
}
// Usage:
forgen(1, { $0 <= high - low }, { 2 * $0 }) { m in
print(m)
}
// Almost like in C/C++ code
Here is a solution using for:
let n = Int(log(Double(high - low))/log(2.0))
var m = 1
for p in 1...n {
print("m =", m)
...
m = m << 1
}
(Supposing that high - low is greater than 2)
I am building a calculator and want it to automatically convert every decimal into a fraction. So if the user calculates an expression for which the answer is "0.333333...", it would return "1/3". For "0.25" it would return "1/4". Using GCD, as found here (Decimal to fraction conversion), I have figured out how to convert any rational, terminating decimal into a decimal, but this does not work on any decimal that repeats (like .333333).
Every other function for this on stack overflow is in Objective-C. But I need a function in my swift app! So a translated version of this (https://stackoverflow.com/a/13430237/5700898) would be nice!
Any ideas or solutions on how to convert a rational or repeating/irrational decimal to a fraction (i.e. convert "0.1764705882..." to 3/17) would be great!
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)
So a little late here, but I had a similar problem and ended up building Swift FractionFormatter. This works because most of the irrational numbers you care about are part of the set of vulgar, or common fractions and are easy to validate proper transformation. The rest may or may not round, but you get very close on any reasonable fraction your user might generate. It is designed to be a drop in replacement for NumberFormatter.
As Martin R said, the Only way to have (99.99%)exact calculations, is to calculate everything with rational numbers, from beginning to the end.
the reason behind the creation of this class was also the fact that i needed to have very accurate calculations, and that was not possible with the swift-provided types. so i created my own type.
here is the code, i'll explain it below.
class Rational {
var alpha = 0
var beta = 0
init(_ a: Int, _ b: Int) {
if (a > 0 && b > 0) || (a < 0 && b < 0) {
simplifier(a,b,"+")
}
else {
simplifier(a,b,"-")
}
}
init(_ double: Double, accuracy: Int = -1) {
exponent(double, accuracy)
}
func exponent(_ double: Double, _ accuracy: Int) {
//Converts a double to a rational number, in which the denominator is of power of 10.
var exp = 1
var double = double
if accuracy != -1 {
double = Double(NSString(format: "%.\(accuracy)f" as NSString, double) as String)!
}
while (double*Double(exp)).remainder(dividingBy: 1) != 0 {
exp *= 10
}
if double > 0 {
simplifier(Int(double*Double(exp)), exp, "+")
}
else {
simplifier(Int(double*Double(exp)), exp, "-")
}
}
func gcd(_ alpha: Int, _ beta: Int) -> Int {
// Calculates 'Greatest Common Divisor'
var inti: [Int] = []
var multi = 1
var a = Swift.min(alpha,beta)
var b = Swift.max(alpha,beta)
for idx in 2...a {
if idx != 1 {
while (a%idx == 0 && b%idx == 0) {
a = a/idx
b = b/idx
inti.append(idx)
}
}
}
inti.map{ multi *= $0 }
return multi
}
func simplifier(_ alpha: Int, _ beta: Int, _ posOrNeg: String) {
//Simplifies nominator and denominator (alpha and beta) so they are 'prime' to one another.
let alpha = alpha > 0 ? alpha : -alpha
let beta = beta > 0 ? beta : -beta
let greatestCommonDivisor = gcd(alpha,beta)
self.alpha = posOrNeg == "+" ? alpha/greatestCommonDivisor : -alpha/greatestCommonDivisor
self.beta = beta/greatestCommonDivisor
}
}
typealias Rnl = Rational
func *(a: Rational, b: Rational) -> Rational {
let aa = a.alpha*b.alpha
let bb = a.beta*b.beta
return Rational(aa, bb)
}
func /(a: Rational, b: Rational) -> Rational {
let aa = a.alpha*b.beta
let bb = a.beta*b.alpha
return Rational(aa, bb)
}
func +(a: Rational, b: Rational) -> Rational {
let aa = a.alpha*b.beta + a.beta*b.alpha
let bb = a.beta*b.beta
return Rational(aa, bb)
}
func -(a: Rational, b: Rational) -> Rational {
let aa = a.alpha*b.beta - a.beta*b.alpha
let bb = a.beta*b.beta
return Rational(aa, bb)
}
extension Rational {
func value() -> Double {
return Double(self.alpha) / Double(self.beta)
}
}
extension Rational {
func rnlValue() -> String {
if self.beta == 1 {
return "\(self.alpha)"
}
else if self.alpha == 0 {
return "0"
}
else {
return "\(self.alpha) / \(self.beta)"
}
}
}
// examples:
let first = Rnl(120,45)
let second = Rnl(36,88)
let third = Rnl(2.33435, accuracy: 2)
let forth = Rnl(2.33435)
print(first.alpha, first.beta, first.value(), first.rnlValue()) // prints 8 3 2.6666666666666665 8 / 3
print((first*second).rnlValue()) // prints 12 / 11
print((first+second).rnlValue()) // prints 203 / 66
print(third.value(), forth.value()) // prints 2.33 2.33435
First of all, we have the class itself. the class can be initialised in two ways:
in the Rational class, alpha ~= nominator & beta ~= denominator
The First way is initialising the class using two integers, first of with is the nominator, and the second one is the denominator. the class gets those two integers, and then reduces them to the least numbers possible. e.g reduces (10,5) to (2,1) or as another example, reduces (144, 60) to (12,5). this way, always the simplest numbers are stored.
this is possible using the gcd (greatest common divisor) function and simplifier function, which are not hard to understand from the code.
the only thing is that the class faces some issues with negative numbers, so it always saves whether the final rational number is negative or positive, and if its negative it makes the nominator negative.
The Second way to initialise the class, is with a double, and with an optional parameter called 'accuracy'. the class gets the double, and also the accuracy of how much numbers after decimal point you need, and converts the double to a nominator/denominator form, in which the denominator is of power of 10. e.g 2.334 will be 2334/1000 or 342.57 will be 34257/100. then tries to simplify the rational numbers using the same method which was explained in the #1 way.
After the class definition, there is type-alias 'Rnl', which you can obviously change it as you wish.
Then there are 4 functions, for the 4 main actions of math: * / + -, which i defined so e.g. you can easily multiply two numbers of type Rational.
After that, there are 2 extensions to Rational type, first of which ('value') gives you the double value of a Rational number, the second one ('rnlValue') gives you the the Rational number in form of a human-readable string: "nominator / denominator"
At last, you can see some examples of how all these work.
I've got two Int values (they have to be Ints) and I want them to round off to the nearest value when in an equation;
var Example = Int()
var secondExample = Int()
Example = (secondExample / 7000)
This equation makes the variable Example always round down to the lowest value. Say for example that the numbers are the following;
var Example = Int()
var secondExample : Int = 20000
Example = (20000 / 7000)
20000 / 7000 equals 2.857... But the variable Example displays 2.
How can I make Example round off to closest number without changing it to a Double
For nonnegative integers, the following function gives
the desired result in pure integer arithmetic :
func divideAndRound(numerator: Int, _ denominator: Int) -> Int {
return (2 * numerator + denominator)/(2 * denominator)
}
Examples:
print(20000.0/7000.0) // 2.85714285714286
print(divideAndRound(20000, 7000)) // 3 (rounded up)
print(10000.0/7000.0) // 1.42857142857143
print(divideAndRound(10000, 7000)) // 1 (rounded down)
The idea is that
a 1 2 * a + b
- + - = ---------
b 2 2 * b
And here is a possible implementation for arbitrarily signed
integers which also does not overflow:
func divideAndRound(num: Int, _ den: Int) -> Int {
return num / den + (num % den) / (den / 2 + den % 2)
}
(Based on #user3441734's updated solution, so we have a reference
cycle between our answers now :)
There is also a ldiv function which computes both quotient
and remainder of a division, so the last function could also be
implemented as
func divideAndRound(num: Int, _ den: Int) -> Int {
let div = ldiv(num, den)
let div2 = ldiv(den, 2)
return div.quot + div.rem / (div2.quot + div2.rem)
}
(I did not test which version is faster.)
see Martin's answer! his idea is great, so i extend his solution for negative numbers
func divideAndRound(n: Int, _ d: Int) -> Int {
let sn = n < 0 ? -1 : 1
let sd = d < 0 ? -1 : 1
let s = sn * sd
let n = n * sn
let d = d * sd
return (2 * n + d)/(2 * d) * s
}
divideAndRound(1, 2) // 1
divideAndRound(1, 3) // 0
divideAndRound(-1, 2) // -1
divideAndRound(-1, 3) // 0
divideAndRound(1, -2) // -1
divideAndRound(1, -3) // 0
the only trouble is that (2 * n + d) can overflow and code will crash.
UPDATE! with help of mathematics for children
func divr(a: Int, _ b: Int)->Int {
return (a % b) * 2 / b + a / b
}
it should work for any Int :-) except 0 denominator.
I want to solve this
Write a function applyKTimes(f:(Float -> Float),x:Float,k:Int) -> Float
that takes a function f and a float x and applies f to x k times
for example the function f is
func square (number:Float) -> Float
{
return number * number
}
is there a short solution using Swift?
If you don't want to use recursion:
func applyKTimes(f:(Float -> Float), x: Float, k: Int) -> Float {
var result = x
for _ in 0..k {
result = f(result)
}
return result
}
A recursive solution:
func applyKTimes(f:(Float -> Float), x:Float, k:Int) -> Float
{
return k > 0 ? applyKTimes(f, f(x), k - 1) : x
}
I have not checked if Swift guarantees tail call optimization. If it does, than the recursive solution is what you really want. However if it does not you need to unroll it by hand to avoid problems with the stack.
The same, but unrolled by hand:
func applyKTimes(f:(Float -> Float), x:Float, k:Int) -> Float
{
var result = x
for _ in 0..k {
result = f(result)
}
return result
}