Factorials in Swift

How to calculate the 21! (21 factorial) in swift?

Unsigned 64 bit integer has a maximum value of 18,446,744,073,709,551,615. While 21! = 51,090,942,171,709,440,000. For this kind of case, you need a Big Integer type. I found a question about Big Integer in Swift. There's a library for Big Integer in that link.

BigInteger equivalent in Swift?

Factorials in Swift

As others have said you can use libraries that support larger numbers, or just don't allow values that are too big.

Note that if you want to handle very large values you might need to use a loop rather than a recursive algorithm because recursion can cause a Stack Overflow. Yes, that's right, an SO "eponymous crash".

To prevent numbers that are too big, figure out the largest number that doesn't crash, and do input checking and reject numbers bigger than that.

You can figure out the number that crashes it by going in the opposite direction and logging the count and the result every 10 steps:

1 * 2 * 3 * 4 * 5 * 6...

When you crash, go back to the previous largest logged value, start from there plugging in your previously logged factorial result, and step 1 at a time until you crash. Then only allow n that's 1 less than your crash value.

In Swift 3, how to calculate the factorial when the result becomes too high?

Here's an approach that will let you find very large factorials.

Represent large numbers as an array of digits. For instance 987 would be [9, 8, 7]. Multiplying that number by an integer n would require two steps.

1. Multiply each value in that array by n.
2. Perform a carry operation to return a result that is again single digits.

For example 987 * 2:

let arr = [9, 8, 7]
let arr2 = arr.map { $0 * 2 }
print(arr2)  // [18, 16, 14]

Now, perform the carry operation. Starting at the one's digit, 14 is too big, so keep the 4 and carry the 1. Add the 1 to 16 to get 17.

[18, 17, 4]

Repeat with the ten's place:

[19, 7, 4]

And then with the hundred's place:

[1, 9, 7, 4]

Finally, for printing, you could convert this back to a string:

let arr = [1, 9, 7, 4]
print(arr.map(String.init).joined())

1974

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Applying that technique, here is a carryAll function that performs the carry operation, and a factorial that uses it to calculate very large factorials:

func carryAll(_ arr: [Int]) -> [Int] {
    var result = [Int]()

    var carry = 0
    for val in arr.reversed() {
        let total = val + carry
        let digit = total % 10
        carry = total / 10
        result.append(digit)
    }

    while carry > 0 {
        let digit = carry % 10
        carry = carry / 10
        result.append(digit)
    }

    return result.reversed()
}

func factorial(_ n: Int) -> String {
    var result = [1]
    for i in 2...n {
        result = result.map { $0 * i }
        result = carryAll(result)
    }

    return result.map(String.init).joined()
}

print(factorial(1000))

402387260077093773543702433923003985719374864210714632543799910429938512398629020592044208486969404800479988610197196058631666872994808558901323829669944590997424504087073759918823627727188732519779505950995276120874975462497043601418278094646496291056393887437886487337119181045825783647849977012476632889835955735432513185323958463075557409114262417474349347553428646576611667797396668820291207379143853719588249808126867838374559731746136085379534524221586593201928090878297308431392844403281231558611036976801357304216168747609675871348312025478589320767169132448426236131412508780208000261683151027341827977704784635868170164365024153691398281264810213092761244896359928705114964975419909342221566832572080821333186116811553615836546984046708975602900950537616475847728421889679646244945160765353408198901385442487984959953319101723355556602139450399736280750137837615307127761926849034352625200015888535147331611702103968175921510907788019393178114194545257223865541461062892187960223838971476088506276862967146674697562911234082439208160153780889893964518263243671616762179168909779911903754031274622289988005195444414282012187361745992642956581746628302955570299024324153181617210465832036786906117260158783520751516284225540265170483304226143974286933061690897968482590125458327168226458066526769958652682272807075781391858178889652208164348344825993266043367660176999612831860788386150279465955131156552036093988180612138558600301435694527224206344631797460594682573103790084024432438465657245014402821885252470935190620929023136493273497565513958720559654228749774011413346962715422845862377387538230483865688976461927383814900140767310446640259899490222221765904339901886018566526485061799702356193897017860040811889729918311021171229845901641921068884387121855646124960798722908519296819372388642614839657382291123125024186649353143970137428531926649875337218940694281434118520158014123344828015051399694290153483077644569099073152433278288269864602789864321139083506217095002597389863554277196742822248757586765752344220207573630569498825087968928162753848863396909959826280956121450994871701244516461260379029309120889086942028510640182154399457156805941872748998094254742173582401063677404595741785160829230135358081840096996372524230560855903700624271243416909004153690105933983835777939410970027753472000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

Factorial with intermediate results - Swift playgrounds - index out of range error

results is an empty array but you try to access values without appending values first.

The simplest solution might be to pre-populate your array with zeros.

var results: [Int] = Array(repeating: 0, count: n)

Swift has error : thread return -x to return to the state before expression evaluation

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
}

How handle different threads in recursive function in swift?

My recommendation would be to split it into synchronous and asynchronous functions:

func factorial(of number: Double, completion: @escaping (Double) -> ()) {

    DispatchQueue.global(qos: .background).async {
        let result = factorial(of: number)
        DispatchQueue.main.sync { //I don't know what thread you're intending to use, so I picked main :)
            completion(result)
        }
    }
}

func factorial(of number: Double) -> Double {

    var result: Double = 0
    if number > 170 {
        return Double.infinity
    }
    if number != Double(Int(number)) || number < 1 {return 0}
    if number == 1 {
        return 1
    }

    result = number * factorial(of: number - 1)

    return result
}

This allows the meet of the work to happen in whichever way you want, without having to worry about potential threading issues.

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