inFullMobile Language Wars: Round 1 - Swift Solution 1 - full code by @mikchmie-ifm - See: https://gist.github.com/rsp/d8bdbafa09f24f99eebc8ed60fe205c8

ifm-lw-r1-swift-solution-1-full.swift

import Foundation

func inc(_ num: Int) -> Int {
    return num + 1
}

func count(_ foo: @escaping (@escaping (Int) -> Int) -> (Int) -> Int) -> Int {
    return foo(inc(_:))(0)
}

func f(_ foo: @escaping (@escaping (Int) -> Int) -> (Int) -> Int) -> (@escaping (Int) -> Int) -> (Int) -> Int {
    return { op in
        return { num in
            let times = foo({ x in x + 1 })(0) - 1
            return (0..<times).reduce(num, { result, _ in op(result) })
        }
    }
}

// Applies x() 1 time:
func input1(_ x: @escaping (Int) -> Int) -> (Int) -> Int {
    return { y in
        return x(y)
    }
}

// Applies x() 2 times:
func input2(_ x: @escaping (Int) -> Int) -> (Int) -> Int {
    return { y in
        return x(x(y))
    }
}

// Applies x() 10 times:
func input3(_ x: @escaping (Int) -> Int) -> (Int) -> Int {
    return { y in
        return x(x(x(x(x(x(x(x(x(x(y))))))))))
    }
}

// Applies x() 100 times:
func input4(_ x: @escaping (Int) -> Int) -> (Int) -> Int {
    return input3(input3(x));
}

// Applies x() 1000 times:
func input5(_ x: @escaping (Int) -> Int) -> (Int) -> Int {
    return input3(input4(x));
}

count(input1(_:)) == 1
count(input2(_:)) == 2
count(input3(_:)) == 10
count(input4(_:)) == 100
count(input5(_:)) == 1000

count(input1(_:)) == count(f(input1(_:))) + 1
count(input2(_:)) == count(f(input2(_:))) + 1
count(input3(_:)) == count(f(input3(_:))) + 1
count(input4(_:)) == count(f(input4(_:))) + 1
count(input5(_:)) == count(f(input5(_:))) + 1