Fibonacci numbers generator in Swift 3

Question

The following Q&A covers a few methods of generating Fibonacci numbers in Swift, but it's quite outdated (Swift 1.2?):

• Sum of Fibonacci term using Functional Swift

Question: How could we generate Fibonacci numbers neatly using modern Swift (Swift >= 3)? Preferably methods avoiding explicit recursion.

Answer 1

An alternative for Swift 3.0 would be to use the helper function

public func sequence<T>(first: T, while condition: @escaping (T)-> Bool, next: @escaping (T) -> T) -> UnfoldSequence<T, T> {
    let nextState = { (state: inout T) -> T? in
        // Return `nil` if condition is no longer satisfied:
        guard condition(state) else { return nil }
        // Update current value _after_ returning from this call:
        defer { state = next(state) }
        // Return current value:
        return state
    }
    return sequence(state: first, next: nextState)
}

from Express for loops in swift with dynamic range:

for f in sequence(first: (0, 1), while: { $1 <= 50 }, next: { ($1, $0 + $1)}) {
    print(f.1)
}
// 1 1 2 3 5 8 13 21 34

Note that in order to include zero in the resulting sequence, it suffices to replace the initial value (0, 1) by (1, 0):

for f in sequence(first: (1, 0), while: { $1 <= 50 }, next: { ($1, $0 + $1)}) {
    print(f.1)
}
// 0 1 1 2 3 5 8 13 21 34

That makes the "artificial" check

if pair.1 == 0 { pair.1 = 1; return 0 }

redundant. The underlying reason is that the Fibonacci numbers can be generalized to negative indices (https://en.wikipedia.org/wiki/Generalizations_of_Fibonacci_numbers):

 ... -8, 5, -3, 2, -1, 1, 0, 1, 1, 2, 3, 5, 8, ...