How to print the Fibonacci sequence in Swift Playground using recursion - swift

Question

I am trying to use recursion in Swift to print out the Fibonacci sequence for a number "n" iterations. However, I keep getting the same error.
I have already tried doing it without recursion and was able to do it. However, I am now trying to do in a more complex and "computer scientisty" way by using recursion.
func fibonacciSequence (n: Int) -> [Int] {
// Consumes a number "n", which is the number of iterations to go through with the Fibonacci formula and prints such sequence.
var fibonacciArray = [Int]()
for n in 0 ... n {
if n == 0 {
fibonacciArray.append(0)
}
else if n == 1 {
fibonacciArray.append(1)
}
else {
fibonacciArray.append (fibonacciSequence(n: (n - 1)) +
fibonacciSequence(n: (n-2)))
}
}
return fibonacciArray
I expect to call the function with a number n and for the function to print out the Fibonacci sequence. Example: if n = 5, I expect the console to print 0, 1, 1, 2, 3, 5. The error I get is this: (Cannot convert value of type '[Int]' to expected argument type 'Int').

Answer 1

As pointed out above, the return value is causing an error when summed. A possible way (but not recursive) of fixing the code would be to simply change the else statement:
func fibonacciSequence (n: Int) -> [Int] {
// Consumes a number "n", which is the number of iterations to go through with the Fibonacci formula and prints such sequence.
var fibonacciArray = [Int]()
for n in 0 ... n {
if n == 0 {
fibonacciArray.append(0)
}
else if n == 1 {
fibonacciArray.append(1)
}
else {
fibonacciArray.append (fibonacciArray[n-1] + fibonacciArray[n-2] )
}
}
return fibonacciArray
}
A recursive solution would be the following:
func fibonacciSequence (n: Int, sumOne: Int, sumTwo: Int, counter: Int, start: Bool) {
if start {
print(0)
print(1)
}
if counter == -1 {
print(1)
}
if (counter == n - 2) {
return
}
let sum = sumOne + sumTwo
print(sum)
fibonacciSequence(n: n, sumOne: sumTwo , sumTwo: sum, counter: counter + 1, start: false)
}
fibonacciSequence(n: 8, sumOne: 0, sumTwo: 1, counter: 0, start: true)
There is probably a "nicer" way, but I hope it helps. Cheers.

Answer 2

These is my solution for fabonacci series in swift 5 playground
func fibonacci(n: Int) {
var num1 = 0
var num2 = 1
var nextNum = Int()
let i = 1
var array = [Int]()
array.append(num1)
array.append(num2)
for _ in i...n {
nextNum = num1 + num2
num1 = num2
num2 = nextNum
array.append(num2)
print(array)
}
print("result = \(num2)")
}
print(fibonacci(n: 5))

Answer 3

let fibonacci = sequence(state: (0, 1)) {(state: inout (Int, Int)) -> Int? in
defer { state = (state.1, state.0 + state.1) }
return state.0
}
//limit 10
for number in fibonacci.prefix(10) {
print(number)
}

Answer 4

// MARK: - Function
func fibonacciSeries(_ num1 : Int,_ num2 : Int,_ term : Int,_ termCount : Int) -> Void{
if termCount != term{
print(num1)
fibonacciSeries(num2, num2+num1, term, termCount + 1)
}
}
// MARK: - Calling Of Function fibonacciSeries(0, 1, 5, 0)
// MARK: - out Put 0 1 1 2 3
Note Need to Change only No Of term for fibonacci Series.

Answer 5

func fibonacci(n: Int) {
var seq: [Int] = n == 0 ? [0] : [0, 1]
var curNum = 2
while curNum < n{
seq.append(seq[curNum - 1] + seq[curNum - 2])
curNum += 1 }
print(seq) }

Answer 6

Recursive way of fabonacci -> Solutions
func fibo( n: Int) -> Int {
guard n > 1 else { return n }
return fibo(n: n-1) + fibo(n: n-2)
}