Fibonacci sequence in Ruby (recursion)

Question

I\'m trying to implement the following function, but it keeps giving me the stack level too deep (SystemStackError) error.

Any ideas what the problem mi

Answer 1

If you want to write the fastest functonal algorithem for fib, it will not be recursive. This is one of the few times were the functional way to write a solution is slower. Because the stack repeats its self if you use somethingn like

fibonacci( n - 1 ) + fibonacci( n - 2 ) 

eventually n-1 and n-2 will create the same number thus repeats will be made in the calculation. The fastest way to to this is iteratvily

def fib(num)
# first 5 in the sequence 0,1,1,2,3
fib1 = 1 #3
fib2 = 2 #4
i = 5 #start at 5 or 4 depending on wheather you want to include 0 as the first number
while i <= num
    temp = fib2
    fib2 = fib2 + fib1
    fib1 = temp
    i += 1
end
p fib2
end
fib(500)

Answer 2

Try this

def fibonacci( n )
  return  n  if ( 0..1 ).include? n
  ( fibonacci( n - 1 ) + fibonacci( n - 2 ) )
end
puts fibonacci( 5 )
# => 5

check this post too Fibonacci One-Liner

and more .. https://web.archive.org/web/20120427224512/http://en.literateprograms.org/Fibonacci_numbers_(Ruby)

You have now been bombarded with many solutions :)

regarding problem in ur solution

you should return n if its 0 or 1

and add last two numbers not last and next

New Modified version

def fibonacci( n )
    return  n  if n <= 1 
    fibonacci( n - 1 ) + fibonacci( n - 2 )
end 
puts fibonacci( 10 )
# => 55

One liner

def fibonacci(n)
   n <= 1 ? n :  fibonacci( n - 1 ) + fibonacci( n - 2 ) 
end
puts fibonacci( 10 )
# => 55

Answer 3

It's been a while, but you can write a fairly elegant and simple one line function:

def fib(n)
  n > 1 ? fib(n-1) + fib(n-2) : n
end

Answer 4

Another approach of calculating fibonacci numbers taking the advantage of memoization:

$FIB_ARRAY = [0,1]
def fib(n)
  return n if $FIB_ARRAY.include? n
  ($FIB_ARRAY[n-1] ||= fib(n-1)) + ($FIB_ARRAY[n-2] ||= fib(n-2))
end

This ensures that each fibonacci number is being calculated only once reducing the number of calls to fib method greatly.

Answer 5

1) Example, where max element < 100

def fibonachi_to(max_value)
  fib = [0, 1]
  loop do
    value = fib[-1] + fib[-2]
    break if value >= max_value
    fib << value
  end
  fib
end

puts fibonachi_to(100)

Output:

0
1
1
2
3
5
8
13
21
34
55
89

2) Example, where 10 elements

def fibonachi_of(numbers)
  fib = [0, 1]
  (2..numbers-1).each { fib << fib[-1] + fib[-2] }
  fib
end

puts fibonachi_of(10)

Output:

0
1
1
2
3
5
8
13
21
34

Answer 6

Here is one in Scala:

object Fib {
  def fib(n: Int) {
    var a = 1: Int
    var b = 0: Int
    var i = 0: Int
    var f = 0: Int
    while(i < n) {
      println(s"f(${i+1}) -> $f")
      f = a+b
      a = b
      b = f
      i += 1
    }
  }

  def main(args: Array[String]) {
    fib(10)
  }
}