Ruby Fibonacci algorithm

Question

The following is a method I wrote to calculate a value in the Fibonacci sequence:

def fib(n)

    if n == 0
        return 0
    end
    if n == 1
        return 1
    end

    if n >= 2
        return fib(n-1) + (fib(n-2))
    end

end

It works uptil n = 14, but after that I get a message saying the program is taking too long to respond (I'm using repl.it). Anyone know why this is happening?

Solution 1

Naive fibonacci makes a lot of repeat calculations - in fib(14) fib(4) is calculated many times.

You can add memoization to your algorithm to make it a lot faster:

def fib(n, memo = {})
  if n == 0 || n == 1
    return n
  end
  memo[n] ||= fib(n-1, memo) + fib(n-2, memo)
end
fib 14
# => 377
fib 24
# => 46368
fib 124
# => 36726740705505779255899443

Solution 2

As others have pointed out, your implementation's run time grows exponentially in n. There are much cleaner implementations.

Constructive [O(n) run time, O(1) storage]:

def fib(n)
  raise "fib not defined for negative numbers" if n < 0
  new, old = 1, 0
  n.times {new, old = new + old, new}
  old
end

Memoized recursion [O(n) run time, O(n) storage]:

def fib_memo(n, memo)
  memo[n] ||= fib_memo(n-1, memo) + fib_memo(n-2, memo)
end

def fib(n)
  raise "fib not defined for negative numbers" if n < 0
  fib_memo(n, [0, 1])
end

Recursive powers of a matrix multiplication using squared halving of the power for when you just gotta know really big factorials like 1_000_000.fib [O(log n) run time and storage (on stack)]:

def matrix_fib(n)
  if n == 1
    [0,1]
  else
    f = matrix_fib(n/2)
    c = f[0] * f[0] + f[1] * f[1]
    d = f[1] * (f[1] + 2 * f[0])
    n.even? ? [c,d] : [d,c+d]
  end
end

def fib(n)
  raise "fib not defined for negative numbers" if n < 0
  n.zero? ? n : matrix_fib(n)[1]
end

Solution 3

Your program has exponential runtime due to the recursion you use.

Expanding only the recursive calls a few levels to show you why:

fib(14) = fib(13) + fib(12)
        = (fib(12) + fib(11)) + (fib(11) + fib (10))
        = (((fib(11) + fib(10)) + (fib(10) + fib(9))) (((fib(10) + fib(9)) + (fib(9) + fib(8)))
        = ... //a ton more calls

The terrible runtime might be causing your program to hang, as increasing fib(n) by 1 means you have a TON more recursive calls

Solution 4

your program overflows as Kevin L explained.

instead, you can use an iterative algorithm like this:

def fib (n)
  return 0 if n == 0
  return 1 if n == 1 or n == 2

  x = 0
  y = 1

  (2..n).each do
    z = (x + y)
    x = y
    y = z
  end

  return y
end

(0..14).map { |n| fib(n) }
# [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377]

Solution 5

I tried comparing the run time of two fibonacci methods on repl.it

require 'benchmark'

def fib_memo(n, memo = {})
  if n == 0 || n == 1
    return n
  end
  memo[n] ||= fib_memo(n-1, memo) + fib_memo(n-2, memo)
end


def fib_naive(n)
  if n == 0 || n == 1
    return n
  end
  fib_naive(n-1) + fib_naive(n-2)
end

def time(&block) 
  puts Benchmark.measure(&block) 
end

time {fib_memo(14)}
time {fib_naive(14)}

Output

0.000000   0.000000   0.000000 (  0.000008)
0.000000   0.000000   0.000000 (  0.000099)

As you can see, the runtime is quite different. As @Uri Agassi suggested, use memoization. The concept is explained quite well here https://stackoverflow.com/a/1988826/5256509

View more solutions

Comments

[Casimir et Hippolyte]

I have dreamed about that and Ruby makes it possible.

[Daniël Knippers]

Your code does not run by the way, there should not be an end after the if before the else, and range is not valid either. I will edit your code to make it work, revert / adjust if you disagree.

[sertsedat]

well, I just wrote it with a sleepy head, I would appreciate the edit.

[pjs]

Very elegant, nice job.

[Anatoly]

this called not caching but memoization instead

[JCC]

This method seems to return the number after the one you want. fib(5) should return 3 and not 5 which is the number after.

[Uri Agassi]

@JCC - like everything in computing - this was implemented as a zero based vector, meaning the first element is fib(0). This means that fib(5) retrieves the 6th element in the list...

[JCC]

@UriAgassi What would we have to do to make it start on fib(1)?

[Uri Agassi]

@JCC - that's a great question - give it a try! see if you can make the minimal change to enable this - remember to check rigorously (including edge cases!)

[Pablo]

The Fibonacci series is: 0, 1, 1, 2, 3, 5, 8, 13, 21.... Your algorithm is not producing the correct result. For example, for the 20th element in the fibonacci series it returns 10946 instead of 6765; it's doing an extra iteration. Also, it should return 1 when the input is 2. To fix these issues, the range shoud be (2..n) and you should add the line return 1 if n == 2. I like your algorithm though.