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
        re         


        

Answer 1

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

Answer 2

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

Answer 3

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]

Answer 4

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

Answer 5

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