Ruby Fibonacci Algorithm

Question

The following is the method I wrote to calculate the 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 that the program has been reacting for too long (I use repl.it). Does anyone know why this is happening?

Answer 1

A naive Fibonacci makes many repeated calculations - it is fib(14) fib(4)calculated many times.

You can add memoization to your algorithm to make it much 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 2

As others have pointed out, the runtime of your implementation exponentially grows at n. There is a much cleaner implementation.

O (n), 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

[O (n) , 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

, , 1_000_000.fib [O (log n) ( )]:

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

Answer 3

- .

, , :

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

, fib(n) 1 , TON

Answer 4

, .

:

def fib (n)
  return 0 if n == 0

  x = 0
  y = 1

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

  return y
end

(0..14).map { |n| fib(n) }
# [0, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610]

Answer 5

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)}

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

, . @Uri Agassi, memoization. fooobar.com/questions/25184/...