Implementing fibonacci series using greedy approach?

Question

I have implemented fibonacci series using recursion:

def fibonacci(n): 
    if n==0: 
        return 0
    elif n==1: 
        return 1
    else: 
        return fibonacci(n-1) + fibonacci(n-2)

I have also implemented it using dynamic programming:

def fibonacci(n): 
    result = [0, 1]

    if n > 1:
        for i in range(2, n+1):
            result.append(result[i-1] + result[i-2])
    return result[n]

I want to implement it using greedy approach. I am unable to think of it in greedy terms. Please provide a greedy approach for this problem.

Answer 1

I didn't understand what you wanted to say by saying the word 'greedy'. But these are ways:

Example 1: Using looping technique

 def fib(n):
     a,b = 1,1
     for i in range(n-1):
      a,b = b,a+b
     return a
    print fib(5)

Example 2: Using recursion

def fibR(n):
 if n==1 or n==2:
  return 1
 return fibR(n-1)+fibR(n-2)
print fibR(5)

Example 3: Using generators

a,b = 0,1
def fibI():
 global a,b
 while True:
  a,b = b, a+b
  yield a
f=fibI()
f.next()
f.next()
f.next()
f.next()
print f.next()

Example 4: Using memoization

def memoize(fn, arg):
 memo = {}
 if arg not in memo:
  memo[arg] = fn(arg)
  return memo[arg]

fib() as written in example 1.

fibm = memoize(fib,5)
print fibm

Example 5: Using memoization as the decorator

class Memoize:
 def __init__(self, fn):
  self.fn = fn
  self.memo = {}
 def __call__(self, arg):
  if arg not in self.memo:
   self.memo[arg] = self.fn(arg)
   return self.memo[arg]

@Memoize
def fib(n):
 a,b = 1,1
 for i in range(n-1):
  a,b = b,a+b
 return a
print fib(5)