Fibonacci number(Recursive, DP and Decorator)

I found a fatastic video about Dynamic Programming, it can be used to solve Fibonacci number problem efficiently. Decorater can also be used to make code elegent ^o^

Fibonacci number: 1, 1, 2, 3, 5, 8, 13…

Original recursive implementation

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def fib(n):
if n<=2: f = 1
else: f = fib(n-1) + fib(n-2)

return f

Improvement: memoization

  • Reuse solutions to sub-problems to solve the problem
  • So time = #sub-problems * O(sub-problem)
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    #!/usr/bin/python3
    from collections import defaultdict
    import sys

    n = int(sys.argv[1])
    mem = defaultdict(lambda: 0)

    def fib(n):
    global mem
    # print(n)
    if mem[n]:
    return mem[n]

    if n<=2: f = 1
    else: f = fib(n-1) + fib(n-2)

    mem[n] = f
    print(f)
    return f


    def fib_con(n):
    mem = {}
    for k in range(1, 1+n):
    if k<=2:
    mem[k] = 1
    else:
    mem[k] = mem[k-1] + mem[k-2]

    print(mem[k])
    return mem[n]

    fib(n)
    fib_con(n)

Video link: https://www.youtube.com/watch?v=OQ5jsbhAv_M
DP = recursion + memorization + guessing

Problem of the improvement above:

The disadvantage of this method is that the clarity and the beauty of the original recursive implementation is lost.
So we use a helper function to handle the fib() function, the idea of .

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def memoize(f):
memo = {}

def helper(x):
if x not in memo:
memo[x] = f(x)
return memo[x]

return helper


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


fib = memoize(fib)
print(fib(40))

That is Decorator: @memoize
U can check this website for more information:
http://www.python-course.eu/python3_memoization.php

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@memoize
def fib(n):
...

print(fib(40))

Generator version

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def fib(n):
a, b = 0, 1
for _ in range(n):
yield a
a, b = b, a+b

for i in fib(10):
print(i)