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Python2017_zadania/labs05/tools/fib.py
2017-12-16 06:21:44 +01:00

39 lines
975 B
Python

#!/usr/bin/env python
# -*- coding: utf-8 -*-
"""
Obliczenie n-tego wyrazu ciągu fibonacciego na dwa sposoby.
1. Naiwna rekurencja: podstawienie do wzoru.
2. Wersja z cachem: każdy wyraz jest obliczany dokładnie raz.
"""
def naive_fibonacci(n):
if n <= 0:
return 0
if n in [1,2]:
return 1
return naive_fibonacci(n-1) + naive_fibonacci(n-2)
def cache_fibonacci(n, cache=None):
if cache is None:
cache = [None for i in range(n+1)]
cache[0] = 0
cache[1] = cache[2] = 1
return cache_fibonacci(n, cache)
else:
if cache[n] is not None:
return cache[n]
else:
cache[n] = cache_fibonacci(n-1, cache) + cache_fibonacci(n-2, cache)
return cache[n]
def non_reccurent_fibonacci(n):
cache = [None for i in range(n+1)]
cache[0] = 0
cache[1] = cache[2] = 1
for i in range(2, n + 1):
cache[i] = cache[i-1] + cache[i-2]
return cache[n]