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python的递归算法学习（2）：具体实现：斐波那契和其中的陷阱
1.斐波那契
什么是斐波那契，斐波那契额就是一个序列的整数的排序，其定义如下；
Fn = Fn-1 + Fn-2
with F0 = 0 and F1 = 1
也就是，0，1，1，2，3，5，8，13.。。。。
递归实现：
def fib(n): if n == 0: return 0 elif n == 1: return 1 else: return fib(n-1) + fib(n-2)
非递归实现：
def fibi(n): a, b = 0, 1 for i in range(n): a, b = b, a + b return a
在这里，我们如果仔细调试，会发现，递归实现，会消耗更多的时间，这里测试如下：
from timeit import Timer from fibo import fib t1 = Timer("fib(10)","from fibo import fib") for i in range(1,41): s = "fib(" + str(i) + ")" t1 = Timer(s,"from fibo import fib") time1 = t1.timeit(3) s = "fibi(" + str(i) + ")" t2 = Timer(s,"from fibo import fibi") time2 = t2.timeit(3) print("n=%2d, fib: %8.6f, fibi: %7.6f, percent: %10.2f" % (i, time1, time2, time1/time2))
结果如下；
C:\Python35\python.exe C:\pylearn\bottlelearn\fibnaqie.py n= 1, fib: 0.000002, fibi: 0.000003, percent: 0.55 n= 2, fib: 0.000002, fibi: 0.000003, percent: 0.73 n= 3, fib: 0.000003, fibi: 0.000003, percent: 1.20 n= 4, fib: 0.000005, fibi: 0.000003, percent: 1.80 n= 5, fib: 0.000007, fibi: 0.000003, percent: 2.45 n= 6, fib: 0.000012, fibi: 0.000004, percent: 3.31 n= 7, fib: 0.000022, fibi: 0.000003, percent: 6.50 n= 8, fib: 0.000030, fibi: 0.000004, percent: 8.38 n= 9, fib: 0.000049, fibi: 0.000003, percent: 14.58 n=10, fib: 0.000078, fibi: 0.000004, percent: 20.07 n=11, fib: 0.000126, fibi: 0.000004, percent: 35.00 n=12, fib: 0.000203, fibi: 0.000004, percent: 52.29 n=13, fib: 0.000330, fibi: 0.000004, percent: 79.20 n=14, fib: 0.000537, fibi: 0.000004, percent: 120.94 n=15, fib: 0.000925, fibi: 0.000005, percent: 196.18 n=16, fib: 0.001693, fibi: 0.000007, percent: 244.16 n=17, fib: 0.007242, fibi: 0.000011, percent: 652.70 n=18, fib: 0.004422, fibi: 0.000007, percent: 637.64 n=19, fib: 0.010322, fibi: 0.000008, percent: 1240.40 n=20, fib: 0.010813, fibi: 0.000007, percent: 1443.78 n=21, fib: 0.025547, fibi: 0.000011, percent: 2246.24 n=22, fib: 0.035421, fibi: 0.000011, percent: 3192.22 n=23, fib: 0.060877, fibi: 0.000008, percent: 7837.86 n=24, fib: 0.090561, fibi: 0.000007, percent: 12091.41 n=25, fib: 0.132058, fibi: 0.000008, percent: 16415.97 n=26, fib: 0.227813, fibi: 0.000008, percent: 29330.54 n=27, fib: 0.557644, fibi: 0.000014, percent: 38659.17 n=28, fib: 0.578976, fibi: 0.000009, percent: 65224.25 n=29, fib: 1.133326, fibi: 0.000008, percent: 140882.03 n=30, fib: 1.454107, fibi: 0.000009, percent: 169095.94 n=31, fib: 2.274395, fibi: 0.000009, percent: 264486.10 n=32, fib: 3.817956, fibi: 0.000009, percent: 430110.09 n=33, fib: 5.923710, fibi: 0.000009, percent: 688859.58 n=34, fib: 9.629423, fibi: 0.000009, percent: 1020986.44 n=35, fib: 15.910085, fibi: 0.000009, percent: 1686911.18 n=36, fib: 26.556680, fibi: 0.000009, percent: 2901071.88 n=37, fib: 43.014073, fibi: 0.000016, percent: 2673506.31
为什么会这样呢？我们思考一下，递归过程的计算数，具体如下；
从树中可以看到，有重复计算的情况,(f(1)就计算了很多次。
在这里，尝试做一个改进，引入一个临时的字典，把计算过的内容做一个保存：
memo = {0:0, 1:1}
def fibm(n):
if not n in memo:
memo[n] = fibm(n-1) + fibm(n-2)
return memo[n]
重新进行时间的测试：
from timeit import Timer from fibo import fib t1 = Timer("fib(10)","from fibo import fib") for i in range(1,41): s = "fibm(" + str(i) + ")" t1 = Timer(s,"from fibo import fibm") time1 = t1.timeit(3) s = "fibi(" + str(i) + ")" t2 = Timer(s,"from fibo import fibi") time2 = t2.timeit(3) print("n=%2d, fib: %8.6f, fibi: %7.6f, percent: %10.2f" % (i, time1, time2, time1/time2))
n= 1, fib: 0.000011, fibi: 0.000015, percent: 0.73 n= 2, fib: 0.000011, fibi: 0.000013, percent: 0.85 n= 3, fib: 0.000012, fibi: 0.000014, percent: 0.86 n= 4, fib: 0.000012, fibi: 0.000015, percent: 0.79 n= 5, fib: 0.000012, fibi: 0.000016, percent: 0.75 n= 6, fib: 0.000011, fibi: 0.000017, percent: 0.65 n= 7, fib: 0.000012, fibi: 0.000017, percent: 0.72 n= 8, fib: 0.000011, fibi: 0.000018, percent: 0.61 n= 9, fib: 0.000011, fibi: 0.000018, percent: 0.61 n=10, fib: 0.000010, fibi: 0.000020, percent: 0.50 n=11, fib: 0.000011, fibi: 0.000020, percent: 0.55 n=12, fib: 0.000004, fibi: 0.000007, percent: 0.59 n=13, fib: 0.000004, fibi: 0.000007, percent: 0.57 n=14, fib: 0.000004, fibi: 0.000008, percent: 0.52 n=15, fib: 0.000004, fibi: 0.000008, percent: 0.50 n=16, fib: 0.000003, fibi: 0.000008, percent: 0.39 n=17, fib: 0.000004, fibi: 0.000009, percent: 0.45 n=18, fib: 0.000004, fibi: 0.000009, percent: 0.45 n=19, fib: 0.000004, fibi: 0.000009, percent: 0.45 n=20, fib: 0.000003, fibi: 0.000010, percent: 0.29 n=21, fib: 0.000004, fibi: 0.000009, percent: 0.45 n=22, fib: 0.000004, fibi: 0.000010, percent: 0.40 n=23, fib: 0.000004, fibi: 0.000010, percent: 0.40 n=24, fib: 0.000004, fibi: 0.000011, percent: 0.35 n=25, fib: 0.000004, fibi: 0.000012, percent: 0.33 n=26, fib: 0.000004, fibi: 0.000011, percent: 0.34 n=27, fib: 0.000004, fibi: 0.000011, percent: 0.35 n=28, fib: 0.000004, fibi: 0.000012, percent: 0.32 n=29, fib: 0.000004, fibi: 0.000012, percent: 0.33 n=30, fib: 0.000004, fibi: 0.000013, percent: 0.31 n=31, fib: 0.000004, fibi: 0.000012, percent: 0.34 n=32, fib: 0.000004, fibi: 0.000012, percent: 0.33 n=33, fib: 0.000004, fibi: 0.000013, percent: 0.30 n=34, fib: 0.000004, fibi: 0.000012, percent: 0.34 n=35, fib: 0.000004, fibi: 0.000013, percent: 0.31 n=36, fib: 0.000004, fibi: 0.000013, percent: 0.31 n=37, fib: 0.000004, fibi: 0.000014, percent: 0.29 n=38, fib: 0.000004, fibi: 0.000014, percent: 0.29 n=39, fib: 0.000004, fibi: 0.000013, percent: 0.31 n=40, fib: 0.000004, fibi: 0.000014, percent: 0.29
python的递归算法学习（2）：具体实现：斐波那契和其中的陷阱
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