python 2.7.6和3.3.3导致不同的递归

Question

我写的递归一个简单的分类代码,我测试它Python 2.7.6和Python 3.3.3.但我得到了两个不同的结果.代码如下.

import math, copy, random, sys
call_count = 0;    # Keep track of number of times of stooge_sort() get called
swap_count = 0;    # Keep track of number of times swapping.
def stooge_sort(a, origin):
    global call_count, swap_count;
    call_count += 1;
    n = len(a);
    m = int(math.ceil(n*2/3));
    if(n == 2 and a[0] > a[1]):
        a[0], a[1] = a[1], a[0];
        swap_count += 1;
    elif (n > 2):
        first = copy.deepcopy(a[0:m]);
        a[0:m] = stooge_sort(first, origin);
        second = copy.deepcopy(a[n-m:]);
        a[n-m:] = stooge_sort(second, origin);
        first = copy.deepcopy(a[0:m]);
        a[0:m] = stooge_sort(first, origin);
    return a;

a = [int(random.random() * 100) for i in range(10)];
stooge_sort(a, a);
print("sort function call count = " + str(call_count))
print("swap count = " + str(swap_count));

1)如果我跑进来Python 2.7.6,我会得到错误的排序和

sort function call count = 40
swap count = 2

2)如果我跑进来Python 3.3.3,我得到正确的排序和

sort function call count = 364
swap count = 18

所以我想知道哪个部分出了问题Python 2.7.6？

Answer 1

这完全是因为这条线

m = int(math.ceil(n*2/3));

在Python 2中,n*2/3给出一个比实际值小1的值,因为浮点值被截断(因为/在Python 2.x中进行了分区),但是在Python 3中,它将进行适当的浮点除法.

要使程序表现一致,只需确保使用浮点数

m = int(math.ceil(n*2.0/3.0))