如何测试洗牌程序 | 酷 壳 - CoolShell.cn


如何测试洗牌程序 | 酷 壳 - CoolShell.cn
下面,我们来看看性能高且正确的算法—— Fisher_Yates算法
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void ShuffleArray_Fisher_Yates(char* arr, int len)
{
    int i = len, j;
    char temp;
 
    if ( i == 0 ) return;
    while ( --i ) {
        j = rand() % (i+1);
        temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
    }
}
在做测试之前,我们还需要了解一下一个基本知识——PC机上是做不出真随机数的,只能做出伪随机数。真随机数需要硬件支持。但是不是这样我们就无法测试了呢,不是的。我们依然可以测试。
我们知道,洗牌洗得好不好,主要是看是不是够随机。那么如何测试随机性呢?
试想,我们有个随机函数rand()返回1到10中的一个数,如果够随机的话,每个数返回的概率都应该是一样的,也就是说每个数都应该有10分之1的概率会被返回。
一到概率问题,我们只有一个方法来做测试,那就是用统计的方式。也就是说,你调用rand()函数100次,其中,每个数出现的次数大约都在10次左右。(注意:我用了左右,这说明概率并不是很准确的)不应该有一个数出现了15次以上,另一个在5次以下,要是这样的话,这个函数就是错的。
举一反三,测试洗牌程序也一样,需要通过概率的方式来做统计,是不是每张牌出现在第一个位置的次数都是差不多的。
于是,这样一来上面的程序就可以很容易做测试了。
下面是测试结果(测试样本1000次——列是每个位置出现的次数,行是各个字符的统计,出现概率应该是1/10,也就是100次):
其实,以上的测试只是测试了牌在各个位置的概率。这个还不足够好。因为还可能会现在有Patten的情况。如:每次洗牌出来的都是一个循环顺序数组。这完全可以满足我上面的测试条件。但是那明显是错的。所以,还需要统计每种排列的出现的次数,看看是不是均匀。但是,如果这些排列又是以某种规律出现的呢?看来,这没完没了了。

之前忘贴了一个模拟我们玩牌洗牌的算法,现补充如下:
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void ShuffleArray_Manual(char* arr, int len)
{
    int mid = len / 2;
 
    for (int n=0; n<5; n++){
 
        //两手洗牌
        for (int i=1; i<mid; i+=2){
            char tmp = arr[i];
            arr[i] = arr[mid+i];
            arr[mid+i] = tmp;
        }
 
        //随机切牌
        char *buf = (char*)malloc(sizeof(char)*len);
 
        for(int j=0; j<5; j++) {
            int start= rand() % (len-1) + 1;
            int numCards= rand()% (len/2) + 1;
 
            if (start + numCards > len ){
                numCards = len - start;
            }
 
            memset(buf, 0, len);
            strncpy(buf, arr, start);
            strncpy(arr, arr+start, numCards);
            strncpy(arr+numCards, buf, start);
        }
        free(buf);
 
    }
}
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