算法 - 如何使用2n / 3比较对0/1数组进行排序？ [英] algorithm - How to sort a 0/1 array with 2n/3 comparisons?

问题描述

在算法设计手册中，有这样一个消费者

4-26考虑使用
比较排序n 0和1的序列的问题。对于两个值x和y的每个比较，算法
学习x < y，x = y或x> y成立。

（a）给出一种算法，在最坏的情况下进行n-1个比较。
显示您的算法是最优的。

（b）给出一个算法在平均值为
的2n / 3比较中排序（假设每个的n个输入是0或1等概率）。显示
，您的算法是最佳的。

对于（a），我认为这是相当容易的。我可以选择一个[n-1]作为枢轴，然后在快速分区中进行类似的操作，扫描0到n - 2，找到左侧全部为0的中间点，右侧全部为1，这取n - 1个比较

但是（b），我不能得到一个线索。它说n个输入中的每一个都是0或1，具有相同的概率，所以我想我可以假设0和1的数相等吗？但是如何获得与1/3相关的结果？将整个数组分成3组？

谢谢

解决方案

0或1等概率是平均情况的条件。提示1：2/3 = 1/2 + 1/8 + 1/32 + 1/128 + ...

p>

提示2：将序列作为对的序列，并比较每对中的项目。一半会平等回报;一半不会。在不平等的一半中，你知道该对中的哪个项目是0，哪个是1，所以这些不需要比较。

In Algorithm Design Manual, there is such an excise

4-26 Consider the problem of sorting a sequence of n 0’s and 1’s using comparisons. For each comparison of two values x and y, the algorithm learns which of x < y, x = y, or x > y holds.

(a) Give an algorithm to sort in n − 1 comparisons in the worst case. Show that your algorithm is optimal.

(b) Give an algorithm to sort in 2n/3 comparisons in the average case (assuming each of the n inputs is 0 or 1 with equal probability). Show that your algorithm is optimal.

For (a), I think it is fairly easy. I can choose a[n-1] as pivot, then do something like in quicksort partition, scan 0 to n - 2, find the middle point where left side is all 0 and right side is all 1, this take n - 1 comparisons.

But for (b), I can't get a clue. It says "each of the n inputs is 0 or 1 with equal probability", so I guess I can assume the numbers of 0 and 1 equal? But how can I get a result which is related to 1/3? divide the whole array into 3 groups?

Thanks

解决方案

"0 or 1 with equal probability" is the condition for "average" case. Other cases may have worse timing.

Hint 1: 2/3 = 1/2 + 1/8 + 1/32 + 1/128 + ...

Hint 2: Consider the sequence as a sequence of pairs and compare the items in each pair. Half will return equal; half will not. Of the half that are unequal you know which item in the pair is 0 and which is 1, so those need no more comparisons.

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