DSAA之排序下界分析（一）

1. 回顾

• In this section, we prove that any algorithm for sorting that uses only comparisons requires $\Omega(n log n)$ comparisons (and hence time) in the worst case, so that mergesort and heapsort are optimal to within a constant factor.
• The proof can be extended to show that $\Omega(n log n)$ comparisons are required, even on average, for any sorting algorithm that uses only comparisons, which means that quicksort is optimal on average to within a constant factor.
  　　

　　这里就应该回顾以前的知识了，上文强调了仅仅使用swap手段的排序算法，在最坏和平均情况下都需要$\Omega(n log n)$次比较，这是个下界，也就是说至少需要$nlogn$次比较。
　　再次回顾，根据以前的记录篇，总结一下：

排序方法	最差情况	平均情况	最好情况
Quicksort	$O(n^2)$	$O(nlogn)$	$O(nlogn)$
Mergesort	$O(nlogn)$	$O(nlogn)$	$O(nlogn)$
Heapsort	$O(nlogn)$	$O(nlogn)$	$O(nlogn)$
Insertsort	$O(n^2)$	$O(n^2)$	$O(n)$
Shellsort	$O(n^2)$	$o(n^2)$
Radixsort	$O(n)$	$O(n)$	$O(n)$空间换时间

　　以上表格，总结的了目前为止，笔者记录的几种排序的时间复杂度情况。希尔排序有些特殊，一些总结和笔者上面的时间复杂度有出入，可能是选用的增量不同的原因。

2. Decision Trees

Any sorting algorithm that uses only comparisons requires $[log n!]$comparisons in the worst case and $log n!$ comparisons on average.

　　以下DSAA的篇幅都在证明以上结论：

Each node represents a set of possible orderings, consistent with comparisons that have been made, among the elements. The results of the comparisons are the tree edges.如下图所示，决策树的概念比较简单。

　　使用决策树的抽象概念是为了得到以下的结论：

• Every algorithm that sorts by using only comparisons can be represented by a decision tree.
• The number of comparisons used by the sorting algorithm is equal to the depth of the deepest leaf.最深树叶的深度其实就是整个树的高度
• The average number of comparisons used is equal to the average depth of the leaves.平均比较次数，就是树的所有树叶的平均深度。
• Let T be a binary tree of depth d. Then T has at most $2^d$ leaves.DSAA树的深度或者高度的是以路径长计数的，区别其他定义。
• A binary tree with L leaves must have depth at least $logL$ .上一点的反推，如果有L个树叶，那么至少应该有logL的深度。可以简单的假设下，2个叶节点至少树高为1。
• Any binary tree with L leaves has an average depth of at least $logL$.
• Any sorting algorithm that uses only comparisons between elements requires at least $logn!$ comparisons in the worst case.
  
  • A decision tree to sort n elements must have $n!$ leaves.
    所以n!的树叶，至少树高为logn!,那么最深树叶的高度至少为logn!,所以最少的比较次数为logn!
• Any sorting algorithm that uses only comparisons between elements requires $\Omega(n log n)$ comparisons.根据上面的几点，无论是平均还是最坏情况，比较次数的下界都是logn