# 8.16. Lower Bounds for Sorting¶

## 8.16.1. Lower Bounds for Sorting¶

By now you have seen many analyses for algorithms.
These analyses generally define the upper and lower bounds for
algorithms in their worst and average cases.
For many of the algorithms presented so far, analysis has been easy.
This module considers a more difficult task: An analysis for
the cost of a *problem* as opposed to an *algorithm*.
The upper bound for a problem can be defined as the asymptotic cost of
the fastest known algorithm.
The lower bound defines the best possible cost for *any*
algorithm that solves the problem, including algorithms not yet
invented.
Once the upper and lower bounds for the problem meet, we know that no
future algorithm can possibly be (asymptotically) more efficient.

A simple estimate for a problem’s lower bound can be obtained by
measuring the size of the input that must be read and the output
that must be written.
Certainly no algorithm can be more efficient than the problem’s
I/O time.
From this we see that the sorting problem cannot be solved by
*any* algorithm in less than \(\Omega(n)\) time because it
takes at least \(n\) steps to read and write the \(n\) values
to be sorted.
Alternatively, any sorting algorithm must at least look at every input
value to recognize whether the input values are in sorted order.
So, based on our current knowledge of sorting algorithms and the
size of the input, we know that the *problem* of sorting is
bounded by \(\Omega(n)\) and \(O(n \log n)\).

Computer scientists have spent much time devising efficient general-purpose sorting algorithms, but no one has ever found one that is faster than \(O(n \log n)\) in the worst or average cases. Should we keep searching for a faster sorting algorithm? Or can we prove that there is no faster sorting algorithm by finding a tighter lower bound?

This section presents one of the most important and most useful proofs in computer science: No sorting algorithm based on key comparisons can possibly be faster than \(\Omega(n \log n)\) in the worst case. This proof is important for three reasons. First, knowing that widely used sorting algorithms are asymptotically optimal is reassuring. In particular, it means that you need not bang your head against the wall searching for an \(O(n)\) sorting algorithm. (Or at least not one that is in any way based on key comparisons. But it is hard to imagine how to sort without any comparisons. Even Radix Sort is does comparisons, though in quite a different way.) Second, this proof is one of the few non-trivial lower-bounds proofs that we have for any problem; that is, this proof provides one of the relatively few instances where our lower bound is tighter than simply measuring the size of the input and output. As such, it provides a useful model for proving lower bounds on other problems. Finally, knowing a lower bound for sorting gives us a lower bound in turn for other problems whose solution could be made to work as the basis for a sorting algorithm. The process of deriving asymptotic bounds for one problem from the asymptotic bounds of another is called a reduction.

Except for the Radix Sort and Binsort, all of the sorting algorithms we have studied make decisions based on the direct comparison of two key values. For example, Insertion Sort sequentially compares the value to be inserted into the sorted list until a comparison against the next value in the list fails. In contrast, Radix Sort has no direct comparison of key values. All decisions are based on the value of specific digits in the key value, so it is possible to take approaches to sorting that do not involve direct key comparisons. Of course, Radix Sort in the end does not provide a more efficient sorting algorithm than comparison-based sorting. Thus, empirical evidence suggests that comparison-based sorting is a good approach.

(Actually, the truth is stronger than this statement implies. In reality, Radix Sort relies on comparisons as well and so can be modeled by the technique used in this section. The result is an \(\Omega(n \log n)\) bound in the general case even for algorithms that look like Radix Sort.)

The proof that any comparison sort requires \(\Omega(n \log n)\) comparisons in the worst case is structured as follows. First, comparison-based decisions can be modeled as the branches in a tree. This means that any sorting algorithm based on comparisons between records can be viewed as a binary tree whose nodes correspond to the comparisons, and whose branches correspond to the possible outcomes. Next, the minimum number of leaves in the resulting tree is shown to be the factorial of \(n\). Finally, the minimum depth of a tree with \(n!\) leaves is shown to be in \(\Omega(n \log n)\).

Before presenting the proof of an \(\Omega(n \log n)\) lower bound for sorting, we first must define the concept of a decision tree. A decision tree is a binary tree that can model the processing for any algorithm that makes binary decisions. Each (binary) decision is represented by a branch in the tree. For the purpose of modeling sorting algorithms, we count all comparisons of key values as decisions. If two keys are compared and the first is less than the second, then this is modeled as a left branch in the decision tree. In the case where the first value is greater than the second, the algorithm takes the right branch.

Here is a Visualization that illustrates decision trees and the sorting lower bound proof.

Any sorting algorithm requiring \(\Omega(n \log n)\) comparisons in the worst case requires \(\Omega(n \log n)\) running time in the worst case. Because any sorting algorithm requires \(\Omega(n \log n)\) running time, the problem of sorting also requires \(\Omega(n \log n)\) time. We already know of sorting algorithms with \(O(n \log n)\) running time, so we can conclude that the problem of sorting requires \(\Theta(n \log n)\) time. As a corollary, we know that no comparison-based sorting algorithm can improve on existing \(\Theta(n \log n)\) time sorting algorithms by more than a constant factor.

Here are some review questions to check that you understand this proof.