Lecture 18

Comparison sorts

MCS 275 Spring 2022
David Dumas

Lecture 18: Comparison sorts

Course bulletins:

• Homework 6 due at Noon tomorrow.
• Project 2 due 6pm Friday. Autograder open.
• I cannot promise any availability between 1pm and 6pm on Friday. Plan accordingly.
• Worksheet 7 coming soon.
• Starting new topic (trees) on Wednesday.

Growth rates

Let's look at the functions $n$, $n \log(n)$, and $n^2$ as $n$ grows.

Mergesort recursion tree

Efficiency

Theorem: If you measure the time cost of mergesort in any of these terms

• Number of comparisons made
• Number of assignments (e.g. L[i] = x counts as 1)
• Number of Python statements executed

then the cost to sort a list of length $n$ is less than $C n \log(n)$, for some constant $C$ that only depends on which expense measure you chose.

Looking back on quicksort

It ought to be called partitionsort because the algorithm is simply:

• Partition the list
• Quicksort the part before the pivot
• Quicksort the part after the pivot

Other partition strategories

We used the last element of the list as a pivot. Other popular choices:

• The first element, L[start]
• A random element of L[start:end]
• The element L[(start+end)//2]
• An element near the median of L[start:end] (more complicated to find!)

How to choose?

Knowing something about your starting data may guide choice of partition strategy (or even the choice to use something other than quicksort).

Almost-sorted data is a common special case where first or last pivots are bad.

Efficiency

Theorem: If you measure the time cost of quicksort in any of these terms

• Number of comparisons made
• Number of swaps or assignments
• Number of Python statements executed

then the cost to sort a list of length $n$ is less than $C n^2$, for some constant $C$.

But if you average over all possible orders of the input data, the result is less than $C n \log(n)$.

Quicksort recursion tree

Bad case

What if we ask our version of quicksort to sort a list that is already sorted?

Recursion depth is $n$ (whereas if the pivot is always the median it would be $\approx \log_2 n$).

Number of comparisons $\approx C n^2$. Very slow!

Stability

A sort is called stable if items that compare as equal stay in the same relative order after sorting.

This could be important if the items are more complex objects we want to sort by one attribute (e.g. sort alphabetized employee records by hiring year).

As we implemented them:

• Mergesort is stable
• Quicksort is not stable

Other comparison sorts

• Insertion sort — Convert the beginning of the list to a sorted list, starting with one element and growing by one element at a time.
• Bubble sort — Process the list from left to right. Any time two adjacent elements are in the wrong order, switch them. Repeat $n$ times.

Closing thoughts on sorting

Mergesort is rarely a bad choice. It is stable and sorts in $C n \log(n)$ time. Nearly sorted input is not a pathological case. Its main weakness is its use of memory proportional to the input size.

Heapsort, which we may discuss later, has $C n \log(n)$ running time and uses constant space, but it is not stable.

There are stable comparison sorts with $C n \log(n)$ running time and constant space (best in every category!) though they tend to be more complex.

If swaps and comparisons have very different cost, it may be important to select an algorithm that minimizes one of them. Python's list.sort assumes that comparisons are expensive, and uses Timsort.

Quadratic danger

Algorithms that take time proportional to $n^2$ are a big source of real-world trouble. They are often fast enough in small-scale tests to not be noticed as a problem, yet are slow enough for large inputs to disable the fastest computers.