Dmitry Peter Borzov /posts/quicksort/

Analysis of randomized algorithms (Part I)

Estimating the quicksort’s time performance

Abstract: It is not that hard to build skills for analysing randomized algorithms. Here I list some of the basic tricks that work for me using quicksort as an example

Randomized algorithms are fascinating to me. Many classic algorithms are somewhat obvious in the hindsight: you can see right away their value and what makes them good. That is not the case for Randomized algorithms. They can surprise you with how efficient they can be on average. I remember being amazed when I learned that the randomized algorithm for finding a median element in an unsorted graph has asymptotic performance being linearly proportional to the size of the array, for example. but being able to actually understand what gives them their properties can be pretty hard at times.

Specifically, deriving the execution time estimation can get pretty challenging in general case, in a sharp contrast to the deterministic algorithms. For the average execution time, for example, we have to consider each of the possible random outcomes along with its probability and then average out over all of them. Not to mention that just the average time may not enough to characterize the alorithm and we may have to investigate metrics beyond just the average. The mathematical expressions behind those easily gets so hairy that they tend to be omitted at all in many educational materials and textbooks.

However, that complexity is also a bit deceptive in practice. The key is that we are usually only interested in asymptotics, that is the behaviour in some limit, such as the limit of large dataset. That simplifies the problem immensely and means that, with a bit of experience behind our belt, we can learn to arrive at the answer pretty easily. It all comes down to carefully studying what terms are important in our case of interest and what are not.

Let’s illustrate this with an example. We will look at the asymptotics of what is probably the most famous random-based algorithm: the quicksort.

Heads on

We will be interested in the asymptotic limit of large arrays: that is, when the number of elements n in the array we need to sort gets pretty big. Let’s denote this execution time for quicksort with f(n).

The challenge here is that we don’t know how lucky we are going to get with our random choice of the pivot element: it could be the median which means we divided our problem into two smaller ones perfectly, or it could be the smallest element and we may have shifted the whole array to only “sort” one element.

Let’s assume all the elements are different for simplicity and the pivot happened to be the m the element of the array. We know the execution time for the given value m: at each iteration, we would have to go through the entire array, then sort the both subsets with times f(m) and f(n-m) for each:

` f(m) + f(n-m) + c_{1} + c_{2} n`

The pivot element choice being random means that the execution time is the average over all possible values m, from 1 up to n:

`f(n) = \frac{1}{n} \sum_{m=1...n} [ f(m) + f(n-m) + c_{1} + c_{2} n ] `

This expression does not look very friendly.

Let’s start with building some intuition for what the solution might look like.

What does the solution may look like?

Let’s see what kind of solution we may expect here.

Why do we need to go through the derivation at all? Unless you are working in alorithms research, it is unlikely that we would ever . Numerical simulation is good enough. My reason is that it helps build a proper understanding of what is going on: see which decisions in our design matter and which are not.

That is why the best thing is to look into the custom cases what kind of answer we may reasonably expect.

Let’s look at the worst case O(n): this case unlikiest ever: every time we manage to make the worst pivot element ever: the smaller subset only has the pivot element, and the other one contains all the other ones, n-1. We then we have to make the recursive call on that set.

`f_{\text{worst}}(n) = f_{\text{worst}}(n-1) + n`

We can just unfold the expression recursively:

`f_{\text{worst}}(n) = f_{\text{worst}}(n-1) + n = f_{\text{worst}}(n-2) + (n-1) + n = f_{\text{worst}}(1) + \sum_{x=1..n} x`

Since we are only interested in asymptotics we don’t even need to be careful with the arithmetic sum expression 1 + 2 + 3 + .. + n. We can just apply the fact that with the paring up of the elements, the average is about x/2 and we get x of them, so

`f_{\text{worst}}(n) = n^2`

In the worst case, quicksort looks like the naive sort implementation: we keep selecting the smallest (or the biggest) element of the array, insert it into the very beginning (or end).

Now let’s look at the best case: when every pivot we choose turns out to be the median element of the array! The two subsets are then exactly half in size:

`f_{\text{best}}(n) = 2 \cdot f_{\text{best}}(\frac{n}{2}) + n`

Continuing iteration, we get

`f_{\text{best}}(n) = 2 \cdot f_{\text{best}}(\frac{n}{2}) + n = 4 \cdot f_{\text{best}}(\frac{n}{4}) + 3*n`

This recursive expression looks similar to the one we for the worst case, except. Let’s make it the iterative with rewriting this relation in terms of a new variable x ~ log n:

`n=2^{x} \text{ or } x= \{log}_{2} n` Now we can express the same relation in terms of `g(x)=f(n(x))`:

`g(x) = 2 g(x-1) + 2^{x}`

Solving this relation gets us

`g(x-1) = 2 g(x-2) + \frac{1}{2} 2^{x}`

`g(x) = 2 g(x-1) + 2^{x} = 4 g(x-2) + 2 \cdot \frac{1}{2} 2^{x} + 2^{x} = .. = g(1) + x \cdot 2^{x}`

so the solution is f(n) ~ n log (n).

We learned the upper boundaries and the lower boundaries for what we might expect as the solution.

Building up some intuition

The second thing that I do is I work through some custom cases. We looked at the absolutely worst case and the perfect choice, but in practice it is likely to be somewhere in between on every step. Let’s go through the case of where somehow we happen to split the array into 1/10n and 9/10n on every step.

Then the performance time would be of order:

`f_{\text{sample}}(n) = f_{\text{best}}(\frac{9 n}{10}) + f_{\text{sample}}(\frac{n}{10}) + n`

The expression may not look very trivial on the first glance. This recursive expression has multiplication instead of addition in the arguments of its function and that complicates things. However it is easy to notice that it is pretty similar to the one we already tackled, the one for the best case scenario. We were able to reduce it to something obvious by rewriting it in terms of a different variable. Let’s can apply the same trick here:

`n = 10^{x}, x = \log_{10} n \text{ and } g(x) = f(x(n))` `g(x) = g(x - \log_{10}\frac{10}{9}) + g(x-1) + 10^{x}`

lg (10/9) = lg(1 + 1/9) = 0.04.


[1] “Algorithms”, Dasgupta, C. H. Papadimitriou, and U. V. Vazirani, ยง2.4 “Medians”, pp.64-66