Theta(n^2) running time of Quicksort Algorithm

ATroelstein

New member
I am trying to prove the following scenario of Quicksort has a running time of Theta(n^2). Initially, we only have arrays of size n, where n = ij. The idea is that at every partition step of Quicksort, you end up with two sub-arrays where one is of size i and the other is of size i(j-1). i in this case is an integer constant greater than 0. I have drawn out the recursive tree of some examples and understand why this is a worst-case scenario and that the running time will be Theta(n^2). To prove this, I've used repeated substitution to solve the recurrence equation:

T(n) = T(ij) = m if j = 1, where m just denotes a constant
T(n) = T(ij) = T(i) + T(i(j-1)) + cn if j > 1

T(i) = m
T(2i) = m + m + c*2i = 2m + 2ci
T(3i) = m + 2m + 2ci + 3ci = 3m + 5ci

So it looks like the recurrence is:

$$T(n) = jm + (ci\sum_{k=1}^{j}k) - 1$$

At this point, I'm a bit lost as to what to do. It looks the summation at the end will result in j^2 if expanded out, but I need to show that it somehow equals n^2. Any explanation on how to continue with this would be appreciated.

CaptainBlack

Well-known member
I am trying to prove the following scenario of Quicksort has a running time of Theta(n^2). Initially, we only have arrays of size n, where n = ij. The idea is that at every partition step of Quicksort, you end up with two sub-arrays where one is of size i and the other is of size i(j-1). i in this case is an integer constant greater than 0. I have drawn out the recursive tree of some examples and understand why this is a worst-case scenario and that the running time will be Theta(n^2). To prove this, I've used repeated substitution to solve the recurrence equation:

T(n) = T(ij) = m if j = 1, where m just denotes a constant
T(n) = T(ij) = T(i) + T(i(j-1)) + cn if j > 1

T(i) = m
T(2i) = m + m + c*2i = 2m + 2ci
T(3i) = m + 2m + 2ci + 3ci = 3m + 5ci

So it looks like the recurrence is:

$$T(n) = jm + (ci\sum_{k=1}^{j}k) - 1$$

At this point, I'm a bit lost as to what to do. It looks the summation at the end will result in j^2 if expanded out, but I need to show that it somehow equals n^2. Any explanation on how to continue with this would be appreciated.
Why are you factorising $$n$$?

In the worst case the pivot splits the array of size $$n$$ into arrays of size $$0$$, and size $$n-1$$. This spliting requires $$n-1$$ compartisons, so:

$T_{wc}(n)= k.(n-1)+T_{wc}(n-1)=k\sum_{k=1}^{n-1} n=k \frac{n.(n-1)}{2}=K(n^2-n)$

which implies that $$T_{wx}\in \Theta(n^2)$$

CB

Last edited:

ATroelstein

New member
Why are you factorising $$n$$?

In the worst case the pivot splits the array of size $$n$$ into arrays of size $$0$$, and size $$n-1$$. This spliting requires $$n-1$$ compartisons, so:

$T_{wc}(n)= k.(n-1)+T_{wc}(n-1)=k\sum_{k=1}^{n-1} n=k \frac{n.(n-1)}{2}=K(n^2-n)$

which implies that $$T_{wx}\in \Theta(n^2)$$

CB

Thank you for your reply CaptainBlack. I am factorizing n because the specific problem I am working on is trying to show that while the case you described is the absolute worst-case scenario, where you have a sub-arrays of size 0 and n-1, that the scenario I gave also has a running time of theta(n^2). I am able to find many examples of the scenario you gave in textbooks and online, but am unfortunately am unable to find any examples of the case that I have given above and that's why I've become so stuck at this point. Thanks.

CaptainBlack

Well-known member
Thank you for your reply CaptainBlack. I am factorizing n because the specific problem I am working on is trying to show that while the case you described is the absolute worst-case scenario, where you have a sub-arrays of size 0 and n-1, that the scenario I gave also has a running time of theta(n^2). I am able to find many examples of the scenario you gave in textbooks and online, but am unfortunately am unable to find any examples of the case that I have given above and that's why I've become so stuck at this point. Thanks.
Can you tell us exactly what the case you are considering is?

In particular how are the i's and j's behaving asymtotically as n becomes large?

CB

ATroelstein

New member
Since i is an integer constant, its asymptotic behavior won't depend on n. The asymptotic behavior of j however will be increasing as n increases, since n = ij. To make my scenario a little more clear, we could look at the case where i = 1. This falls into my scenario and is just one step away from the worst case you described, as now we have sub-arrays of size 1 and n-2. The next case is my scenario would be where i = 2 and we have sub-arrays of 2 and n-3, and so on. So now I'm trying to show that no matter what the integer constant i is, the running time will still be theta(n^2) as j is becoming large as n becomes large, but i does not. Thank you.

CaptainBlack

Well-known member
Since i is an integer constant, its asymptotic behavior won't depend on n. The asymptotic behavior of j however will be increasing as n increases, since n = ij. To make my scenario a little more clear, we could look at the case where i = 1. This falls into my scenario and is just one step away from the worst case you described, as now we have sub-arrays of size 1 and n-2. The next case is my scenario would be where i = 2 and we have sub-arrays of 2 and n-3, and so on. So now I'm trying to show that no matter what the integer constant i is, the running time will still be theta(n^2) as j is becoming large as n becomes large, but i does not. Thank you.
If you can show that:

$T(ij)=\Theta(j^2)$

you are done, since in this context $$i$$ is a constant and so $$\Theta(j^2)=\Theta(i^2j^2)=\Theta(n^2)$$

CB