In the book

Introduction to Algorithms (3th edition) chapter 7

the recurrence of the running time of quicksorts partitioning is given by $$T(n) = T(n-1) + \Theta(n)$$ as the worst-case happens when the partitioning has one subproblem with $n-1$ elements and one with $0$ elements.

Now, they they guess $T(n)=\Theta(n^2)$ for the substitution method. I don't understand this, why choose $\Theta(n^2)$? I personally would have chosen $\Theta(n)$, which is wrong.


With the help from a fellow friend I came to the following conclusion:

We know that we have to make $$n-1,n-2, \dots, 2,1 \text{ comparisons}$$

So when looking at the arithmetic series $$\sum_{i=1}^n k = \Theta(n^2)$$

We can conclude that $T(n) = \Theta(n^2)$.

Please comment if I can improve my answer in any way.


With the substitution method, you add up $\Theta(n) + \Theta(n-1) = ... + \Theta(1)$, roughly n times $\Theta(n)$, therefore about $n^2$.

Note that this is for an initial guess - to help you make a guess your maths doesn't have to be strict. To prove it, you need to be strict.

If you have a sum that isn't as simple as this one here, you can just give a rought upper and lower bound: There are in the worst case n/2 partitions each taking n/2 or more comparisons (lower bound $n^2/4$) but also n partitions taking at most n comparisons (upper bound $n^2$).


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