Problem 7.3

Alternative quicksort analysis

An alternative analysis of the running time of randomized quicksort focuses on the expected running time of each individual recursive call to RANDOMIZED-QUICKSORT, rather than on the number of comparisons performed.

1. Argue that, given an array of size $n$, the probability that any particular element is chosen as the pivot is $1/n$. Use this to define indicator random variables $X_i = I\{i\text{th smallest element is chosen as the pivot}\}$. What is $\E[X_i]$?
2. Let $T(n)$ be a random variable denoting the running time of quicksort on an array of size $n$. Argue that $$\E[T(n)] = \E\bigg[\sum_{q=1}^nX_q(T(q-1) + T(n-q) + \Theta(n))\bigg] \tag{7.5}$$
3. Show that we can rewrite equation (7.5) as $$\E[T(n)] = \frac{2}{n}\sum_{q=2}^{n-1}\E[T(q)] + \Theta(n) \tag{7.6}$$
4. Show that $$\sum_{k=2}^{n-1}k\lg{k} \le \frac{1}{2}n^2\lg{n} - \frac{1}{8}n^2 \tag{7.7}$$ (Hint: Split the summation into two parts, one for $k = 2, 3, \ldots, \lceil n/2 \rceil - 1$ and one for $k = \lceil n/2 \rceil, \ldots, n - 1$.
5. Using the bound from equation (7.7), show that the recurrence in equation (7.6) has the solution $\E[T(n)] = \Theta(n\lg{n})$. (Hint: Show, by substitution, that $\E[T(n)] \le an\lg{n}$ for sufficiently large $n$ and for some positive constant $a$.)

Choosing a pivot

PARTITION is equally likely to pick any element as a pivot. Since there are $n$ elements, the probability of one being picked is $1/n$. For the same reason, $\E[X_i] = 1/n$.

Running time of quicksort

Let the $q$th smallest element be the pivot. There are $n$ possible choices for it, each with chance $X_q$. Each will solve the problem by breaking it down in two parts of size $q - 1$ and $n - q$ and adding a linear factor. The formula in (7.5) follows by the definition of expectation.

The rewrite

\begin{aligned} \E[T(n)] &= \E\bigg[\sum_{q=1}^nX_q(T(q-1) + T(n-q) + \Theta(n))\bigg] \\ &= \sum_{q=1}^n\frac{1}{n}(\E[T(q-1)] + \E[T(n-q)] + \Theta(n))\bigg] \\ &= \frac{1}{n}\sum_{q=1}^n\E[T(q-1)] + \frac{1}{n}\sum_{q=1}^n\E[T(n - q)] + \frac{1}{n}\sum_{q=1}^n\Theta(n) \\ &= \frac{1}{n}\sum_{q=0}^{n-1}\E[T(q)] + \frac{1}{n}\sum_{q=0}^{n-1}\E[T(n - q + 1)] + \Theta(n) \\ &= \frac{1}{n}\sum_{q=0}^{n-1}\E[T(q)] + \frac{1}{n}\sum_{q=0}^{n-1}\E[T(q)] + \Theta(n) \\ &= \frac{2}{n}\sum_{q=0}^{n-1}\E[T(q)] + \Theta(n) \\ &= \frac{2}{n}\sum_{q=2}^{n-1}\E[T(q)] + \frac{2\E[T(0)]}{n} + \frac{2\E[T(1)]}{n} + \Theta(n) \\ &= \frac{2}{n}\sum_{q=2}^{n-1}\E[T(q)] + \Theta(n) \end{aligned}

The bound

\begin{aligned} \sum_{k=2}^{n-1}k\lg{k} &= \sum_{k=2}^{\lceil n/2 \rceil - 1}k\lg{k} + \sum_{k=\lceil n/2 \rceil}^{n - 1}k\lg{k} \\ &\le \sum_{k=2}^{n/2}k\lg{k} + \sum_{k=n/2 + 1}^{n}k\lg{k} \\ &\le \sum_{k=2}^{n/2}k\lg(n/2) + \sum_{k=n/2 + 1}^{n}k\lg{n} \\ &= \lg(n/2)\sum_{k=2}^{n/2}k\ + \lg{n}\sum_{k=n/2 + 1}^{n}k \\ &= (\lg{n} - \lg{2})\bigg(\frac{(n/2)(n/2 + 1)}{2}\bigg) + \lg{n}\bigg(\frac{n(n+1)}{2} - \frac{(n/2)(n/2 + 1)}{2}\bigg) \\ &= \lg{n}\frac{n(n+1)}{2} - \frac{(n/2)(n/2 + 1)}{2} \\ &= \frac{1}{2}\lg{n}(n^2 + 2n + 1) - \frac{1}{8}(n^2 + 2n + 1/8) \\ &= \frac{1}{2}n^2\lg{n} - \frac{1}{8}n^2 - \frac{8n\lg{n} + 4\lg{n} - 2n - 1/8}{8} \\ &\le \frac{1}{2}n^2\lg{n} - \frac{1}{8}n^2 \end{aligned}

The solution

We guess $\E[T(n)] \le an\lg{n}$:

\begin{aligned} \E[T(n)] &= \frac{2}{n}\sum_{q=2}^{n-1}\E[T(q)] + \Theta(n) \\ &\le \frac{2}{n}\sum_{q=2}^{n-1}an\lg{n} + \Theta(n) & \text{(by the guess)} \\ &\le \frac{2a}{n}\bigg(\frac{1}{2}n^2\lg{n} - \frac{1}{8}n^2\bigg) + \Theta(n) & \text{(by 7.7)} \\ &= an\lg{n} - \frac{a}{4}n + \Theta(n) & \text{(by }\Theta\text{-notation)} \\ &\le an\lg{n} \end{aligned}

Note that $\Theta$-notation allows us to pick $a$ and $n$ such that the last derivation is possible.