Randomized Algorithms (Spring 2010)/Introduction

Randomized Quicksort

For an input set S with an arbitrary order, the Quicksort algorithm sorts [math]\displaystyle{ S }[/math] as described in following psuedocode:

• if [math]\displaystyle{ |S|\gt 1 }[/math] do:
  • pick an element [math]\displaystyle{ x }[/math] from [math]\displaystyle{ S }[/math] as the pivot;
  • partition [math]\displaystyle{ S }[/math] into [math]\displaystyle{ S_1 }[/math], [math]\displaystyle{ \{x\} }[/math], and [math]\displaystyle{ S_2 }[/math], where all elements in [math]\displaystyle{ S_1 }[/math] are smaller than [math]\displaystyle{ x }[/math] and all elements in [math]\displaystyle{ S_2 }[/math] are larger than [math]\displaystyle{ x }[/math];
  • recursively sort [math]\displaystyle{ S_1 }[/math] and [math]\displaystyle{ S_2 }[/math];

Let us measure the time complexity of this sorting algorithm by the number of comparisons.

For the deterministic quicksort algorithm, the pivot element is chosen deterministically (usually the first one in the sequence [math]\displaystyle{ S }[/math]). This will make the worst-case time complexity [math]\displaystyle{ \Omega(n^2) }[/math], which means there exists a bad case [math]\displaystyle{ S }[/math], sorting which will cost us [math]\displaystyle{ \Omega(n^2) }[/math] comparisons, every time!

It is just so unfair to have an unbeatable case for this brilliant algorithm. So we tweak the algorithm a little bit:

Algorithm: RandQSort

• if [math]\displaystyle{ |S|\gt 1 }[/math] do:
  • uniformly pick a random element [math]\displaystyle{ x }[/math] from [math]\displaystyle{ S }[/math] as the pivot;
  • partition [math]\displaystyle{ S }[/math] into [math]\displaystyle{ S_1 }[/math], [math]\displaystyle{ \{x\} }[/math], and [math]\displaystyle{ S_2 }[/math], where all elements in [math]\displaystyle{ S_1 }[/math] are smaller than [math]\displaystyle{ x }[/math] and all elements in [math]\displaystyle{ S_2 }[/math] are larger than [math]\displaystyle{ x }[/math];
  • recursively sort [math]\displaystyle{ S_1 }[/math] and [math]\displaystyle{ S_2 }[/math];

Analysis of RandQSort

Our goal is to analyze the expected number of comparisons during an execution of RandQSort with an arbitrary input [math]\displaystyle{ S }[/math]. We achieve this by measuring the chance that each pair of elements are compared, and summing all of them up due to Linearity of Expectation.

Let [math]\displaystyle{ a_i }[/math] denote the [math]\displaystyle{ i }[/math]th smallest element in [math]\displaystyle{ S }[/math]. Let [math]\displaystyle{ X_{ij}\in\{0,1\} }[/math] be the random variable which indicates whether [math]\displaystyle{ a_i }[/math] and [math]\displaystyle{ a_j }[/math] are compared during the execution of RandQSort. That is:

[math]\displaystyle{ \begin{align} X_{ij} &= \begin{cases} 1 & a_i\mbox{ and }a_j\mbox{ are compared}\\ 0 & \mbox{otherwise} \end{cases}. \end{align} }[/math]

Elements [math]\displaystyle{ a_i }[/math] and [math]\displaystyle{ a_j }[/math] are compared only if one of them is chosen as pivot. After comparison they are separated (thus are never compared again). So we have the following observation:

Claim 1: Every pair of [math]\displaystyle{ a_i }[/math] and [math]\displaystyle{ a_j }[/math] are compared at most once.

Therefore the sum of [math]\displaystyle{ X_{ij} }[/math] for all pair [math]\displaystyle{ \{i,j\} }[/math] gives the total number of comparisons. The expected number of comparisons is [math]\displaystyle{ \mathbb{E}\left[\sum_{i=1}^n\sum_{j\gt i}X_{ij}\right] }[/math]. Due to Linearity of Expectation, [math]\displaystyle{ \mathbb{E}\left[\sum_{i=1}^n\sum_{j\gt i}X_{ij}\right] = \sum_{i=1}^n\sum_{j\gt i}\mathbb{E}\left[X_{ij}\right] }[/math]. Our next step is to analyze [math]\displaystyle{ \mathbb{E}\left[X_{ij}\right] }[/math] for each [math]\displaystyle{ \{i,j\} }[/math].

By the definition of expectation and [math]\displaystyle{ X_{ij} }[/math], it holds that

[math]\displaystyle{ \begin{align} \mathbb{E}\left[X_{ij}\right] &= 1\cdot \Pr[a_i\mbox{ and }a_j\mbox{ are compared}] + 0\cdot \Pr[a_i\mbox{ and }a_j\mbox{ are not compared}]\\ &= \Pr[a_i\mbox{ and }a_j\mbox{ are compared}]. \end{align} }[/math]

We are going to bound this probability.

Claim 2: [math]\displaystyle{ a_i }[/math] and [math]\displaystyle{ a_j }[/math] are compared if and only if one of them is chosen as pivot when they are still in the same subset.

This is easy to verify: just check the algorithm. The next one is a bit complicated.

Claim 3: If [math]\displaystyle{ a_i }[/math] and [math]\displaystyle{ a_j }[/math] are still in the same subset then all [math]\displaystyle{ \{a_i,a_{i+1},\ldots,a_{j-1},a_{j}\} }[/math] are in the same subset.

We can verify this by induction. Initially, [math]\displaystyle{ S }[/math] itself has the property described above; and partitioning any [math]\displaystyle{ S }[/math] with the property into [math]\displaystyle{ S_1 }[/math] and [math]\displaystyle{ S_2 }[/math] will preserve the property for both [math]\displaystyle{ S_1 }[/math] and [math]\displaystyle{ S_2 }[/math]. Therefore Claim 3 holds.

Combining Claim 2 and 3, we have:

Claim 4: [math]\displaystyle{ a_i }[/math] and [math]\displaystyle{ a_j }[/math] are compared only if one of [math]\displaystyle{ \{a_i, a_j\} }[/math] is chosen from [math]\displaystyle{ \{a_i,a_{i+1},\ldots,a_{j-1},a_{j}\} }[/math].

And apparently,

Claim 5: Every one of [math]\displaystyle{ \{a_i,a_{i+1},\ldots,a_{j-1},a_{j}\} }[/math] is chosen with the same probability.

This is because our RandQSort chooses the pivot uniformly at random.

Claim 4 and Claim 5 together give:

[math]\displaystyle{ \begin{align} \Pr[a_i\mbox{ and }a_j\mbox{ are compared}] &\le \frac{2}{j-i+1}. \end{align} }[/math]

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Perhaps you feel confused about the above argument. You may ask: "Many instead of one pivot are chosen during the execution. Why in the above argument, it looks like the pivot is chosen only once?" Good question. Let's see what really is happening here by looking a bit closer.

For any pair [math]\displaystyle{ a_i }[/math] and [math]\displaystyle{ a_j }[/math], initially [math]\displaystyle{ \{a_i,a_{i+1},\ldots,a_{j-1},a_{j}\} }[/math] are all in the same set [math]\displaystyle{ S }[/math] (obviously!). During the execution of the algorithm, the set which containing [math]\displaystyle{ \{a_i,a_{i+1},\ldots,a_{j-1},a_{j}\} }[/math] are shrinking, till one of [math]\displaystyle{ \{a_i,a_{i+1},\ldots,a_{j-1},a_{j}\} }[/math] is chosen. And we ask