随机算法 (Spring 2014)/Random Recurrence

Random Quicksort

Given as input a set [math]\displaystyle{ S }[/math] of [math]\displaystyle{ n }[/math] numbers, we want to sort the numbers in [math]\displaystyle{ S }[/math] in increasing order. One of the most famous algorithm for this problem is the Quicksort algorithm.

• if [math]\displaystyle{ |S|\gt 1 }[/math] do:
  • pick an [math]\displaystyle{ x\in 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 numbers in [math]\displaystyle{ S_1 }[/math] are smaller than [math]\displaystyle{ x }[/math] and all numbers 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];

The time complexity of this sorting algorithm is measured by the number of comparisons.

For the deterministic quicksort algorithm, the pivot is picked from a fixed position (e.g. the first number in the array). The worst-case time complexity in terms of number of comparisons is [math]\displaystyle{ \Theta(n^2) }[/math].

We consider the following randomized version of the quicksort.

• if [math]\displaystyle{ |S|\gt 1 }[/math] do:
  • uniformly pick a random [math]\displaystyle{ x\in 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 numbers in [math]\displaystyle{ S_1 }[/math] are smaller than [math]\displaystyle{ x }[/math] and all numbers 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 Random Quicksort

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 observations:

Observation 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{ \mathbf{E}\left[\sum_{i=1}^n\sum_{j\gt i}X_{ij}\right] }[/math]. Due to Linearity of Expectation, [math]\displaystyle{ \mathbf{E}\left[\sum_{i=1}^n\sum_{j\gt i}X_{ij}\right] = \sum_{i=1}^n\sum_{j\gt i}\mathbf{E}\left[X_{ij}\right] }[/math]. Our next step is to analyze [math]\displaystyle{ \mathbf{E}\left[X_{ij}\right] }[/math] for each [math]\displaystyle{ \{i, j\} }[/math].

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

[math]\displaystyle{ \begin{align} \mathbf{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.

Observation 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.

Observation 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 Observation 2 and 3, we have:

Observation 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,

Observation 5: Every one of [math]\displaystyle{ \{a_i, a_{i+1}, \ldots, a_{j-1}, a_{j}\} }[/math] is chosen equal-probably.

This is because the Random Quicksort chooses the pivot uniformly at random.

Observation 4 and 5 together imply:

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

Remark: Perhaps you feel confused about the above argument. You may ask: "The algorithm chooses pivots for many times during the execution. Why in the above argument, it looks like the pivot is chosen only once?" Good question! Let's see what really happens by looking closely. 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 (due to the pivoting), until one of [math]\displaystyle{ \{a_i, a_{i+1}, \ldots, a_{j-1}, a_{j}\} }[/math] is chosen, and the set is partitioned into different subsets. We ask for the probability that the chosen one is among [math]\displaystyle{ \{a_i, a_j\} }[/math]. So we really care about "the last" pivoting before [math]\displaystyle{ \{a_i, a_{i+1}, \ldots, a_{j-1}, a_{j}\} }[/math] is split. Formally, let [math]\displaystyle{ Y }[/math] be the random variable denoting the pivot element. We know that for each [math]\displaystyle{ a_k\in\{a_i, a_{i+1}, \ldots, a_{j-1}, a_{j}\} }[/math], [math]\displaystyle{ Y=a_k }[/math] with the same probability, and [math]\displaystyle{ Y\not\in\{a_i, a_{i+1}, \ldots, a_{j-1}, a_{j}\} }[/math] with an unknown probability (remember that there might be other elements in the same subset with [math]\displaystyle{ \{a_i, a_{i+1}, \ldots, a_{j-1}, a_{j}\} }[/math]). The probability we are looking for is actually [math]\displaystyle{ \Pr[Y\in \{a_i, a_j\}\mid Y\in\{a_i, a_{i+1}, \ldots, a_{j-1}, a_{j}\}] }[/math], which is always [math]\displaystyle{ \frac{2}{j-i+1} }[/math], provided that [math]\displaystyle{ Y }[/math] is uniform over [math]\displaystyle{ \{a_i, a_{i+1}, \ldots, a_{j-1}, a_{j}\} }[/math]. The conditional probability rules out the irrelevant events in a probabilistic argument.

Summing all up:

[math]\displaystyle{ \begin{align} \mathbf{E}\left[\sum_{i=1}^n\sum_{j\gt i}X_{ij}\right] &= \sum_{i=1}^n\sum_{j\gt i}\mathbf{E}\left[X_{ij}\right]\\ &\le \sum_{i=1}^n\sum_{j\gt i}\frac{2}{j-i+1}\\ &= \sum_{i=1}^n\sum_{k=2}^{n-i+1}\frac{2}{k} & & (\mbox{Let }k=j-i+1)\\ &\le \sum_{i=1}^n\sum_{k=1}^{n}\frac{2}{k}\\ &= 2n\sum_{k=1}^{n}\frac{1}{k}\\ &= 2n H(n). \end{align} }[/math]

[math]\displaystyle{ H(n) }[/math] is the [math]\displaystyle{ n }[/math]th Harmonic number. It holds that

[math]\displaystyle{ \begin{align}H(n) = \ln n+O(1)\end{align} }[/math].

Therefore, for an arbitrary input [math]\displaystyle{ S }[/math] of [math]\displaystyle{ n }[/math] numbers, the expected number of comparisons taken by RandQSort to sort [math]\displaystyle{ S }[/math] is [math]\displaystyle{ \mathrm{O}(n\log n) }[/math].

Random Recurrence

We consider the selection problem: given as input a set [math]\displaystyle{ S }[/math] of [math]\displaystyle{ n }[/math] distinct numbers and an integer [math]\displaystyle{ 1\le k\le n }[/math], find the [math]\displaystyle{ k }[/math]-th smallest number in [math]\displaystyle{ S }[/math].

We know that this problem can be solved deterministically by the median-of-median algorithm or randomly by the LazySelect algorithm, both in linear time. In the same spirit of random QuickSort, we propose the following random QuickSelection algorithm, called the RandomQS.

RandomQS([math]\displaystyle{ S }[/math],[math]\displaystyle{ k }[/math])

pick a uniform random pivot [math]\displaystyle{ x\in S }[/math]; construct [math]\displaystyle{ S_1=\{y\in S\mid y\lt x\} }[/math] and [math]\displaystyle{ S_2=\{y\in S\mid y\gt x\} }[/math]; if [math]\displaystyle{ |S_1|=k-1 }[/math]: return [math]\displaystyle{ x }[/math]; if [math]\displaystyle{ |S_1|\gt k-1 }[/math]: return RandomQS[math]\displaystyle{ (S_1,k) }[/math]; if [math]\displaystyle{ |S_1|\lt k-1 }[/math]: return RandomQS[math]\displaystyle{ (S_2,k-|S_1|-1) }[/math];

It can be verified as an exercise problem that the expected number of comparisons of RandomQS is [math]\displaystyle{ O(n) }[/math]. Alternatively, we ask the following question:

• How many recursive calls to RandomQS in expectation in a running of the algorithm?

Let [math]\displaystyle{ T(n) }[/math] denote the number of recursive calls in a running of RandomQS. Obviously [math]\displaystyle{ T(n) }[/math] is a random variable and it satisfies the following recurrence:

[math]\displaystyle{ T(n) = \begin{cases} 1+T(n-X(n)) & n\gt 0\\ 0 & n=0 \end{cases} }[/math]

where [math]\displaystyle{ X(n) }[/math] is a random variable representing the number of elements in the input [math]\displaystyle{ S }[/math] which are thrown away in the current call. Specifically, [math]\displaystyle{ X(n)=|S_2|+1 }[/math] when [math]\displaystyle{ |S_1|\gt k-1 }[/math], [math]\displaystyle{ X(n)=|S_1|+1 }[/math] when [math]\displaystyle{ |S_1|\lt k-1 }[/math] and [math]\displaystyle{ X(n)=n-1 }[/math] when [math]\displaystyle{ |S_1|=k-1 }[/math]. We are looking for an upper bound on the expectation [math]\displaystyle{ \mathbf{E}[T(n)] }[/math].

Karp-Upfal-Wigderson bound

Theorem (Karp-Upfal-Wigderson 1988)

Consider the following recurrence [math]\displaystyle{ \begin{align} T(n) &= \begin{cases} 1+T(n-X(n)) & n\gt c\\ 0 & n=c \end{cases} \end{align} }[/math] where [math]\displaystyle{ c }[/math] is a constant and each [math]\displaystyle{ X(n) }[/math] is an independent drawn of a nonnegative integral random variable [math]\displaystyle{ X_n }[/math] such that [math]\displaystyle{ 0\le X_n\le n-c }[/math]. If there exists a positive-valued non-degreasing function [math]\displaystyle{ \,\mu(x) }[/math] satisfying that [math]\displaystyle{ \mathbf{E}[X_n]\ge\mu(n) }[/math] for all [math]\displaystyle{ n }[/math], then [math]\displaystyle{ \mathbf{E}[T(n)] \le \int_{c}^n\frac{1}{\mu(x)}\,\mathrm{d} x. }[/math]

Proof. We prove the theorem by applying an induction on [math]\displaystyle{ n }[/math]. The bound is trivially true when [math]\displaystyle{ n=c }[/math]. For [math]\displaystyle{ n\gt c }[/math], assume the induction hypothesis for all smaller [math]\displaystyle{ n }[/math]. Denote that [math]\displaystyle{ X=X(n) }[/math]. Due to the recursion [math]\displaystyle{ T(n)=1+T(n-X(n)) }[/math], we have [math]\displaystyle{ \begin{align} \mathbf{E}[T(n)] &= 1+\mathbf{E}[T(n-X)]\\ &= 1+\mathbf{E}[T(n-X)\mid X=0]\Pr[X_n=0]+\mathbf{E}[T(n-X)\mid X\gt 0]\Pr[X\gt 0]. \end{align} }[/math] Denote that [math]\displaystyle{ p=\Pr[X=0] }[/math]. We have [math]\displaystyle{ \Pr[X\gt 0]=1-p }[/math] because [math]\displaystyle{ X=X(n) }[/math] is nonnegative. Thus the above equation becomes [math]\displaystyle{ \begin{align} \mathbf{E}[T(n)] &= 1+p\mathbf{E}[T(n)]+(1-p)\mathbf{E}[T(n-X)\mid X\gt 0], \end{align} }[/math] which gives us that [math]\displaystyle{ \mathbf{E}[T(n)]=\frac{1}{1-p}+\mathbf{E}[T(n-X)\mid X\gt 0] }[/math]. For the conditional expectation, we have [math]\displaystyle{ \begin{align} &\mathbf{E}[T(n-X)\mid X\gt 0]\\ = &\mathbf{E}[\,\mathbf{E}[T(n-X)\mid X]\mid X\gt 0]\\ \le &\mathbf{E}\left[\int_{c}^{n-X}\frac{1}{\mu(x)}\,\mathrm{d}x \,\Big|\, X\gt 0\right] & \text{(induction hypothesis)}\\ = &\mathbf{E}\left[\int_{c}^{n}\frac{1}{\mu(x)}\,\mathrm{d}x - \int_{n-X}^{n}\frac{1}{\mu(x)}\,\mathrm{d}x\,\Big|\, X\gt 0\right] \\ = &\int_{c}^{n}\frac{1}{\mu(x)}\,\mathrm{d}x-\mathbf{E}\left[\int_{n-X}^{n}\frac{1}{\mu(x)}\,\mathrm{d}x\,\Big|\, X\gt 0\right] \\ \le &\int_{c}^{n}\frac{1}{\mu(x)}\,\mathrm{d}x-\mathbf{E}\left[\int_{n-X}^{n}\frac{1}{\mu(n)}\,\mathrm{d}x\,\Big|\, X\gt 0\right] &\text{(}\mu(x)\text{ is non-degreasing)}\\ = &\int_{c}^{n}\frac{1}{\mu(x)}\,\mathrm{d}x-\frac{1}{\mu(n)}\mathbf{E}[X\mid X\gt 0]. \end{align} }[/math] On the other hand, due to the total expectation, we have [math]\displaystyle{ \mathbf{E}[X_n] =\mathbf{E}[X_n\mid X_n=0]\Pr[X_n=0]+\mathbf{E}[X_n\mid X_n\gt 0]\Pr[X_n\gt 0]=(1-p)\mathbf{E}[X_n\mid X_n\gt 0], }[/math] which gives us that [math]\displaystyle{ \mathbf{E}[X\mid X\gt 0]=\frac{\mathbf{E}[X_n]}{1-p}\ge\frac{\mu(n)}{1-p} }[/math]. Substituting this in the above inequality gives us that [math]\displaystyle{ \begin{align} \mathbf{E}[T(n)] &= \frac{1}{1-p}+\mathbf{E}[T(n-X)\mid X\gt 0]\\ &\le \frac{1}{1-p}+\int_{c}^{n}\frac{1}{\mu(x)}\,\mathrm{d}x-\frac{1}{\mu(n)}\cdot\frac{\mu(n)}{1-p}\\ &= \int_{c}^{n}\frac{1}{\mu(x)}\,\mathrm{d}x. \end{align} }[/math] [math]\displaystyle{ \square }[/math]