Randomized Algorithms (Spring 2010)/Hashing, limited independence

Limited Independence

k-wise independence

Recall the definition of independence between events:

Definition (Independent events): Events [math]\displaystyle{ \mathcal{E}_1, \mathcal{E}_2, \ldots, \mathcal{E}_n }[/math] are mutually independent if, for any subset [math]\displaystyle{ I\subseteq\{1,2,\ldots,n\} }[/math], [math]\displaystyle{ \begin{align} \Pr\left[\bigwedge_{i\in I}\mathcal{E}_i\right] &= \prod_{i\in I}\Pr[\mathcal{E}_i]. \end{align} }[/math]

Similarly, we can define independence between random variables:

Definition (Independent variables): Random variables [math]\displaystyle{ X_1, X_2, \ldots, X_n }[/math] are mutually independent if, for any subset [math]\displaystyle{ I\subseteq\{1,2,\ldots,n\} }[/math] and any values [math]\displaystyle{ x_i }[/math], where [math]\displaystyle{ i\in I }[/math], [math]\displaystyle{ \begin{align} \Pr\left[\bigwedge_{i\in I}(X_i=x_i)\right] &= \prod_{i\in I}\Pr[X_i=x_i]. \end{align} }[/math]

Mutual independence is an ideal condition of independence. The limited notion of independence is usually defined by the k-wise independence.

Definition (k-wise Independenc): 1. Events [math]\displaystyle{ \mathcal{E}_1, \mathcal{E}_2, \ldots, \mathcal{E}_n }[/math] are k-wise independent if, for any subset [math]\displaystyle{ I\subseteq\{1,2,\ldots,n\} }[/math] with [math]\displaystyle{ |I|\le k }[/math] [math]\displaystyle{ \begin{align} \Pr\left[\bigwedge_{i\in I}\mathcal{E}_i\right] &= \prod_{i\in I}\Pr[\mathcal{E}_i]. \end{align} }[/math] 2. Random variables [math]\displaystyle{ X_1, X_2, \ldots, X_n }[/math] are k-wise independent if, for any subset [math]\displaystyle{ I\subseteq\{1,2,\ldots,n\} }[/math] with [math]\displaystyle{ |I|\le k }[/math] and any values [math]\displaystyle{ x_i }[/math], where [math]\displaystyle{ i\in I }[/math], [math]\displaystyle{ \begin{align} \Pr\left[\bigwedge_{i\in I}(X_i=x_i)\right] &= \prod_{i\in I}\Pr[X_i=x_i]. \end{align} }[/math]

A very common case is pairwise independence, i.e. the 2-wise independence.

Definition (pairwise Independent random variables): Random variables [math]\displaystyle{ X_1, X_2, \ldots, X_n }[/math] are pairwise independent if, for any [math]\displaystyle{ X_i,X_j }[/math] where [math]\displaystyle{ i\neq j }[/math] and any values [math]\displaystyle{ a,b }[/math] [math]\displaystyle{ \begin{align} \Pr\left[X_i=a\wedge X_j=b\right] &= \Pr[X_i=a]\cdot\Pr[X_j=b]. \end{align} }[/math]

Note that the definition of k-wise independence is hereditary:

• If [math]\displaystyle{ X_1, X_2, \ldots, X_n }[/math] are k-wise independent, then they are also [math]\displaystyle{ \ell }[/math]-wise independent for any [math]\displaystyle{ \ell\lt k }[/math].
• If [math]\displaystyle{ X_1, X_2, \ldots, X_n }[/math] are NOT k-wise independent, then they cannot be [math]\displaystyle{ \ell }[/math]-wise independent for any [math]\displaystyle{ \ell\gt k }[/math].

Construction via XOR

Suppose we have [math]\displaystyle{ m }[/math] mutually independent and uniform random bits [math]\displaystyle{ X_1,\ldots, X_m }[/math]. We are going to extract [math]\displaystyle{ n=2^m-1 }[/math] pairwise independent bits from these [math]\displaystyle{ m }[/math] mutually independent bits.

Enumerate all the nonempty subsets of [math]\displaystyle{ \{1,2,\ldots,m\} }[/math] in some order. Let [math]\displaystyle{ S_j }[/math] be the [math]\displaystyle{ j }[/math]th subset. Let

[math]\displaystyle{ Y_j=\bigoplus_{i\in S_j} X_i, }[/math]

where [math]\displaystyle{ \oplus }[/math] is the exclusive-or, whose truth table is as follows.

[math]\displaystyle{ a }[/math]	[math]\displaystyle{ b }[/math]	[math]\displaystyle{ a }[/math][math]\displaystyle{ \oplus }[/math][math]\displaystyle{ b }[/math]
0	0	0
0	1	1
1	0	1
1	1	0

There are [math]\displaystyle{ n=2^m-1 }[/math] such [math]\displaystyle{ Y_j }[/math], because there are [math]\displaystyle{ 2^m-1 }[/math] nonempty subsets of [math]\displaystyle{ \{1,2,\ldots,m\} }[/math]. An equivalent definition of [math]\displaystyle{ Y_j }[/math] is

[math]\displaystyle{ Y_j=\left(\sum_{i\in S_j}X_i\right)\bmod 2 }[/math].

Sometimes, [math]\displaystyle{ Y_j }[/math] is called the parity of the bits in [math]\displaystyle{ S_j }[/math].

We claim that [math]\displaystyle{ Y_j }[/math] are pairwise independent and uniform.

Theorem: For any [math]\displaystyle{ Y_j }[/math] and any [math]\displaystyle{ b\in\{0,1\} }[/math], [math]\displaystyle{ \begin{align} \Pr\left[Y_j=b\right] &= \frac{1}{2}. \end{align} }[/math] For any [math]\displaystyle{ Y_j,Y_\ell }[/math] that [math]\displaystyle{ j\neq\ell }[/math] and any [math]\displaystyle{ a,b\in\{0,1\} }[/math], [math]\displaystyle{ \begin{align} \Pr\left[Y_j=a\wedge Y_\ell=b\right] &= \frac{1}{4}. \end{align} }[/math]

The proof is left for your exercise.

Therefore, we extract exponentially many pairwise independent uniform random bits from a sequence of mutually independent uniform random bits.

Note that [math]\displaystyle{ Y_j }[/math] are not 3-wise independent. For example, consider the subsets [math]\displaystyle{ S_1=\{1\},S_2=\{2\},S_3=\{1,2\} }[/math] and the corresponding random bits [math]\displaystyle{ Y_1,Y_2,Y_3 }[/math]. Any two of [math]\displaystyle{ Y_1,Y_2,Y_3 }[/math] would decide the value of the third one.

Construction via modulo a prime

We now consider constructing pairwise independent random variables ranging over [math]\displaystyle{ [p]=\{0,1,2,\ldots,p-1\} }[/math] for some prime [math]\displaystyle{ p }[/math]. Unlike the above construction, now we only need two independent random sources [math]\displaystyle{ X_0,X_1 }[/math], which are uniformly and independently distributed over [math]\displaystyle{ [p] }[/math].

Let [math]\displaystyle{ Y_0,Y_1,\ldots, Y_{p-1} }[/math] be defined as:

[math]\displaystyle{ \begin{align} Y_i=(X_0+i\cdot X_i)\bmod p &\quad \mbox{for }i\in[p]. \end{align} }[/math]

Theorem: The random variables [math]\displaystyle{ Y_0,Y_1,\ldots, Y_{p-1} }[/math] are pairwise independent uniform random variables over [math]\displaystyle{ [p] }[/math].

Proof: We first show that [math]\displaystyle{ Y_i }[/math] are uniform. That is, we will show that for any [math]\displaystyle{ i,a\in[p] }[/math],

[math]\displaystyle{ \begin{align} \Pr\left[(X_0+i\cdot X_1)\bmod p=a\right] &= \frac{1}{p}. \end{align} }[/math]

Due to the law of total probability,

[math]\displaystyle{ \begin{align} \Pr\left[(X_0+i\cdot X_1)\bmod p=a\right] &= \sum_{j\in[p]}\Pr[X_1=j]\cdot\Pr\left[(X_0+ij)\bmod p=a\right]\\ &=\frac{1}{p}\sum_{j\in[p]}\Pr\left[X_0\equiv(a-ij)\pmod{p}\right]. \end{align} }[/math]

For prime [math]\displaystyle{ p }[/math], for any [math]\displaystyle{ i,j,a\in[p] }[/math], there is exact one value in [math]\displaystyle{ [p] }[/math] of [math]\displaystyle{ X_0 }[/math] satisfying [math]\displaystyle{ X_0\equiv(a-ij)\pmod{p} }[/math]. Thus, [math]\displaystyle{ \Pr\left[X_0\equiv(a-ij)\pmod{p}\right]=1/p }[/math] and the above probability is [math]\displaystyle{ \frac{1}{p} }[/math].

We then show that [math]\displaystyle{ Y_i }[/math] are pairwise independent, i.e. we will show that for any [math]\displaystyle{ Y_i,Y_j }[/math] that [math]\displaystyle{ i\neq j }[/math] and any [math]\displaystyle{ a,b\in[p] }[/math],

[math]\displaystyle{ \begin{align} \Pr\left[Y_i=a\wedge Y_j=b\right] &= \frac{1}{p^2}. \end{align} }[/math]

The event [math]\displaystyle{ Y_i=a\wedge Y_j=b }[/math] is equivalent to that

[math]\displaystyle{ \begin{cases} (X_0+iX_1)\equiv a\pmod{p}\\ (X_0+jX_1)\equiv b\pmod{p} \end{cases} }[/math]

Due to the Chinese remainder theorem, there exists a unique solution of [math]\displaystyle{ X_0 }[/math] and [math]\displaystyle{ X_1 }[/math] in [math]\displaystyle{ [p] }[/math] to the above linear congruential system. Thus the probability of the event is [math]\displaystyle{ \frac{1}{p^2} }[/math].

[math]\displaystyle{ \square }[/math]

Chebyshev's inequality for pairwise independent random variables

For random viables with limited independence, we are not able to directly use the probability tools which rely on the independence of random variables, such as the Chernoff bounds. On the positive side, there are tools that require less independence.

In lecture 4, we show the following theorem of linearity of variance for pairwise independent random variables.

Theorem: For pairwise independent random variables [math]\displaystyle{ X_1,X_2,\ldots,X_n }[/math], [math]\displaystyle{ \begin{align} \mathbf{Var}\left[\sum_{i=1}^n X_i\right]=\sum_{i=1}^n\mathbf{Var}[X_i]. \end{align} }[/math]

We proved the theorem by showing that the covariances of pairwise independent random variables are 0. The theorem is actually a consequence of a more general statement.

Theorem 1: Let [math]\displaystyle{ X_1,X_2,\ldots,X_n }[/math] be mutually independent random variables, [math]\displaystyle{ Y_1,Y_2,\ldots,Y_n }[/math] be k-wise independent random variables, and [math]\displaystyle{ Pr[X_i=z]=\Pr[Y_i=z] }[/math] for every [math]\displaystyle{ 1\le i\le n }[/math] and any [math]\displaystyle{ z }[/math]. Let [math]\displaystyle{ f:\mathbb{R}^n\rightarrow\mathbb{R} }[/math] be a multivariate polynomial of degree at most [math]\displaystyle{ k }[/math]. Then [math]\displaystyle{ \begin{align} \mathbf{E}\left[f(X_1,X_2,\ldots,X_n)\right]=\mathbf{E}[f(Y_1,Y_2,\ldots,Y_n)]. \end{align} }[/math]

This phenomenon is sometimes called that the k-degree polynomials are fooled by k-wise independence. In other words, a k-degree polynomial behaves the same on the k-wise independent random variables as on the mutual independent random variables.

This theorem is implied by the following lemma.

Lemma: Let [math]\displaystyle{ X_1,X_2,\ldots,X_k }[/math] be [math]\displaystyle{ k }[/math] mutually independent random variables. Then [math]\displaystyle{ \begin{align} \mathbf{E}\left[\prod_{i=1}^k X_i\right]=\prod_{i=1}^k\mathbf{E}[X_i]. \end{align} }[/math]

The lemma can be proved by directly compute the expectation. We omit the detailed proof. By the linearity of expectation, the expectation of a polynomial is reduced to the sum of the expectations of terms. For a k-degree polynomial, each term has at most [math]\displaystyle{ k }[/math] variables. Due to the above lemma, with k-wise independence, the expectation of each term behaves exactly the same as mutual independence. Theorem 1 is proved.

Since the [math]\displaystyle{ k }[/math]th moment is the expectation of a k-degree polynomial of random variables, the tools based on the [math]\displaystyle{ k }[/math]th moment can be safely used for the k-wise independence. In particular, Chebyshev's inequality for pairwise independent random variables:

Chebyshev's inequality: Let [math]\displaystyle{ X=\sum_{i=1}^n X_i }[/math], where [math]\displaystyle{ X_1, X_2, \ldots, X_n }[/math] are pairwise independent Poisson trials. Let [math]\displaystyle{ \mu=\mathbf{E}[X] }[/math]. Then [math]\displaystyle{ \Pr[|X-\mu|\ge t]\le\frac{\mathbf{Var}[X]}{t^2}=\frac{\sum_{i=1}^n\mathbf{Var}[X_i]}{t^2}. }[/math]

Two-point sampling

Consider a Monte Carlo randomized algorithm with one-sided error for a decision problem [math]\displaystyle{ f }[/math]. We formulate the algorithm as a deterministic algorithm [math]\displaystyle{ A }[/math] that takes as input [math]\displaystyle{ x }[/math] and a uniform random number [math]\displaystyle{ r\in[p] }[/math] where [math]\displaystyle{ p }[/math] is a prime, such that for any input [math]\displaystyle{ x }[/math]:

• If [math]\displaystyle{ f(x)=1 }[/math], then [math]\displaystyle{ \Pr[A(x,r)=1]\ge\frac{1}{2} }[/math], where the probability is taken over the random choice of [math]\displaystyle{ r }[/math].
• If [math]\displaystyle{ f(x)=0 }[/math], then [math]\displaystyle{ A(x,r)=0 }[/math] for any [math]\displaystyle{ r }[/math].

We call [math]\displaystyle{ r }[/math] the random source for the algorithm.

For the [math]\displaystyle{ x }[/math] that [math]\displaystyle{ f(x)=1 }[/math], we call the [math]\displaystyle{ r }[/math] that makes [math]\displaystyle{ A(x,r)=1 }[/math] a witness for [math]\displaystyle{ x }[/math]. For a positive [math]\displaystyle{ x }[/math], at least half of [math]\displaystyle{ [p] }[/math] are witnesses. The random source [math]\displaystyle{ r }[/math] has polynomial number of bits, which means that [math]\displaystyle{ p }[/math] is exponentially large, thus it is infeasible to find the witness for an input [math]\displaystyle{ x }[/math] by exhaustive search. Deterministic overcomes this by having sophisticated deterministic rules for efficiently searching for a witness. Randomization, on the other hard, reduce this to a bit of luck, by randomly choosing an [math]\displaystyle{ r }[/math] and winning with a probability of 1/2.

We can boost the accuracy (equivalently, reduce the error) of any Monte Carlo randomized algorithm with one-sided error by running the algorithm for a number of times.

Suppose that we sample [math]\displaystyle{ t }[/math] values [math]\displaystyle{ r_1,r_2,\ldots,r_t }[/math] uniformly and independently from [math]\displaystyle{ [p] }[/math], and run the following scheme:

[math]\displaystyle{ B(x,r_1,r_2,\ldots,r_t): }[/math]

return [math]\displaystyle{ \bigvee_{i=1}^t A(x,r_i) }[/math];

That is, return 1 if any instance of [math]\displaystyle{ A(x,r_i)=1 }[/math]. For any [math]\displaystyle{ x }[/math] that [math]\displaystyle{ f(x)=1 }[/math], due to the independence of [math]\displaystyle{ r_1,r_2,\ldots,r_t }[/math], the probability that [math]\displaystyle{ B(x,r_1,r_2,\ldots,r_t) }[/math] returns an incorrect result is at most [math]\displaystyle{ 2^{-t} }[/math]. On the other hand, [math]\displaystyle{ B }[/math] never makes mistakes for the [math]\displaystyle{ x }[/math] that [math]\displaystyle{ f(x)=0 }[/math] since [math]\displaystyle{ A }[/math] has no false positives. Thus, the error of the Monte Carlo algorithm is reduced to [math]\displaystyle{ 2^{-t} }[/math].

Sampling [math]\displaystyle{ t }[/math] mutually independent random numbers from [math]\displaystyle{ [p] }[/math] can be quite expensive since it requires [math]\displaystyle{ \Omega(t\log p) }[/math] random bits. Suppose that we can only afford [math]\displaystyle{ O(\log p) }[/math] random bits. In particular, we sample two independent uniform random number [math]\displaystyle{ a }[/math] and [math]\displaystyle{ b }[/math] from [math]\displaystyle{ [p] }[/math]. If we use [math]\displaystyle{ a }[/math] and [math]\displaystyle{ b }[/math] directly bu running two independent instances [math]\displaystyle{ A(x,a) }[/math] and [math]\displaystyle{ A(x,b) }[/math], we only get an error upper bound of 1/4.

The following scheme reduces the error significantly with the same number of random bits:

Choose two independent uniform random number [math]\displaystyle{ a }[/math] and [math]\displaystyle{ b }[/math] from [math]\displaystyle{ [p] }[/math]. Construct [math]\displaystyle{ t }[/math] random number [math]\displaystyle{ r_1,r_2,\ldots,r_t }[/math] by: [math]\displaystyle{ \begin{align} \forall 1\le i\le t, &\quad \mbox{let }r_i = (a\cdot i+b)\bmod p. \end{align} }[/math] Run [math]\displaystyle{ B(x,r_1,r_2,\ldots,r_t): }[/math].

Due to the discussion in the last section, we know that for [math]\displaystyle{ t\le p }[/math], [math]\displaystyle{ r_1,r_2,\ldots,r_t }[/math] are pairwise independent and uniform over [math]\displaystyle{ [p] }[/math]. Let [math]\displaystyle{ X_i=A(x,r_i) }[/math] and [math]\displaystyle{ X=\sum_{i=1}^tX_i }[/math]. Due to the uniformity of [math]\displaystyle{ r_i }[/math] and our definition of [math]\displaystyle{ A }[/math], for any [math]\displaystyle{ x }[/math] that [math]\displaystyle{ f(x)=1 }[/math], it holds that

[math]\displaystyle{ \Pr[X_i=1]=\Pr[A(x,r_i)=1]\ge\frac{1}{2}. }[/math]

By the linearity of expectations,

[math]\displaystyle{ \mathbf{E}[X]=\sum_{i=1}^t\mathbf{E}[X_i]=\sum_{i=1}^t\Pr[X_i=1]\ge\frac{t}{2}. }[/math]

Since [math]\displaystyle{ X_i }[/math] is Bernoulli trial with a probability of success at least [math]\displaystyle{ p=1/2 }[/math]. We can estimate the variance of each [math]\displaystyle{ X_i }[/math] as follows.

[math]\displaystyle{ \mathbf{Var}[X_i]=p(1-p)\le\frac{1}{4}. }[/math]

Applying Chebyshev's inequality, we have

[math]\displaystyle{ \begin{align} \Pr\left[\bigvee_{i=1}^t A(x,r_i)\right] &= \Pr[X=0]\\ &\le \Pr[|X-\mathbf{E}[X]|\ge \mathbf{E}[X]]\\ &\le \Pr\left[|X-\mathbf{E}[X]|\ge \frac{t}{2}\right]\\ &\le \frac{4}{t^2}\sum_{i=1}^t\mathbf{Var}[X_i]\\ &\le \frac{1}{t}. \end{align} }[/math]

Collision number

By seeing the lodas of bins as a vector of random variables, called load vector, the expectation of the maximum load is the expected [math]\displaystyle{ L_\infty }[/math]-norm of this load vector. Since there are [math]\displaystyle{ m }[/math] balls, the [math]\displaystyle{ L_1 }[/math]-norm of the load vector is definitely [math]\displaystyle{ m }[/math]. We ask about something between these two extremes, specifically, the sum of the squares of the loads. We will see that this quantity approximately gives the expected number of collision pairs in the balls-into-bins game.

For any two balls, we say that there is a collision between them if they are thrown into the same bin. Let [math]\displaystyle{ Y_{ij} }[/math] indicates whether ball [math]\displaystyle{ i }[/math] and ball[math]\displaystyle{ j }[/math] collides, i.e.

[math]\displaystyle{ Y_{ij} = \begin{cases} 1 & \text{if ball }i\text{ and ball }j\text{ are thrown into the same bin},\\ 0 & \text{otherwise.} \end{cases} }[/math]

The total number of collision pairs is [math]\displaystyle{ Y=\sum_{i\lt j} Y_{ij} }[/math]. Since each ball is uniformly and independently thrown into one of the [math]\displaystyle{ n }[/math] bins, for any particular [math]\displaystyle{ i\neq j }[/math], the probability that ball [math]\displaystyle{ i }[/math] and ball[math]\displaystyle{ j }[/math] are thrown into the same bin is

[math]\displaystyle{ \begin{align} \Pr[Y_{ij}=1] &= \Pr[\text{ball }i\text{ and ball }j\text{ are in the same bin}]\\ &= \sum_{k=1}^n\Pr[\text{ball }i\text{ is in bin }k]\cdot\Pr[\text{ball }i\text{ and ball }j\text{ are in the same bin}\mid \text{ball }i\text{ is in bin }k]\\ &=n\cdot\frac{1}{n}\cdot\frac{1}{n}\\ &= \frac{1}{n}. \end{align} }[/math]

Therefore,

[math]\displaystyle{ \mathbf{E}[Y]=\mathbf{E}\left[\sum_{i\lt j}Y_{ij}\right]=\sum_{i\lt j}\mathbf{E}[Y_{ij}]=\sum_{i\lt j}\Pr[Y_{ij}=1]={m\choose 2}\frac{1}{n}. }[/math]

For [math]\displaystyle{ 1\le k\le n }[/math], let [math]\displaystyle{ X_k }[/math] be the load of the [math]\displaystyle{ k }[/math]-th bin. The number of collision pairs in the [math]\displaystyle{ k }[/math]-th bin can be computed as [math]\displaystyle{ {X_k\choose 2} }[/math], therefore the total number of collision pairs is also given by

[math]\displaystyle{ Y=\sum_{i=1}^n {X_i\choose 2}. }[/math]

The sum of the squares of the loads is

[math]\displaystyle{ \sum_{i=1}^n X_{i}^2 =\sum_{i=1}^n \left(X_i+X_i(X_i-1)\right) =m+2\sum_{i=1}^n{X_i\choose 2} =m+2Y. }[/math]

Its expectation is

[math]\displaystyle{ \mathbf{E}\left[\sum_{i=1}^nX_i^2\right]=m+2\mathbf{E}[Y]=m+2{m\choose 2}\frac{1}{n}=m+\frac{m(m-1)}{n}. }[/math]

Hashing

Perfect hashing

In a hash table, [math]\displaystyle{ m }[/math] keys are stored in [math]\displaystyle{ n }[/math] slots, and the keys are mapped to slots by a hash function. A collision occurs if two keys are mapped to the same slots. There are various strategies for resolving collisions, such as by chaining, or by "open addressing" techniques like linear probing or double hashing. However, we could also just wish there is no collision.

A hash function [math]\displaystyle{ h:U\rightarrow[n] }[/math] is perfect for a set [math]\displaystyle{ S\subseteq U }[/math] of keys if [math]\displaystyle{ h }[/math] maps all keys in [math]\displaystyle{ S }[/math] to different values, i.e. there is no collision.

For hash tables, the hash function is a random mapping from keys to values. To simplify the analysis, we assume that the hash function is uniformly random function [math]\displaystyle{ h:U\rightarrow[n] }[/math]. This assumption is called the Simple Uniform Hash Assumption (SUHA or UHA), which is a standard assumption used in the analysis of hashing

With UHA, the [math]\displaystyle{ m }[/math] different keys are uniformly and independently assigned to [math]\displaystyle{ n }[/math] slots. Due to our analysis of the birthday problem, for some [math]\displaystyle{ m=O(\sqrt{n}) }[/math], with good chance (a constant probability), the hash function is perfect. However, as [math]\displaystyle{ m }[/math] grows larger, it becomes quite unlikely that the hash function is perfect.