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 uniform and pairwise independent random bits from a sequence of uniform and mutually independent 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]

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.