概率论与数理统计 (Spring 2023)/Polynomial identity testing: Difference between revisions

Latest revision as of 03:05, 12 March 2023

In the Polynomial Identity Testing (PIT) problem, we are given two polynomials, and we want to determine whether they are identical.

Polynomial Identity Testing

Input: two [math]\displaystyle{ n }[/math]-variate polynomials [math]\displaystyle{ f, g\in\mathbb{F}[x_1,x_2,\ldots,x_n] }[/math] of degree [math]\displaystyle{ d }[/math].
Determine whether [math]\displaystyle{ f\equiv g }[/math].

The [math]\displaystyle{ \mathbb{F}[x_1,x_2,\ldots,x_n] }[/math] denotes the ring of multivariate polynomials over field [math]\displaystyle{ \mathbb{F} }[/math]. An [math]\displaystyle{ n }[/math]-variate polynomial of degree [math]\displaystyle{ d }[/math], written as a sum of monomials, is:

[math]\displaystyle{ f(x_1,x_2,\ldots,x_n)=\sum_{i_1,i_2,\ldots,i_n\ge 0\atop i_1+i_2+\cdots+i_n\le d}a_{i_1,i_2,\ldots,i_n}x_{1}^{i_1}x_2^{i_2}\cdots x_{n}^{i_n} }[/math].

The degree or total degree of a monomial [math]\displaystyle{ a_{i_1,i_2,\ldots,i_n}x_{1}^{i_1}x_2^{i_2}\cdots x_{n}^{i_n} }[/math] is given by [math]\displaystyle{ i_1+i_2+\cdots+i_n }[/math] and the degree of a polynomial [math]\displaystyle{ f }[/math] is the maximum degree of monomials of nonzero coefficients.

This problem is equivalent to the following problem of checking whether [math]\displaystyle{ f-g }[/math] (which is a polynomial of degree at most [math]\displaystyle{ d }[/math]) is the zero-polynomial.

Polynomial Identity Testing (equivalent version)

Input: a polynomial [math]\displaystyle{ f\in\mathbb{F}[x_1,x_2,\ldots,x_n] }[/math] of degree [math]\displaystyle{ d }[/math].
Determine whether [math]\displaystyle{ f\equiv 0 }[/math].

The problem seems to be solvable by checking whether all coefficients of monomials in [math]\displaystyle{ f }[/math] are zeros. And there are at most [math]\displaystyle{ (n+d)^{d} }[/math] coefficients since [math]\displaystyle{ f }[/math] is an [math]\displaystyle{ n }[/math]-variate polynomial of degree at most [math]\displaystyle{ d }[/math]. However, this relies on that all coefficients of [math]\displaystyle{ f }[/math] are explicitly given, which may not always be the case.

For example, a multivariate polynomial [math]\displaystyle{ f }[/math] may be expressed in a product form.

Example

The Vandermonde matrix [math]\displaystyle{ M=M(x_1,x_2,\ldots,x_n) }[/math] is defined as that [math]\displaystyle{ M_{ij}=x_i^{j-1} }[/math], that is

[math]\displaystyle{ M=\begin{bmatrix} 1 & x_1 & x_1^2 & \dots & x_1^{n-1}\\ 1 & x_2 & x_2^2 & \dots & x_2^{n-1}\\ 1 & x_3 & x_3^2 & \dots & x_3^{n-1}\\ \vdots & \vdots & \vdots & \ddots &\vdots \\ 1 & x_n & x_n^2 & \dots & x_n^{n-1} \end{bmatrix} }[/math].

Let [math]\displaystyle{ f }[/math] be the polynomial defined as

[math]\displaystyle{ f(x_1,\ldots,x_n)=\det(M)=\prod_{j\lt i}(x_i-x_j). }[/math]

For polynomials [math]\displaystyle{ f }[/math] in product form, it is quite possible that evaluating the polynomial [math]\displaystyle{ f(x_1,\ldots,x_n) }[/math] at any [math]\displaystyle{ (x_1,\ldots,x_n)\in\mathbb{F}^n }[/math] is always efficient but symbolically expanding the polynomial to calculate the coefficients may be very expensive.

We assume that the input polynomial [math]\displaystyle{ f }[/math] is given as a block-box or in product form, so that the evaluation of [math]\displaystyle{ f(x_1,\ldots,x_n) }[/math] at any specific [math]\displaystyle{ (x_1,\ldots,x_n)\in\mathbb{F}^n }[/math] is efficient.

Polynomial Interpolation

First, let's consider the univariate case where [math]\displaystyle{ n=1 }[/math]. In such case [math]\displaystyle{ f(x) }[/math] is a univariate polynomial of degree at most [math]\displaystyle{ d }[/math].

A straightforward deterministic algorithm is to evaluate [math]\displaystyle{ f(x_1),f(x_2),\ldots,f(x_{d+1}) }[/math] over [math]\displaystyle{ d+1 }[/math] distinct elements [math]\displaystyle{ x_1,x_2,\ldots,x_{d+1} }[/math] from the field [math]\displaystyle{ \mathbb{F} }[/math] and check whether they are all zero. By the fundamental theorem of algebra, also known as polynomial interpolation, this guarantees to answer correctly whether a degree-[math]\displaystyle{ d }[/math] univariate polynomial [math]\displaystyle{ f\equiv 0 }[/math].

Fundamental Theorem of Algebra

Any non-zero univariate polynomial of degree [math]\displaystyle{ d }[/math] has at most [math]\displaystyle{ d }[/math] roots.

The reason for this fundamental theorem holding generally over any field [math]\displaystyle{ \mathbb{F} }[/math] is that any univariate polynomial of degree [math]\displaystyle{ d }[/math] factors uniquely into at most [math]\displaystyle{ d }[/math] irreducible polynomials, each of which has at most one root.

The following simple randomized algorithm is natural:

Randomized algorithm for univariate PIT

suppose we have a finite subset [math]\displaystyle{ S\subseteq\mathbb{F} }[/math] (to be specified later);
pick [math]\displaystyle{ r\in S }[/math] uniformly at random;
if [math]\displaystyle{ f(r) = 0 }[/math] then return “yes” else return “no”;

This algorithm evaluates [math]\displaystyle{ f }[/math] at one point chosen uniformly at random from a finite subset [math]\displaystyle{ S\subseteq\mathbb{F} }[/math]. It is easy to see the followings:

If [math]\displaystyle{ f\equiv 0 }[/math], the algorithm always returns "yes", so it is always correct.
If [math]\displaystyle{ f\not\equiv 0 }[/math], the algorithm may wrongly return "yes" (a false positive). But this happens only when the random [math]\displaystyle{ r }[/math] is a root of [math]\displaystyle{ f }[/math]. By the fundamental theorem of algebra, [math]\displaystyle{ f }[/math] has at most [math]\displaystyle{ d }[/math] roots, so the probability that the algorithm is wrong is bounded as

[math]\displaystyle{ \Pr[f(r)=0]\le\frac{d}{|S|}. }[/math]

By fixing [math]\displaystyle{ S\subseteq\mathbb{F} }[/math] to be an arbitrary subset of size [math]\displaystyle{ |S|=2d }[/math], this probability of false positive is at most [math]\displaystyle{ 1/2 }[/math]. We can reduce it to an arbitrarily small constant [math]\displaystyle{ \delta }[/math] by repeat the above testing independently for [math]\displaystyle{ \log_2 \frac{1}{\delta} }[/math] times, since the error probability decays geometrically as we repeat the algorithm independently.

Schwartz-Zippel Lemma

Now let's consider the general multivariate case, where [math]\displaystyle{ f\in\mathbb{F}[x_1,x_2,\ldots,x_n] }[/math] is an [math]\displaystyle{ n }[/math]-variate polynomial of degree [math]\displaystyle{ d }[/math].

The following is a simple randomized algorithm for testing identity of multivariate polynomials:

Randomized algorithm for multivariate PIT

suppose we have a finite subset [math]\displaystyle{ S\subseteq\mathbb{F} }[/math] (to be specified later);
pick [math]\displaystyle{ r_1,r_2,\ldots,r_n\in S }[/math] uniformly and independently at random;
if [math]\displaystyle{ f(\vec{r})=f(r_1,r_2,\ldots,r_n) = 0 }[/math] then return “yes” else return “no”;

This algorithm evaluates [math]\displaystyle{ f }[/math] at one point chosen uniformly from an [math]\displaystyle{ n }[/math]-dimensional cube [math]\displaystyle{ S^n }[/math], where [math]\displaystyle{ S\subseteq\mathbb{F} }[/math] is a finite subset. And:

If [math]\displaystyle{ f\equiv 0 }[/math], the algorithm always returns "yes", so it is always correct.
If [math]\displaystyle{ f\not\equiv 0 }[/math], the algorithm may wrongly return "yes" (a false positive). But this happens only when the random [math]\displaystyle{ \vec{r}=(r_1,r_2,\ldots,r_n) }[/math] is a root of [math]\displaystyle{ f }[/math]. The probability of this bad event is upper bounded by the following famous result due to Schwartz (1980) and Zippel (1979).

Schwartz-Zippel Lemma

Let [math]\displaystyle{ f\in\mathbb{F}[x_1,x_2,\ldots,x_n] }[/math] be an [math]\displaystyle{ n }[/math]-variate polynomial of degree [math]\displaystyle{ d }[/math] over a field [math]\displaystyle{ \mathbb{F} }[/math]. If [math]\displaystyle{ f\not\equiv 0 }[/math], then for any finite set [math]\displaystyle{ S\subset\mathbb{F} }[/math], and [math]\displaystyle{ r_1,r_2\ldots,r_n\in S }[/math] chosen uniformly and independently at random,

[math]\displaystyle{ \Pr[f(r_1,r_2,\ldots,r_n)=0]\le\frac{d}{|S|}. }[/math]

The Schwartz-Zippel lemma states that for any nonzero [math]\displaystyle{ n }[/math]-variate polynomial of degree at most [math]\displaystyle{ d }[/math], the number of roots in any cube [math]\displaystyle{ S^n }[/math] is at most [math]\displaystyle{ d\cdot |S|^{n-1} }[/math]. And such algebraic proposition enjoins a probabilistic proof which uses no more than the law of total probability.

A naive thought for proving the Schwartz-Zippel lemma is to express the multivariate polynomial [math]\displaystyle{ f(r_1,r_2,\ldots,r_n)=g_{r_1,r_2,\ldots,r_{n-1}}(r_n) }[/math] as a univariate polynomial [math]\displaystyle{ g_{r_1,r_2,\ldots,r_{n-1}} }[/math] on the [math]\displaystyle{ {n} }[/math]-th variable [math]\displaystyle{ r_{n} }[/math], where [math]\displaystyle{ g_{r_1,\ldots,r_{n-1}}(\cdot)=f(r_1,\ldots,r_{n-1},\cdot) }[/math] is a univariate polynomial "parameterized" by the random values [math]\displaystyle{ r_1,r_2,\ldots,r_{n-1} }[/math].

Then, the argument for the univariate case can apply, and easily we have

[math]\displaystyle{ \Pr[f(r_1,r_2,\ldots,r_n)=0]=\Pr[g_{r_1,r_2,\ldots,r_{n-1}}(r_n)=0]\le\frac{d}{|S|}. }[/math]

A fatal flaw of this "proof" is that when the first [math]\displaystyle{ {n-1} }[/math] variables are assigned with random values [math]\displaystyle{ r_1,r_2,\ldots,r_{n-1} }[/math], the univariate polynomial [math]\displaystyle{ g_{r_1,\ldots,r_{n-1}}(\cdot)=f(r_1,\ldots,r_{n-1},\cdot) }[/math] may no longer always satisfy that [math]\displaystyle{ g_{r_1,\ldots,r_{n-1}}\not\equiv 0 }[/math], which is required by polynomial interpolation.

And this is precisely where the law of total probability kicks in, to separate the case where [math]\displaystyle{ g_{r_1,\ldots,r_{n-1}}\not\equiv 0 }[/math] from the case where [math]\displaystyle{ g_{r_1,\ldots,r_{n-1}}\equiv 0 }[/math].

Proof.

The theorem is proved by induction on [math]\displaystyle{ n }[/math].

Induction basis:

For [math]\displaystyle{ n=1 }[/math], this is the univariate case. Assume that [math]\displaystyle{ f\not\equiv 0 }[/math]. Due to the fundamental theorem of algebra, any polynomial [math]\displaystyle{ f(x) }[/math] of degree at most [math]\displaystyle{ d }[/math] must have at most [math]\displaystyle{ d }[/math] roots, thus

[math]\displaystyle{ \Pr[f(r)=0]\le\frac{d}{|S|}. }[/math]

Induction hypothesis:

Assume the theorem holds for any [math]\displaystyle{ m }[/math]-variate polynomials for all [math]\displaystyle{ m\lt n }[/math].

Induction step:

For any [math]\displaystyle{ n }[/math]-variate polynomial [math]\displaystyle{ f(x_1,x_2,\ldots,x_n) }[/math] of degree at most [math]\displaystyle{ d }[/math], write [math]\displaystyle{ f }[/math] as

[math]\displaystyle{ f(x_1,x_2,\ldots,x_n)=\sum_{i=0}^kx_n^{i}f_i(x_1,x_2,\ldots,x_{n-1}) }[/math],

where [math]\displaystyle{ k }[/math] is the highest degree of [math]\displaystyle{ x_n }[/math]. Note that the degree of [math]\displaystyle{ f_k }[/math] is at most [math]\displaystyle{ d-k }[/math] and [math]\displaystyle{ f_k\not\equiv 0 }[/math] (otherwise [math]\displaystyle{ k }[/math] wouldn't be the highest degree of [math]\displaystyle{ x_n }[/math]).

For short, express [math]\displaystyle{ f }[/math] as:

[math]\displaystyle{ f(x_1,x_2,\ldots,x_n)=x_n^k f_k(x_1,x_2,\ldots,x_{n-1})+\bar{f}(x_1,x_2,\ldots,x_n) }[/math],

where both [math]\displaystyle{ f_k }[/math] and [math]\displaystyle{ \bar{f} }[/math] are polynomials, such that

[math]\displaystyle{ f_k\not\equiv 0 }[/math] and has degree at most [math]\displaystyle{ d-k }[/math];
[math]\displaystyle{ \bar{f}(x_1,x_2,\ldots,x_n)=\sum_{i=0}^{k-1}x_n^i f_i(x_1,x_2,\ldots,x_{n-1}) }[/math], thus [math]\displaystyle{ \bar{f}(x_1,x_2,\ldots,x_n) }[/math] has no [math]\displaystyle{ x_n^{k} }[/math] factor in any term.

By the law of total probability,

[math]\displaystyle{ \begin{align} \Pr[f(r_1,r_2,\ldots,r_n)=0] = &\Pr[f(\vec{r})=0\mid f_k(r_1,r_2,\ldots,r_{n-1})=0]\cdot\Pr[f_k(r_1,r_2,\ldots,r_{n-1})=0]\\ &+\Pr[f(\vec{r})=0\mid f_k(r_1,r_2,\ldots,r_{n-1})\neq0]\cdot\Pr[f_k(r_1,r_2,\ldots,r_{n-1})\neq0]. \end{align} }[/math]

Since [math]\displaystyle{ f_k(r_1,r_2,\ldots,r_{n-1}) }[/math] is an [math]\displaystyle{ (n-1) }[/math]-variate polynomial of degree at most [math]\displaystyle{ d-k }[/math] satisfying that [math]\displaystyle{ f_k\not\equiv 0 }[/math], by the induction hypothesis,

[math]\displaystyle{ \begin{align} (\text{case-I}) &\qquad &\Pr[f_k(r_1,r_2,\ldots,r_{n-1})=0]\le\frac{d-k}{|S|}. \end{align} }[/math]

Now look at the case with condition [math]\displaystyle{ f_k(r_1,r_2,\ldots,r_{n-1})\neq0 }[/math]. Recall that [math]\displaystyle{ \bar{f}(x_1,\ldots,x_n) }[/math] has no [math]\displaystyle{ x_n^k }[/math] factor in any term, thus the condition [math]\displaystyle{ f_k(r_1,r_2,\ldots,r_{n-1})\neq0 }[/math] implies that

[math]\displaystyle{ f(r_1,\ldots,r_{n-1},x_n)=x_n^k f_k(r_1,r_2,\ldots,r_{n-1})+\bar{f}(r_1,r_2,\ldots,r_{n-1},x_n)=g_{r_1,\ldots,r_{n-1}}(x_n) }[/math]

when being interpreted as a univariate polynomial [math]\displaystyle{ g_{r_1,\ldots,r_{n-1}}(x_n) }[/math] on variable [math]\displaystyle{ x_n }[/math], always has that [math]\displaystyle{ g_{r_1,\ldots,r_{n-1}}\not\equiv 0 }[/math] and the degree of [math]\displaystyle{ g_{r_1,\ldots,r_{n-1}}(x_n) }[/math] is at most [math]\displaystyle{ k }[/math], and thus due to the argument for the univariate case, the probability [math]\displaystyle{ g_{r_1,\ldots,r_{n-1}}(r_n)=0 }[/math] is at most [math]\displaystyle{ \frac{k}{|S|} }[/math].

Therefore,

[math]\displaystyle{ \begin{align} (\text{case-II}) &\qquad &\Pr[f(\vec{r})=0\mid f_k(r_1,r_2,\ldots,r_{n-1})\neq0]=\Pr[g_{r_1,\ldots,r_{n-1}}(r_n)=0\mid f_k(r_1,r_2,\ldots,r_{n-1})\neq0]\le\frac{k}{|S|} \end{align} }[/math].

Applying the inequalities of (case-I) and (case-II) in the total probability above, gives

[math]\displaystyle{ \Pr[f(r_1,r_2,\ldots,r_n)=0] \le\frac{d-k}{|S|}+\frac{k}{|S|}=\frac{d}{|S|}. }[/math]

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

@@ Line 1: / Line 1: @@
-=Polynomial Identity Testing (PIT) =
+In the '''Polynomial Identity Testing (PIT)''' problem, we are given two polynomials, and we want to determine whether they are identical.
-The  '''Polynomial Identity Testing (PIT)''' is such a problem: given as input two polynomials, determine whether they are identical. It plays a fundamental role in ''Identity Testing'' problems.
+{{Theorem|Polynomial Identity Testing|
-First, let's consider the univariate ("one variable") case:
-* '''Input:''' two polynomials <math>f, g\in\mathbb{F}[x]</math> of degree <math>d</math>.
-* Determine whether <math>f\equiv g</math> (<math>f</math> and <math>g</math> are identical).
-Here the <math>\mathbb{F}[x]</math> denotes the [http://en.wikipedia.org/wiki/Polynomial_ring ring of univariate polynomials] on a field <math>\mathbb{F}</math>. More precisely, a polynomial <math>f\in\mathbb{F}[x]</math> is
-:<math>f(x)=\sum_{i=0}^\infty a_ix^i</math>,
-where the coefficients <math>a_i</math> are taken from the field <math>\mathbb{F}</math>, and the addition and multiplication are also defined over the field <math>\mathbb{F}</math>.
-And:
-* the '''degree''' of <math>f</math> is the highest <math>i</math> with non-zero <math>a_i</math>;
-* a polynomial <math>f</math> is a '''zero-polynomial''', denoted as <math>f\equiv 0</math>, if all coefficients <math>a_i=0</math>.
-Alternatively, we can consider the following equivalent problem by comparing the polynomial <math>f-g</math> (whose degree is at most <math>d</math>) with the zero-polynomial:
-* '''Input:''' a polynomial <math>f\in\mathbb{F}[x]</math> of degree <math>d</math>.
-* Determine whether <math>f\equiv 0</math> (<math>f</math> is the 0 polynomial).
-The problem is trivial if the input polynomial <math>f</math> is given explicitly: one can trivially solve the problem by checking whether all <math>d+1</math> coefficients are <math>0</math>. To make the problem nontrivial, we assume that the input polynomial is given implicitly as a ''black box'' (also called an ''oracle''): the only way the algorithm can access to <math>f</math> is to evaluate <math>f(x)</math> over some <math>x</math> from the field <math>\mathbb{F}</math>, <math>x</math> chosen by the algorithm.
-A straightforward deterministic algorithm is to evaluate <math>f(x_1),f(x_2),\ldots,f(x_{d+1})</math> over <math>d+1</math> ''distinct'' elements <math>x_1,x_2,\ldots,x_{d+1}</math> from the field <math>\mathbb{F}</math> and check whether they are all zero. By the [https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra fundamental theorem of algebra], also known as polynomial interpolations, this guarantees to verify whether a degree-<math>d</math> univariate polynomial <math>f\equiv 0</math>.
-{{Theorem|Fundamental Theorem of Algebra|
-:Any non-zero univariate polynomial of degree <math>d</math> has at most <math>d</math> roots.
-}}
-The reason for this fundamental theorem holding generally over any field <math>\mathbb{F}</math> is that any univariate polynomial of degree <math>d</math> factors uniquely into at most <math>d</math> irreducible polynomials, each of which has at most one root.
-The following simple randomized algorithm is natural:
-{{Theorem|Algorithm for PIT|
-*suppose we have a finite subset <math>S\subseteq\mathbb{F}</math> (to be specified later);
-*pick <math>r\in S</math> ''uniformly'' at random;
-*if <math>f(r) = 0</math> then return “yes” else return “no”;
-}}
-This algorithm evaluates <math>f</math> at one point chosen uniformly at random from a finite subset <math>S\subseteq\mathbb{F}</math>. It is easy to see the followings:
-* If <math>f\equiv 0</math>, the algorithm always returns "yes", so it is always correct.
-* If <math>f\not\equiv 0</math>, the algorithm may wrongly return "yes" (a '''false positive'''). But this happens only when the random <math>r</math> is a root of <math>f</math>. By the fundamental theorem of algebra, <math>f</math> has at most <math>d</math> roots, so the probability that the algorithm is wrong is bounded as
-::<math>\Pr[f(r)=0]\le\frac{d}{|S|}.</math>
-By fixing <math>S\subseteq\mathbb{F}</math> to be an arbitrary subset of size <math>|S|=2d</math>, this probability of false positive is at most <math>1/2</math>. We can reduce it to an arbitrarily small constant <math>\delta</math> by repeat the above testing ''independently'' for <math>\log_2 \frac{1}{\delta}</math> times, since the error probability decays geometrically as we repeat the algorithm independently.
-== Communication Complexity of Equality ==
-The [http://en.wikipedia.org/wiki/Communication_complexity communication complexity] is introduced by Andrew Chi-Chih Yao as a model of computation with more than one entities, each with partial information about the input.
-Assume that there are two entities, say Alice and Bob. Alice has a private input <math>a</math> and Bob has a private input <math>b</math>. Together they want to compute a function <math>f(a,b)</math> by communicating with each other. The communication follows a predefined '''communication protocol''' (the "algorithm" in this model). The complexity of a communication protocol is measured by the number of bits communicated between Alice and Bob in the worst case.
-The problem of checking identity is formally defined by the function EQ as follows: <math>\mathrm{EQ}:\{0,1\}^n\times\{0,1\}^n\rightarrow\{0,1\}</math> and for any <math>a,b\in\{0,1\}^n</math>,
-:<math>
-\mathrm{EQ}(a,b)=
-\begin{cases}
-& \mbox{if } a=b,\\
-& \mbox{otherwise.}
-\end{cases}</math>
-A trivial way to solve EQ is to let Bob send his entire input string <math>b</math> to Alice and let Alice check whether <math>a=b</math>. This costs <math>n</math> bits of communications.
-It is known that for deterministic communication protocols, this is the best we can get for computing EQ.
-{{Theorem|Theorem (Yao 1979)|
-:Any deterministic communication protocol computing EQ on two <math>n</math>-bit strings costs <math>n</math> bits of communication in the worst-case.
-}}
-This theorem is much more nontrivial to prove than it looks, because Alice and Bob are allowed to interact with each other in arbitrary ways. The proof of this theorem is in Yao's [http://math.ucla.edu/~znorwood/290d.2.14s/papers/yao.pdf  celebrated paper in 1979 with a humble title]. It pioneered the field of communication complexity.
-If we allow randomness in protocols, and also tolerate a small probabilistic error, the problem can be solved with significantly less communications. To present this randomized protocol, we need a few preparations:
-* We represent the inputs  <math>a,b \in\{0,1\}^{n}</math> of Alice and Bob as two univariate polynomials of degree at most <math>n-1</math>, respectively
-::<math>f(x)=\sum_{i=0}^{n-1}a_ix^{i}</math> and <math>g(x)=\sum_{i=0}^{n-1}b_ix^{i}</math>.
-* The two polynomials <math>f</math> and <math>g</math> are defined over finite field <math>\mathbb{Z}_p=\{0,1,\ldots,p-1\}</math> for some suitable prime <math>p</math> (to be specified later), which means the additions and multiplications are modulo <math>p</math>.
-The randomized communication protocol is then as follows:
-{{Theorem|A randomized protocol for EQ|
-'''Bob does''':
-:* pick <math>r\in\mathbb{Z}_p</math> uniformly at random;
-:* send <math>r</math> and <math>g(r)</math> to Alice;
-'''Upon receiving''' <math>r</math> and <math>g(r)</math> '''Alice does''':
-:* compute <math>f(r)</math>;
-:* If <math>f(r)= g(r)</math> return "'''yes'''"; else return "'''no'''".
-}}
-The communication complexity of the protocol is given by the number of bits used to represent the values of <math>r</math> and <math>g(r)</math>.
-Since the polynomials are defined over finite field <math>\mathbb{Z}_p</math> and the random number <math>f</math> is also chosen from <math>\mathbb{Z}_p</math>, this is bounded by <math>O(\log p)</math>.
-On the other hand the protocol makes mistakes only when <math>a\neq b</math> but wrongly answers "yes". This happens only when <math>f\not\equiv g</math> but <math>f(r)=g(r)</math>. The degrees of <math>f, g</math> are at most <math>n-1</math>, and <math>r</math> is chosen among <math>p</math> distinct values, we have
-:<math>\Pr[f(r)=g(r)]\le \frac{n-1}{p}</math>.
-By choosing <math>p</math> to be a prime in the interval <math>[n^2, 2n^2]</math> (by [https://en.wikipedia.org/wiki/Bertrand%27s_postulate Chebyshev's theorem], such prime <math>p</math> always exists), the above randomized communication protocol solves the Equality function EQ with an error probability of false positive at most <math>O(1/n)</math>, with communication complexity <math>O(\log n)</math>, an EXPONENTIAL improvement to ANY deterministic communication protocol!
-== Schwartz-Zippel Theorem ==
-Now let's see the the true form of '''Polynomial Identity Testing (PIT)''', for multivariate polynomials:
 * '''Input:''' two <math>n</math>-variate polynomials <math>f, g\in\mathbb{F}[x_1,x_2,\ldots,x_n]</math> of degree <math>d</math>.
 * Determine whether <math>f\equiv g</math>.
-The <math>\mathbb{F}[x_1,x_2,\ldots,x_n]</math> is the [http://en.wikipedia.org/wiki/Polynomial_ring#The_polynomial_ring_in_several_variables ring of multivariate polynomials] over field <math>\mathbb{F}</math>. An <math>n</math>-variate polynomial of degree <math>d</math>, written as a sum of monomials, is:
+}}
+The <math>\mathbb{F}[x_1,x_2,\ldots,x_n]</math> denotes the [http://en.wikipedia.org/wiki/Polynomial_ring#The_polynomial_ring_in_several_variables ring of multivariate polynomials] over field <math>\mathbb{F}</math>. An <math>n</math>-variate polynomial of degree <math>d</math>, written as a sum of monomials, is:
 :<math>f(x_1,x_2,\ldots,x_n)=\sum_{i_1,i_2,\ldots,i_n\ge 0\atop i_1+i_2+\cdots+i_n\le d}a_{i_1,i_2,\ldots,i_n}x_{1}^{i_1}x_2^{i_2}\cdots x_{n}^{i_n}</math>.
 The '''degree''' or '''total degree''' of a monomial <math>a_{i_1,i_2,\ldots,i_n}x_{1}^{i_1}x_2^{i_2}\cdots x_{n}^{i_n}</math> is given by <math>i_1+i_2+\cdots+i_n</math> and the degree of a polynomial <math>f</math> is the maximum degree of monomials of nonzero coefficients.
-As before, we also consider the following equivalent problem:
+This problem is equivalent to the following problem of checking whether <math>f-g</math> (which is a polynomial of degree at most <math>d</math>) is the zero-polynomial.
+{{Theorem|Polynomial Identity Testing (equivalent version)|
 * '''Input:''' a polynomial <math>f\in\mathbb{F}[x_1,x_2,\ldots,x_n]</math> of degree <math>d</math>.
 * Determine whether <math>f\equiv 0</math>.
+}}
-If <math>f</math> is written explicitly as a sum of monomials, then the problem can be solved by checking whether all coefficients, and there at most <math>{n+d\choose d}\le (n+d)^{d}</math> coefficients in an <math>n</math>-variate polynomial of degree at most <math>d</math>.
+The problem seems to be solvable by checking whether all coefficients of monomials in <math>f</math> are zeros. And there are at most <math>(n+d)^{d}</math> coefficients since <math>f</math> is an <math>n</math>-variate polynomial of degree at most <math>d</math>. However, this relies on that all coefficients of <math>f</math> are explicitly given, which may not always be the case.
-A multivariate polynomial <math>f</math> can also be presented in its '''product form''', for example:
+For example, a multivariate polynomial <math>f</math> may be expressed in a '''product form'''.
 {{Theorem|Example|
 The [http://en.wikipedia.org/wiki/Vandermonde_matrix Vandermonde matrix] <math>M=M(x_1,x_2,\ldots,x_n)</math> is defined as that <math>M_{ij}=x_i^{j-1}</math>, that is
@@ Line 111: / Line 31: @@
 </math>
 }}
-For polynomials in product form, it is quite efficient to '''evaluate''' the polynomial at any specific point from the field over which the polynomial is defined, however, '''expanding''' the polynomial to a sum of monomials can be very expensive.
+For polynomials  <math>f</math> in product form, it is quite possible that '''evaluating''' the polynomial  <math>f(x_1,\ldots,x_n)</math> at any <math>(x_1,\ldots,x_n)\in\mathbb{F}^n</math> is always efficient but symbolically '''expanding''' the polynomial to calculate the coefficients may be very expensive.
+We assume that the input polynomial <math>f</math> is given as a '''block-box''' or in '''product form''', so that the evaluation of <math>f(x_1,\ldots,x_n)</math> at any specific <math>(x_1,\ldots,x_n)\in\mathbb{F}^n</math> is efficient.
+== Polynomial Interpolation ==
+First, let's consider the '''univariate''' case where <math>n=1</math>. In such case <math>f(x)</math> is a univariate polynomial of degree at most <math>d</math>.
+A straightforward deterministic algorithm is to evaluate <math>f(x_1),f(x_2),\ldots,f(x_{d+1})</math> over <math>d+1</math> ''distinct'' elements <math>x_1,x_2,\ldots,x_{d+1}</math> from the field <math>\mathbb{F}</math> and check whether they are all zero. By the [https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra fundamental theorem of algebra], also known as '''polynomial interpolation''', this guarantees to answer correctly whether a degree-<math>d</math> univariate polynomial <math>f\equiv 0</math>.
+{{Theorem|Fundamental Theorem of Algebra|
+:Any non-zero univariate polynomial of degree <math>d</math> has at most <math>d</math> roots.
+}}
+The reason for this fundamental theorem holding generally over any field <math>\mathbb{F}</math> is that any univariate polynomial of degree <math>d</math> factors uniquely into at most <math>d</math> irreducible polynomials, each of which has at most one root.
+The following simple randomized algorithm is natural:
+{{Theorem|Randomized algorithm for univariate PIT|
+*suppose we have a finite subset <math>S\subseteq\mathbb{F}</math> (to be specified later);
+*pick <math>r\in S</math> ''uniformly'' at random;
+*if <math>f(r) = 0</math> then return “yes” else return “no”;
+}}
+This algorithm evaluates <math>f</math> at one point chosen uniformly at random from a finite subset <math>S\subseteq\mathbb{F}</math>. It is easy to see the followings:
+* If <math>f\equiv 0</math>, the algorithm always returns "yes", so it is always correct.
+* If <math>f\not\equiv 0</math>, the algorithm may wrongly return "yes" (a '''false positive'''). But this happens only when the random <math>r</math> is a root of <math>f</math>. By the fundamental theorem of algebra, <math>f</math> has at most <math>d</math> roots, so the probability that the algorithm is wrong is bounded as
+::<math>\Pr[f(r)=0]\le\frac{d}{|S|}.</math>
+By fixing <math>S\subseteq\mathbb{F}</math> to be an arbitrary subset of size <math>|S|=2d</math>, this probability of false positive is at most <math>1/2</math>. We can reduce it to an arbitrarily small constant <math>\delta</math> by repeat the above testing ''independently'' for <math>\log_2 \frac{1}{\delta}</math> times, since the error probability decays geometrically as we repeat the algorithm independently.
+== Schwartz-Zippel Lemma ==
+Now let's consider the general multivariate case, where <math>f\in\mathbb{F}[x_1,x_2,\ldots,x_n]</math> is an <math>n</math>-variate polynomial of degree <math>d</math>.
 The following is a simple randomized algorithm for testing identity of multivariate polynomials:
@@ Line 124: / Line 72: @@
 * If <math>f\not\equiv 0</math>, the algorithm may wrongly return "yes" (a '''false positive'''). But this happens only when the random <math>\vec{r}=(r_1,r_2,\ldots,r_n)</math> is a root of <math>f</math>. The probability of this bad event is upper bounded by the following famous result due to [https://pdfs.semanticscholar.org/b913/cf330852035f49b4ec5fe2db86c47d8a98fd.pdf Schwartz (1980)] and [http://www.cecm.sfu.ca/~monaganm/teaching/TopicsinCA15/zippel79.pdf Zippel (1979)].
-{{Theorem|Schwartz-Zippel Theorem|
+{{Theorem|Schwartz-Zippel Lemma|
-: Let <math>f\in\mathbb{F}[x_1,x_2,\ldots,x_n]</math> be a multivariate polynomial of degree <math>d</math> over a field <math>\mathbb{F}</math>. If <math>f\not\equiv 0</math>, then for any finite set <math>S\subset\mathbb{F}</math>, and <math>r_1,r_2\ldots,r_n\in S</math> chosen uniformly and independently at random,
+: Let <math>f\in\mathbb{F}[x_1,x_2,\ldots,x_n]</math> be an <math>n</math>-variate polynomial of degree <math>d</math> over a field <math>\mathbb{F}</math>. If <math>f\not\equiv 0</math>, then for any finite set <math>S\subset\mathbb{F}</math>, and <math>r_1,r_2\ldots,r_n\in S</math> chosen uniformly and independently at random,
 ::<math>\Pr[f(r_1,r_2,\ldots,r_n)=0]\le\frac{d}{|S|}.</math>
 }}
-The Schwartz-Zippel Theorem states that for any nonzero <math>n</math>-variate polynomial of degree at most <math>d</math>, the number of roots in any cube <math>S^n</math> is at most <math>d\cdot |S|^{n-1}</math>.
+The Schwartz-Zippel lemma states that for any nonzero <math>n</math>-variate polynomial of degree at most <math>d</math>, the number of roots in any cube <math>S^n</math> is at most <math>d\cdot |S|^{n-1}</math>. And such algebraic proposition enjoins a probabilistic proof which uses no more than the '''law of total probability'''.
+A naive thought for proving the Schwartz-Zippel lemma is to express the multivariate polynomial <math>f(r_1,r_2,\ldots,r_n)=g_{r_1,r_2,\ldots,r_{n-1}}(r_n)</math> as a univariate polynomial <math>g_{r_1,r_2,\ldots,r_{n-1}}</math> on the <math>{n}</math>-th variable <math>r_{n}</math>, where <math>g_{r_1,\ldots,r_{n-1}}(\cdot)=f(r_1,\ldots,r_{n-1},\cdot)</math> is a univariate polynomial "parameterized" by the random values <math>r_1,r_2,\ldots,r_{n-1}</math>.
+Then, the argument for the univariate case can apply, and easily we have
+:<math>\Pr[f(r_1,r_2,\ldots,r_n)=0]=\Pr[g_{r_1,r_2,\ldots,r_{n-1}}(r_n)=0]\le\frac{d}{|S|}.</math>
+A fatal flaw of this "proof" is that when the first <math>{n-1}</math> variables are assigned with random values <math>r_1,r_2,\ldots,r_{n-1}</math>, the univariate polynomial <math>g_{r_1,\ldots,r_{n-1}}(\cdot)=f(r_1,\ldots,r_{n-1},\cdot)</math> may no longer always satisfy that <math>g_{r_1,\ldots,r_{n-1}}\not\equiv 0</math>, which is required by polynomial interpolation.
-Dana Moshkovitz gave a surprisingly simply and elegant [http://eccc.hpi-web.de/report/2010/096/ proof] of Schwartz-Zippel Theorem, using some advanced ideas. Now we introduce the standard proof by induction.
+And this is precisely where the law of total probability kicks in, to separate the case where <math>g_{r_1,\ldots,r_{n-1}}\not\equiv 0</math> from the case where <math>g_{r_1,\ldots,r_{n-1}}\equiv 0</math>.
 {{Proof|
@@ Line 144: / Line 99: @@
 ;Induction step''':'''
-For any <math>n</math>-variate polynomial <math>f(x_1,x_2,\ldots,x_n)</math>  of degree at most <math>d</math>, we write <math>f</math> as
+For any <math>n</math>-variate polynomial <math>f(x_1,x_2,\ldots,x_n)</math>  of degree at most <math>d</math>, write <math>f</math> as
 :<math>f(x_1,x_2,\ldots,x_n)=\sum_{i=0}^kx_n^{i}f_i(x_1,x_2,\ldots,x_{n-1})</math>,
-where <math>k</math> is the highest degree of <math>x_n</math>, which means the degree of <math>f_k</math> is at most <math>d-k</math> and <math>f_k\not\equiv 0</math>.
+where <math>k</math> is the highest degree of <math>x_n</math>. Note that the degree of <math>f_k</math> is at most <math>d-k</math> and <math>f_k\not\equiv 0</math> (otherwise <math>k</math> wouldn't be the highest degree of <math>x_n</math>).
-In particular, we write <math>f</math> as a sum of two parts:
+For short, express <math>f</math> as:
 :<math>f(x_1,x_2,\ldots,x_n)=x_n^k f_k(x_1,x_2,\ldots,x_{n-1})+\bar{f}(x_1,x_2,\ldots,x_n)</math>,
 where both <math>f_k</math> and <math>\bar{f}</math> are polynomials, such that
-* <math>f_k\not\equiv 0</math> is as above, whose degree is  at most <math>d-k</math>;
+* <math>f_k\not\equiv 0</math> and has degree at most <math>d-k</math>;
 * <math>\bar{f}(x_1,x_2,\ldots,x_n)=\sum_{i=0}^{k-1}x_n^i f_i(x_1,x_2,\ldots,x_{n-1})</math>, thus <math>\bar{f}(x_1,x_2,\ldots,x_n)</math> has no <math>x_n^{k}</math> factor in any term.
-By the law of total probability, we have
+By the '''law of total probability''',
 :<math>
 \begin{align}
-&\Pr[f(r_1,r_2,\ldots,r_n)=0]\\
+\Pr[f(r_1,r_2,\ldots,r_n)=0]
 =
 &\Pr[f(\vec{r})=0\mid f_k(r_1,r_2,\ldots,r_{n-1})=0]\cdot\Pr[f_k(r_1,r_2,\ldots,r_{n-1})=0]\\
@@ Line 163: / Line 118: @@
 \end{align}
 </math>
-Note that <math>f_k(r_1,r_2,\ldots,r_{n-1})</math> is a polynomial on <math>n-1</math> variables of degree <math>d-k</math> such that <math>f_k\not\equiv 0</math>.
+Since <math>f_k(r_1,r_2,\ldots,r_{n-1})</math> is an <math>(n-1)</math>-variate polynomial of degree at most <math>d-k</math> satisfying that <math>f_k\not\equiv 0</math>,
-By the induction hypothesis, we have
+by the induction hypothesis,
 :<math>
 \begin{align}
-(*)
+(\text{case-I})
 &\qquad
 &\Pr[f_k(r_1,r_2,\ldots,r_{n-1})=0]\le\frac{d-k}{|S|}.
@@ Line 173: / Line 128: @@
 </math>
-Now we look at the case conditioning on <math>f_k(r_1,r_2,\ldots,r_{n-1})\neq0</math>. Recall that <math>\bar{f}(x_1,\ldots,x_n)</math> has no <math>x_n^k</math> factor in any term, thus the condition <math>f_k(r_1,r_2,\ldots,r_{n-1})\neq0</math> guarantees that
+Now look at the case with condition <math>f_k(r_1,r_2,\ldots,r_{n-1})\neq0</math>.
+Recall that <math>\bar{f}(x_1,\ldots,x_n)</math> has no <math>x_n^k</math> factor in any term, thus the condition <math>f_k(r_1,r_2,\ldots,r_{n-1})\neq0</math> implies that
 :<math>f(r_1,\ldots,r_{n-1},x_n)=x_n^k f_k(r_1,r_2,\ldots,r_{n-1})+\bar{f}(r_1,r_2,\ldots,r_{n-1},x_n)=g_{r_1,\ldots,r_{n-1}}(x_n)</math>
-is a nonzero univariate polynomial of <math>x_n</math> such that the degree of <math>g_{r_1,\ldots,r_{n-1}}(x_n)</math> is <math>k</math> and <math>g_{r_1,\ldots,r_{n-1}}\not\equiv 0</math>, for which we already known that the probability <math>g_{r_1,\ldots,r_{n-1}}(r_n)=0</math> is at most <math>\frac{k}{|S|}</math>.
+when being interpreted as a univariate polynomial <math>g_{r_1,\ldots,r_{n-1}}(x_n)</math> on variable <math>x_n</math>, always has that <math>g_{r_1,\ldots,r_{n-1}}\not\equiv 0</math> and the degree of <math>g_{r_1,\ldots,r_{n-1}}(x_n)</math> is at most <math>k</math>,
+and thus due to the argument for the univariate case, the probability <math>g_{r_1,\ldots,r_{n-1}}(r_n)=0</math> is at most <math>\frac{k}{|S|}</math>.
 Therefore,
 :<math>
 \begin{align}
-(**)
+(\text{case-II})
 &\qquad
 &\Pr[f(\vec{r})=0\mid f_k(r_1,r_2,\ldots,r_{n-1})\neq0]=\Pr[g_{r_1,\ldots,r_{n-1}}(r_n)=0\mid f_k(r_1,r_2,\ldots,r_{n-1})\neq0]\le\frac{k}{|S|}
 \end{align}
 </math>.
-Substituting both <math>(*)</math> and <math>(**)</math> back in the total probability, we have
+Applying the inequalities of (case-I) and (case-II) in the total probability above, gives
 :<math>
 \Pr[f(r_1,r_2,\ldots,r_n)=0]
-\le\frac{d-k}{|S|}+\frac{k}{|S|}=\frac{d}{|S|},
+\le\frac{d-k}{|S|}+\frac{k}{|S|}=\frac{d}{|S|}.
-</math>
-which proves the theorem.
-}}
-= Fingerprinting =
-The polynomial identity testing algorithm in the Schwartz-Zippel theorem can be abstracted as the following framework:
-Suppose we want to compare two objects <math>X</math> and <math>Y</math>. Instead of comparing them directly, we compute random '''fingerprints''' <math>\mathrm{FING}(X)</math> and <math>\mathrm{FING}(Y)</math> of them and compare the fingerprints.
-The fingerprints has the following properties:
-* <math>\mathrm{FING}(\cdot)</math> is a function, meaning that if <math>X= Y</math> then <math>\mathrm{FING}(X)=\mathrm{FING}(Y)</math>.
-* It is much easier to compute and compare the fingerprints.
-* Ideally, the domain of fingerprints is much smaller than the domain of original objects, so storing and comparing fingerprints are easy. This means the fingerprint function <math>\mathrm{FING}(\cdot)</math> cannot be an injection (one-to-one mapping), so it's possible that different <math>X</math> and <math>Y</math> are mapped to the same fingerprint. We resolve this by making fingerprint function ''randomized'', and for <math>X\neq Y</math>, we want the probability <math>\Pr[\mathrm{FING}(X)=\mathrm{FING}(Y)]</math> to be small.
-In Schwartz-Zippel theorem, the objects to compare are polynomials from <math>\mathbb{F}[x_1,\ldots,x_n]</math>. Given a polynomial <math>f\in \mathbb{F}[x_1,\ldots,x_n]</math>, its fingerprint is computed as <math>\mathrm{FING}(f)=f(r_1,\ldots,r_n)</math> for <math>r_i</math> chosen independently and uniformly at random from some fixed set <math>S\subseteq\mathbb{F}</math>.
-With this generic framework, for various identity testing problems, we may design different fingerprints <math>\mathrm{FING}(\cdot)</math>.
-== Communication protocols for Equality by fingerprinting==
-Now consider again the communication model where the two players Alice with a private input <math>x\in\{0,1\}^n</math> and Bob with a private input <math>y\in\{0,1\}^n</math> together compute a function <math>f(x,y)</math> by running a communication protocol.
-We still consider the communication protocols for the equality function EQ
-:<math>
-\mathrm{EQ}(x,y)=
-\begin{cases}
-& \mbox{if } x=y,\\
-& \mbox{otherwise.}
-\end{cases}</math>
-With the language of fingerprinting, this communication problem can be solved by the following generic scheme:
-{{Theorem|Communication protocol for EQ by fingerprinting|
-'''Bob does''':
-:* choose a random fingerprint function <math>\mathrm{FING}(\cdot)</math> and compute the fingerprint of her input <math>\mathrm{FING}(y)</math>;
-:* sends both the description of <math>\mathrm{FING}(\cdot)</math> and the value of <math>\mathrm{FING}(y)</math> to Alice;
-'''Upon receiving''' the description of <math>\mathrm{FING}(\cdot)</math> and the value of <math>\mathrm{FING}(y)</math>, '''Alice does''':
-:* computes <math>\mathrm{FING}(x)</math> and check whether <math>\mathrm{FING}(x)=\mathrm{FING}(y)</math>.
-}}
-In this way we have a randomized communication protocol for the equality function EQ with false positive. The communication cost as well as the error probability are reduced to the question of how to design this random fingerprint function <math>\mathrm{FING}(\cdot)</math> to guarantee:
-# A random fingerprint function <math>\mathrm{FING}(\cdot)</math> can be described succinctly.
-# The range of <math>\mathrm{FING}(\cdot)</math> is small, so the fingerprints are succinct.
-# If <math>x\neq y</math>, the probability <math>\Pr[\mathrm{FING}(x)=\mathrm{FING}(y)]</math> is small.
-=== Fingerprinting by PIT===
-As before, we can define the fingerprint function as: for any bit-string <math>x\in\{0,1\}^n</math>,  its random fingerprint is <math>\mathrm{FING}(x)=\sum_{i=1}^n x_i r^{i}</math>, where the additions and multiplications are defined over a finite field <math>\mathbb{Z}_p</math>, and <math>r</math> is chosen uniformly at random from <math>\mathbb{Z}_p</math>, where <math>p</math> is some suitable prime which can be represented in <math>\Theta(\log n)</math> bits. More specifically, we can choose <math>p</math> to be any prime from the interval <math>[n^2, 2n^2]</math>. Due to Chebyshev's theorem, such prime must exist.
-As we have shown before, it takes <math>O(\log p)=O(\log n)</math> bits to represent <math>\mathrm{FING}(y)</math> and to describe the random function <math>\mathrm{FING}(\cdot)</math> (since it a random function <math>\mathrm{FING}(\cdot)</math> from this family is uniquely identified by a random <math>r\in\mathbb{Z}_p</math>, which can be represented within <math>\log p=O(\log n)</math> bits). And it follows easily from the fundamental theorem of algebra that for any distinct <math>x, y\in\{0,1\}^n</math>,
-:<math>
-\Pr[\mathrm{FING}(x)=\mathrm{FING}(y)] \le \frac{n-1}{p}\le \frac{1}{n}.
-</math>
-=== Fingerprinting by randomized checksum===
-Now we consider a new fingerprint function: We treat each input string <math>x\in\{0,1\}^n</math> as the binary representation of a number, and let <math>\mathrm{FING}(x)=x\bmod p</math> for some random prime <math>p</math> chosen from <math>[k]=\{0,1,\ldots,k-1\}</math>, for some <math>k</math> to be specified later.
-Now a random fingerprint function <math>\mathrm{FING}(\cdot)</math> can be uniquely identified by this random prime <math>p</math>. The new communication protocol for EQ with this fingerprint is as follows:
-{{Theorem|Communication protocol for EQ by random checksum|
-'''Bob does''':
-:for some parameter <math>k</math> (to be specified),
-:* choose a prime <math>p\in[k]</math> uniformly at random;
-:* send <math>p</math> and <math>x\bmod p</math> to Alice;
-'''Upon receiving''' <math>p</math> and <math>x\bmod p</math>, '''Alice does''':
-:* check whether <math>x\bmod p=y\bmod p</math>.
-}}
-The number of bits to be communicated is obviously <math>O(\log k)</math>. When <math>x\neq y</math>, we want to upper bound the error probability <math>\Pr[x\bmod p=y\bmod p]</math>.
-Suppose without loss of generality <math>x>y</math>. Let <math>z=x-y</math>. Then <math>z<2^n</math> since <math>x,y\in[2^n]</math>, and
-<math>z\neq 0</math> for <math>x\neq y</math>.  It holds that <math>x\equiv y\pmod p</math> if and only if <math>p\mid z</math>. Therefore, we only need to upper bound the probability
-:<math>\Pr[z\bmod p=0]</math>
-for an arbitrarily fixed <math>0<z<2^n</math>, and a uniform random prime <math>p\in[k]</math>.
-The probability <math>\Pr[z\bmod p=0]</math> is computed directly as
-:<math>\Pr[z\bmod p=0]\le\frac{\mbox{the number of prime divisors of }z}{\mbox{the number of primes in }[k]}</math>.
-For the numerator, any positive <math>z<2^n</math> has at most <math>n</math> prime factors. To see this, by contradiction assume that <math>z</math> has more than <math>n</math> prime factors. Note that any prime number is at least 2. Then <math>z</math> must be greater than <math>2^n</math>, contradicting the fact that <math>z<2^n</math>.
-For the denominator, we need to lower bound the number of primes in <math>[k]</math>. This is given by the celebrated [http://en.wikipedia.org/wiki/Prime_number_theorem '''Prime Number Theorem (PNT)'''].
-{{Theorem|Prime Number Theorem|
-:Let <math>\pi(k)</math> denote the number of primes less than <math>k</math>. Then <math>\pi(k)\sim\frac{k}{\ln k}</math> as <math>k\rightarrow\infty</math>.
-}}
-Therefore, by choosing <math>k=2n^2\ln n</math>, we have that for a <math>0<z<2^n</math>, and a random prime <math>p\in[k]</math>,
-:<math>\Pr[z\bmod p=0]\le\frac{n}{\pi(k)}\sim\frac{1}{n}</math>,
-which means the for any inputs <math>x,y\in\{0,1\}^n</math>, if <math>x\neq y</math>, then the false positive is bounded as
-:<math>\Pr[\mathrm{FING}(x)=\mathrm{FING}(y)]\le\Pr[|x-y|\bmod p=0]\le \frac{1}{n}</math>.
-Moreover, by this choice of parameter <math>k=2n^2\ln n</math>, the communication complexity of the protocol is bounded by <math>O(\log k)=O(\log n)</math>.
-= Checking distinctness =
-Consider the following problem of '''checking distinctness''':
-*Given a sequence <math>x_1,x_2,\ldots,x_n\in\{1,2,\ldots,n\}</math>, check whether every element of <math>\{1,2,\ldots,n\}</math> appears '''exactly''' once.
-Obviously this problem can be solved in linear time and linear space (in addition to the space for storing the input) by maintaining a <math>n</math>-bit vector that indicates which numbers among <math>\{1,2,\ldots,n\}</math> have appeared.
-When this <math>n</math> is enormously large, <math>\Omega(n)</math> space cost is too expensive. We wonder whether we could solve this problem with a space cost (in addition to the space for storing the input) much less than <math>O(n)</math>. This can be done by fingerprinting if we tolerate a certain degree of inaccuracy.
-== Fingerprinting multisets ==
-We consider the following more generalized problem, '''checking identity of multisets''':
-* '''Input:''' two multisets <math>A=\{a_1,a_2,\ldots, a_n\}</math> and <math>B=\{b_1,b_2,\ldots, b_n\}</math> where <math>a_1,a_2,\ldots,b_1,b_2,\ldots,b_n\in \{1,2,\ldots,n\}</math>.
-* Determine whether <math>A=B</math> (multiset equivalence).
-Here for a '''multiset''' <math>A=\{a_1,a_2,\ldots, a_n\}</math>, its elements <math>a_i</math> are not necessarily distinct. The '''multiplicity''' of an element <math>a_i</math> in a multiset <math>A</math> is the number of times <math>a_i</math> appears in <math>A</math>. Two multisets <math>A</math> and <math>B</math> are equivalent if they contain the same set of elements and the multiplicities of every element in <math>A</math> and <math>B</math> are equal.
-Obviously the above problem of checking distinctness can be treated as a special case of checking identity of multisets: by checking the identity of the multiset <math>A</math> and set <math>\{1,2,\ldots, n\}</math>.
-The following fingerprinting function for multisets was introduced by Lipton for solving multiset identity testing.
-{{Theorem|Fingerprint for multiset|
-:Let <math>p</math> be a uniform random prime chosen from the interval <math>[(n\log n)^2,2(n\log n)^2]</math>. By Chebyshev's theorem, such prime must exist. And consider the the finite field <math>\mathbb{Z}_p=[p]</math>.
-:Given a multiset <math>A=\{a_1,a_2,\ldots,a_n\}</math>, we define a univariate polynomial <math>f_A\in\mathbb{Z}_p[x]</math> over the finite field <math>\mathbb{Z}_p</math> as follows:
-::<math>f_A(x)=\prod_{i=1}^n(x-a_i)</math>,
-:where <math>+</math> and <math>\cdot</math> are defined over the finite field <math>\mathbb{Z}_p</math>.
-:We then define the random fingerprinting function as:
-::<math>\mathrm{FING}(A)=f_A(r)=\prod_{i=1}^n(r-a_i)</math>,
-:where <math>r</math> is chosen uniformly at random from <math>\mathbb{Z}_p</math>.
-}}
-Since all computations of <math>\mathrm{FING}(A)=\prod_{i=1}^n(r-a_i)</math> are over the finite field <math>\mathbb{Z}_p</math>, the space cost for computing the fingerprint <math>\mathrm{FING}(A)</math> is only <math>O(\log p)=O(\log n)</math>.
-Moreover, the fingerprinting function <math>\mathrm{FING}(A)</math> is invariant under permutation of elements of the multiset <math>A=\{a_1,a_2,\ldots,a_n\}</math>, thus it is indeed a function of multisets (meaning every multiset has only one fingerprint). Therefore, if <math>A=B</math> then <math>\mathrm{FING}(A)=\mathrm{FING}(B)</math>.
-For two distinct multisets <math>A\neq B</math>, it is possible that <math>\mathrm{FING}(A)=\mathrm{FING}(B)</math>, but the following theorem due to Lipton bounds this error probability of false positive.
-{{Theorem|Theorem (Lipton 1989)|
-:Let <math>A=\{a_1,a_2,\ldots,a_n\}</math> and <math>B=\{b_1,b_2,\ldots,b_n\}</math> be two multisets whose elements are from <math>\{1,2,\ldots,n\}</math>. If <math>A\neq B</math>, then
-::<math>\Pr[\mathrm{FING}(A)= \mathrm{FING}(B)]=O\left(\frac{1}{n}\right)</math>.
-}}
-{{Proof|
-Let <math>\tilde{f}_A(x)=\prod_{i=1}^n(x-a_i)</math> and <math>\tilde{f}_B(x)=\prod_{i=1}^n(x-b_i)</math> be two univariate polynomials defined over reals <math>\mathbb{R}</math>. Note that in contrast to <math>f_A(x)</math> and <math>f_B(x)</math>,  the <math>+</math> and <math>\cdot</math> in <math>\tilde{f}_A(x), \tilde{f}_B(x)</math> do not modulo <math>p</math>. It is easy to verify that the polynomials <math>\tilde{f}_A(x), \tilde{f}_B(x)</math> have the following properties:
-*<math>\tilde{f}_A\equiv \tilde{f}_B</math> if and only if <math>A=B</math>. Here <math>A=B</math> means the multiset equivalence.
-*By the properties of finite field, for any value <math>r\in\mathbb{Z}_p</math>, it holds that <math>f_A(r)=\tilde{f}_A(r)\bmod p</math> and <math>f_B(r)=\tilde{f}_B(r)\bmod p</math>.
-Therefore, assuming that <math>A\neq B</math>, we must have <math>\tilde{f}_A(x)\not\equiv \tilde{f}_B(x)</math>. Then by the law of total probability:
-:<math>
-\begin{align}
-\Pr[\mathrm{FING}(A)= \mathrm{FING}(B)]
-&=
-\Pr\left[f_A(r)=f_B(r)\mid f_A\not\equiv f_B\right]\Pr[f_A\not\equiv f_B]\\
-&\quad\,\,+\Pr\left[f_A(r)=f_B(r)\mid f_A\equiv f_B\right]\Pr[f_A\equiv f_B]\\
-&\le \Pr\left[f_A(r)=f_B(r)\mid f_A\not\equiv f_B\right]+\Pr[f_A\equiv f_B].
-\end{align}
-</math>
-Note that the degrees of <math>f_A,f_B</math> are at most <math>n</math> and <math>r</math> is chosen uniformly from <math>[p]</math>. By the Schwartz-Zippel theorem for univariate polynomials, the first probability
-:<math>
-\Pr\left[f_A(r)=f_B(r)\mid f_A\not\equiv f_B\right]\le \frac{n}{p}=o\left(\frac{1}{n}\right),
-</math>
-since <math>p</math> is chosen from the interval <math>[(n\log n)^2,2(n\log n)^2]</math>.
-For the second probability <math>\Pr[f_A\equiv f_B]</math>, recall that <math>\tilde{f}_A\not\equiv \tilde{f}_B</math>, therefore there is at least a non-zero coefficient <math>c\le n^n</math> in <math>\tilde{f}_A-\tilde{f}_B</math>. The event <math>f_A\equiv f_B</math> occurs only if <math>c\bmod p=0</math>, which means
-:<math>
-\begin{align}
-\Pr[f_A\equiv f_B]
-&\le \Pr[c\bmod p=0]\\
-&=\frac{\text{number of prime factors of }c}{\text{number of primes in }[(n\log n)^2,2(n\log n)^2]}\\
-&\le \frac{n\log_2n}{\pi(2(n\log n)^2)-\pi((n\log n)^2)}.
-\end{align}
-</math>
-By the prime number theorem, <math>\pi(N)\rightarrow \frac{N}{\ln N}</math> as <math>N\to\infty</math>. Therefore,
-:<math>
-\Pr[f_A\equiv f_B]=O\left(\frac{n\log n}{n^2\log n}\right)=O\left(\frac{1}{n}\right).
 </math>
-Combining everything together, we have
-<math>\Pr[\mathrm{FING}(A)= \mathrm{FING}(B)]=O\left(\frac{1}{n}\right)</math>.
 }}

概率论与数理统计 (Spring 2023)/Polynomial identity testing: Difference between revisions

Latest revision as of 03:05, 12 March 2023

Polynomial Interpolation

Schwartz-Zippel Lemma

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