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| =Introduction= | | == Problem 1== |
| This course will study ''Randomized Algorithms'', the algorithms that use randomness in computation.
| | Prove the following identity: |
| ;Why do we use randomness in computation?
| | *<math>\sum_{k=1}^n k{n\choose k}= n2^{n-1}</math>. |
| * Randomized algorithms can be simpler than deterministic ones.
| |
| :(median selection, load balancing, etc.) | |
| * Randomized algorithms can be faster than the best known deterministic algorithms. | |
| :(min-cut, checking matrix multiplication, primality testing, etc.)
| |
| * Randomized algorithms can do things that deterministic algorithms cannot do.
| |
| :(routing, volume estimation, communication complexity, data streams, etc.)
| |
| * Randomized algorithms may lead us to smart deterministic algorithms.
| |
| :(hashing, derandomization, SL=L, Lovász Local Lemma, etc.)
| |
| * Randomness is presented in the input.
| |
| :(average-case analysis, smoothed analysis, learning, etc.)
| |
| * Some deterministic problems are random in nature.
| |
| :(counting, inference, etc.)
| |
| * ...
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|
| |
|
| ;How is randomness used in computation?
| | (Hint: Use double counting.) |
| * To hit a witness/certificate.
| |
| :(identity testing, fingerprinting, primality testing, etc.)
| |
| * To avoid worst case or to deal with adversaries.
| |
| :(randomized quick sort, perfect hashing, etc.)
| |
| * To simulate random samples.
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| :(random walk, Markov chain Monte Carlo, approximate counting etc.) | |
| * To enumerate/construct solutions.
| |
| :(the probabilistic method, min-cut, etc.)
| |
| * To break symmetry.
| |
| :(mutual exclusion, leader election, parrallel/distributed algorithms, etc.)
| |
| * ...
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|
| |
|
| == Principles in probability theory == | | == Problem 2 == |
| The course is organized by the advancedness of the probabilistic tools. We do this for two reasons: First, for randomized algorithms, analysis is usually more difficult and involved than the algorithm itself; and second, getting familiar with these probability principles will help you understand the true reasons for which the smart algorithms are designed.
| | (Erdős-Spencer 1974) |
| * '''Basics of probability theory''': probability space, events, the union bound, independence, conditional probability.
| |
| * '''Moments and deviations''': random variables, expectation, linearity of expectation, Markov's inequality, variance, second moment method.
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| * '''The probabilistic method''': averaging principle, threshold phenomena, Lovász Local Lemma.
| |
| * '''Concentrations''': Chernoff-Hoeffding bound, martingales, Azuma's inequality, bounded difference method.
| |
| * '''Markov chains and random walks''': Markov chians, random walks, hitting/cover time, mixing time, coupling, conductance.
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|
| =Probability Space=
| | Let <math>n</math> coins of weights 0 and 1 be given. We are also given a scale with which we may weigh any subset of the coins. Our goal is to determine the weights of coins (that is, to known which coins are 0 and which are 1) with the minimal number of weighings. |
| The axiom foundation of probability theory is laid by [http://en.wikipedia.org/wiki/Andrey_Kolmogorov Kolmogorov], one of the greatest mathematician of the 20th century, who advanced various very different fields of mathematics.
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|
| {{Theorem|Definition (Probability Space)|
| | This problem can be formalized as follows: A collection <math>S_1,S_1,\ldots,S_m\subseteq [n]</math> is called '''determining''' if an arbitrary subset <math>T\subseteq[n]</math> can be uniquely determined by the cardinalities <math>|S_i\cap T|, 1\le i\le m</math>. |
| A '''probability space''' is a triple <math>(\Omega,\Sigma,\Pr)</math>. | |
| *<math>\Omega</math> is a set, called the '''sample space'''.
| |
| *<math>\Sigma\subseteq 2^{\Omega}</math> is the set of all '''events''', satisfying:
| |
| *:(K1). <math>\Omega\in\Sigma</math> and <math>\empty\in\Sigma</math>. (The ''certain'' event and the ''impossible'' event.)
| |
| *:(K2). If <math>A,B\in\Sigma</math>, then <math>A\cap B, A\cup B, A-B\in\Sigma</math>. (Intersection, union, and diference of two events are events).
| |
| * A '''probability measure''' <math>\Pr:\Sigma\rightarrow\mathbb{R}</math> is a function that maps each event to a nonnegative real number, satisfying
| |
| *:(K3). <math>\Pr(\Omega)=1</math>.
| |
| *:(K4). If <math>A\cap B=\emptyset</math> (such events are call ''disjoint'' events), then <math>\Pr(A\cup B)=\Pr(A)+\Pr(B)</math>.
| |
| *:(K5*). For a decreasing sequence of events <math>A_1\supset A_2\supset \cdots\supset A_n\supset\cdots</math> of events with <math>\bigcap_n A_n=\emptyset</math>, it holds that <math>\lim_{n\rightarrow \infty}\Pr(A_n)=0</math>.
| |
| }}
| |
|
| |
|
| ;Remark
| | * Prove that if there is a determining collection <math>S_1,S_1,\ldots,S_m\subseteq [n]</math>, then there is a way to determine the weights of <math>n</math> coins with <math>m</math> weighings. |
| * In general, the set <math>\Omega</math> may be continuous, but we only consider '''discrete''' probability in this lecture, thus we assume that <math>\Omega</math> is either finite or countably infinite. | | * Use pigeonhole principle to show that if a collection <math>S_1,S_1,\ldots,S_m\subseteq [n]</math> is determining, then it must hold that <math>m\ge \frac{n}{\log_2(n+1)}</math>. |
| * Sometimes it is convenient to assume <math>\Sigma=2^{\Omega}</math>, i.e. the events enumerates all subsets of <math>\Omega</math>. But in general, a probability space is well-defined by any <math>\Sigma</math> satisfying (K1) and (K2). Such <math>\Sigma</math> is called a <math>\sigma</math>-algebra defined on <math>\Omega</math>.
| |
| * The last axiom (K5*) is redundant if <math>\Sigma</math> is finite, thus it is only essential when there are infinitely many events. The role of axiom (K5*) in probability theory is like [http://en.wikipedia.org/wiki/Zorn's_lemma Zorn's Lemma] (or equivalently the [http://en.wikipedia.org/wiki/Axiom_of_choice Axiom of Choice]) in axiomatic set theory.
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|
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| Useful laws for probability can be deduced from the ''axioms'' (K1)-(K5).
| | (This gives a lower bound for the number of weighings required to determine the weights of <math>n</math> coins.) |
| {{Theorem|Proposition|
| |
| # Let <math>\bar{A}=\Omega\setminus A</math>. It holds that <math>\Pr(\bar{A})=1-\Pr(A)</math>.
| |
| # If <math>A\subseteq B</math> then <math>\Pr(A)\le\Pr(B)</math>.
| |
| }}
| |
| {{Proof|
| |
| # The events <math>\bar{A}</math> and <math>A</math> are disjoint and <math>\bar{A}\cup A=\Omega</math>. Due to Axiom (K4) and (K3), <math>\Pr(\bar{A})+\Pr(A)=\Pr(\Omega)=1</math>.
| |
| # The events <math>A</math> and <math>B\setminus A</math> are disjoint and <math>A\cup(B\setminus A)=B</math> since <math>A\subseteq B</math>. Due to Axiom (K4), <math>\Pr(A)+\Pr(B\setminus A)=\Pr(B)</math>, thus <math>\Pr(A)\le\Pr(B)</math>.
| |
| }}
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|
| |
|
| ;Notation
| |
| An event <math>A\subseteq\Omega</math> can be represented as <math>A=\{a\in\Omega\mid \mathcal{E}(a)\}</math> with a predicate <math>\mathcal{E}</math>.
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| The predicate notation of probability is
| | == Problem 3 == |
| :<math>\Pr[\mathcal{E}]=\Pr(\{a\in\Omega\mid \mathcal{E}(a)\})</math>.
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|
| During the lecture, we mostly use the predicate notation instead of subset notation.
| | A set of vertices <math>D\subseteq V</math> of graph <math>G(V,E)</math> is a [http://en.wikipedia.org/wiki/Dominating_set ''dominating set''] if for every <math>v\in V</math>, it holds that <math>v\in D</math> or <math>v</math> is adjacent to a vertex in <math>D</math>. The problem of computing minimum dominating set is NP-hard. |
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| == Independence ==
| | * Prove that for every <math>d</math>-regular graph with <math>n</math> vertices, there exists a dominating set with size at most <math>\frac{n(1+\ln(d+1))}{d+1}</math>. |
| {{Theorem
| |
| |Definition (Independent events)|
| |
| :Two events <math>\mathcal{E}_1</math> and <math>\mathcal{E}_2</math> are '''independent''' if and only if
| |
| ::<math>\begin{align}
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| \Pr\left[\mathcal{E}_1 \wedge \mathcal{E}_2\right]
| |
| &=
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| \Pr[\mathcal{E}_1]\cdot\Pr[\mathcal{E}_2].
| |
| \end{align}</math> | |
| }}
| |
| This definition can be generalized to any number of events:
| |
| {{Theorem
| |
| |Definition (Independent events)|
| |
| :Events <math>\mathcal{E}_1, \mathcal{E}_2, \ldots, \mathcal{E}_n</math> are '''mutually independent''' if and only if, for any subset <math>I\subseteq\{1,2,\ldots,n\}</math>,
| |
| ::<math>\begin{align}
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| \Pr\left[\bigwedge_{i\in I}\mathcal{E}_i\right]
| |
| &=
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| \prod_{i\in I}\Pr[\mathcal{E}_i].
| |
| \end{align}</math>
| |
| }}
| |
|
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|
| Note that in probability theory, the "mutual independence" is <font color="red">not</font> equivalent with "pair-wise independence", which we will learn in the future.
| | * Try to obtain an upper bound for the size of dominating set using Lovász Local Lemma. Is it better or worse than previous one? |
|
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|
| = Model of Computation = | | == Problem 4 == |
| Our model of computation extends the standard model ([http://en.wikipedia.org/wiki/Turing_machine Turing machine] or [http://en.wikipedia.org/wiki/Random-access_machine random-access machine]) with access to uniform and independent random bits (fair coin flips). On a fixed input, the behavior of the algorithm is random. To be specific, the output or running time of the algorithm may be random.
| | Let <math>H(W,F)</math> be a graph and <math>n>|W|</math> be an integer. It is known that for some graph <math>G(V,E)</math> such that <math>|V|=n</math>, <math>|E|=m</math>, <math>G</math> does not contain <math>H</math> as a subgraph. Prove that for <math>k>\frac{n^2\ln n}{m}</math>, there is an edge <math>k</math>-coloring for <math>K_n</math> that <math>K_n</math> contains no monochromatic <math>H</math>. |
|
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|
| == Monte Carlo algorithms == | | Remark: Let <math>E=\binom{V}{2}</math> be the edge set of <math>K_n</math>. "An edge <math>k</math>-coloring for <math>K_n</math>" is a mapping <math>f:E\to[k]</math>. |
| [http://en.wikipedia.org/wiki/Monte_carlo_algorithm Monte Carlo algorithms] always returns in finite steps but may output the wrong answer. For decision problems (problems with two answers "yes" and "no"), the Monte Carlo algorithms are further divided into those with ''one-sided errors'' and ''two-sided errors''.
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|
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|
| ;Monte Carlo algorithms with one-sided errors
| | == Problem 5 == |
| :These algorithms only make errors in one direction, which may be further divided into two cases:
| |
| :'''False positive''': If the true answer is "yes" then the algorithm returns "yes" with probability 1, and if the true answer is "no" then the algorithm returns "no" with probability at least <math>\epsilon</math>, where <math>0<\epsilon< 1</math> is the ''confidence''. The algorithm may return a wrong "yes" while the true answer is "no".
| |
| :The one-sided error can be reduced by independent repetitions. Run the algorithm independently for <math>t</math> times, output "yes" if all running instances return "yes", and output "no" if otherwise. If the true answer is "yes" this new algorithm returns "yes" since all running instances are guaranteed to do so, and if the true answer is "no" the new algorithm returns "yes" only if all running instances return "yes", whose probability is bounded by <math>(1-\epsilon)^t</math>, which can be reduced to any <math>\delta\in(0,1)</math> by setting <math>t=O\left(\frac{1}{\epsilon}\log\frac{1}{\delta}\right)</math>.
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|
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|
| :'''False negative''': If the true answer is "yes" then the algorithm returns "yes" with probability at least <math>\epsilon</math>, and if the true answer is "no" then the algorithm returns "no" with probability 1. The algorithm may return a wrong "no" while the true answer is "yes". The error can be reduced in the same way.
| | Let <math>G(V,E)</math> be a cycle of length <math>k\cdot n</math> and let <math>V=V_1\cup V_2\cup\dots V_n</math> be a partition of its <math>k\cdot n</math> vertices into <math>n</math> pairwise disjoint subsets, each of cardinality <math>k</math>. |
| | For <math>k\ge 11</math>, show that there must be an independent set of <math>G</math> containing precisely one vertex from each <math>V_i</math>. |
|
| |
|
| | | Hint: you can use Lovász Local Lemma. |
| ;Monte Carlo algorithms with two-sided errors
| |
| :These algorithms make errors in both directions. If the true answer is "yes" then the algorithm returns "yes" with probability at least <math>\frac{1}{2}+\epsilon</math>, and if the true answer is "no" then the algorithm returns "no" with probability at least <math>\frac{1}{2}+\epsilon</math>, where <math>\epsilon\in \left(0,\frac{1}{2}\right)</math> is a bias. | |
| :The error can be reduced by repetitions and majority vote. Run the algorithm independently for <math>t</math> times, output "yes" if over half running instances return "yes", and output "no" if otherwise. The numbers of "yes"s and "no"s in the <math>t</math> trials follow the Binomial distribution. For each <math>0\le i\le t</math>, the probability that there are precisely <math>i</math> correct answers in <math>t</math> trials is given by
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| ::<math>
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| {t\choose i}\left(\frac{1}{2}+\epsilon\right)^i\left(\frac{1}{2}-\epsilon\right)^{t-i},
| |
| </math>
| |
| :and the new algorithm returns a wrong answer only if at most <math>\lfloor t/2\rfloor</math> correct answers in <math>t</math> trials, which is given by
| |
| ::<math>
| |
| \begin{align}
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| \sum_{i=0}^{\lfloor t/2\rfloor} {t\choose i}\left(\frac{1}{2}+\epsilon\right)^i\left(\frac{1}{2}-\epsilon\right)^{t-i}
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| &\le
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| \sum_{i=0}^{\lfloor t/2\rfloor} {t\choose i}\left(\frac{1}{2}+\epsilon\right)^{t/2}\left(\frac{1}{2}-\epsilon\right)^{t/2}\\
| |
| &=
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| \left(\frac{1}{4}-\epsilon^2\right)^{t/2}\sum_{i=0}^{\lfloor t/2\rfloor}{t\choose i}\\
| |
| &\le
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| \left(\frac{1}{4}-\epsilon^2\right)^{t/2}2^t\\
| |
| &=(1-4\epsilon^2)^{t/2},
| |
| \end{align}
| |
| </math>
| |
| :which can be reduced to any <math>\delta\in(0,1)</math> by setting <math>t=O\left(\frac{1}{\epsilon^2}\log\frac{1}{\delta}\right)</math>.
| |
| | |
| == Las Vegas algorithms ==
| |
| [http://en.wikipedia.org/wiki/Las_Vegas_algorithm Las Vegas algorithms] always output correct answers but the running time is random. The time complexity of a Las Vegas algorithm is measure by the expected running time. The concept of Las Vegas algorithm is introduced by Babai in 1979 in his seminal work on graph isomorphsm testing.
| |
| | |
| A Las Vegas algorithm can be converted to a Monte Carlo algorithm by truncating. The error of the resulting Monte Carlo algorithm can be made one-sided and can be bounded by Markov's inequality. There is no general way known to convert a Monte Carlo algorithm to a Las Vegas algorithm.
| |
| | |
| = Checking Matrix Multiplication=
| |
| [[File: matrix_multiplication.png|thumb|360px|right|The evolution of time complexity <math>O(n^{\omega})</math> for matrix multiplication.]]
| |
| | |
| Let <math>\mathbb{F}</math> be a feild (you may think of it as the filed <math>\mathbb{Q}</math> of rational numbers, or the finite field <math>\mathbb{Z}_p</math> of integers modulo prime <math>p</math>). We suppose that each field operation (addition, subtraction, multiplication, division) has unit cost. This model is called the '''unit-cost RAM''' model, which is an ideal abstraction of a computer.
| |
| | |
| Consider the following problem:
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| * '''Input''': Three <math>n\times n</math> matrices <math>A</math>, <math>B</math>, and <math>C</math> over the field <math>\mathbb{F}</math>.
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| * '''Output''': "yes" if <math>C=AB</math> and "no" if otherwise.
| |
| | |
| A naive way to solve this is to multiply <math>A</math> and <math>B</math> and compare the result with <math>C</math>.
| |
| The straightforward algorithm for matrix multiplication takes <math>O(n^3)</math> time, assuming that each arithmetic operation takes unit time.
| |
| The [http://en.wikipedia.org/wiki/Strassen_algorithm Strassen's algorithm] discovered in 1969 now implemented by many numerical libraries runs in time <math>O(n^{\log_2 7})\approx O(n^{2.81})</math>. Strassen's algorithm starts the search for fast matrix multiplication algorithms. The [http://en.wikipedia.org/wiki/Coppersmith%E2%80%93Winograd_algorithm Coppersmith–Winograd algorithm] discovered in 1987 runs in time <math>O(n^{2.376})</math> but is only faster than Strassens' algorithm on extremely large matrices due to the very large constant coefficient. This has been the best known for decades, until recently Stothers got an <math>O(n^{2.3737})</math> algorithm in his PhD thesis in 2010, and independently Vassilevska Williams got an <math>O(n^{2.3727})</math> algorithm in 2012. Both these improvements are based on generalization of Coppersmith–Winograd algorithm. It is unknown whether the matrix multiplication can be done in time <math>O(n^{2+o(1)})</math>.
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| | |
| == Freivalds Algorithm ==
| |
| The following is a very simple randomized algorithm due to Freivalds, running in <math>O(n^2)</math> time:
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| {{Theorem|Algorithm (Freivalds, 1979)|
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| *pick a vector <math>r \in\{0, 1\}^n</math> uniformly at random;
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| *if <math>A(Br) = Cr</math> then return "yes" else return "no";
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| }}
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| The product <math>A(Br)</math> is computed by first multiplying <math>Br</math> and then <math>A(Br)</math>.
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| The running time of Freivalds algorithm is <math>O(n^2)</math> because the algorithm computes 3 matrix-vector multiplications.
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| If <math>AB=C</math> then <math>A(Br) = Cr</math> for any <math>r \in\{0, 1\}^n</math>, thus the algorithm will return a "yes" for any positive instance (<math>AB=C</math>).
| |
| But if <math>AB \neq C</math> then the algorithm will make a mistake if it chooses such an <math>r</math> that <math>ABr = Cr</math>. However, the following lemma states that the probability of this event is bounded.
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| | |
| {{Theorem|Lemma|
| |
| :If <math>AB\neq C</math> then for a uniformly random <math>r \in\{0, 1\}^n</math>,
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| ::<math>\Pr[ABr = Cr]\le \frac{1}{2}</math>.
| |
| }}
| |
| {{Proof| Let <math>D=AB-C</math>. The event <math>ABr=Cr</math> is equivalent to that <math>Dr=0</math>. It is then sufficient to show that for a <math>D\neq \boldsymbol{0}</math>, it holds that <math>\Pr[Dr = \boldsymbol{0}]\le \frac{1}{2}</math>.
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| Since <math>D\neq \boldsymbol{0}</math>, it must have at least one non-zero entry. Suppose that <math>D_{ij}\neq 0</math>.
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| We assume the event that <math>Dr=\boldsymbol{0}</math>. In particular, the <math>i</math>-th entry of <math>Dr</math> is
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| :<math>(Dr)_{i}=\sum_{k=1}^n D_{ik}r_k=0.</math>
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| The <math>r_j</math> can be calculated by
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| :<math>r_j=-\frac{1}{D_{ij}}\sum_{k\neq j}^n D_{ik}r_k.</math>
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| Once all other entries <math>r_k</math> with <math>k\neq j</math> are fixed, there is a unique solution of <math>r_j</math>. Therefore, the number of <math>r\in\{0,1\}^n</math> satisfying <math>Dr=\boldsymbol{0}</math> is at most <math>2^{n-1}</math>. The probability that <math>ABr=Cr</math> is bounded as
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| :<math>\Pr[ABr=Cr]=\Pr[Dr=\boldsymbol{0}]\le\frac{2^{n-1}}{2^n}=\frac{1}{2}</math>.
| |
| }}
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| When <math>AB=C</math>, Freivalds algorithm always returns "yes"; and when <math>AB\neq C</math>, Freivalds algorithm returns "no" with probability at least 1/2.
| |
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| To improve its accuracy, we can run Freivalds algorithm for <math>k</math> times, each time with an ''independent'' <math>r\in\{0,1\}^n</math>, and return "yes" if and only if all running instances returns "yes".
| |
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| {{Theorem|Freivalds' Algorithm (multi-round)|
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| *pick <math>k</math> vectors <math>r_1,r_2,\ldots,r_k \in\{0, 1\}^n</math> uniformly and independently at random;
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| *if <math>A(Br_i) = Cr_i</math> for all <math>i=1,\ldots,k</math> then return "yes" else return "no";
| |
| }}
| |
| | |
| If <math>AB=C</math>, then the algorithm returns a "yes" with probability 1. If <math>AB\neq C</math>, then due to the independence, the probability that all <math>r_i</math> have <math>ABr_i=C_i</math> is at most <math>2^{-k}</math>, so the algorithm returns "no" with probability at least <math>1-2^{-k}</math>. For any <math>0<\epsilon<1</math>, choose <math>k=\log_2 \frac{1}{\epsilon}</math>. The algorithm runs in time <math>O(n^2\log_2\frac{1}{\epsilon})</math> and has a one-sided error (false positive) bounded by <math>\epsilon</math>.
| |
| | |
| =Polynomial Identity Testing (PIT) =
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| Consider the following problem of '''Polynomial Identity Testing (PIT)''':
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| * '''Input:''' two polynomials <math>P_1, P_2\in\mathbb{F}[x]</math> of degree <math>d</math>.
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| * '''Output:''' "yes" if two polynomials are identical, i.e. <math>P_1\equiv P_2</math>, and "no" if otherwise.
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| The <math>\mathbb{F}[x]</math> denote the [http://en.wikipedia.org/wiki/Polynomial_ring ring of polynomials] over field <math>\mathbb{F}</math>.
| |
| | |
| Alternatively, we can consider the following equivalent problem:
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| * '''Input:''' a polynomial <math>P\in\mathbb{F}[x]</math> of degree <math>d</math>.
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| * '''Output:''' "yes" if <math>P\equiv 0</math>, and "no" if otherwise.
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| | |
| The probalem is trivial if <math>P</math> is presented in its explicit form <math>P(x)=\sum_{i=0}^d a_ix^i</math>. But we assume that <math>P</math> is given in product form or as black box.
| |
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| A straightforward deterministic algorithm that solves PIT is to query <math>d+1</math> points <math>P(1),P(2),\ldots,P(d+1)</math> and check whether thay are all zero. This can determine whether <math>P\equiv 0</math> by interpolation.
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| We now introduce a simple randomized algorithm for the problem.
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| {{Theorem|Algorithm for PIT|
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| *pick <math>x\in\{1,2,\ldots,2d\}</math> uniformly at random;
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| *if <math>P(x) = 0</math> then return “yes” else return “no”;
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| }}
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| This algorithm requires only the evaluation of <math>P</math> at a single point. And if <math>P\equiv 0</math> it is always correct.
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| And if <math>P\not\equiv 0</math> then the probability that the algorithm wrongly returns "yes" is bounded as follows.
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| | |
| {{Theorem|Theorem|
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| : Let <math>P\in\mathbb{F}[x]</math> be a polynomial of degree <math>d</math> over the field <math>\mathbb{F}</math>. Let <math>S\subset\mathbb{F}</math> be an arbitrary set and <math>x\in S</math> is chosen uniformly at random from <math>S</math>. If <math>P\not\equiv 0</math> then
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| ::<math>\Pr[P(x)=0]\le\frac{d}{|S|}.</math>
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| }}
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| {{Proof|
| |
| A non-zero <math>d</math>-degree polynomial <math>P</math> has at most <math>d</math> distinct roots, thus at most <math>d</math> members <math>x</math> of <math>S</math> satisfy that <math>P(x)=0</math>. Therefore, <math>\Pr[P(x)=0]\le\frac{d}{|S|}</math>.
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| }}
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| By the theorem, the algorithm can distinguish a non-zero polynomial from 0 with probability at least <math>1/2</math>. This is achieved by evaluation of the polynomial at only one point and <math>1+\log_2 d</math> many random bits.
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| == Communication Complexity of Identity ==
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| The [http://en.wikipedia.org/wiki/Communication_complexity communication complexity] is introduced by Andrew Chi-Chih Yao as a model of computation which involves multiple participants, each with partial information of the input.
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| 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) whose logics depends only on the problem <math>f</math> but not on the inputs. The complexity of a communication protocol is measured by the number of bits communicated between Alice and Bob in the worst case.
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| 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>,
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| :<math>
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| \mathrm{EQ}(a,b)=
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| \begin{cases}
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| 1& \mbox{if } a=b,\\
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| 0& \mbox{otherwise.}
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| \end{cases}</math>
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| 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.
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| It is known that for deterministic communication protocols, this is the best we can get for computing EQ.
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| {{Theorem|Theorem (Yao 1979)|
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| :Any deterministic communication protocol computing EQ on two <math>n</math>-bit strings costs <math>n</math> bits of communication in the worst-case.
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| }}
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| 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 in Yao's 1979 paper initiates the field of communication complexity.
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| If the randomness is allowed, we can solve this problem up to a tolerable probabilistic error with significantly less communications. The inputs <math>a,b\in\{0,1\}^{n}</math> are two strings <math>a=a_0a_1\cdots a_{n-1}, b=b_0b_1\cdots b_{n-1}</math> of <math>n</math> bits. Let <math>k=\lceil\log_2 (2n)\rceil</math> and <math>p\in[2^k,2^{k+1}]</math> be an arbitrary prime number. (Such a prime <math>p</math> always exists.) The input strings <math>a,b</math> can be respectively represented as two polynomials <math>f,g\in\mathbb{Z}_p[x]</math> such that <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> of degree <math>n-1</math>, where all additions and multiplications are modulo <math>p</math>. The randomized communication protocol is given as follows:
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| {{Theorem|A randomized protocol for EQ|
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| '''Alice does''':
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| :* pick <math>x\in[p]</math> uniformly at random;
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| :* send <math>x</math> and <math>f(x)</math> to Bob;
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| '''Upon receiving''' <math>x</math> and <math>f(x)</math> '''Bob does''':
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| :* If <math>f(x)= g(x)</math> return "'''yes'''"; else return "'''no'''".
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| }}
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| Repeat this protocol for 100 times.
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| The total number of bits to communicate is bounded by <math>200\log_2p=O(\log n)</math>. Due to the analysis of the randomized algorithm for PIT, if <math>a=b</math> the protocol is always correct and if <math>a\neq b</math> the protocol fails to report a difference with probability less than <math>2^{-100}</math>.
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| =Polynomial Identity Testing (PIT)=
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| Consider the following problem of '''Polynomial Identity Testing (PIT)''':
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| * '''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>.
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| * '''Output:''' "yes" if <math>f\equiv g</math>, and "no" if otherwise.
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| 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 of multi-variate polynomials] over field <math>\mathbb{F}</math>. The most natural way to represent an <math>n</math>-variate polynomial of degree <math>d</math> is to write it as a sum of monomials:
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| :<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>.
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| 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.
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| Alternatively, we can consider the following equivalent problem:
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| * '''Input:''' a polynomial <math>f\in\mathbb{F}[x_1,x_2,\ldots,x_n]</math> of degree <math>d</math>.
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| * '''Output:''' "yes" if <math>f\equiv 0</math>, and "no" if otherwise.
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| If <math>f</math> is written explicitly as a sum of monomials, then the problem is trivial. Again we allow <math>f</math> to be represented in product form.
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| {{Theorem|Example|
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| 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
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| :<math>M=\begin{bmatrix}
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| 1 & x_1 & x_1^2 & \dots & x_1^{n-1}\\
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| 1 & x_2 & x_2^2 & \dots & x_2^{n-1}\\
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| 1 & x_3 & x_3^2 & \dots & x_3^{n-1}\\
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| \vdots & \vdots & \vdots & \ddots &\vdots \\
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| 1 & x_n & x_n^2 & \dots & x_n^{n-1}
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| \end{bmatrix}</math>.
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| Let <math>f</math> be the polynomial defined as
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| :<math>
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| f(x_1,\ldots,x_n)=\det(M)=\prod_{j<i}(x_i-x_j).
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| </math>
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| It is pretty easy to evaluate <math>f(x_1,x_2,\ldots,x_n)</math> on any particular <math>x_1,x_2,\ldots,x_n</math>, however it is prohibitively expensive to symbolically expand <math>f(x_1,\ldots,x_n)</math> to its sum-of-monomial form.
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| }}
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| | |
| == Schwartz-Zippel Theorem==
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| Here is a very simple randomized algorithm, due to Schwartz and Zippel.
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| {{Theorem|Randomized algorithm for multi-variate PIT|
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| * fix an arbitrary set <math>S\subseteq \mathbb{F}</math> whose size to be fixed;
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| * pick <math>r_1,r_2,\ldots,r_n\in S</math> uniformly and independently at random;
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| * if <math>f(\vec{r})=f(r_1,r_2,\ldots,r_n) = 0</math> then return “yes” else return “no”;
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| }}
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| This algorithm requires only the evaluation of <math>f</math> at a single point <math>\vec{r}</math>. And if <math>f\equiv 0</math> it is always correct.
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| In the Theorem below, we’ll see that if <math>f\not\equiv 0</math> then the algorithm is incorrect with probability at most <math>\frac{d}{|S|}</math>, where <math>d</math> is the degree of the polynomial <math>f</math>.
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| {{Theorem|Schwartz-Zippel Theorem|
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| : 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> such that <math>f\not\equiv 0</math>. Fix any finite set <math>S\subset\mathbb{F}</math>, and let <math>r_1,r_2\ldots,r_n</math> be chosen uniformly and independently at random from <math>S</math>. Then
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| ::<math>\Pr[f(r_1,r_2,\ldots,r_n)=0]\le\frac{d}{|S|}.</math>
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| }}
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| {{Proof|
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| We prove by induction on <math>n</math> the number of variables.
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| For <math>n=1</math>, assuming that <math>f\not\equiv 0</math>, due to the fundamental theorem of algebra, the degree-<math>d</math> polynomial <math>f(x)</math> has at most <math>d</math> roots, thus
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| :<math>\Pr[f(r)=0]\le\frac{d}{|S|}.
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| </math>
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| Assume the induction hypothesis for a multi-variate polynomial up to <math>n-1</math> variable.
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| An <math>n</math>-variate polynomial <math>f(x_1,x_2,\ldots,x_n)</math> can be represented as
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| :<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>,
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| where <math>k</math> is the largest power of <math>x_n</math>, which means that the degree of <math>f_k</math> is at most <math>d-k</math> and <math>f_k\not\equiv 0</math>.
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| In particular, we write <math>f</math> as a sum of two parts:
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| :<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>,
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| where both <math>f_k</math> and <math>\bar{f}</math> are polynomials, such that
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| * as argued above, the degree of <math>f_k</math> is at most <math>d-k</math> and <math>f_k\not\equiv 0</math>;
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| * <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.
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| By the law of total probability, it holds that
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| :<math>
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| \begin{align}
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| &\Pr[f(r_1,r_2,\ldots,r_n)=0]\\
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| =
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| &\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]\\
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| &+\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].
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| \end{align}
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| </math>
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| 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>.
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| By the induction hypothesis, we have
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| :<math>
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| \begin{align}
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| (*)
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| &\qquad
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| &\Pr[f_k(r_1,r_2,\ldots,r_{n-1})=0]\le\frac{d-k}{|S|}.
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| \end{align}
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| </math>
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| For the second case, 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
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| :<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)=g_{r_1,\ldots,r_{n-1}}(x_n)</math>
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| is a single-variate polynomial 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>.
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| Therefore,
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| :<math>
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| \begin{align}
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| (**)
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| &\qquad
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| &\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|}
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| \end{align}
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| </math>.
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| Substituting both <math>(*)</math> and <math>(**)</math> back in the total probability, we have
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| :<math>
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| \Pr[f(r_1,r_2,\ldots,r_n)=0]
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| \le\frac{d-k}{|S|}+\frac{k}{|S|}=\frac{d}{|S|},
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| </math>
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| which proves the theorem.
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| ------
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| In above proof, for the second case that <math>f_k(r_1,\ldots,r_{n-1})\neq 0</math>, we use an "probabilistic arguement" to deal with the random choices in the condition. Here we give a more rigorous proof by enumerating all elementary events in applying the law of total probability. You make your own judgement which proof is better.
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| By the law of total probability,
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| :<math>
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| \begin{align}
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| &\Pr[f(\vec{r})=0]\\
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| =
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| &\sum_{x_1,\ldots,x_{n-1}\in S}\Pr[f(\vec{r})=0\mid \forall i<n, r_i=x_i]\cdot\Pr[\forall i<n, r_i=x_i]\\
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| =
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| &\sum_{x_1,\ldots,x_{n-1}\in S\atop f_k(x_1,\ldots,x_{n-1})=0}\Pr[f(\vec{r})=0\mid \forall i<n, r_i=x_i]\cdot\Pr[\forall i<n, r_i=x_i]\\
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| &+\sum_{x_1,\ldots,x_{n-1}\in S\atop f_k(x_1,\ldots,x_{n-1})\neq0}\Pr[f(\vec{r})=0\mid \forall i<n, r_i=x_i]\cdot\Pr[\forall i<n, r_i=x_i]\\
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| \le
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| &\sum_{x_1,\ldots,x_{n-1}\in S\atop f_k(x_1,\ldots,x_{n-1})=0}\Pr[\forall i<n, r_i=x_i]\\
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| &+\sum_{x_1,\ldots,x_{n-1}\in S\atop f_k(x_1,\ldots,x_{n-1})\neq 0}\Pr[f(x_1,\ldots,x_{n-1},r_n)=0\mid \forall i<n, r_i=x_i]\cdot\Pr[\forall i<n, r_i=x_i]\\
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| =
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| &\Pr[f_k(r_1,\ldots,r_{n-1})=0]+\sum_{x_1,\ldots,x_{n-1}\in S\atop f_k(x_1,\ldots,x_{n-1})\neq 0}\Pr[f(x_1,\ldots,x_{n-1},r_n)=0]\cdot\Pr[\forall i<n, r_i=x_i].
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| \end{align}
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| </math>
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| We have argued that <math>f_k\not\equiv 0</math> and the degree of <math>f_k</math> is <math>d-k</math>. By the induction hypothesis, we have
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| :<math>
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| \Pr[f_k(r_1,\ldots,r_{n-1})=0]\le\frac{d-k}{|S|}.
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| </math>
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| And for every fixed <math>x_1,\ldots,x_{n-1}\in S</math> such that <math>f_k(x_1,\ldots,x_{n-1})\neq 0</math>, we have argued that <math>f(x_1,\ldots,x_{n-1},x_n)</math> is a polynomial in <math>x_n</math> of degree <math>k</math>, thus
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| :<math>
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| \Pr[f(x_1,\ldots,x_{n-1},r_n)=0]\le\frac{k}{|S|},
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| </math>
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| which holds for all <math>x_1,\ldots,x_{n-1}\in S</math> such that <math>f_k(x_1,\ldots,x_{n-1})\neq 0</math>, therefore the weighted average
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| :<math>
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| \sum_{x_1,\ldots,x_{n-1}\in S\atop f_k(x_1,\ldots,x_{n-1})\neq 0}\Pr[f(x_1,\ldots,x_{n-1},r_n)=0]\cdot\Pr[\forall i<n, r_i=x_i]
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| \le\frac{k}{|S|}.
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| </math>
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| Substituting these inequalities back to the total probability, we have
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| <math>
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| \Pr[f(\vec{r})=0]
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| \le\frac{d-k}{|S|}+\frac{k}{|S|}
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| =\frac{d}{|S|}.
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| </math>
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| }}
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