imported>Etone |
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| = Metric Embedding= | | == Problem 1 == |
| A '''metric space''' is a pair <math>(X,d)</math>, where <math>X</math> is a set and <math>d</math> is a '''metric''' (or '''distance''') on <math>X</math>, i.e., a function
| | (Matching vs. Star) |
| :<math>d:X^2\to\mathbb{R}_{\ge 0}</math>
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| such that for any <math>x,y,z\in X</math>, the following axioms hold:
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| # (identity of indiscernibles) <math>d(x,y)=0\Leftrightarrow x=y</math>
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| # (symmetry) <math>d(x,y)=d(y,x)</math>
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| # (triangle inequality) <math>d(x,z)\le d(x,y)+d(y,z)</math>
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| Let <math>(X,d_X)</math> and <math>(Y,d_Y)</math> be two metric spaces. A mapping
| | Given a graph <math>G(V,E)</math>, a ''matching'' is a subset <math>M\subseteq E</math> of edges such that there are no two edges in <math>M</math> sharing a vertex, and a ''star'' is a subset <math>S\subseteq E</math> of edges such that every pair <math>e_1,e_2\in S</math> of distinct edges in <math>S</math> share the same vertex <math>v</math>. |
| :<math>\phi:X\to Y</math>
| |
| is called an '''embedding''' of metric space <math>X</math> into <math>Y</math>. The embedding is said to be with '''distortion''' <math>\alpha\ge1</math> if for any <math>x,y\in X</math> it holds that
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| :<math>\frac{1}{\alpha}\cdot d(x,y)\le d(\phi(x),\phi(y))\le \alpha\cdot d(x,y)</math>.
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| In Computer Science, a typical scenario for the metric embedding is as follows. We want to solve some difficult computation problem on a metric space <math>(X,d)</math>. Instead of solving this problem directly on the original metric space, we embed the metric into a new metric space <math>(Y,d_Y)</math> (with low distortion) where the computation problem is much easier to solve.
| | Prove that any graph <math>G</math> containing more than <math>2(k-1)^2</math> edges either contains a matching of size <math>k</math> or a star of size <math>k</math>. |
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| One particular important case for the metric embedding is to embed a high-dimensional metric space to a new metric space whose dimension is much lower. This is called dimension reduction. This can be very helpful because various very common computation tasks can be very hard to solve on high-dimensional space due to the [https://en.wikipedia.org/wiki/Curse_of_dimensionality curse of dimensionality].
| | (Hint: Learn from the proof of Erdos-Rado's sunflower lemma.) |
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| = The Johnson-Lindenstrauss Theorem = | | == Problem 2 == |
| The '''Johnson-Lindenstrauss Theorem''' or '''Johnson-Lindenstrauss Transformation''' (both shorten as '''JLT''') is a fundamental result for dimension reduction in Euclidian space.
| | (Frankl 1986) |
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| Recall that in Euclidian space <math>\mathbf{R}^d</math>, for any two points <math>x,y\in\mathbf{R}^d</math>, the Euclidian distance between them is given by <math>\|x-y\|=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2+\cdots +(x_d-y_d)^2}</math>, where <math>\|\cdot\|=\|\cdot\|_2</math> denotes the Euclidian norm (a.k.a. the <math>\ell_2</math>-norm).
| | Let <math>\mathcal{F}\subseteq {[n]\choose k}</math> be a <math>k</math>-uniform family, and suppose that it satisfies that <math>A\cap B \not\subset C</math> for any <math>A,B,C\in\mathcal{F}</math>. |
| | * Fix any <math>B\in\mathcal{F}</math>. Show that the family <math>\{A\cap B\mid A\in\mathcal{F}, A\neq B\}</math> is an anti chain. |
| | * Show that <math>|\mathcal{F}|\le 1+{k\choose \lfloor k/2\rfloor}</math>. |
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| The JLT says that in Euclidian space, it is always possible to embed a set of <math>n</math> points in ''arbitrary'' dimension to <math>O(\log n)</math> dimension with constant distortion. The theorem itself is stated formally as follows.
| | ==Problem 3 == |
| | An <math>n</math>-player tournament (竞赛图) <math>T([n],E)</math> is said to be '''transitive''', if there exists a permutation <math>\pi</math> of <math>[n]</math> such that <math>\pi_i<\pi_j</math> for every <math>(i,j)\in E</math>. |
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| {{Theorem
| | Show that for any <math>k\ge 3</math>, there exists a finite <math>N(k)</math> such that every tournament of <math>n\ge N(k)</math> players contains a transitive sub-tournament of <math>k</math> players. Express <math>N(k)</math> in terms of Ramsey number. |
| |Johnson-Lindenstrauss Theorem, 1984|
| |
| :For any <math>0<\epsilon<1/2</math> and any positive integer <math>n</math>, there is a positive integer <math>k=O(\epsilon^{-2}\log n)</math> such that the following holds:
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| :For any set <math>S\subset\mathbf{R}^d</math> with <math>|S|=n</math>, where <math>d</math> is arbitrary, there is an embedding <math>\phi:\mathbf{R}^d\rightarrow\mathbf{R}^k</math> such that
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| ::<math>\forall x,y\in S,\quad (1-\epsilon)\|x-y\|^2\le\|\phi(x)-\phi(y)\|^2\le(1+\epsilon)\|x-y\|^2</math>.
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| }}
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| The Johnson-Lindenstrauss Theorem is usually stated for the <math>\ell_2^2</math>-norm <math>\|\cdot\|^2</math> instead of the Euclidian norm <math>\|\cdot\|</math> itself. Note that this does not change anything other than the constant faction in <math>k=O(\epsilon^{-2}\log n)</math> because <math>(1\pm\epsilon)^{\frac{1}{2}}=1\pm\Theta(\epsilon)</math>. The reason for stating the theorem in <math>\ell_2^2</math>-norm is because <math>\|\cdot\|^2</math> is a sum (rather than a square root) which is easier to analyze.
| | ==Problem 4== |
| | | We color each non-empty subset of <math>[n]=\{1,2,\ldots,n\}</math> with one of the <math>r</math> colors in <math>[r]</math>. Show that for any finite <math>r</math> there is a finite <math>N</math> such that for all <math>n\ge </math>$, for any <math>r</math>-coloring of all non-empty subsets of <math>[n]</math>, there always exist <math>1\le i<j<k\le n</math> such that the intervals <math>[i,j)=\{i,i+1,\ldots, j-1\}</math>, <math>[j,k)=\{j,j+1,\ldots, k-1\}</math> and <math>[i,k)=\{i,i+1,\ldots, k-1\}</math> are all assigned with the same color by the <math>r</math>-coloring. |
| In fact, the embedding <math>\phi:\mathbf{R}^d\rightarrow\mathbf{R}^k</math> can be as simple as a linear transformation <math>A\in\mathbf{R}^{k\times d}</math> so that <math>\phi(x)=Ax</math> for any <math>x\in\mathbf{R}^{d}</math>.
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| Therefore, the above theorem can be stated more precisely as follows.
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| {{Theorem
| |
| |Johnson-Lindenstrauss Theorem (linear embedding)|
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| :For any <math>0<\epsilon<1/2</math> and any positive integer <math>n</math>, there is a positive integer <math>k=O(\epsilon^{-2}\log n)</math> such that the following holds:
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| :For any set <math>S\subset\mathbf{R}^d</math> with <math>|S|=n</math>, where <math>d</math> is arbitrary, there is a linear transformation <math>A\in\mathbf{R}^{k\times d}</math> such that
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| ::<math>\forall x,y\in S,\quad (1-\epsilon)\|x-y\|^2\le\|Ax-Ay\|^2\le(1+\epsilon)\|x-y\|^2</math>.
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| }}
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| The theorem is proved by the probabilistic method. Specifically, we construct a random matrix <math>A\in\mathbf{R}^{k\times d}</math> and show that with high probability (<math>1-O(1/n)</math>) it is a good embedding satisfying:
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| :<math>\forall x,y\in S,\quad (1-\epsilon)\|x-y\|\le\|Ax-Ay\|\le(1+\epsilon)\|x-y\|</math>.
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| Therefore, if such random matrix <math>A</math> is efficient to construct, it immediately gives us an efficient randomized algorithm for dimension reduction in the Euclidian space.
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| There are several such constructions of the random matrix <math>A\in\mathbf{R}^{k\times d}</math>, including:
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| * projection onto uniform random <math>k</math>-dimensional subspace of <math>\mathbf{R}^{d}</math>; (the original construction of Johnson and Lindenstrauss in 1984; a simplified analysis due to [https://cseweb.ucsd.edu/~dasgupta/papers/jl.pdf Dasgupta and Gupta in 1999])
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| * random matrix with i.i.d. Gaussian entries; (due to [https://www.cs.princeton.edu/courses/archive/spr04/cos598B/bib/IndykM-curse.pdf Indyk and Motwani in 1998])
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| * random matrix with i.i.d. -1/+1 entries; (due to [https://www.sciencedirect.com/science/article/pii/S0022000003000254?via%3Dihub#BIB8 Achlioptas in 2003])
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| It was proved by [https://arxiv.org/abs/1609.02094 Kasper Green Larsen and Jelani Nelson in 2016] that the JLT is optimal: there is a set of <math>n</math> points from <math>\mathbf{R}^d</math> such that any embedding <math>\phi:\mathbf{R}^d\rightarrow\mathbf{R}^k</math> having the distortion asserted in the JLT must have dimension <math>k</math> of the target space satisfy <math>k=\Omega(\epsilon^{-2}\log n)</math>.
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| ==JLT via Gaussian matrix==
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| Here we prove the Johnson-Lindenstrauss theorem with the second construction, by the random matrix <math>A\in\mathbf{R}^{k\times d}</math> with i.i.d. Gaussian entries. The construction of <math>A</math> is very simple:
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| :{|border="2" width="100%" cellspacing="4" cellpadding="3" rules="all" style="margin:1em 1em 1em 0; border:solid 1px #AAAAAA; border-collapse:collapse;empty-cells:show;"
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| *Each entry of <math>A\in\mathbf{R}^{k\times d}</math> is drawn independently from the Gaussian distribution <math>\mathcal{N}(0,1/k)</math>.
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| |}
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| Recall that a [https://en.wikipedia.org/wiki/Normal_distribution Gaussian distribution] <math>\mathcal{N}(\mu,\sigma^2)</math> is specified by its mean <math>\mu</math> and standard deviation <math>\sigma</math> such that for a random variable <math>X</math> distributed as <math>\mathcal{N}(\mu,\sigma^2)</math>, we have
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| :<math>\mathbf{E}[X]=\mu</math> and <math>\mathbf{Var}[X]=\sigma^2</math>,
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| and the probability density function is given by <math>p(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\mathrm{e}^{-\frac{(x-\mu)^2}{2\sigma^2}}</math>, therefore
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| :<math>\Pr[X\le t]=\int_{-\infty}^t\frac{1}{\sqrt{2\pi\sigma^2}}\mathrm{e}^{-\frac{(x-\mu)^2}{2\sigma^2}}\,\mathrm{d}x</math>.
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| Fix any two points <math>x,y</math> out of the <math>n</math> points in <math>S\subset \mathbf{R}^d</math>, if we can show that
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| :<math>\Pr\left[(1-\epsilon)\|x-y\|^2\le\|Ax-Ay\|^2\le(1+\epsilon)\|x-y\|^2\right]\ge 1-\frac{1}{n^3}</math>,
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| then by the union bound, the following event
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| :<math>\forall x,y\in S,\quad (1-\epsilon)\|x-y\|^2\le\|Ax-Ay\|^2\le(1+\epsilon)\|x-y\|^2</math>
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| holds with probability <math>1-O(1/n)</math>. The Johnson-Lindenstrauss theorem follows.
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| Furthermore, dividing both sides of the inequalities <math>(1-\epsilon)\|x-y\|^2\le\|Ax-Ay\|^2\le(1+\epsilon)\|x-y\|^2</math> by the factor <math>\|x-y\|^2</math> gives us
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| :<math>(1-\epsilon)\le\frac{\|Ax-Ay\|^2}{\|x-y\|^2}=\left\|A\frac{(x-y)}{\|x-y\|^2}\right\|^2\le(1+\epsilon)</math>,
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| where <math>\frac{(x-y)}{\|x-y\|^2}</math> is a unit vector.
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| Therefore, the Johnson-Lindenstrauss theorem is proved once the following theorem on the unit vector is proved.
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| {{Theorem
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| |Theorem (JLT on unit vector)|
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| :For any positive <math>\epsilon,\delta<1/2</math> there is a positive integer <math>k=O\left(\epsilon^{-2}\log \frac{1}{\delta}\right)</math> such that for the random matrix <math>A\in\mathbf{R}^{k\times d}</math> with each entry drawn independently from the Gaussian distribution <math>\mathcal{N}(0,1/k)</math>, for any unit vector <math>u\in\mathbf{R}^d</math> with <math>\|u\|=1</math>,
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| ::<math>\Pr\left[\left|\|Au\|^2-1\right|>\epsilon\right]<\delta</math>.
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| }}
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| For any <math>u\in\mathbf{R}^d</math>, we have <math>\|Au\|^2=\sum_{i=1}^k(Au)_i^2</math>, where the <math>(Au)_i</math>'s are independent of each other, because each
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| :<math>(Au)_i=\langle A_{i\cdot},u\rangle=\sum_{j=1}^dA_{ij}u_j</math>
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| is determined by a distinct row vector <math>A_{i\cdot}</math> of the matrix <math>A</math> with independent entries.
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| Moreover, since each entry <math>A_{ij}\sim\mathcal{N}(0,1/k)</math> independently and recall that for any two independent Gaussian random variables <math>X_1\sim \mathcal{N}(\mu_1,\sigma_1^2)</math> and <math>X_2\sim \mathcal{N}(\mu_2,\sigma_2^2)</math>, their weighted sum <math>aX_1+bX_2</math> follows the Gaussian distribution <math>\mathcal{N}(a\mu_1+b\mu_2,a^2\sigma_1^2+b^2\sigma_2^2)</math>, therefore <math>(Au)_i=\langle A_{i\cdot},u\rangle=\sum_{j=1}^dA_{ij}u_j</math> follows the Gaussian distribution
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| :<math>(Au)_i\sim\mathcal{N}\left(0,\sum_{j=1}^d\frac{u_j^2}{k}\right)</math>,
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| which is precisely <math>\mathcal{N}\left(0,\frac{1}{k}\right)</math> if <math>u\in\mathbf{R}^d</math> is a unit vector, i.e. <math>\|u\|^2=\sum_{j=1}^du_j^2=1</math>.
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| In summary, the random variable <math>\|Au\|^2</math> that we are interested in, can be represented as a sum-of-square form
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| :<math>\|Au\|^2=\sum_{i=1}^kY_i^2</math>,
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| where each <math>Y_i=(Au)_i</math> independently follows the Gaussian distribution <math>\mathcal{N}\left(0,\frac{1}{k}\right)</math>.
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| ==Concentration of <math>\chi^2</math>-distribution ==
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| By the above argument, the Johnson-Lindenstrauss theorem is proved by the following concentration inequality for the sum of the squares of <math>k</math> Gaussian distributions.
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| {{Theorem
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| |Chernoff bound for sum-of-squares of Gaussian distributions|
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| :For any positive <math>\epsilon,\delta<1/2</math> there is a positive integer <math>k=O\left(\epsilon^{-2}\log \frac{1}{\delta}\right)</math> such that for i.i.d. Gaussian random variables <math>Y_1,Y_2,\ldots,Y_k\sim\mathcal{N}(0,1/k)</math>,
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| ::<math>\Pr\left[\left|\sum_{i=1}^kY_i^2-1\right|>\epsilon\right]<\delta</math>.
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| }}
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| First, we see that this is indeed a concentration inequality that bounds the deviation of a random variable from its mean.
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| For each <math>Y_i\sim\mathcal{N}\left(0,\frac{1}{k}\right)</math>, we know that <math>\mathbf{E}[Y_i]=0</math> and <math>\mathbf{Var}[Y_i]=\mathbf{E}[Y_i^2]-\mathbf{E}[Y_i]^2=\frac{1}{k}</math>, thus
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| :<math>\mathbf{E}[Y_i^2]=\mathbf{Var}[Y_i]+\mathbf{E}[Y_i]^2=\frac{1}{k}</math>.
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| By linearity of expectation, it holds that
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| :<math>\mathbf{E}\left[\sum_{i=1}^kY_i^2\right]=\sum_{i=1}^k\mathbf{E}\left[Y_i^2\right]=1</math>.
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| We prove an equivalent concentration bound stated for the [https://en.wikipedia.org/wiki/Chi-squared_distribution '''<math>\chi^2</math>-distribution'''] (sum of the squares of <math>k</math> '''standard''' Gaussian distributions).
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| {{Theorem
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| |Chernoff bound for the <math>\chi^2</math>-distribution|
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| :For i.i.d. standard Gaussian random variables <math>X_1,X_2,\ldots,X_k\sim\mathcal{N}(0,1)</math>, for <math>0<\epsilon<1</math>,
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| :*<math>\Pr\left[\sum_{i=1}^kX_i^2>(1+\epsilon)k\right]<\mathrm{e}^{-\epsilon^2k/8}</math>,
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| :*<math>\Pr\left[\sum_{i=1}^kX_i^2<(1-\epsilon)k\right]<\mathrm{e}^{-\epsilon^2k/8}</math>.
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| }}
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| Note that this indeed implies the above concentration bound by setting <math>X_i=\sqrt{k}\cdot Y_i</math> and <math>\delta\ge\mathrm{e}^{-\epsilon^2k/8}</math>.
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| {{Proof|
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| We first prove the upper tail. Let <math>\lambda>0</math> be a parameter to be determined.
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| :<math>
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| \begin{align}
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| \Pr\left[\sum_{i=1}^kX_i^2>(1+\epsilon)k\right]
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| &=\Pr\left[\mathrm{e}^{\lambda \sum_{i=1}^kX_i^2}>\mathrm{e}^{(1+\epsilon)\lambda k}\right]\\
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| &< \mathrm{e}^{-(1+\epsilon)\lambda k}\cdot \mathbf{E}\left[\mathrm{e}^{\lambda \sum_{i=1}^kX_i^2}\right] && \text{(Markov's inequality)}\\
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| &= \mathrm{e}^{-(1+\epsilon)\lambda k}\cdot \prod_{i=1}^k \mathbf{E}\left[\mathrm{e}^{\lambda X_i^2}\right]. && \text{(independence)}
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| \end{align}
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| </math>
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| {{Theorem|Proposition|
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| :For standard Gaussian <math>X\sim\mathcal{N}(0,1)</math>, <math>\mathbf{E}\left[\mathrm{e}^{\lambda X^2}\right]=\frac{1}{\sqrt{1-2\lambda}}</math>.
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| }}
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| {{Proof|
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| :<math>
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| \begin{align}
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| \mathbf{E}\left[\mathrm{e}^{\lambda X^2}\right]
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| &=
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| \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\mathrm{e}^{\lambda x^2}\mathrm{e}^{-x^2/2}\,\mathrm{d}x\\
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| &=
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| \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\mathrm{e}^{-(1-2\lambda)x^2/2}\,\mathrm{d}x\\
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| &=
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| \frac{1}{\sqrt{1-2\lambda}}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\mathrm{e}^{-y^2/2}\,\mathrm{d}y &&(\text{set }y=\sqrt{1-2\lambda}x)\\
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| &=\frac{1}{\sqrt{1-2\lambda}}.
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| \end{align}
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| </math>
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| The last equation is because <math>\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\mathrm{e}^{-y^2/2}\,\mathrm{d}y</math> is the total probability mass of a standard Gaussian distribution (which is obviously 1).
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| }}
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| Then continue the above calculation:
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| :<math>
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| \begin{align}
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| \Pr\left[\sum_{i=1}^kX_i^2>(1+\epsilon)k\right]
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| &<
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| \mathrm{e}^{-(1+\epsilon)\lambda k}\cdot \left(\frac{1}{\sqrt{1-2\lambda}}\right)^k\\
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| &=
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| \mathrm{e}^{-\epsilon\lambda k}\cdot \left(\frac{\mathrm{e}^{-\lambda}}{\sqrt{1-2\lambda}}\right)^k\\
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| &\le
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| \mathrm{e}^{-\epsilon\lambda k + 2\lambda^2 k} && (\text{for }\lambda<1/4)\\
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| &=
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| \mathrm{e}^{-\epsilon k/8}. && (\text{by setting the stationary point }\lambda=\epsilon/4)
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| \end{align}
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| </math>
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| This proved the upper tail. The lower tail can be symmetrically proved by optimizing over a parameter <math>\lambda<0</math>.
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| }}
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| As we argued above, this finishes the proof of the Johnson-Lindenstrauss Theorem.
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| = Nearest Neighbor Search (NNS)=
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| {{Theorem|Nearest Neighbor Search (NNS)|
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| *'''Data''': a set <math>S</math> of <math>n</math> points <math>y_1,y_2,\ldots,y_n\in X</math> from a metric space <math>(X,\mathrm{dist})</math>;
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| *'''Query''': a point <math>x\in X</math>;
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| *'''Answer''': a nearest neighbor of the query point <math>x</math> among all data points, i.e. a data point <math>y_{i^*}\in S</math> such that <math>\mathrm{dist}(x,y_{i^*})=\min_{y_i\in S}\mathrm{dist}(x,y_i)</math>.
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| }}
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| {{Theorem|<math>c</math>-ANN (Approximate Nearest Neighbor)|
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| :<math>c>1</math> is an approximate ratio.
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| *'''Data''': a set <math>S</math> of <math>n</math> points <math>y_1,y_2,\ldots,y_n\in X</math> from a metric space <math>(X,\mathrm{dist})</math>;
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| *'''Query''': a point <math>x\in X</math>;
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| *'''Answer''': a <math>c</math>-approximate nearest neighbor of the query point <math>x</math> among all data points, i.e. a data point <math>y_{i^*}\in S</math> such that <math>\mathrm{dist}(x,y_{i^*})\le c\, \min_{y_i\in S}\mathrm{dist}(x,y_i)</math>.
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| }}
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| {{Theorem|<math>(c,r)</math>-ANN (Approximate ''Near'' Neighbor)|
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| :<math>c>1</math> is an approximate ratio; <math>r>0</math> is a radius;
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| *'''Data''': a set <math>S</math> of <math>n</math> points <math>y_1,y_2,\ldots,y_n\in X</math> from a metric space <math>(X,\mathrm{dist})</math>;
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| *'''Query''': a point <math>x\in X</math>;
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| *'''Answer''': <math>
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| \begin{cases}
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| \text{a data point }y_{i^*}\in S\text{ such that }\mathrm{dist}(x,y_{i^*})\le c\cdot r & \text{if }\exists y_i\in S, \mathrm{dist}(x,y_{i})\le r,\\
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| \bot & \text{if }\forall y_i\in S, \mathrm{dist}(x,y_{i})>c\cdot r,\\
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| \text{arbitrary} & \text{otherwise}.
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| \end{cases}
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| </math>
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| }}
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| {{Theorem|Theorem|
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| :For any instance <math>S=\{y_1,y_2,\ldots,y_n\}</math> of the nearest neighbor problem, let <math>D_{\min}\triangleq\min_{y_i,y_j\in S}\mathrm{dist}(y_i,y_j)</math>, <math>D_{\max}\triangleq\max_{y_i,y_j\in S}\mathrm{dist}(y_i,y_j)</math> and <math>R\triangleq \frac{D_{\max}}{D_{\min}}</math>.
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| :If for any <math>r>0</math> there always exists a data structure for the <math>(c,r)</math>-ANN problem, using space at most <math>s</math>, answering each query in time <math>t</math>, correctly with probability <math>1-\delta</math>,
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| :then there exists a data structure for the <math>c</math>-ANN problem, using space <math>O(s\log_c R)</math>, answering each query in time <math>O(s\log_c R)</math>, correctly with probability <math>1-O(\delta\log\log_c R)</math>.
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| }}
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| = Locality-Sensitive Hashing (LSH)=
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