Foundations of Data Science

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10.4.2 A Representation Cannot be Sparse in Both Time and Frequency Domains We now show that there is an uncertainty principle that states that a time signal cannot be sparse in both the time domain and the frequency domain. If the signal is of length n, then the product of the number of nonzero coordinates in the time domain and the number of nonzero coordinates in the frequency domain must be at least n. We first prove two technical lemmas. In dealing with the Fourier transform it is convenient for indices to run from 0 to n−1 rather than from 1 to n. Let x 0 , x 1 , . . . , x n−1 be a sequence and let f 0 , f 1 , . . . , f n−1 be its discrete Fourier transform. Let i = √ −1. Then f j = √ 1 n−1 ∑ n x k e − 2πijk n , j = 0, . . . , n−1. In matrix form f = Zx where z jk = e − 2πi n jk . k=0 ⎛ ⎜ ⎝ ⎞ f 0 f 1 ⎟ . f n−1 ⎛ ⎠ = √ 1 n ⎜ ⎝ 1 e − 2πi n e − 2πi2 n · · · e − 2πi n (n − 1)2 1 1 1 · · · 1 ⎞ ⎛ 1 e − 2πi n e − 2πi2 n · · · e − 2πi (n − 1) n . . . ⎟ ⎜ ⎠ ⎝ ⎞ x 0 x 1 ⎟ . ⎠ x n−1 If some of the elements of x are zero, delete the zero elements of x and the corresponding columns of the matrix. To maintain a square matrix, let n x be the number of nonzero elements in x and select n x consecutive rows of the matrix. Normalize the columns of the resulting submatrix by dividing each element in a column by the column element in the first row. The resulting submatrix is a Vandermonde matrix that looks like ⎛ ⎞ 1 1 1 1 ⎜ a b c d ⎟ ⎝ a 2 b 2 c 2 d 2 ⎠ a 3 b 3 c 3 d 3 and is nonsingular. Lemma 10.6 If x 0 , x 1 , . . . , x n−1 has n x nonzero elements, then f 0 , f 1 , . . . , f n−1 cannot have n x consecutive zeros. Proof: Let i 1 , i 2 , . . . , i nx be the indices of the nonzero elements of x. Then the elements of the Fourier transform in the range k = m + 1, m + 2, . . . , m + n x are ∑ f k = √ 1 n x n j=1 x ij e −2πiki n j Note the use of i as √ −1 and the multiplication of the exponent by i j to account for the actual location of the element in the sequence. Normally, if every element in the sequence 340
was included, we would just multiply by the index of summation. Convert the equation to matrix form by defining z kj = 1 √ n exp(− 2πi n ki j) and write f = Zx. Actually instead of x, write the vector consisting of the nonzero elements of x. By its definition, x ≠ 0. To prove the lemma we need to show that f is nonzero. This will be true provided Z is nonsingular. If we rescale Z by dividing each column by its leading entry we get the Vandermonde determinant which is nonsingular. Theorem 10.7 Let n x be the number of nonzero elements in x and let n f be the number of nonzero elements in the Fourier transform of x. Let n x divide n. Then n x n f ≥ n. Proof: If x has n x nonzero elements, f cannot have a consecutive block of n x zeros. Since n x divides n there are n n x blocks each containing at least one nonzero element. Thus, the product of nonzero elements in x and f is at least n. Fourier transform of spikes prove that above bound is tight To show that the bound in Theorem 10.7 is tight we show that the Fourier transform of the sequence of length n consisting of √ n ones, each one separated by √ n − 1 zeros, is the sequence itself. For example, the Fourier transform of the sequence 100100100 is 100100100. Thus, for this class of sequences, n x n f = n. Theorem 10.8 Let S ( √ n, √ n) be the sequence of 1’s and 0’s with √ n 1’s spaced √ n apart. The Fourier transform of S ( √ n, √ n) is itself. Proof: Consider the columns 0, √ n, 2 √ n, . . . , ( √ n − 1) √ n. These are the columns for which S ( √ n, √ n) has value 1. The element of the matrix Z in the row j √ n of column k √ n, 0 ≤ k < √ n is z nkj = 1. Thus, for these rows Z times the vector S ( √ n, √ n) = √ n and the 1/ √ n normalization yields f j √ n = 1. For rows whose index is not of the form j √ n, the row b, b ≠ j √ n, j ∈ {0, √ n, . . . , √ n − 1}, the elements in row b in the columns 0, √ n, 2 √ n, . . . , ( √ n − 1) √ n are 1, z b , z 2b , . . . , z (√ n−1)b ) and thus f b = √ 1 n (1 + z b + z 2b · · · + z (√ n−1)b = √ 1 z √ nb −1 n = 0 since z b√n = 1 and z ≠ 1 z−1 . Uniqueness of l 1 optimization Consider a redundant representation for a sequence. One such representation would be representing a sequence as the concatenation of two sequences, one specified by its coordinates and the other by its Fourier transform. Suppose some sequence could be represented as a sequence of coordinates and Fourier coefficients sparsely in two different ways. Then by subtraction, the zero sequence could be represented by a sparse sequence. The representation of the zero sequence cannot be solely coordinates or Fourier coefficients. If y is the coordinate sequence in the representation of the zero sequence, then the Fourier portion of the representation must represent −y. Thus y and its Fourier transform would have sparse representations contradicting n x n f ≥ n. Notice that a factor of two comes in 341
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Foundations of Data Science ∗ Avr
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4.4 Branching Processes . . . . . .
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8.2.2 Structural properties of the
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12.4.7 Median . . . . . . . . . . .
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densities, discrete optimization, e
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2 High-Dimensional Space 2.1 Introd
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Also, if x and y are independent, t
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1 1 − ɛ Annulus of width 1 d Fig
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integral gives ∫ ∞ 0 e −r2 r
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The volume of the hemisphere below
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points onto the circle. In higher d
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Using the substitution 2z = y 2 , |
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For the nearest neighbor problem, i
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√ √ 2d+O(1) ≤ 2d + ∆2 −O(
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2.10 Bibliographic Notes The word v
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Exercise 2.9 A 3-dimensional cube h
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Exercise 2.26 Explain how the volum
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Exercise 2.40 Consider a non orthog
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1 Exercise 2.50 Use the probability
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This decomposition of A can be view
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3.3 Singular Vectors We now define
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orthonormal basis w 1 , w 2 , . . .
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A n × d = U n × r D r × r V T r
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3.6 Left Singular Vectors Theorem 3
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Lemma 3.10 (Analog of eigenvalues a
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Proof: Let A = r∑ σ i u i vi T i
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a i , let dist(a i , l) denote its
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such models. Mixture models are a v
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1. The best fit 1-dimension subspac
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This leads to the following theorem
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Thus, the maximum cut problem can b
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and this meets the claimed error bo
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(0,3) (1,1) (3,0) ⎛ M = ⎝ 1 1 0
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Exercise 3.18 Modify the power meth
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2. What percent of the Forbenius no
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City Bei- Tian- Shang- Chong- Hoh-
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5 10 15 20 25 30 35 40 4 9 14 19 24
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Prob(vertex has degree k) = ( ) n
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and thus k k ≤ n. Since k! ≤ k
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For x to be equal to zero, it must
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No items E(x) ≥ 0.1 At least one
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Our first task is to figure out wha
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√ Thus, the probability for all s
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Figure 4.7: A degree three vertex w
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ftp://ftp.cs.rochester.edu/pub/u/jo
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Size of frontier 0 ln n nθ n Numbe
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For a small number i of steps, the
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same or are disjoint, ⎛( n∑ )
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are O(ln n) in size and the expecte
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f(x) q p 0 f(f(x)) f(x) x Figure 4.
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may be infinite. That is, ∞∑ i=
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Property cycles 1/n giant component
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Keep in mind that the leading terms
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4.6 Phase Transitions for Increasin
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p(n) A symmetric argument shows tha
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simple. We supply some intuition be
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Proof: Say that t is a “busy time
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Consider a graph in which half of t
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problem contributes a minuscule err
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one. On the other hand, for δ = 1,
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for all δ greater than δ critical
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expected degree d i = 1, 2, 3 √ t
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Destination Figure 4.17: For d < 2,
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For at least half of the pairs of v
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S i+1 is (1 − 1 ) |Si| ≤ 1 −
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Exercise 4.7 Let f (n) be a functio
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Exercise 4.19 (Birthday problem) Wh
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Exercise 4.38 Consider a model of a
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Exercise 4.53 Consider graph 3-colo
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5 Random Walks and Markov Chains A
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A B C (a) A B C (b) Figure 5.1: (a)
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in time polynomial in n. A quantity
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5.2 Markov Chain Monte Carlo The Ma
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p(a) = 1 2 p(b) = 1 4 p(c) = 1 8 p(
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5 8 7 12 1 3 1 6 1 8 1 3 3,1 3,2 3,
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Figure 5.4: A network with a constr
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There is a simple interpretation of
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Let γ i = π 1 + π 2 + · · · +
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5.4.1 Using Normalized Conductance
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The Gaussian distribution on the in
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6 6 1 8 1 5 8 4 5 5 Graph with boun
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and then reaching a from y before r
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every vertex? Hitting time The hitt
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clique of size n/2 x y } {{ } n/2 F
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i ↑ ↓ ↑ ↓ ↑ j ↑ =⇒ i
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Theorem 5.13 Let G be an undirected
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0 1 2 3 4 12 20 Number of resistors
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1 2 4 Figure 5.11: Paths obtained f
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whereas, with k self-loops, the equ
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or p[I − (1 − α)A] = α (1, 1,
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Exercise 5.10 Let p be a probabilit
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i 1 i 2 R 1 R 2 R 3 Figure 5.13: An
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(a) 1 2 3 4 (b) 1 2 3 4 (c) 1 2 3 4
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Exercise 5.40 Suppose that the cliq
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Exercise 5.55 Using a web browser b
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will allow us to make mathematical
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Not spam { }} { { Spam }} { x 1 x 2
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1 ∨ x 8 = 1}, or more succinctly,
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described using O(k log d) bits: lo
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In fact, we can show this bound is
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y definition of w ∗ . Similarly,
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y x φ Figure 6.4: Data that is not
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Theorem 6.11 (Online to Batch via R
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function of concept class H, will g
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Corollary 6.16 (VC-dimension sample
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A sphere in d-dimensions is a set o
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then with probability ≥ 1 − δ,
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work on a variety of measures. One
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Since weight(0) = n, the total weig
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Stochastic Gradient Descent: Given:
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sleeping experts problem. Combining
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⎫ ⎪⎬ ⎪⎭ Each gate is conn
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W 1 W 2 W 1 W 2 W 3 (a) (b) Figure
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convolution pooling Image Convoluti
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discard separators in advance that
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6.14.2 Active learning Active learn
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6.16 Exercises Exercise 6.1 (Sectio
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√ L log(L(T )) |S|) will be at mo
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7 Algorithms for Massive Data Probl
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important to know if some popular s
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We now give an example of a 2-unive
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estimate of m. To obtain a coin tha
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expected value of the sum will be z
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δ have values in ±1. There are ex
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mean. (ii) There is “no free lunc
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⎡ ⎢ ⎣ A m × n ⎤ ⎡ ⎥
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One uses the primitive in Section ?
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would require s ≥ n, which clearl
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its that capture sufficient informa
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7.6 Exercises Algorithms for Massiv
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Counting Frequent Elements The Majo
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Exercise 7.30 Blast: Given a long s
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Preliminaries: We will follow the s
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B A Figure 8.1: Example where the n
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Lloyd’s algorithm: Start with k c
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8.2.5 k-means clustering on the lin
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a clustering that is ɛ-close to C
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Theorem 8.5 Assume A satisfies (c,
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probability is q (where, q < p). 32
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8.6.4 Spectral Clustering Algorithm
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But for l ≠ l ′ , by hypothesis
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4. the σ-interior of the clusters
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Figure 8.4: Example of a bipartite
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are likely to be balanced given tha
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Page 325 and 326: Now consider a very tall tree. If t
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Page 329 and 330: 10 Other Topics 10.1 Rankings Ranki
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Page 333 and 334: In the discrete case, x = [x 0 , x
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Page 375 and 376: 12 Appendix 12.1 Asymptotic Notatio
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Page 379 and 380: Gaussian and related integrals To v
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12.4.7 Median One often calculates
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The density of y is the unit varian
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Generation of random numbers accord
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Example: Consider flipping a coin 1
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Setting λ = ln(1 + δ) Prob ( s >
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12.5 Bounds on Tail Probability Aft
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Collect terms of the summation with
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The first integral is just the stan
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of a symmetric matrix, counting mul
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capture this situation. Now AA T =
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The above theorem tells us that the
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Important special cases are |x| 0 t
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Lemma 12.23 Let A be a symmetric ma
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etween vertices in different blocks
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∞∑ ∑ ix i = ∞ i=0 i=0 x d d
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Generating functions are useful for
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Thus, u n = (n − 1) u n−2 . The
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f(x) a c b Figure 12.3: Illustratio
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need to argue that no other sequenc
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1. How large must δ be if we wish
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Exercise 12.40 We are given the pro
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Index 2-universal, 240 4-way indepe
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Lagrange, 423 Law of large numbers,
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References [AK] Sanjeev Arora and R
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[MR95b] [MU05] [MV10] Rajeev Motwan
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