Foundations of Data Science

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The Hare system implements this method by asking each voter to rank all the candidates. Then one counts how many voters ranked each candidate as number one. If no candidate receives a majority, the candidate with the fewest number one votes is dropped from each voters ranking. If the dropped candidate was number one on some voters list, then the number two candidate becomes that voter’s number one choice. The process of counting the number one rankings is then repeated. Although the Hare system is widely used it fails to satisfy Arrow’ axioms as all voting systems must. Consider the following situation in which there are 21 voters that fall into four categories. Voters within a category rank individuals in the same order. Category Number of voters in category Preference order 1 7 abcd 2 6 bacd 3 5 cbad 4 3 dcba The Hare system would first eliminate d since d gets only three rank one votes. Then it would eliminate b since b gets only six rank one votes whereas a gets seven and c gets eight. At this point a is declared the winner since a has thirteen votes to c’s eight votes. Now assume that Category 4 voters who prefer b to a move a up to first place. Then the election proceeds as follows. In round one, d is eliminated since it gets no rank one votes. Then c with five votes is eliminated and b is declared the winner with 11 votes. Note that by moving a up, category 4 voters were able to deny a the election and get b to win, whom they prefer over a. 10.3 Compressed Sensing and Sparse Vectors Given a function x(t), one can represent the function by the composition of sinusoidal functions. Basically one is representing the time function by its frequency components. The transformation from the time representation of a function to it frequency representation is accomplished by a Fourier transform. The Fourier transform of a function x(t) is given by ∫ f(ω) = x(t)e −2πωt dt Converting the frequency representation back to the time representation is done by the inverse Fourier transformation ∫ x(t) = f(ω)e −2πωt dω 332
In the discrete case, x = [x 0 , x 1 , . . . , x n−1 ] and f = [f 0 , f 1 , . . . , f n−1 ]. The Fourier transform and its inverse are f = Ax with a ij = ω ij where ω is the principle n th root of unity. There are many other transforms such as the Laplace, wavelets, chirplets, etc. In fact, any nonsingular n × n matrix can be used as a transform. If one has a discrete time sequence x of length n, the Nyquist theorem states that n coefficients in the frequency domain are needed to represent the signal x. However, if the signal x has only s nonzero elements, even though one does not know which elements they are, one can recover the signal by randomly selecting a small subset of the coefficients in the frequency domain. It turns out that one can reconstruct sparse signals with far fewer samples than one might suspect and an area called compressed sampling has emerged with important applications. Motivation Let A be an n × d matrix with n much smaller than d whose elements are generated by independent, zero mean, unit variance, Gaussian processes. Let x be a sparse d-dimensional vector with at most s nonzero coordinates, s
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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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Page 325 and 326: Now consider a very tall tree. If t
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Page 379 and 380: Gaussian and related integrals To v
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Thus, n! ≤ n n e −n√ ne. For
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from which the current inequality f
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Example: Let f (x) = x k for k an e
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Random variables x 1 , x 2 , . . .
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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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