Foundations of Data Science

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4.8 Nonuniform and Growth Models of Random Graphs 4.8.1 Nonuniform Models So far we have considered the random graph G(n, p) in which all vertices have the same expected degree and showed that the degree is concentrated close to its expectation. However, large graphs occurring in the real world tend to have power law degree distributions. For a power law degree distribution, the number f(d) of vertices of degree d plotted as a function of d satisfies f(d) ≤ c/d α , where α and c are constants. To generate such graphs, we stipulate that there are f(d) vertices of degree d and choose uniformly at random from the set of graphs with this degree distribution. Clearly, in this model the graph edges are not independent and this makes these random graphs harder to analyze. But the question of when phase transitions occur in random graphs with arbitrary degree distributions is still of interest. In this section, we consider when a random graph with a nonuniform degree distribution has a giant component. Our treatment in this section, and subsequent ones, will be more intuitive without providing rigorous proofs. 4.8.2 Giant Component in Random Graphs with Given Degree Distribution Molloy and Reed address the issue of when a random graph with a nonuniform degree distribution has a giant component. Let λ i be the fraction of vertices of degree i. There ∑ will be a giant component if and only if ∞ i(i − 2)λ i > 0. i=0 To see intuitively that this is the correct formula, consider exploring a component of a graph starting from a given seed vertex. Degree zero vertices do not occur except in the case where the vertex is the seed. If a degree one vertex is encountered, then that terminates the expansion along the edge into the vertex. Thus, we do not want to encounter too many degree one vertices. A degree two vertex is neutral in that the vertex is entered by one edge and left by the other. There is no net increase in the size of the frontier. Vertices of degree i greater than two increase the frontier by i − 2 vertices. The vertex is entered by one of its edges and thus there are i − 1 edges to new vertices in the frontier for a net gain of i − 2. The iλ i in i (i − 2) λ i is proportional to the probability of reaching a degree i vertex and the i − 2 accounts for the increase or decrease in size of the frontier when a degree i vertex is reached. Example: Consider applying the Molloy Reed conditions to the G(n, p) model, and use p i to denote the probability that a vertex has degree i, i.e., to analog to λ i . It turns out that the summation ∑ n i=0 i(i − 2)p i gives value zero precisely when p = 1/n, the point at which the phase transition occurs. At p = 1/n, the average degree of each vertex is one and there are n/2 edges. However, the actual degree distribution of the vertices is binomial, where the probability that a vertex is of degree i is given by p i = ( n i) p i (1−p) n−i . n∑ We now show that lim i(i − 2)p i = 0 for p i = ( n i) p i (1 − p) n−i when p = 1/n. n→∞ i=0 114
Consider a graph in which half of the vertices are degree one and half are degree two. If a vertex is selected at random, it is equally likely to be degree one or degree two. However, if we select an edge at random and walk to a random endpoint, the vertex is twice as likely to be degree two as degree one. In many graph algorithms, a vertex is reached by randomly selecting an edge and traversing the edge to reach an endpoint. In this case, the probability of reaching a degree i vertex is proportional to iλ i where λ i is the fraction of vertices that are degree i. Figure 4.13: Probability of encountering a degree d vertex when following a path in a graph. lim n→∞ n∑ i=0 ( ) ( i ( n 1 i(i − 2) 1 − i n) 1 ) n−i n = lim n→∞ n∑ i=0 = 1 e lim n→∞ ≤ ∞∑ i=0 n(n − 1) · · · (n − i + 1) i(i − 2) n∑ i=0 i(i − 2) . i! i! n i (1 − 1 n ( n(n − 1) · · · (n − i + 1) n i(i − 2) i! n i n − 1 ) n ( 1 − 1 ) −i n ) i To see that ∞ ∑ i=0 i(i−2) i! = 0, note that ∞∑ i=0 ∞ i i! = ∑ i=1 ∞ i i! = ∑ i=1 ∞ 1 (i − 1)! = ∑ i=0 1 i! and Thus, ∞∑ i=0 i 2 ∞ i! = ∑ i=1 ∞ i (i − 1)! = ∑ ∞∑ i=0 i=0 i(i−2) i! = ∞ ∑ i + 1 i! i=0 = ∞∑ i=0 ∑ i 2 − 2 ∞ i! i=0 ∞ i i! + ∑ i = 0. i! i=0 ∞ 1 i! = 2 ∑ i=0 1 i! . 115
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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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Page 67 and 68: Exercise 3.18 Modify the power meth
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Page 75 and 76: Prob(vertex has degree k) = ( ) n
Page 77 and 78: and thus k k ≤ n. Since k! ≤ k
Page 79 and 80: For x to be equal to zero, it must
Page 81 and 82: No items E(x) ≥ 0.1 At least one
Page 83 and 84: Our first task is to figure out wha
Page 85 and 86: √ Thus, the probability for all s
Page 87 and 88: Figure 4.7: A degree three vertex w
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Page 91 and 92: Size of frontier 0 ln n nθ n Numbe
Page 93 and 94: For a small number i of steps, the
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Page 97 and 98: are O(ln n) in size and the expecte
Page 99 and 100: f(x) q p 0 f(f(x)) f(x) x Figure 4.
Page 101 and 102: may be infinite. That is, ∞∑ i=
Page 103 and 104: Property cycles 1/n giant component
Page 105 and 106: Keep in mind that the leading terms
Page 107 and 108: 4.6 Phase Transitions for Increasin
Page 109 and 110: p(n) A symmetric argument shows tha
Page 111 and 112: simple. We supply some intuition be
Page 113: Proof: Say that t is a “busy time
Page 117 and 118: problem contributes a minuscule err
Page 119 and 120: one. On the other hand, for δ = 1,
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Page 123 and 124: expected degree d i = 1, 2, 3 √ t
Page 125 and 126: Destination Figure 4.17: For d < 2,
Page 127 and 128: For at least half of the pairs of v
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Page 131 and 132: Exercise 4.7 Let f (n) be a functio
Page 133 and 134: Exercise 4.19 (Birthday problem) Wh
Page 135 and 136: Exercise 4.38 Consider a model of a
Page 137 and 138: Exercise 4.53 Consider graph 3-colo
Page 139 and 140: 5 Random Walks and Markov Chains A
Page 141 and 142: A B C (a) A B C (b) Figure 5.1: (a)
Page 143 and 144: in time polynomial in n. A quantity
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Page 151 and 152: Figure 5.4: A network with a constr
Page 153 and 154: There is a simple interpretation of
Page 155 and 156: Let γ i = π 1 + π 2 + · · · +
Page 157 and 158: 5.4.1 Using Normalized Conductance
Page 159 and 160: The Gaussian distribution on the in
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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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s 1 ∞ ∞ λ t ∞ edges and vert
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A clustering algorithm is consisten
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less than n, then richness is not r
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8.13 Exercises Exercise 8.1 Constru
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A B Figure 8.8: insert caption Exer
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9 Topic Models, Hidden Markov Proce
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Nonnegative matrix factorization (N
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This is a linear program. As we rem
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for t = 1 to T Prob(O 0 O 1 · ·
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a ij transition probability from st
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C 1 S 2 C 2 causes D 1 D 2 diseases
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x 1 + x 2 + x 3 x 1 + x 2 x 1 + x 3
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x 1 x 1 + x 2 + x 3 x 3 + x 4 + x 5
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the messages coming to a variable n
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y 2 y 3 x 2 x 3 y 1 x 1 x 4 y 4 y n
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two pieces of information, the valu
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graph. The vertices of Π are denot
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present and ask how correlated this
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Now consider a very tall tree. If t
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9.14 Exercises Exercise 9.1 Find a
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10 Other Topics 10.1 Rankings Ranki
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b . a . a b b b b . . b b a a b b .
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In the discrete case, x = [x 0 , x
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Figure 10.3: Some subgradients for
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Suppose ˜x 0 were another minimum.
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A T S A S is invertible. We will pr
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was included, we would just multipl
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position on genome trees = Phenotyp
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form a matrix, called the Hessian,
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The simplex algorithm is a classica
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This problem is NP-hard. One way to
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10.9 Exercises Exercise 10.1 Select
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Exercise 10.18 Repeat the above exe
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Lemma 11.1 If a dilation equation i
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The Haar Wavelet ⎧ ⎨ 1 0 ≤ x
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11.3 Wavelet Systems So far we have
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11.5 Conditions on the Dilation Equ
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the support of both sides of the eq
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and the wavelet functions are ortho
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11.7 Sufficient Conditions for the
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Example of orthogonality when wavel
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11.9 Designing a Wavelet System In
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3. f(x) = 1 2 f(2x) + 1 2 f(2x −
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12 Appendix 12.1 Asymptotic Notatio
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these equalities by derivatives.
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Gaussian and related integrals To v
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1 + x ≤ e x for all real x (1 −
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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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