Foundations of Data Science

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∑ j c 1j = 1 Constraint forcing b 2 to have exactly one neighbor c 32 w 12 c 12 c 42 ∑ c 12 i c i2 = 1 ← → β(1, 2) α(1, 2) Constraint forcing c n2 a 1 to have exactly one neighbor Figure 9.7: Portion of factor graph for the maximum weight matching problem. ∑ j c 1j = 1 c 11 c 13 c 1n ∑ j c i2 = 1 ∑ i c i2 = 1 c 12 c 22 c 32 c n2 Figure 9.8: Tree for MWM problem. 320
graph. The vertices of Π are denoted by dotted circles in the above tree. Consider a set of trees where each tree has a root that corresponds to one of the constraints. If the constraint at each root is satisfied by the edge of the MWM, then we have found the MWM. Suppose that the matching at the root in one of the trees disagrees with the MWM. Then there is an alternating path of vertices of length 2k consisting of vertices corresponding to edges in Π and edges in Λ. Map this path onto the bipartite graph. In the bipartite graph the path will consist of a number of cycles plus a simple path. If k is large enough there will be a large number of cycles since no cycle can be of length more than 2n. Let m be the number of cycles. Then m ≥ 2k 2n = k n . Let π ∗ be the MWM in the bipartite graph. Take one of the cycles and use it as an alternating path to convert the MWM to another matching. Assuming that the MWM is unique and that the next closest matching is ε less, W π∗ − W π > ε where π is the new matching. Consider the tree matching. Modify the tree matching by using the alternating path of all cycles and the left over simple path. The simple path is converted to a cycle by adding two edges. The cost of the two edges is at most 2w* where w* is the weight of the maximum weight edge. Each time we modify Λ by an alternating cycle, we increase the cost of the matching by at least ε. When we modify Λ by the left over simple path, we increase the cost of the tree matching by ε − 2w∗ since the two edges that were used to create a cycle in the bipartite graph are not used. Thus weight of Λ - weight of Λ ′ ≥ k n ε − 2w∗ which must be negative since Λ ′ is optimal for the tree. However, if k is large enough this becomes positive, an impossibility since Λ ′ is the best possible. Since we have a tree, there can be no cycles, as messages are passed up the tree, each sub tree is optimal and hence the total tree is optimal. Thus the message passing algorithm must find the maximum weight matching in the weighted complete bipartite graph assuming that the maximum weight matching is unique. Note that applying one of the cycles that makes up the alternating path decreased the bipartite graph match but increases the value of the tree. However, it does not give a higher tree matching, which is not possible since we already have the maximum tree matching. The reason for this is that the application of a single cycle does not result in a valid tree matching. One must apply the entire alternating path to go from one matching to another. 9.12 Warning Propagation Significant progress has been made using methods similar to belief propagation in finding satisfying assignments for 3-CNF formulas. Thus, we include a section on a version of belief propagation, called warning propagation, that is quite effective in finding assignments. Consider a factor graph for a SAT problem. Index the variables by i, j, and 321
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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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Page 319: two pieces of information, the valu
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Page 353 and 354: Exercise 10.18 Repeat the above exe
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Page 359 and 360: 11.3 Wavelet Systems So far we have
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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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