COMPSCI 514: Algorithms for Data Science
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1 COMPSCI 514: Algorithms for Data Science Arya Mazumdar University of Massachusetts at Amherst Fall 2018
2 Lecture 4 Markov Chain & Pagerank
3 Homework Announcement Show your work in the homework Write the full names and ids of all the people in the group The writing should be in latex/legible. If the graders have trouble reading what you wrote you may be penalized. For the coding part, attach an explanation file with results, pseudocode and description of what is done. The best code is going to be distributed in the class as a sample. The graders are instructed to deduct points if you do not adhere to these rules Announce your full group name in Piazza (refer to the that TAs sent) by Friday.
4 Markov Chain A Markov Chain consists of a set of states For each pair of states i and j, there is a transition probability p ji of going from state i to state j For each i, j p ji = 1. P = {p ij } probability transition matrix Suppose, q(t) = (q 1 (t), q 2 (t),..., q n (t)) T is a vector containing the probabilities of states at time t At time t + 1, q(t + 1) = Pq(t).
5 Markov Chain The probably transition matrix induces a directed graph on the states The chain is connected (or the graph is strongly connected) if there is a directed path from any vertex to any other vertex A B C (a) A B C (b) Figure 4.1: (a) A directed graph with vertices having no out out edges and a strongly connected component A with no in edges. (b) A directed graph with three strongly connected components. courtesy: textbook
6 Markov Chain: Stationary distribution q(t) probability of the states after t steps Long term average: a(t) = 1 t (q(0) + q(1) + + q(t 1)) (Fundamental Theorem of Markov Chains) For a connected Markov chain there is a unique probability vector p satisfying Pp = p. Moreover, for any starting distribution, lim t a(t) exists and equals p
7 Markov Chain: Stationary distribution How? First, a(t) is a valid probability vector. Pa(t) a(t) = 1 ( ) P(q(0) + q(1) + + q(t 1)) (q(0) + q(1) + + q(t 1)) t = 1 ( ) (q(1) + q(2) + + q(t)) (q(0) + q(1) + + q(t 1)) t = 1 (q(t) q(0)). t Therefore, Pa(t) a(t) 1 = 1 t q(t) q(0) 1 1 t ( q(t) 1+ q(0) 1 ) = 2 t 0 as t.
8 Markov Chain: Stationary distribution Pa(t) a(t) 1 = 1 t q(t) q(0) 1 1 t ( q(t) 1+ q(0) 1 ) = 2 t 0 as t. lim t a(t) is an eigenvector of P. Still need to show that this is the unique eigenvector with eigenvalue 1. Prove that at home (also in the textbook).
9 Pagerank: A ranking method for the pages in the web Content can be manipulated Reputation cannot! This is the main reason to create a data matrix that depends on the link-structure Incoming links are votes for a webpage Links from important pages count more Each incoming link s vote is proportional to the importance of its source page
10 Pagerank 166 CHAPTER 5. LINK ANALYSIS by modifications that are necessary for dealing with some real-world problems concerning the structure of the Web. If page i with importance v i has d out-links, each link gets v i Think of the Web as a directed graph, where pages are the nodes, and there d votes is an arc from page p 1 to page p 2 if there are one or more links from p 1 to p 2. Figure 5.1 is an example of a tiny version of the Web, where there are only four Page pages. i s Pageown A hasimportance links to each ofis the the other sum threeof pages; thepage votes B has onlinks its to in-links A and D only; page C has a link only to A, andpaged has links to B and C only. A B C D Figure 5.1: A hypothetical example of the Web v A = v B /2 + v C ; v B = v A /3 + v D /2; v C = v A /3 + v D /2; v D = v A /3 + v B /2 courtesy: mmds.org Suppose a random surfer starts at page A in Fig There are links to B, C, andd, so this surfer will next be at each of those pages with probability 1/3, and has zero probability of being at A. ArandomsurferatB has, at the next step, probability 1/2 of being at A, 1/2 of being at D, and 0 of being at B or C.
11 by modifications that are necessary for dealing with some real-world problems concerning the structure of the Pagerank Web. Think of the Web as a directed graph, where pages are the nodes, and there is an arc from page p 1 to page p 2 if there are one or more links from p 1 to p 2. The behavior of a random surfer indicates which pages users of the Figure 5.1 is an example of a tiny version of the Web, where there are only four Web pages. are likely Page Atohas visit. links to Users each of are the more other three likely pages; topage visit B useful has linkspages to A and D only; C has a link only to A, andpaged has links to B and C than useless pages. only. A B C D The transition matrix Figure 5.1: A hypothetical example of the Web Suppose a random surfer starts at page 1 A in Fig There are links to B, C, andd, so this surfer will next 0be at2 each1 of those 0 pages with probability 1/3, and has zero probability of being 1 at A. ArandomsurferatB 1 has, at the next step, probability 1/2P of= being at A, 1/2 of being 2 at D, and 0 of being at B or C In general, we can define the transition 1 1 matrix of the Web to describe what happens to random surfers after one 3 step This matrix M has n rows and columns, if there are n pages. The element m ij in row i and column j has value
12 Pagerank Network flow equation: Pv = v If a random surfer starts with probability vector v 0 (probabilities of being in any website, say uniform) then v 1 = Pv 0 v 2 = Pv 1 = P 2 v 0 v 3 = Pv 2 = P 3 v 0 v will converge to the unique stationary distribution of the Markov chain, if the graph is strongly connected (no dead ends, etc). Can be computed by the power method. Which webpage do you guess will have the highest probability of a random surfer being there?
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