pagerank - Googles PageRank Algorithm Notes for Math 204...

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Google’s PageRank Algorithm Notes for Math 204 October 8, 2010 PageRank The original papers by Brin and Page [1, 2] describe the PageRank algorithm. The problem is to decide which pages are authoritative sources of information given certain terms. The PageRank algorithm assigns a a positive real number π i to each page with the properties that π i 0 and π 1 + ··· + π n = 1, where n is the number of webpages. In more advanced language the PageRank is a probability distribution over webpages. There are two ways to think of PageRank. We begin with the original definition due to Brin and Page. Let G = ( V,E ) be the webpage graph, where the vertices V are the individual webpages and E the edges. There is an edge from the vertex u to the vertex v if there is a hyperlink from u to v . Let B u be the set of pages that have links to u (back links), let F u be the pages to which u has a link (forward links), and let N u be the size of F u , which is the number of pages to which u links. Each page is assigned a PageRank R ( u ) by the following equation R ( u ) = X v B u R ( v ) N v . (1) A simple interpretation is that each page distributes its PageRank equally amongst all pages to which it points. In order to make this a probability distribution we normalize so that v V R ( v ) = 1 and R ( v ) 0. We are immediately faced with two problems. First it appears as though in order to compute the PageRank we need to know the PageRank, since this quantity appears on both sides of the equation. Let us number the vertices from 1 to n and denote the PageRank of the k th page by R ( k ). Let W be the matrix defined by w i,j = ( 1 /N i if i j 0 otherwise (2) Then the equation in (1) can be written R ( j ) = n X i =1 R ( i ) N i = n X i =1 R ( i ) w i,j . If we let R be the row vector with entries R (1) ,...,R ( n ), then the above equation can be written R = RW , where W is an n × n matrix. This can be further rewritten as R ( I n - W ) = 0. This of course is a linear system and therefore has at least one solution R = 0. This solution does not satisfy any of the conditions we need. The hard part is determine why there should be a non-zero solution with non-negative entries. Before
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This document was uploaded on 10/29/2011 for the course MATH 204 at Vanderbilt.

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pagerank - Googles PageRank Algorithm Notes for Math 204...

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