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Unformatted text preview: Review Problems 4 iCME and MS&E Refresher Course Thursday, 16 September, 2010 The problems in this problem set concern a mouse randomly walking around a maze with n rooms. At every room in the maze, the mouse chooses one of the doors in the room uniformly at random and proceeds to the next room (so if a room has m doors, the probability of using any given door is m 1 ). Assume that every room is reachable in at most n steps from any other. 1. Describe the maze as a graph. What are the nodes? What are the edges? The adjacency matrix of a graph has a ij = 1 if and only if there is an edge between node i and node j . Solution :The set of nodes will be the rooms; V = { 1 , 2 , ··· ,n } . ( i,j ) is an edge if there is a door between room i and j . Thus the doors are represented by edges. The adjacency matrix a ij has nonzero entries when there is a door from room i to room j . 2. Let P = { p ij } be a matrix of probabilities so that p ij is the probability of going from room i to room j . Construct P from A , the adjacency matrix of the graph. Show e is an eigenvector of P . Solution :The total number of doors in room i is given by N i = n X j =1 a ij ; the probability of selecting a given door in room i is thus ( ∑ n j =1 a ij ) 1 . Thus the probability of going from room i to room j is given by: p ij = a ij ∑ n j =1 a ij . Define the diagonal matrix D with diagonal elements d i = ( ∑ n j =1 a ij ) 1 . Then the probability transi tion matrix P is given by P = DA. The i th, j th element of P is p ij . To show that e is an eigenvector of P , consider the following: ( Pe ) i = n X j =1 p ij · 1 = n X j =1 a ij ∑ n j =1 a ij = ∑ n j =1 a ij ∑ n j =1 a ij = 1 = ( e ) i ....
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This note was uploaded on 10/01/2011 for the course EE 221 taught by Professor Ee221a during the Spring '08 term at Berkeley.
 Spring '08
 ee221a

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