Random WalksRandom Walks on GraphsRandom Walks on GraphsRandom Walks on GraphsRandom Walks on GraphsRandom Walks on GraphsRandom Walks on GraphsRandom walk on a lineRandom walk on a lineRandom walk on a lineRandom walk on a lineAnother way of looking at itRandom walks and electrical networksRandom walks and electrical networksElectrical networks save the day…Random walks and electrical networksGetting back homeGetting back homeCover timesWe will eventually get homeAn averaging argumentMarkov’s InequalityAn averaging argumentso let’s walk some more!The power of independence“3-regular” citiesGuidebookGuidebookGuidebookGuidebookUniversal Guidebooksdegree=2 n=3 graphsdegree=2 n=3 graphsdegree=2 n=3 graphsdegree=2 n=3 graphsUniversal Traversal sequencesProofProofProof (continued)Univeral Traversal SequencesBut here’s a randomized procedureAsideElectrical Networks againElectrical Networks againElectrical Networks againElectrical Networks againSuppose G is the graphPick a spanning tree of GRandom Walk on a lineUnbiased Random WalkUnbiased Random WalkSimple ClaimWe will return…How about a 2-d grid?How about a 2-d grid?How about a 2-d grid?How about a 2-d grid?How about a 2-d grid?in the 2-d walkWe will return (again!)…But in 3-dGreat Theoretical Ideas In Computer ScienceSteven Rudich, Anupam Gupta CS 15-251 Spring 2005Lecture 24 April 7, 2005 Carnegie Mellon UniversityRandom WalksRandom Walks on Graphs -Random Walks on Graphs -At any node, go to one of the neighbors of the node with equal probability.Random Walks on Graphs -At any node, go to one of the neighbors of the node with equal probability.Random Walks on Graphs -At any node, go to one of the neighbors of the node with equal probability.Random Walks on Graphs -At any node, go to one of the neighbors of the node with equal probability.Random Walks on Graphs -At any node, go to one of the neighbors of the node with equal probability.Let’s start simple…We’ll just walk in a straight line.Random walk on a lineRandom walk on a lineYou go into a casino with $k, and at each time step,you bet $1 on a fair game. You leave when you are broke or have $n.Question 1:what is your expected amount of money at time t?Let Xtbe a R.V. for the amount of money at time t.0nkRandom walk on a lineRandom walk on a lineYou go into a casino with $k, and at each time step,you bet $1 on a fair game. You leave when you are broke or have $n.0nXtXt= k + δ1+ δ2+ ... + δt,(δiis a RV for the change in your money at time i.)E[δi] = 0, since E[δi|A] = 0 for all situations A at time i.So, E[Xt] = k.Random walk on a lineRandom walk on a lineYou go into a casino with $k, and at each time step,you bet $1 on a fair game. You leave when you are broke or have $n.Question 2:what is the probability that you leave with $n ?0nkRandom walk on a lineRandom walk on a lineQuestion 2:what is the probability that you leave with $n ?E[Xt] = k.E[Xt] = E[Xt| Xt= 0] × Pr(Xt= 0) + E[Xt| Xt= n] × Pr(Xt= n) + E[ Xt| neither] × Pr(neither)As t →∞, Pr(neither) → 0, also somethingt< nHence Pr(Xt= n) → k/n.0+ n × Pr(Xt= n) + (somethingt× Pr(neither))Another way of looking at itAnother way of looking at itYou go into a casino with $k, and at each time step,you bet $1 on a fair game. You leave when you are broke or have $n.Question 2:what is the probability that you leave with $n ?= the probability that I hit green before I hit red.0nkRandom walks and electrical networksWhat is chance I reach green before red?-Same as voltage if edges are resistors and we put 1-volt battery between green and red.Random walks and electrical networks-•px= Pr(reach green first starting from x)•pgreen= 1, pred= 0• and for the rest px = Averagey2 Nbr(x)(py)Same as equations for voltage if edges all have same resistance!Electrical networks save the day…Electrical networks save the day…You go into a casino with $k, and at each time step,you bet $1 on a fair game. You leave when you are broke or have $n.Question 2:what is the probability that you leave with $n ?voltage(k) = k/n = Pr[ hitting n before 0 starting at k] !!!0nk1 volt0 voltsRandom walks and electrical networksWhat is chance I reach green before red?-Of course, it holds for general graphs as well…Let’s move on tosome other questions on general graphsGetting back home-Lost in a city, you want to get back to your hotel.How should you do this?Depth First Search:requires a good memory and a piece of chalkGetting back home-Lost in a city, you want to get back to your hotel.How should you do this?How about walking randomly?no memory, no chalk, just coins…Will this work?When will I get home?I have a curfew of 10 PM!Will this work?Is Pr[ reach home ] = 1?When will I get home?What is E[ time to reach home ]?I have a curfew of 10 PM!Relax, Bonzo!Yes,Pr[ will reach home ] = 1Furthermore:If the graph has n nodes and m edges, then E[ time to visit allnodes ] ≤ 2m × (n-1)E[ time to reach home ] is at most thisCover timesLet us define a couple of useful things:Cover time (from u)Cu= E [ time to visit all vertices | start at u ]Cover time of the graph:C(G) = maxu { Cu}Cover Time TheoremIf the graph G has n nodes and m edges, thenthe cover time of G isC(G) ≤ 2m (n – 1)Any graph on n vertices has < n2/2 edges.Hence C(G) < n3for all graphs G.First, let’s prove thatPr[ eventually get home ] = 1We will eventually get homeLook at the first n steps.There is a non-zero chance p1that we get home.Suppose we fail.Then, wherever we are, there a chance p2> 0that we hit home in the next n steps from there.Probability of failing to reach home by time kn= (1 – p1)(1- p2) … (1 – pk) → 0 as k → ∞In factPr[ we don’t get home by 2k C(G) steps ] ≤ (½)kRecall: C(G) = cover time of G ≤ 2m(n-1)An averaging argumentSuppose I start at u.E[ time to hit all vertices | start at u ] ≤ C(G)Hence,Pr[ time to hit all vertices > 2C(G) | start at u ] ≤ ½.Why?Else this average would be higher.(called Markov’s inequality.)Markov’s InequalityRandom variable X has expectation A = E[X].A = E[X] = E[X | X > 2A ] Pr[X > 2A]+ E[X | X ≤ 2A ] Pr[X ≤ 2A]≥ E[X | X > 2A ] Pr[X > 2A]Also, E[X | X > 2A]> 2A⇒ A ≥ 2A × Pr[X > 2A] ⇒ ½ ≥ Pr[X > 2A]Pr[ X exceeds k × expectation ] ≤ 1/k.An averaging argumentSuppose I start at u.E[ time to hit all vertices | start at u ] ≤ C(G)Hence, by Markov’s InequalityPr[ time to hit all vertices > 2C(G) | start at u ] ≤ ½.so let’s walk some more!Pr [ time to hit
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