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Random walks on undirected graphs and a little bit about Markov Chains

Explore the world of random walks on graphs and delve into Markov Chains. Understand 2-SAT problems, algorithms, and Markov inequality. Discover fast random walks on undirected graphs and the fundamental theorem of Markov chains.

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Random walks on undirected graphs and a little bit about Markov Chains

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  1. Random walks on undirected graphs and a little bit about Markov Chains Guy

  2. One dimensional Random walk • A random walk on a line. • A line is n 0 1 2 3 4 5 If the walk is on 0 it goes into 1. Else it goes to i+1 or toi-1 with probability 1/2 What is the expected number of steps to go to n?

  3. The expected time function • T(n)=0 • T(i)=1+(T(i+1)+T(i-1))/2, i0 • T(0)=1+T(1) • Add allequation gives T(n-1)=2n-1. • From that we get:T(n-2)=4n-4 • T(i)=2(n-i)n-(n-i)2 • T(0)=n2

  4. 2-SAT: definition • A 2-CNF formula: • (x+ ¬y)& (¬ x+ ¬ z) &(x+y) &(z+¬ x) • (x+y) and the others are called CLAUSES • CNF means that we want to satisfy all of them • (x+ ¬y) is satisfied if X=T or y=F. • The question: is there a satisfying assignment? 2n is not a number the computer can run even for n=70

  5. Remark: the case of two literals very special • If three literals per clause: NPC • Not only that, but can not be approximated better than 7/8 • Obviously for (x+¬y+¬z) if we draw a random value the probability that the clause is satisfied is 1-1/8=7/8. Best approximation possible • Also hard 3-SAT(5)

  6. The random algorithm for 2-SAT • 2-SAT : Start with an arbitrary assignment • Let C be a non satisfied clause. Choose at random one of the two literals of C and flip its value. • We know that if the variables are x1 and x2 the optimum disagrees with us on x1 or on x2. • Distance to OPT: with probability ½ smaller by 1 and with probability ½ larger by 1 (worst case). Thus E(RW)≤n2

  7. RP algorithm (can make a mistake) • If you do these changes 2n2 times the probability that we do not get a truth assigment is ½ if one exists • You can do n4 and the probability that if there is a truth assignment we don’t find it is 1/n2 • What we use here is called the Markov inequality that states there are at most 1/3 of the numbers in a collection of numbers that are at least 3 times the average • There are several deterministic algorithms for 2-SAT

  8. Shuffling cards • Take the top card and place it with a random place (including first place). One of n. • A simple claim: if a card is in a random place it will be in a random place after next move. • We know Pr(card is in place i)=1/n • For the card to be in i after next step three possibilities

  9. One possibility: it is not the first card and it is in place i. Chooses one of i-1 first places. Second possibility (disjoint events) its at place 1 and exchanged with i. Third possibility: it is in place i+1 and the first card is placed in one of the places after i+1 1/n*(i-1)/n+1/n*1/n+1/n*1/(n-i)=1/n Probability of the card to be in place i

  10. Stopping time • If all the cards have been upstairs its random • Check the lowes card. To go up by 1G(1/n) thus expectation n. • To go from second last to third last G(2/n) and expectation n/2. • This gives n+n/2+n/3+….= n(ln n+Ө(1)) • FAST

  11. Random Walks on undirected graphs • Given a graph choose a neigbor at random with probability 1/d(v)

  12. Random Walks • Given a graph choose a vertex at random.

  13. Random Walks • Given a graph choose a vertex at random.

  14. Random Walks • Given a graph choose a vertex at random.

  15. Random Walks • Given a graph choose a vertex at random.

  16. Random Walks • Given a graph choose a vertex at random.

  17. Random Walks • Given a graph choose a vertex at random.

  18. Markov chains • Generalization of random walks on undirected graph. • The graph is directed • The sum over the values of outgoing edges is 1 but it does not need to be uniform. • We have a matrix P=(pij) with pij is the probability that on state i it will go to j. • Say that for Π0 =(x1,x2,….,xn)

  19. Changing from state to state • Probability(state=i)=j xj * pji • This is the inner product of the current state and column j of the matrix. • Therefore if we are in distribution Π0 after one step the distribution is at state: Π1= Π0 *P • And after i stages its in Πi= Π0 *Pi • What is a steady state?

  20. Steady state • Steady state is Πso that: Π*P= Π. For any i and any round the probability that it is in i is the same always. • Conditions for convergence: The graph has to be strongly connected. Otherwise there may be many components that have no edges out. No steady state. • hii the time to get back to i from i is finite • Non periodic. Slightly complex for Markov chains. For random walks on undirected graphs: not a bipartite graph.

  21. The bipartite graph example • If the graph is bipartite and V1 and V2 are its sets then if we start with V1 we can not be on V1 vertex in after odd number of transitions. • Therefore a steady state is not possible. • So we need for random walks on graphs that the graph is connected and not bipartite. The other property will follow.

  22. Fundamental theorem of Markov chains • Theorem: Given an aperiodic MC so that hii is not infinite for any i, and non-reducible Markov chain than: 1) There is a unique steady state. Thus to find the steady state just find Π*P= Π 2) hii=1/Πi . Geometric distribution argument. Remark: the mixing time is how fast the chain gets to (very close to) the steady state.

  23. Because its unique you just have to find the correct Π • For random walks in an undirected graph we claim that the steady state is: (2d1/m,2d2/m,……,2dm/m) • It is trivial to show that this is the steady state. Multiply this vector and the i column. The only important ones are neighbors of i. So sum(j,i)E 1/dj*(2dj/m)=2di/m

  24. The expected time to visit all the vertices of a graph • A matrix is doubly stochastic if and only if all columns also sum to 1. • Exercise (very simple): doubly stochastic then {1/n} or uniform is the steady state. • Define a new Markov chain of edges with directions which means 2m states. • The walk is defined naturally. • Exercise: show that the Matrix is doubly stochastic

  25. The time we spend on every edge • By the above, we spend the same time on every edge in the two directions over all edges (of course you have to curve the noise. We are talking on a limit here). • Now, say that I want to bound hijthe expected time we get from i to j.

  26. Showing hij+hji≤ 2m • Assume that we start at i--->j • By the Markov chain of the edges it will take 2m steps until we do this move again • Forget about how you got to j (no memory). • Since we are doing now i---->j again we know that: a) As it was in j, it returned to i. This is half the inequality hji b) Now it goes i----> j. This takes at most hij c) Since this takes at most 2m the claim follows.

  27. Consider a spanning tree and a walk on the spanning tree • Choose any paths that traverses the tree so that the walk goes from a parent to a child once and back once. • Per parent child we have hij+hji≤2m • Thus over the n-1 edges of the graph the cover time is at most 2m(n-1)<n3

  28. Tight example: the clique n/2 vertices u1 un/2 u2

  29. Bridges • For bridges hij+hji=2m (follows from proof). • Say you are the intersection vertex u1 of the clique and of the path. Its harder to go right than left. • Thus it takes about n2 time to go from u1 to u2and the same time to go from u2 to u3 etc. • This gives Ω(n3) • Funny name: Lollipop graph.

  30. Exponential cover time for directed graphs

  31. Universal sequences • Say that the graph is d-regular. • And the edges of every vertex are denoted by 1,2,….,d. • A walks series for d=20 would be 7,14,12,19,3,4,8,6…….. • The question, will it cover all graphs? • We show that if the series is large enough it covers all graphs.

  32. The length of the sequence • Let L=4dn2(dn+2)(log n+1) • What is the probability that a graph is covered? • Let N=4dn2, and think of the above as (dn+2)(ln n+1) different parts of sequences of length N. • By previous proof a graph will be cover in expectation in 2n2d steps. With probability at most ½ it is not covered in N steps.

  33. The number of d regular graphs • From n2 possible edges choose n*d • Thus we have to choose n*d elements and this is bounded by n2n*d graphs. • The probability that a graph is not covered is exp(-nd*lg n) • The union bound implies that with large probability all graphs are covered.

  34. One advise for every n • And with this one advice we can cover all graphs with high probability. • Thus we can find out for example if s and t are in the same connected components. • Later it was proved that with O(log n) space you can check if s and t are in the same connected component!! Even though you can remember O(1) items only.

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