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 1 2 3 4 5
If the walk is on 0 it goes into 1.
Else it goes to i+1 or to i-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 all equation 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. Probability of the card to be in place i
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

10. Stopping time If all the cards have been upstairs its random
Check the lowes card. To go up by 1 G(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 Π so 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 hij the 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 u2 and 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.