Random Walks on Graphs: A Survey - Iowa State Universityorion.math.iastate.edu/rymartin/ISU608EGT/S16/Lazar_TalkSlides.pdf · Random Walk De nition Let G be a graph on n vertices

Random Walks on Graphs: A Survey

Robert Lazar

Department of MathematicsIowa State University

April 26, 2016

Our Mission

I Goals:� Clear understanding of Random Walks including basic

definitions� Find a few bounds for random walk properties� Learn a few concrete examples

I Tool: Random Walks on Graphs: A Survey (1993), L.Lovasz

Robert Lazar (ISU) 2 of 28

Outline

I Random Walk as a Markov Chain� Stationary Distribution� Time-Reversibility� Expected Return Time

I Eigenvalue Connection

I Main Parameters� Bound on Access Time

I Universal Traverse Sequence� Existence for d-regular Graph


Random Walk

Definition

Let G be a graph on n vertices and v0 be a vertex selected bysome probability distribution P0. A random walk on the graph Gwith starting vertex v0 is a sequence of vertices v0 = X0,X1 · · · ,where Xi+1 is uniformly selected from N(Xi ).

Definition

A Markov chain is a sequence of random variables X0,X1, · · ·with state space V , where the probability moving to variable Xi+1

only depends on variable Xi . A Markov chain is said to beirreducible if any state in the state space has a non-zeroprobability of traveling to another state. Let the transition ofprobabilities for the Markov Chain be M.


Random Walks Continued

Example

Let G be K3 with a leaf. Then

M =

0 1

313

13

12 0 1

2 012

12 0 0

1 0 0 0

M is row stochastic as it is the sum of all of its neighbors. We canalso define M in the following way:



Definition

The t-th step of the walk can be represented as Pt = (Mᵀ)tP0,where M = DAG is the matrix of transition probabilities of thisgraph G ,

mij =

{1

d(i) , if ij ∈ E ,

0, otherwise,

D is the diagonal matrix with dii = 1d(i) and P0 is some initial

distribution.

When G is d-regular we have

M = DA =1

dA.



Example

Continuing our example above with G being a K3 with a leaf.

AG =

0 1 1 11 0 1 01 1 0 01 0 0 0

,D =

13 0 0 00 1

2 0 00 0 1

2 00 0 0 1

.Then,

M = DAG =

0 1

313

13

12 0 1

2 012

12 0 0

1 0 0 0



Definition

The probability distribution P0 is stationary for the graph G ifP1 = P0.

In fact, if P0 is stationary, Pt = (Mᵀ)tP0 = P0.

Observation

For any graph G the distribution,

π(v) =d(v)

2m,

is stationary.

P0 is the matrix where∑n

i=1 d(vi ) = 2m and each row is12m (d(v1), d(v2), · · · , d(vn)). Clearly P0 is row stochastic and it isstationary as the entries in each row of M are 0 or 1

d(i) .



Is this stationary distribution unique and is there a way to derive it?

Remark

If the Graph G is connected then the stationary distribution isunique.

Theorem

Perron-Frobenius: For a non-negative irreducible square matrix,there exists a unique distribution π such that Pπ = π and∑π(i) = 1. As well as a real eigenvalue λ which has maximum

absolute value among all eigenvalues and multiplicity 1.

Since G is connected then a random walk corresponds to anirreducible Markov Chain. So by the Perron-Frobenius Theorem,the stationary distribution is unique.



Definition

A random walk is time-reversible if π(i)pij = π(j)pji . In otherwords, walking from i to j is the same as from j to i .

Example

Let M be from the example above with P0 = I ,

M =

0 1

313

13

12 0 1

2 012

12 0 0

1 0 0 0

.Since M is not symmetric this walk is not time-reversible.



Remark

If we require a random walk to be time-reversible then we obtainthe distribution π(i) = d(i)

2m .

Since pij = 1d(i) we have

π(i)

d(i)=π(j)

d(j)= k .

As π is row stochastic and∑

d(i) = 2m,

1 =∑

π(i) =∑

kd(i) = k∑

d(i) = k2m.

Therefore, we obtain k = 12m .



Hence,

π(i) =k

pij=

d(i)

2m

as obtained above. If G is d-regular, then∑

d(i) = dn and

π(i) =1

n,

the uniform distribution.



Remark

1) The expected return rate to an edge is 2m.

2) The expected return rate to a vertex is 2md(i) , if G is d-regular

the expected return rate is n.

Proof.

Let E be the expected value that we return to an edge/vertex withprobability 1

k . Then E = 1k + (1 + E )(1− 1

k ). Solving for E givesus E = k. For 1) we use k = 1

2m as each edge has equally likely

chance of being used. For 2) use k = d(i)2m .

Now, π will represent the unique stationary distribution. We willsee how this relates to eigenvalues.


Eigenvalue Connection

Theorem

For a connected non-bipartite graph G , the stationary distributionis also a limiting distribution.

Proof.

Recall that M = DAG does not always have to be symmetric.Consider

N = D1/2AGD1/2 = D−1/2MD1/2.

D is invertible as graph is connected. Eigenvalues of N are thesame as the eigenvalues of M. Since the matrix is real symmetric,we can order its eigenvalues as λ1 ≥ · · · ≥ λn with correspondingorthonormal eigenvectors v1, · · · , vn.


Eigenvalue Connection Continued

Proof.

By spectral decomposition,

N = V ᵀλV =n∑

i=1

λivᵀi vi .

Let w = (√

d(1), · · · ,√d(n))ᵀ,

Nw = D1/2AGD1/2w = D1/2AG1 = D1/2(d(1), · · · , d(n))ᵀ = 1w .

Therefore, w is an eigenvector of N with eigenvalue 1. ByPerron-Frobenius Theorem, 1 = λ1 > · · · ≥ λn ≥ −1. Hence,v1 = w√

2m, the normalized w eigenvector.

(D1/2v1vᵀ1D−1/2) = π.



Proof.

Then

M = D1/2ND−1/2

= (D1/2v1vᵀ1D−1/2 +

n∑i=2

λiD1/2vᵀi viD

−1/2

= π +n∑

i=2

λiD1/2vᵀi viD

−1/2.

Hence,

Mt = π +n∑

i=2

λtiD1/2vᵀi viD

−1/2.



Proof.

If G is bipartite then its spectrum is symmetric about the origin(Lovasz, Eigenvalues of graphs 2007). If G is bipartite

AG =

[0 BC 0

],

so the sum of the eigenvalues is 0. If G is non-bipartite, then bythe Perron-Frobenius theorem, λn > −1 hence

limt→∞

Mt = π +n∑

i=2

λtiD1/2vᵀi viD

−1/2 = π + 0 = π.


Main Parameters

Definition

The access time Hij is the expected number of steps before nodej is visited, starting from node i .

Definition

The cover time is the expected number of steps to reach everynode. If no starting node is specified we take the worst case.

Definition

The mixing rate is a measure of how fast the random walkconverges to its limiting distribution.


Access Time

Example

Determine the access time for two points of a path on n nodeslabeled 0, · · · , n − 1. Notice that H(k − 1, k) is one less than theexpected return time to vertex k. As calculated before, theexpected return time is

∑d(i)

d(k) . Since we have a path of length k ,∑d(i) = 2(k − 1) + 2 = 2k and d(k) = 1. Hence, the expected

return time is 2k so H(k − 1, k) = 2k − 1. Now,

H(i , k) =k∑

j=i+1

H(i , j) =k∑

j=i+1

2(j)− 1 = k2 − i2.


Cover Time

Example

Let G be the complete graph on n vertices with τi representing thefirst time i vertices have been visited. Then, τi+1 − τi representsthe amount of time for a new vertex to be visited. Now, theprobability we can visit a new vertex is n−i

n−1 since there are n − ivertices we have not visited and n − 1 vertices that are not ourinitial vertex. As above,

E (τi+1 − τi ) =1

n−in−1

=n − 1

n − i.

Therefore, the cover time is

E (τn) =n−1∑i=1

E (τi+1−τi ) =n−1∑i=1

n − 1

n − i= (n−1)

n−1∑i=1

1

i︸︷︷︸HarmonicSeries

≈ n log n


Cover Time Continued

In 1993 Feige was interested in the Coupon Collector Problem. Ifyou want to collect each of n different coupons, and you get arandom coupon every day in the mail how long do you have towait?

Theorem

(Feige 1993). The cover time of a regular graph on n nodes is atmost 2n2.

Remark

The access time is at most 2n2 since we have to cover the wholegraph in that amount of time.


Universal Traverse Sequence

Definition

Let G be a connected d-regular graph and v0 some starting vertex.Label the edges incident to each node as {1, · · · , d}. A traversesequence for the current label and starting vertex v0 is a sequence(h1, · · · , ht) ⊆ {1, · · · , d}t such that if we start a random walk atv0 then at the the i-th step we leave the edge labeled hi , we havevisited every node. A universal traverse sequence for a fixed nand d is a sequence which is a traverse sequence for every d-regulargraph on n vertices, every labeling of it, and every starting point.


Universal Traverse Sequence Continued

Example

K3. (1,2,2) is Universal while (1,1,1) is not. Let 0 represent thelabeling (1,2) and 1 represent the labeling (2,1). Then the 8labelings are (0,0,0),(0,0,1),(0,1,0),(1,0,0),(0,1,1),(1,1,0),(1,0,1)and (1,1,1).


Existence of Universal Traverse Sequence

Theorem

For every d ≥ 2 and n ≥ 3, there exists a universal traversesequence of length O(d2n3 log(n)).

Proof.

I Construct a random sequence

I Calculate probability that sequence is not traverse

I Use Bound on Cover Time

I Use Markov’s Inequality

I Conclude that there is at least one universal traverse sequence


Existence of Universal Traverse Sequence Cont.

Proof.

Let t = 8dn3 log n and H = (h1, · · · , ht) is a random walk. Theprobability that H is not traverse is bounded by the cover time,2n2. Let X be the event that we have not seen all nodes, byMarkov’s Inequality

P(X ≥ 4n2) ≤ E (X )

4n2<

2n2

4n2=

1

2.

We can think of 4n2 as a random walk, in fact we will have t4n2

more random walks that also should not cover the graph, eachhaving an independent probability of 1

2 . The probability that wehave not seen all vertices after t-steps is less than

2−t

4n2 = 2−2dndlog2 ne ≤ n−2dn.


Existence of Universal Traverse Sequence Cont.

Proof.

For a fixed vertex in a d-regular graph on n-vertices, there are(n − 1

d

)≤ nd

choices for its neighbors. So for each vertex we have at most nd

different edges hence there are at most ndn d-regular graphs withdifferent edge labelings. Combining our results, we have ndn

possible labelings, n possible starting vertices and probabilityn−(nd+2) that we have not seen all vertices after t steps.So the probability that H is not a traverse sequence for one ofthese graphs is

nndnn−(nd+2) = n−1 < 1,

as n ≥ 3. So there exists some universal traverse sequence.


Further Work

I The access time can be rewritten in terms of the spectraallowing for a deeper connection with the eigenvalues of thegraphs adjacency matrix

I Calculating the expected number of steps before two RandomWalks starting at different vertices collide

I Applications from electrical networks and statics can beapplied to obtain results about Random Walks

I Applications of the mixing rate


Thank You


Documents

Random Walks on Graphs: A Survey - Iowa State Universityorion.math.iastate.edu/rymartin/ISU608EGT/S16/Lazar_TalkSlides.pdf · Random Walk De nition Let G be a graph on n vertices