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    Random Walks on Graphs

    Sarah Constantin

    April 4, 2012

    1 What is a graph?

    Vertices and edges;

    draw examples;

    Why do we care? Web (whats linked to what? Basis for PageRank) Electronic networksmore generally; Social networks (whos friends with whom?) epidemiology (whos been incontact with whom?)

    For every graph, theres an adjacency matrix that captures all the information of the graph.You can get a graph from its adjacency matrix and vice versa.

    2 Random walks

    An ant moves at random from one node to another neighbor; choice of neighbor is random.Markov chain: new location depends only on present location. No memory of the past.Matrix formulation: probability distribution over each row. Rows sum to 1, all entriesbetween 0 and 1. Transition probability matrix.

    3 Random walks and the heat equation

    Suppose we have a line graph one dimension. Suppose we have the uniform random walk equal probabilities of moving left and right. If we have a distribution u, a function onthe graph, letting the random walk step forward gives

    (Pu)(x) = 1/2u(x + 1) + 1/2u(x 1)

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    where x is a point on the graph, and x + 1 is the point directly to the right and x 1 isthe point directly to the left. If this were a continuous line instead of a line graph, thisequation would be

    (Pu)(x) = 1/2u(x + h) + 1/2u(x h)

    After k timesteps,

    (Pku)(x) = 1/2Pku(x + h) = 1/2Pku(x h)

    1/h(Pk+1u(x) Pku(x)) = 1/2h(Pku(x + h) Pku(x)) + 1/2h(Pku(x h) Pku(x))

    As h 0 we get the one-dimensional heat equation

    ut = 1/2uxx

    The discrete analog is

    du = 1/2(u(x + 1) 2u(x) + u(x 1))

    If we did this in two dimensions, the situation would be the same, except the heat equationwould be

    ut = 1/4u

    the Laplacian,2u

    x2+2u

    y2

    and the discrete analog will be -1 on x, 1/d on its neighbors. Now generalize to *any*graph: -1 on x, 1/d on the neighbors. Note: if the graph approximates the plane, thegraph Laplacian approximates the plane Laplacian. In fact this is also true of graphs that

    approximate other surfaces, but you need a more general Laplacian for surfaces.

    Takeaway: the Laplacian (and the heat equation) emerges from uniform random walks. Inthe continuous case, this gives heat diffusion on surfaces. Keep it discrete and it describesrandom walks on graphs.

    Averaging operator: in the steady-state case where the derivative of u is zero, each point isthe average of its neighbors. A function with zero Laplacian is called harmonic. Example:1/r (unit point charge at the origin). Basically, applying the Laplacian to a function isfMf where M is the averaging operator. Convolving with the Laplacian unsmoothsthe function (in fact, using a discrete Laplacian filter is how they do edge enhancement inimage processing. Its a high pass filter (a sharpener.)

    Definition of graph Laplacian: 1 on the diagonal, 1dudv

    for (u, v) if u and v are adjacent,

    0 otherwise. Its actually the negative of the second difference operator.

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    4 Eigenvalues and Eigenvectors

    Given a matrix A, if

    Av = vthen v is an eigenvector and is an eigenvalue.

    Every non-singular matrix can has n eigenvectors and n eigenvalues (perhaps with repeti-tion.)

    The set of all the eigenvalues is called the spectrum.

    (Stated without proof): every symmetric matrix has real eigenvalues.

    Every positive definite matrix has all positive eigenvalues. Change into eigenbasis:

    yty =

    iy2i

    nonnegative iff all i are nonnegative.

    (Stated without proof): symmetric matrices can be written as UU where is a diagonalmatrix with eigenvalues.

    We can tell a lot about the structure of a graph from the eigenvalues of the graph Lapla-cian.

    5 Properties of graph Laplacians

    Here, i are the eigenvalues of the Graph Laplacian. First note: two disjoint graphs haveblock-wise union of the Laplacians (direct sum.) So the spectrum of the union of disjointgraphs is the union of the spectra of the subgraphs.

    One edge: just (1, -1; -1, 1) around that edge. So only one eigenvalue, namely 2. Everythingelse is 0.

    Laplacian is a quadratic form sum of squared differences of neighbors. So this is positivesemidefinite: xTLx 0.

    xTLx =

    ij(xi xj)

    2

    Fact 0: the null space of the Laplacian is just where

    (xi xj)2 = 0 or all values of the

    function in each connected component are equal. The dimension of the null space is thenumber of connected components. If the graph is connected, the null space has dimensionone, so theres n-1 nonzero eigenvalues, so 2 > 0.

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    Fact 1: the complete graph has all equal eigenvalues of the Laplacian.

    (n 1)vi

    jivj = nvi

    vj = nvi

    Every vector orthogonal to the ones vector has sum of entries equal to 0 (thats whatorthogonal means.) so has eigenvalue n/(n 1).

    Fact 2. For graphs with no self-loops (A(v, v) = 0)

    i = n

    Proof: L has ones on the diagonal; trace is the sum on the diagonal which is n; this is alsothe sum of the eigenvectors. (Trace is multiplicative.)

    Fact 3: for n 2, 1 n

    n1 with equality iff the graph is complete. 0 = 0, so theinequality follows from fact 1. A complete graph has uniform spectrum

    Fact 4. IfG is not complete, then 1 1. Suppose a and b are not neighbors. Its enoughto construct a function f such that

    f(x)dx = 0

    (orthogonal to (d1, d2, . . . dn), the eigenvector associated with 0) and the Rayleigh quo-tient

    R(f) =fLf

    fDf= 1

    Second largest eigenvector. Candidate: f(x) = da if x = b, db if x = a. Clearlyorthogonal,

    R(f) =d2bda + d

    2adb

    d2b

    da + d2a

    db= 1

    Since 1 R(f), the claim follows.

    Fact 6: all the eigenvalues are less than or equal to 2.

    R(f) =

    xy(f(x) f(y))

    2

    f2(x)dx

    xy 2(f

    2(x) + f2(y))

    f2(x)dx

    =2

    x f2(x)dx

    x f2(x)dx

    = 2

    All the eigenvalues are less than or equal to the Rayleigh quotient.Fact 7: The graph G is bipartite iff n1 = 2. If equality exists in the above equation,then f(y) = f(x) for x y, so G is two-colorable, and so bipartite.

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