Laplacian and Random Walks on Graphspeople.math.sc.edu/lu/talks/nankai_2014/spec_nankai_2.pdf ·...

Laplacian and RandomWalks on Graphs

Linyuan Lu

University of South Carolina

Selected Topics on Spectral Graph Theory (II)Nankai University, Tianjin, May 22, 2014

Five talks

Laplacian and Random Walks on Graphs Linyuan Lu – 2 / 59

Selected Topics on Spectral Graph Theory

1. Graphs with Small Spectral RadiusTime: Friday (May 16) 4pm.-5:30p.m.

2. Laplacian and Random Walks on GraphsTime: Thursday (May 22) 4pm.-5:30p.m.

3. Spectra of Random GraphsTime: Thursday (May 29) 4pm.-5:30p.m.

4. Hypergraphs with Small Spectral RadiusTime: Friday (June 6) 4pm.-5:30p.m.

5. Lapalacian of Random HypergraphsTime: Thursday (June 12) 4pm.-5:30p.m.

Backgrounds

Linear Algebra

Graph Theory

Probability Theory

I: Spectral Graph Theory II: Random Graph TheoryIII: Random Matrix Theory

Outline

■ Combinatorial Laplacian

■ Normalized Laplacian

■ An application

Graphs and Matrices

There are several ways to associate a matrix to a graph G.

■ Adjacency matrix

Graphs and Matrices

■ Normalized Laplacian

Basic Graph Notation

■ G = (V,E): a simple connected graph on n vertices

■ G = (V,E): a simple connected graph on n vertices■ A(G): the adjacency matrix

■ G = (V,E): a simple connected graph on n vertices■ A(G): the adjacency matrix■ D(G) = diag(d1, d2, . . . , dn): the diagonal degree matrix

■ G = (V,E): a simple connected graph on n vertices■ A(G): the adjacency matrix■ D(G) = diag(d1, d2, . . . , dn): the diagonal degree matrix■ L = D − A: the combinatorial Laplacian

■ G = (V,E): a simple connected graph on n vertices■ A(G): the adjacency matrix■ D(G) = diag(d1, d2, . . . , dn): the diagonal degree matrix■ L = D − A: the combinatorial Laplacian■ L is semi-definite and 1 is always an eigenvector for the

eigenvalue 0.

① ① ①

L(S4) =

3 −1 −1 −1−1 1 0 0−1 0 1 0−1 0 0 1

eigenvalue 0.

① ① ①

L(S4) =

3 −1 −1 −1−1 1 0 0−1 0 1 0−1 0 0 1

Combinatorial Laplacian eigenvalues of S4: 0, 1, 1, 4.

Matrix-tree Theorem

Kirchhoff’s Matrix-tree Theorem: The (i, j)-cofactor ofD − A equals (−1)i+jt(G), where t(G) is the number ofspanning trees in G.

Matrix-tree Theorem

Kirchhoff’s Matrix-tree Theorem: The (i, j)-cofactor ofD − A equals (−1)i+jt(G), where t(G) is the number ofspanning trees in G.

Proof: Fix an orientation of G, let B be the incidencematrix of the orientation, i.e., bve = 1 if v is the head of thearc e, bve = −1 if i is the tail of e, and bve = 0 otherwise.Let L11 be the sub-matrix obtained from L by deleting thefirst row and first column, and B1 be the matrix obtainedfrom B by deleting the first row. Then L11 = B1B

det(L11) = det(B1B′1)

det(BS)2 By Cauchy-Binet formula

= the number of Spanning Trees. �

An application

Corollary: If G is connected, and λ1, . . . , λn−1 be thenon-zero eigenvalues of L. Then the number of spanningtree is

nλ1λ2 · · ·λn−1.

An application

Corollary: If G is connected, and λ1, . . . , λn−1 be thenon-zero eigenvalues of L. Then the number of spanningtree is

nλ1λ2 · · ·λn−1.

Chung-Yau [1999]: The number of spanning trees in anyd-regular graph on n vertices is at most

(1 + o(1))2 log n

dn log d

(d− 1)d−1

(d2 − 2d)d/2−1

This is best possible within a constant factor.

Normalized Laplacian

Normalized Laplacian: L = I −D−1/2AD−1/2.

■ L is always semi-definite.■ 0 is always an eigenvalue of L with eigenvector

(√d1, . . . ,

√dn)

■ L is always semi-definite.■ 0 is always an eigenvalue of L with eigenvector

(√d1, . . . ,

√dn)

① ① ①

L(S4) =

1 − 1√

3− 1

− 1√

31 0 0

− 1√

30 1 0

− 1√

30 0 1

(Normalized) Laplacian eigenvalues of S4:λ0 = 0, λ1 = λ2 = 1, λ3 = 2.

General properties:

■ The multiplicity of 0 is the number of connectedcomponents.

General properties:

■ Laplacian eigenvalues: λ0, . . . , λn−1

0 = λ0 ≤ λ1 ≤ · · · ≤ λn−1 ≤ 2.

General properties:

0 = λ0 ≤ λ1 ≤ · · · ≤ λn−1 ≤ 2.

■ λn−1 = 2 if and only if G is bipartite.

General properties:

0 = λ0 ≤ λ1 ≤ · · · ≤ λn−1 ≤ 2.

■ λn−1 = 2 if and only if G is bipartite.

■ λ1 > 1 if and only if G is the complete graph.

Rayleigh quotients

The Laplacian eigenvalues can also be computed by Rayleighquotients: for 0 ≤ i ≤ n− 1,

λi = supdim(M)=n−i

inff∈M

x∼y(f(x)− f(y))2∑

x f(x)2dx

Rayleigh quotients

inff∈M

x∼y(f(x)− f(y))2∑

x f(x)2dx

In particular, λ1 can be evaluated by

λ1 = inff⊥D1

x∼y(f(x)− f(y))2∑

x f(x)2dx

Rayleigh quotients

inff∈M

x∼y(f(x)− f(y))2∑

x f(x)2dx

In particular, λ1 can be evaluated by

λ1 = inff⊥D1

x∼y(f(x)− f(y))2∑

x f(x)2dx

An important parameter

λ1 is related to

■ the mixing rate of random walks

λ1 is related to

■ diameter

λ1 is related to

■ diameter

■ neighborhood/edge expansion

λ1 is related to

■ diameter

■ Cheeger constant

λ1 is related to

■ diameter

■ quasi-randomness

λ1 is related to

■ diameter

■ quasi-randomness

■ many other applications.

Random walks

A walk on a graph is a sequence of vertices together asequence of edges:

v0, v1, v2, v3, . . . , vk, vk+1, . . .

v0v1, v1v2, v2v3, . . . , vkvk+1, . . .

Random walks

A walk on a graph is a sequence of vertices together asequence of edges:

v0, v1, v2, v3, . . . , vk, vk+1, . . .

v0v1, v1v2, v2v3, . . . , vkvk+1, . . .

Random walks on a graph G:

fk+1 = fkD−1A.

D−1A ∼ D−1/2AD−1/2 = I−L.λ determines the mixing rate ofrandom walks. ✍✌

✎☞v ✍✌

✎☞

✍✌✎☞

✲��✒✻

Convergence

■ row vector fk: the vertex probability distribution at timek.

fk = f0(D−1A)k.

Convergence

fk = f0(D−1A)k.

■ Stationary distribution π = 1vol(G)(d1, d2, . . . , dn).

π(D−1A) = π.

Convergence

fk = f0(D−1A)k.

π(D−1A) = π.

■ Mixing:

‖(fk − π)D−1/2‖ ≤ λk‖(f0 − π)D−1/2‖.

Convergence

fk = f0(D−1A)k.

π(D−1A) = π.

■ Mixing:

‖(fk − π)D−1/2‖ ≤ λk‖(f0 − π)D−1/2‖.

If G is bipartite, then the random walk does not mix. Inthis case, we will use the α-lazy random walk with thetransition matrix αI + (1− α)D−1A.

Diameter

Suppose that G is not a complete graph. Then the diameterof G satisfies

diam(G) ≤⌈

log(vol(G)/δ)

log λn−1+λ1

λn−1−λ1

where δ is the minimum degree of G.

Edge discrepancy

Let vol(X) =∑

x∈X dx and λ = max{1− λ1, λn−1 − 1}.Then

∣|E(X, Y )| − vol(X)vol(Y )

vol(G)

∣≤ λ

√vol(X)vol(Y )vol(X)vol(Y )

vol(G) .

E(X,Y)

Let 1X and 1Y be the indicated vector of X and Yrespectively. Let α0, α1, . . . , αn−1 be orthogonal uniteigenvectors of L. Write D1/2

1X =∑n−1

i=0 xiαi and

D1/21Y =

∑n−1i=0 yiαi. Then

|E(X, Y )| = 1′XA1Y

= (D1/21X)

′(I − L)D1/21Y

=n−1∑

(1− λi)xiyi.

continue

Note x0 =vol(X)√vol(G)

and y0 =vol(Y )√vol(G)

. Hence,

|E(X, Y )| − vol(X)vol(Y )

vol(G)

(1− λi)xiyi

≤ λ

n−1∑

vol(X)vol(Y )vol(X)vol(Y )

vol(G). �

Cheeger Constant

For a subset S ⊂ V , we define

hG(S) =|E(S, S)|

min(vol(S), vol(S)).

The Cheeger constant hG of a graph G is defined to behG = minS hG(S).

Cheeger Constant

For a subset S ⊂ V , we define

hG(S) =|E(S, S)|

min(vol(S), vol(S)).

The Cheeger constant hG of a graph G is defined to behG = minS hG(S).

Cheeger’s inequality:

2hG ≥ λ1 ≥h2G

d-regular graph

If G is d-regular graph, then adjacency matrix, combinatorialLaplacian, and normalize Laplacian are all equivalent.

d-regular graph

Suppose A has eigenvalues µ1, . . . , µn. Then

■ D − A has eigenvalues d− µ1, . . . , d− µn.

■ I −D−1/2AD−1/2 has eigenvalues1− µ1/d, . . . , 1− µn/d.

d-regular graph

Suppose A has eigenvalues µ1, . . . , µn. Then

■ D − A has eigenvalues d− µ1, . . . , d− µn.

■ I −D−1/2AD−1/2 has eigenvalues1− µ1/d, . . . , 1− µn/d.

The theories of three matrices apply to the d-regular graphs.

An application

Constructing Small Folkman Graphs.

Ramsey’s theorem

For integers k, l ≥ 2, there exists a least positive integerR(k, l) such that no matter how the complete graph KR(k,l)

is two-colored, it will contain a blue subgraph Kk or a redsubgraph Kl.

Ramsey’s theorem

For integers k, l ≥ 2, there exists a least positive integerR(k, l) such that no matter how the complete graph KR(k,l)

is two-colored, it will contain a blue subgraph Kk or a redsubgraph Kl.

R(3, 3) = 6

R(4, 4) = 18

43 ≤ R(5, 5) ≤ 49

102 ≤ R(6, 6) ≤ 165

(1+o(1))

en2n/2 ≤ R(n, n) ≤ (n−1)−C log(n−1)

log log(n−1)

2(n− 1)

n− 1

Spencer[1975] Colon [2009]

Ramsey number R(3, 3) = 6

■ If edges of K6 are 2-colored then there exists amonochromatic triangle.

Ramsey number R(3, 3) = 6

■ If edges of K6 are 2-colored then there exists amonochromatic triangle.

■ There exists a 2-coloring of edges of K5 with nomonochromatic triangle.

Rado’s arrow notation

G → (H): if the edges of G are 2-colored then there exists amonochromatic subgraph of G isomorphic to H.

K6 → (K3)K5 6→ (K3)

Rado’s arrow notation

G → (H): if the edges of G are 2-colored then there exists amonochromatic subgraph of G isomorphic to H.

K6 → (K3)K5 6→ (K3)

Fact: If K6 ⊂ G, then G → (K3).

An Erdos-Hajnal Question

Is there a K6-free graph G with G → (K3)?

An Erdos-Hajnal Question

Is there a K6-free graph G with G → (K3)?

Graham (1968): Yes!

K8 \ C5

Graham’s graph K8 \ C5

Suppose G has no monochromatic triangle.

(b, r)

(r, b)

Label the vertices of C5 by either (r, b) or (b, r).

(b, r)

Label the vertices of C5 by either (r, b) or (b, r).A red triangle is unavoidable since χ(C5) = 3.

K5-free G with G → (K3)

Year Authors |G|1969 Schauble 421971 Graham, Spencer 231973 Irving 181979 Hadziivanov, Nenov 161981 Nenov 15

K5-free G with G → (K3)

Year Authors |G|1969 Schauble 421971 Graham, Spencer 231973 Irving 181979 Hadziivanov, Nenov 161981 Nenov 15

In 1998, Piwakowski, Radziszowski and Urbanski used acomputer-aided exhaustive search to rule out all possiblegraphs on less than 15 vertices.

General results

Folkman’s theorem (1970): For any k2 > k1 ≥ 3, thereexists a Kk2-free graph G with G → (Kk1).

General results

These graphs are called Folkman Graphs.

General results

These graphs are called Folkman Graphs.

Nesetril-Rodl’s theorem (1976): For p ≥ 2 and any

k2 > k1 ≥ 3, there exists a Kk2-free graph G with

G → (Kk1)p.

Here G → (H)p: if the edges of G are p-colored then thereexists a monochromatic subgraph of G isomorphic to H.

f (p, k1, k2)

Let f(p, k1, k2) denote the smallest integer n such that thereexists a Kk2-free graph G on n vertices with G → (Kk1)p.

■ Grahamf(2, 3, 6) = 8.

f (p, k1, k2)

■ Grahamf(2, 3, 6) = 8.

■ Nenov, Piwakowski, Radziszowski and Urbanski

f(2, 3, 5) = 15.

f (p, k1, k2)

■ Grahamf(2, 3, 6) = 8.

■ Nenov, Piwakowski, Radziszowski and Urbanski

f(2, 3, 5) = 15.

■ What about f(2, 3, 4)?

Upper bound of f (2, 3, 4)

■ Folkman, Nesetril-Rodl ’s upper bound is huge.■ Frankl and Rodl (1986)

f(2, 3, 4) ≤ 7× 1011.

■ Erdos set a prize of $100 for the challenge

f(2, 3, 4) ≤ 1010.

f(2, 3, 4) ≤ 7× 1011.

f(2, 3, 4) ≤ 1010.

■ Spencer (1988) claimed the prize.

f(2, 3, 4) ≤ 3× 109.

f(2, 3, 4) ≤ 7× 1011.

f(2, 3, 4) ≤ 1010.

■ Spencer (1988) claimed the prize.

f(2, 3, 4) ≤ 3× 109.

Most wanted Folkman Graph

Difficulty

■ There is no efficient algorithm to test whetherG → (K3).

Difficulty

■ For moderate n, Folkman graphs are very rare among allK4-free graphs on n vertices.

Difficulty

■ For moderate n, Folkman graphs are very rare among allK4-free graphs on n vertices.

■ Probabilistic methods are generally good choices forasymptotic results. However, it is not good for moderatesize n.

Our approach

■ Find a simple and sufficient condition for G → (K3), andan efficient algorithm to verify this condition.

Our approach

■ Search a special class of graphs so that we have a betterchance of finding a Folkman graph.

Our approach

■ Use spectral analysis instead of probabilistic methods.

Our approach

■ Localization and δ-fairness.

Our approach

■ Localization and δ-fairness.

■ Circulant graphs and L(m, s).

My result

I received $100-award by provingTheorem [Lu, 2007]: f(2, 3, 4) ≤ 9697.

My result

Shortly Dudek and Rodl improved it to f(2, 3, 4) ≤ 941.They also received a $50-award.

My result

Shortly Dudek and Rodl improved it to f(2, 3, 4) ≤ 941.They also received a $50-award.

Lange-Radziszowski-Xu [2012+]: f(2, 3, 4) ≤ 786.

Spencer’s Lemma

Notations:

- Gv: the induced graph on the the neighborhood of v.- b(H): the maximum size of edge-cuts for H.

Spencer’s Lemma

Notations:

Lemma (Spencer) If∑

v b(Gv) <23

v |E(Gv)|, thenG → (K3).

Spencer’s Lemma

Notations:

Lemma (Spencer) If∑

v b(Gv) <23

v |E(Gv)|, thenG → (K3).

Localization

For 0 < δ < 12 , a graph H is δ-fair if

b(H) < (1

2+ δ)|E(H)|.

Localization

b(H) < (1

2+ δ)|E(H)|.

G is a Folkman graph if for each v

- Gv is 16-fair.

- Gv is K3-free.

Localization

b(H) < (1

2+ δ)|E(H)|.

G is a Folkman graph if for each v

- Gv is 16-fair.

- Gv is K3-free.

For vertex transitive graph G, all Gv’s are isomorphic.

Spectral lemma

- H: a graph on n vertices- A: the adjacency matrix of H- d = (d1, d2, . . . , dn): degrees of H- Vol(S) =

v∈S dv: the volume of S

- d = Vol(H)n : the average degree

Spectral lemma

Lemma (Lu) If the smallest eigenvalue of

M = A− 1Vol(H)d ·d′ is greater than −2δd, then H is δ-fair.

Spectral lemma

Lemma (Lu) If the smallest eigenvalue of

M = A− 1Vol(H)d ·d′ is greater than −2δd, then H is δ-fair.

Similar results hold for A and L. However, they are weakerthan using M in experiments.

Corollary

Corollary Suppose H is a d-regular graph and the smallest

eigenvalue of its adjacency matrix A is greater than −2δd.Then H is δ-fair.

Corollary

Corollary Suppose H is a d-regular graph and the smallest

eigenvalue of its adjacency matrix A is greater than −2δd.Then H is δ-fair.

Proof: We can replace M by A in the previous lemma.

- 1 is an eigenvector of A with respect to d.- M is the projection of A to the hyperspace 1

⊥.- M and A have the same smallest eigenvalues.

The proof of the Lemma

■ V (H) = X ∪ Y : a partition of the vertex-set.

■ 1X , 1Y : indicated functions of X and Y .

1X + 1Y = 1.

■ We observe M1 = 0.

1X + 1Y = 1.

■ We observe M1 = 0.

■ For each t ∈ (0, 1), let α(t) = (1− t)1X − t1Y . We have

α(t)′ ·M · α(t) = −e(X, Y ) +1

Vol(H)Vol(X)Vol(Y ).

Let ρ be the smallest eigenvalue of M . We have

e(X, Y )− Vol(X)Vol(Y )

Vol(H)≤ −α(t)′ ·M · α(t) ≤ −ρ‖αt‖2.

Choose t = |X|n so that ‖α(t)‖2 reaches its minimum |X||Y |

Vol(H)≤ −α(t)′ ·M · α(t) ≤ −ρ‖αt‖2.

Choose t = |X|n so that ‖α(t)‖2 reaches its minimum |X||Y |

n .We have

e(X, Y ) ≤ Vol(X)Vol(Y )

Vol(H)+ ρ

|X||Y |n

≤ Vol(H)

4− ρ

2+ δ)|E(H)|, since ρ > −2δd. �

Circulant graphs

- Zn = Z/nZ- S: a subset of Zn satisfying −S = S and 0 6∈ S.

We define a circulant graph H by

- V (H) = Zn

- E(H) = {xy | x− y ∈ S}.

Example: A circulant graphwith n = 8 and S = {±1,±3}.

Spectrum of circulant graphs

Lemma: The eigenvalues of the adjacency matrix for the

circulant graph generated by S ⊂ Zn are

s∈Scos

for i = 0, . . . , n− 1.

Spectrum of circulant graphs

Lemma: The eigenvalues of the adjacency matrix for the

circulant graph generated by S ⊂ Zn are

s∈Scos

for i = 0, . . . , n− 1.

Proof: Note A =g(J), where

g(x) =∑

s∈Sxs.

0 1 0 · · · 0 00 0 1 · · · 0 00 0 0 · · · 0 0...

...... . . . . . . ...

0 0 0 · · · 0 11 0 0 · · · 0 0

Proof continues...

Let φ = e2π

−1n denote the primitive n-th unit root.

J has eigenvalues

1, φ, φ2, . . . , φn−1.

Proof continues...

Let φ = e2π

−1n denote the primitive n-th unit root.

J has eigenvalues

1, φ, φ2, . . . , φn−1.

Thus, the eigenvalues of A = g(J) are

g(1), g(φ), . . . , g(φn−1).

For i = 0, 1, 2, . . . , n− 1, we have

g(φi) = ℜ(g(φi)) =∑

s∈Scos

n. �

Graph L(m, s)

Suppose s and m are relatively prime to each other. Let nbe the least positive integer satisfying

sn ≡ 1 mod m.

Graph L(m, s)

sn ≡ 1 mod m.

We define the graph L(m, s) to be a circulant graph on mvertices with

S = {si mod m | i = 0, 1, 2, . . . , n− 1}.

Graph L(m, s)

sn ≡ 1 mod m.

We define the graph L(m, s) to be a circulant graph on mvertices with

S = {si mod m | i = 0, 1, 2, . . . , n− 1}.

Proposition: The local graph Gv of L(m, s) is also acirculant graph.

Algorithm

■ For each L(m, s), compute the local graph Gv.■ If Gv is not triangle-free, reject it and try a new graph

L(m, s).■ If the ratio the smallest eigenvalue verse the largest

eigenvalue of Gv is less than −13 , reject it and try a new

graph L(m, s).■ Output a Folkman graph L(m, s).

Computational results

L(m, s) σL(127, 5) −0.6363 · · ·L(761, 3) −0.5613 · · ·L(785, 53) −0.5404 · · ·L(941, 12) −0.5376 · · ·L(1777, 53) −0.5216 · · ·L(1801, 125) −0.4912 · · ·L(2641, 2) −0.4275 · · ·L(9697, 4) −0.3307 · · ·L(30193, 53) −0.3094 · · ·L(33121, 2) −0.2665 · · ·L(57401, 7) −0.3289 · · ·

■ σ is the ratioof the smallesteigenvalue to thelargest eigenvaluein the local graph.

■ All graphs on theleft are K4-free.

■ Graphs in red areFolkman graphs.

■ Graphs in blackare good candi-dates.

Improvements

Our method has inspired two improvements.

■ Dudek-Rodl [2008]: f(2, 3, 4) ≤ 941.

Improvements

Our method has inspired two improvements.

■ Dudek-Rodl [2008]: f(2, 3, 4) ≤ 941.

■ Lange-Radziszowski-Xu [2012+]: f(2, 3, 4) ≤ 786.

Dudek and Rodl

Given a graph G, a triangle graph HG is defined as

■ V (HG) = E(G)■ e1 ∼ e2 in HG if e1 and e2 belong to the same triangle of

Dudek and Rodl

Given a graph G, a triangle graph HG is defined as

■ V (HG) = E(G)■ e1 ∼ e2 in HG if e1 and e2 belong to the same triangle of

Dudek, Rodl, 2008

■ If b(HG) <23 |E(HG)|, then G → (K3).

■ If HG is 16-fair, then G → (K3).

Dudek and Rodl

Theorem [Dudek, Rodl, 2008]

f(2, 3, 4) ≤ 941.

Dudek and Rodl

f(2, 3, 4) ≤ 941.

Proof: Let G = L(941, 12). Then G is 188 regular. Thetriangle graph H has 941 ∗ 188/2 = 88454 vertices and2122896 edges.

Dudek and Rodl

f(2, 3, 4) ≤ 941.

Proof: Let G = L(941, 12). Then G is 188 regular. Thetriangle graph H has 941 ∗ 188/2 = 88454 vertices and2122896 edges.

Using Matlab, they calculate the least eigenvalue

µn ≥ −15.196 > −(1

So H is 16-fair. Done.

Lange-Radziszowski-Xu

Instead of spectral methods, they use semi-definite program(SDP) to approximate the MAX-CUT problem.

■ First they try the graph G1 obtained from L(941, 12) bydeleting 81 vertices. They showed

3b(HG1) < 1084985 < 1085028 = 2|E(HG1

This implies f(2, 3, 4) ≤ 860.

Lange-Radziszowski-Xu

Instead of spectral methods, they use semi-definite program(SDP) to approximate the MAX-CUT problem.

■ First they try the graph G1 obtained from L(941, 12) bydeleting 81 vertices. They showed

3b(HG1) < 1084985 < 1085028 = 2|E(HG1

■ Second they try the graph G2 obtained from L(785, 53)by one vertex and some 60 edges. They showed

3b(HG2) < 857750 < 857762 = 2|E(HG2

Open questions

■ Exoo conjectured L(127, 5) is a Folkman graph.

■ In 2012 SIAMDM, Ronald Graham announced a $100award for determining if f(2, 3, 4) < 100.

■ A open problem on 3-colors: prove or disprove

f(3, 3, 4) ≤ 334

References

1. Linyuan Lu, Explicit Construction of Small FolkmanGraphs. SIAM Journal on Discrete Mathematics,21(4):1053-1060, January 2008.

2. Andrzej Dudek and Vojtech Rodl. On the FolkmanNumber f(2, 3, 4), Experimental Mathematics,17(1):63-67, 2008.

3. A. Lange, S. Radziszowski, and X. Xu, Use ofMAX-CUT for Ramsey Arrowing of Triangles,http://arxiv.org/pdf/1207.3750.pdf.

Homepage: http://www.math.sc.edu/∼ lu/

Thank You

Laplacian and Random Walks on Graphspeople.math.sc.edu/lu/talks/nankai_2014/spec_nankai_2.pdf ·...

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Spectral analysis of the magnetic Laplacian with vanishing ... · Spectral analysis of the magnetic Laplacian when h →0 17 May 2016, Conference on waveguides (Porquerolles) Spectral

Spectral convergence of the connection Laplacian from ...amits/publications/iaw016.pdf · Spectral convergence of the connection Laplacian from random samples Amit Singer Department

Nonlinear Laplacian Spectral Analysis of Rayleigh-Bénard

Semi-Supervised Learning with Graphszhuxj/tmp/tti.pdf · •semi-supervised kernels by spectral transformation of the graph Laplacian •kernelizing conditional random ﬁelds •combining

Localized Manifold Harmonics for Spectral Shape Analysis€¦ · 03/12/2017 · In particular, Laplacian eigenbases allow generalizing the classical Fourier analysis to manifolds

Nonlinear Laplacian spectral analysis: Capturing ...dimitris/files/GiannakisMajda12_NLSA.pdf · Nonlinear Laplacian spectral analysis: Capturing intermittent and low-frequency spatiotemporal

Spectral properties of the Laplacian on p-forms on the

A Tutorial on Spectral Clusteringcis580/lectures/... · The main tools for spectral clustering are graph Laplacian matrices. There exists a whole ﬁeld dedicated to the study of

Spectral Hashing - People | MIT CSAILpeople.csail.mit.edu/torralba/publications/spectralhashing.pdfin many spectral algorithms including spectral clustering and Laplacian eigenmaps

Research Article The Largest Laplacian Spectral Radius of ...downloads.hindawi.com/journals/jam/2013/563067.pdf · Laplacian spectral radius of . e investigation on the Laplacian

Consistency of spectral clusteringpeople.kyb.tuebingen.mpg.de/ule/publications/publication...of the graph Laplacian matrix constructed from the afÞnity graph between the sam-ple points

On Complementary Distance Signless Laplacian Spectral ...ijmsi.ir/article-1-1017-en.pdfOn Complementary Distance Signless Laplacian Spectral Radius and Energy of Graphs Harishchandra

Research Article Sharp Upper Bounds for the Laplacian ...downloads.hindawi.com/journals/mpe/2013/720854.pdf · the Laplacian spectral radius of irregulargraphs as follows. Let be

MATH-SHU 236 Spectral Clustering, Graph Laplaciansling/MATH-SHU-236-2020-SPRING/MATH-SH… · MATH-SHU 236 Spectral Clustering, Graph Laplacian Shuyang Ling March 11, 2020 1 Limitation

Community Detection with the z-Laplacian · PDF filewe explore the spectral properties of yet another linear operator—the z-Laplacian—which shares spectral ... Figure 1 shows that

Spectral Connectivity Analysis · underlying geometry of the data. In this paper, we describe a version of Laplacian-based spectral methods, called diffusion maps. These techniques

Spectra of Adjacency and Laplaciansco/Spectral/Lesson1Spectra.pdf · Spectra of Laplacian Matrices Properties: Possitive definiteness (all eigenvalues >= 0). The smallest eigenvalue