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Shanghai Jiaotong University On the Generalized Spectral - - PowerPoint PPT Presentation

Introduction Notations and Terminologies A Simple Characterization of DGS Graphs The Method Examples Summaries Shanghai Jiaotong University On the Generalized Spectral Characterization of Graphs (I): The Basics Wei Wang Xian Jiaotong


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Introduction Notations and Terminologies A Simple Characterization of DGS Graphs The Method Examples Summaries

Shanghai Jiaotong University On the Generalized Spectral Characterization of Graphs (I): The Basics

Wei Wang Xi’an Jiaotong University

  • Dec. 2016
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Introduction Notations and Terminologies A Simple Characterization of DGS Graphs The Method Examples Summaries

Outline

1

Introduction

2

Notations and Terminologies

3

A Simple Characterization of DGS Graphs

4

The Method

5

Examples

6

Summaries

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Introduction

Graph Isomorphism Problem (GIP) Given two graphs G and H, determine whether they are isomorphic nor not.

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  • L. Babai (2015)

There is a quasipolynomial time algorithm for all graphs, i.e., one with running time exp((log n)O(1)).a

  • aL. Babai, Graph isomorphism in quasipolynomial time, arXiv:1512.03547v2.

W.X. Du (2016) The running time can be improved to nC log n for some constant C.a

aW.X, Du, On the Automorphism Group of a Graph, arXiv:1607.00547v1.

It remains an unsolved question whether GIP is in P or in NPC.

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Introduction

The spectrum of a graph encodes a lot of information about the given graph, e.g., From the adjacency spectrum, one can deduce

(i) the number of vertices, the number of edges; (ii) the number of triangles ; (iii) the number of closed walks of any fixed length; (iv) bipartiteness; . . .

From the Laplacian spectrum, one can deduce:

(i) the number of spanning trees; (ii) the number of connected components; . . .

Can a graph be determined by the spectrum?

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A pair of cospectral graphs

q q q q q q q q q q ❍❍❍❍❍❍ ✟✟✟✟✟✟

      1 1 1 1 1 1 1 1             1 1 1 1 1 1 1 1       spectrum: -2, 0, 0, 0, 2

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Which Graphs Are DS?

“Which graphs are determined by spectrum(DS for short)?” This is a fundamental question in Spectral Graph Theory that dates back to more than 60 years. In 1956, G¨ unthard and Primas raised the question in a paper that relates the theory of graph spectra to H¨ uckel’s theory from chemistry. Applications:

Graph Isomorphism Problem; The shape and sound of a drum; Energy of hydrocarbon molecules; . . . . . . . . .

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Are almost all graphs DS?

To what extent a graph is DS? Are almost all graph DS? Are Almost all graphs non-DS? or Neither is true? Conjecture (Haemers) Almost all graphs are DS (w.r.t. adjacency spectrum, the Laplacian spectrum etc.) Remark: Formally speaking, the fraction of the DS graphs among all graphs tends to 1 as the order of the graphs tends to infinity.

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Some Evidences Against The Conjecture

Schwenk (1973) Almost every tree has a cospectral mate. a

aA.J. Schwenk, Almost all trees are cospectral, New Directions in The

Theory of Graphs, F. Harary (Ed.), Academic Press, New York (1973), pp. 275-307

Remark: Schwenk’s result holds for the adjacency spectrum, Laplcian spectrum, and etc.

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Schwenk’s Constructions

Cospectrally rooted graph Let G and H be two graphs with root u and v respectively. Then G and H are said to be cospectrally rooted, if Spec(G) = Spec(H) and Spec(G − u) = Spec(H − v). Theorem Let G and H be two cospectrally rooted graphs with root u and v

  • respectively. Let Γ be a graph with root w. Then G • Γ and H • Γ

are cospectral, where the new graphs are obtained from the old

  • nes by identifying the roots u and w, and v with w.
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The Conjecture Is False For Strongly Regular Graphs

16 squares as vertices; adjacent if in same row, same column, or same color. The pair of strongly regular graphs constructed in this way are cospectral. There are lots of Latin squares.

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Cospectral and Non-isomorphic Graphs Can Easily Be Constructed

Godsil and McKay (1982), GM-switching Let the vertex set V of G can be partitioned into V and V2. Suppose G[V1] is regular, and every vertex in V2 is adjacent to non, all, or exactly half number of vertices in V1. Then the new graph obtained by GM-switching and the old one are cospectral.

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Very Small Number of Graphs Are Known to Be DS

1 The complete graph Kn. 2 The regular complete bipartite graph Kn,n. 3 The cycle Cn. 4 The path Pn. 5 The tree Zn. 6 · · · · · · · · ·

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Computer Enumerations

Godsil and McKay (1982); Haemers and Spence (2004); Brouwer and Haemers (2009).

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Our Results on Generalized Spectral Characterization of Graphs

1 A new method; 2 A simple arithmetic criterion; 3 A further development; 4 Some explicit constructions.

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Notations and Terminologies

Two graphs G, H are cospectral w.r.t. the generalized spectrum if Spec(G) = Spec(H) and Spec(¯ G) = Spec(¯ H). E.g.

q q q q q q q ❜❜ ❜ ❜❜ ❜ ✧✧ ✧ ✧✧ ✧ q q q q q q q

❅ ❅ ❅

PG1(λ) = PG2(λ) = λ7 − 6λ5 + 9λ3 − 4λ P¯

G1(λ) = P¯ G2(λ) = λ7 − 15λ5 − 2λ4 + 12λ3 + 24λ2

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DGS Graphs

A graph G is said to be determined by the generalized spectrum (DGS for short), if any graph that is cospectral with G w.r.t. the generalized spectrum is isomorphic to G. In notation, G is DGS if Spec(G) = Spec(H) and Spec(¯ G) = Spec(¯ H) implies H is isomorphic to G for any H.

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DGS Graphs: The basics

The walk-matrix of graph G: W (G) = [e, A(G)e, . . . , A(G)n−1e] where e = (1, 1, . . . , 1)T is the all-one vector. The (i, j)-th entry of W is the number of walks of legth j − 1 starting from the i-th vertex.

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Controllable Graphs

A graph G is called a controllable graph if the corresponding walk-matrix W (G) is non-singular. The set of all controllable graphs of order n is denoted by Gn. It was conjectured (by C.D. Godsil) that almost all graphs are controllable. O

′Rourke and Touri showed recently that this conjecture is

true.

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A Simple Characterization

Theorem 1. [Wang and Xu, 2006] Let G ∈ Gn. Then there exists a graph H such that Spec(G) = Spec(H) and Spec(¯ G) = Spec(¯ H) if and only if there exists a unique rational orthogonal matrix Q such that QTA(G)Q = A(H), and Qe = e, (1) where e is the all-ones vector.a

  • aW. Wang, C. X. Xu, A sufficient condition for a family of graphs being

determined by their generalized spectra, European J. Combin., 27 (2006) 826-840.

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The Proof: Some Lemmas

Lemma Let G be a graph with adjacency matrix A. Let Ak = (a(k)

ij ). Then

a(k)

ij

equals the number of walks of length k in G starting from i and ending at j. Lemma Let Nk(G) be the number of walks of length k in G. Then Nk(G) = eTAe.

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The Proof: Some Lemmas

Lemma Let HG(t) = ∞

k=0 Nk(G)tk be the generating function of Nk(G).

Then HG(t) = 1 t [(−1)n Φ¯

G(− 1+t t )

ΦG( 1

t )

− 1]. Corollary Spec(G) = Spec(H) and Spec(¯ G) = Spec(¯ H) implies W T

G WG = W T H WH.

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The Proof: A Sketch

Proof. “ ⇒ ” Eq. (1)⇒ Q(J − AG − I)Q = J − AH − I ⇒ QTA¯

GQ = A ¯ H.

“ ⇐ ” AGWG = AG[e, AGe, · · · , An

Ge]

= [AGe, A2

G, · · · , An Ge]

= [AGe, A2

Ge, · · · , An−1 G

e, −a1An−1

G

e − a2An−2

G

e − · · · − ane] = WG           · · · −an 1 · · · −an−1 1 · · · −an−2 . . . . . . · · · ... . . . . . . . . . . . . . . . ... −a2 · · · 1           .

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The Proof: A Sketch - contd.

Proof. Similarly, we have AHWH = WH           · · · −an 1 · · · −an−1 1 · · · −an−2 . . . . . . · · · ... . . . . . . . . . . . . . . . ... −a2 · · · 1           . It follows that W −1

G AGWG = W −1 H AHWH. Let Q = WGW −1 H .

Then Q is a rational orthogonal matrix, since W T

G WG = W T H WH,

which implies QTQ = I.

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How to find DGS-Graphs?

Definition Γ(G) = {Q ∈ On(Q)|QTA(G)Q is a (0, 1) − matrix and Qe = e}, whereOn(Q) is the set of all rational orthogonal matrices. Theorem 2. [Wang and Xu, 2006] Let G ∈ Gn. Then G is DGS if and only if Γ(G) contains only permutation matrices. Question: How to find out all Q ∈ Γ(G) explicitly?

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The level of Q

Definition Let Q be a rational orthogonal matrix with Qe = e, the level of Q is the smallest positive integer ℓ such that ℓQ is an integral matrix. If ℓ = 1, then Q is a permutation matrix. Example: 1 2

  • −1

1 1 1 1 −1 1 1 1 1 −1 1 1 1 1 −1

  • , 1

3  

2 1 1 1 1 1 −1 2 −1 1 1 1 −1 −1 2 1 1 1 1 1 1 2 −1 −1 1 1 1 −1 2 −1 1 1 1 −1 −1 2

  .

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The Smith Normal Form

Definition An integral matrix M is called unimodular, if det(M) = ±1. Theorem Let M be an n by n integral matrix with full rank. Then there exist two unimodular matrices U and V such that M = U      d1 d2 ... dn      V , where di|di+1 for i = 1, 2, · · · , n − 1. di is called the i-th elementary divisor of M.

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The level of Q and SNF of W (G)

An Important Fact Let Q ∈ Γ(G) with level ℓ. Then ℓ is always a divisor of the n-th elementary divisor of W (G), i.e. ℓ|dn, and hence ℓ| det(W ). Proof. This is because: QTA(H)Q = A(H), Qe = e

  • =

⇒ QTW (G) = W (H). It follows from W (G) = Udiag(d1, d2, · · · , dn)V that QT = W (H)V −1diag(d−1

1 , d−1 2 , · · · , d−1 n ))U−1 and dnQ is an

integral matrix.

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An Exclusion Principle

Lemma (Wang and Xu, 2006) Let W be the walk-matrix of a graph G ∈ Gn. Let Q ∈ Γ(G) with level ℓ. If prime p|ℓ, then the following system of congruence equations has a non-trivial solution: W Tx ≡ 0, xTx ≡ 0 (mod p). (2) Proof. QTAGQ = AH and Qe = e imply QTWG = WH, i.e., W T

G Q = W T H . Let x be a column of Q such that x ≡ 0 (mod p)

(it always exists by the definition of ℓ). With such an x, we have W Tx ≡ 0 (mod p) and xTx ≡ 0 (mod p).

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Finding DGS-Graphs: Basic Ideas

All the possible prime divisors of ℓ is finite; they are the divisors of dn, and hence are divisors of det(W (G)) Some of the prime divisors of det(W (G)) may not be divisors

  • f ℓ, they can be excluded from further consideration.

If all the prime factors of det(W (G)) are not divisors of ℓ, then we must have ℓ = 1, and hence G is DGS.

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The method

1 Compute det(W ) = pα1

1 pα2 2 · · · pαs s .

2 For every prime pi, if Eq. (2) has no non-trivial solution, then

p ∤ ℓ.

3 If pi ∤ ℓ, ∀i = 1, 2, · · · , s, then ℓ = 1 and G is DGS.

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Excluding Primes p > 2

Let p > 2 be a prime and p|dn. If Eq. (2) has no solution, then p is not a divisor of ℓ. Assume rankp(W (G)) = n − 1. Then the solution to the system of linear equations can be written as x = kξ over finite field Fp. If ξTξ = 0, then p is not a divisor of ℓ. Using this way, the odd prime divisors of dn can be excluded in most cases.

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Examples

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Examples

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The First Graph

It can be computed dn(G1) = 2 × 17 × 67 × 8054231. For p = 17, ξTξ = 12. For p = 67, ξTξ = 25. For p = 8054231, ξTξ = 1492735. Thus, all primes (except p = 2) can be excluded.

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The Second Graph

It can be computed dn(G2) = 2 × 32 × 5 × 197 × 263 × 5821. For p = 3, ξTξ = 1. For p = 5, ξTξ = 0. For p = 197, ξTξ = 139. For p = 263, ξTξ = 101. For p = 5821, ξTξ = 4298. Thus, all primes (except p = 2, 5) can be excluded.

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The prime p = 2

When p = 2, however, the system of linear equations has always non-trivial solutions. Thus, p=2 cannot be excluded using above method. To exclude p = 2, we more intensive exclusion conditions are needed. To conclude, it can be shown that the first graph G1 is DGS. But G2 cannot be shown to be DGS, since p = 5 cannot be excluded by using the existing methods.

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Rational orthogonal matrices with level two

Definition Q is partly decomposable, if there exist two permutation matrix P1 and P2 such that P1QP2 = Q1 O O Q2

  • , where Q1 and Q2 are
  • rthogonal matrices. Otherwise it fully indecomposable.
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Rational orthogonal matrices with level two

Theorem (Wang and Xu, 2006) Let Q be a fully indecomposable rational orthogonal matrix with level two. Then Q can be changed into the following for by permuting rows and columns: (i) 1

2

    −1 1 1 1 1 −1 1 1 1 1 −1 1 1 1 1 −1    . (ii) 1

2

         X O O · · · O Y Y X O · · · O O O Y X · · · O O . . . . . . . . . ... . . . . . . O O O · · · X O O O O · · · Y X          , where X = 1 1 1 1

  • , Y =
  • 1

−1 −1 1

  • , O =
  • .
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Rational orthogonal matrices with level two - contd.

Theorem (Wang and Xu, 2006) (iii) 1

2

          1 1 1 −1 1 1 −1 1 1 1 1 −1 −1 1 1 1 1 −1 1 1 −1 1 1 1 −1 1 1 1          

  • r

1 2

            1 1 1 −1 1 1 −1 1 1 1 1 −1 −1 1 1 1 1 −1 1 1 −1 1 1 1 1 −1 1 1 1 1 1 −1             .

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The case p = 2

Define Hn = {G ∈ Gn|Eq. (2) has only trivial solution for every odd p and the SNF is diag(1, 1, · · · , 1

  • ⌈n/2⌉

, 2, 2, · · · , 2b

  • ⌊n/2⌋

)}. Suppose G ∈ Hn. Then ℓ = 1 or ℓ = 2. If we can eliminate the possibility that ℓ = 2, then ℓ = 1 and G is DGS. But how? ℓ = 2 implies that every column of a fully indecomposable block of Q has four non-zero entry “1”, which satisfies the equation W Tx ≡ 0 (mod 2). Finding the set S of all the solutions to the above equation with exactly four “1” (say, by enumerating all n

2

  • possible

cases). Try to find all Q’s with column vectors from S (modulo 2) and possibly standard unit vectors.

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A Theorem

Theorem Suppose G ∈ Hn. Let S be the set of solutions to the following equations: W Tx ≡ 0 (mod 2), x has exactly four“1”. If S = ∅, then G is DGS. Remark: The larger n is, the easier the above condition holds.

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A Large Example

n = 36, G ∈ Hn and S = ∅. Thus G is DGS.

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A Small Example

A =         1 1 1 1 1 1 1 1 1 1 1 1         , W =         1 1 3 6 16 35 1 3 6 16 35 86 1 2 6 13 32 73 1 3 7 16 38 87 1 2 4 9 20 47 1 1 2 4 9 20         . det(W ) = −23.

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An Example for p = 2

A basis of the solutions to W Tx ≡ 0 (mod2):         1 1 1 1 1 1         ⇒         1 1 1 1 1 1 1 1 1 1 1 1        

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An Example for p = 2

        1 1 1 1         ⇒ Q1 =         2 −1 1 1 1 1 −1 1 1 2 1 1 −1 1 1 1 1 −1        

  • 2

⇒ QT

1 AQ1 =

        1 − 1

2 1 2

1

1 2 1 2

1

1 2 1 2

1

1 2 3 2

− 1

2 1 2 1 2 1 2 1 2 1 2 1 2 3 2

       

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An Example for p = 2

        1 1 1 1         ⇒ Q2 =         −1 1 1 1 1 −1 1 1 2 1 1 −1 1 2 1 1 1 −1         /2 ⇒ QT

2 AQ2 =

       

1 2 1 2

1 1

1 2

−1

1 2

1

1 2 1 2 1 2 1 2

1 1 1 1 1 1        

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An Example for p = 2

        1 1 1 1         ⇒ Q3 =         2 −1 1 1 1 1 −1 1 1 2 1 1 −1 1 1 1 1 −1         /2 ⇒ QT

3 AQ3 =

        1 − 1

2 1 2

1

1 2 1 2

1

1 2 1 2

1

1 2 3 2

− 1

2 1 2 1 2 1 2 1 2 1 2 1 2 3 2

       

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An Example for p = 2

        1 1 1 1 1 1 1 1 1 1 1 1         ⇒ Q4 =         1 −1 1 1 1 −1 1 1 1 1 1 −1 −1 1 1 1 1 1 −1 1 −1 1 1 1         /2 ⇒ QT

4 AQ4 =

        −1

1 2 1 2 1 2 1 2

1

1 2 1 2 1 2 1 2 1 2 1 2

1 1

1 2 1 2

−1 1

1 2 1 2

1

1 2 1 2

1        

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Introduction Notations and Terminologies A Simple Characterization of DGS Graphs The Method Examples Summaries

References

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Introduction Notations and Terminologies A Simple Characterization of DGS Graphs The Method Examples Summaries

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Introduction Notations and Terminologies A Simple Characterization of DGS Graphs The Method Examples Summaries

Thank you! The end!