COUNTING AND ENUMERATINGSPANNING TREES IN (di-) GRAPHS

Xuerong Yong

Version of Feb 5, 9 pm, 06: New York Time

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1 Introduction

• A directed graph (digraph) D is a pair (V, E):V is the vertex set of D, E the edge set ofD. An edge e = (u, v), u, v ∈ V . (u, v) =(v, u)?

• An undirected graph (graph) G is a pair(V, E), where the edge set is of unorderedpair of verties. (u, v) = (v, u)

• Self-loop, Path and Cycle, Degree of aVertex u, Connected Graph, Tree

• Example:

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• A spanning tree in a graph G is a tree hav-ing same vertex set as G.

• A spanning tree in a digraph D is a rootedtree with the same vertex set as D: there isa vertex specified as the root, and from theroot there is a path to any of verties of Dand no cycle.

Multiple edges and self-loops are permitted inany (di-)graph.

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Figure 1: G has 3 spanning trees; D has 4 spanning trees rooted at vertex 1.

Two Examples:

• A graph G and all its spanning trees

• A digraph D and all its (oriented) spanning trees

rooted at vertex 1.

The Main Problem: Given a (di-)graph, how manyspanning trees does it have?

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The number of spanning trees paid much at-

tention.

• Long history, Representatives

– Matrix Tree Theorem (Kirchhoff, 1847)

– Kelman (1965, 1974)

– Boesch (1982, 1991)

– Stanley (1996, 1999)

– Knuth (1997)

– Yamaguchi (2004), Lyons (2005)

– ·Applications

– Analyze the reliability of networks;

– Design electrical circuits;

– Analyze energy of masers.

– ·

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OUTLINE

– Introduction

– Counting the number of spanning trees in a (di-)

graphs

– Graphs with the maximum number of spanning

trees

– Algorithms for enumerating spanning trees

– Possible research trends

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Figure 2: Graph G and its reduced graphs G1 and G2.

2 Counting the numbers of spanning trees

in (di-)graphs

– Basic combinatorial method

Let T (G) be the number of spanning trees in G.

Let e = (u, v) be any edge in G.

Define

G1 = G with e deleted;

G2 = G with u and v contracted together.

Then

T (G) = T (G1) + T (G2).

So what?

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– Algebraically

We need some definitions.

∗ Given G with vertex set {v1, v2, · · · , vn}.The adjacency matrix A of G is a n×n ma-

trix whose (i, j)-entry is aij where aij is the

number of edges between vi and vj.

∗ For matrix A = (aij)n×n, the cofactor of the

(i, j)-entry is (−1)i+j det Aij where Aij is a

(n − 1) × (n − 1) matrix derived from A by

deleting i-th row and j-th column.

Define a diagonal matrix B = diag(d1, d2, · · · , dn)

where di is the degree of vi.

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– Matrix Tree Theorem–a very basic theorem

(Kirchhoff, 1847)

Let H = B −A. Then T (G) is equal to any co-

factor of H . H is usually called Kirchhoff ma-

trix, sometimes, Laplacian matrix or Lucacian

matrix of G.

Its proof follows from the basic combinatorial

idea, the multilinearity of the determinants and

induction on the numbers of edges and vertices

in the graph.

– There is a similar result for digraph.

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Figure 3: Example graph.

– The Kirchhoff matrix H = B − A of G is

H =

4 −1 −1 −1 0 −1

−1 3 0 0 −1 −1

−1 0 3 −1 −1 0

−1 0 −1 3 −1 0

0 −1 −1 −1 4 −1

−1 −1 0 0 −1 3

.

So

T (G) = (−1)1+1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

3 0 0 −1 −1

0 3 −1 −1 0

0 −1 3 −1 0

−1 −1 −1 4 −1

−1 0 0 −1 3

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣= 128.

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– Kelmans and Chelnokov’s method (1974):

The eigenvalues of G are the eigenvalues of its

adjacency matrix.

For G with n vertices, let H be the Kirchhoff

matrix of G and µ1 ≥ µ2 ≥ · · · ≥ µn(= 0) be

the eigenvalues of H . Then

T (G) =1

n

n−1∏j=1

µj.

– Similar formulas hold for digraphs.

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Cont’d

– Sachs’s method (1962): If G is a regular graph of

degree r, then

T (G) =1

n

n−1∏j=1

(r − λj),

where λ1 ≤ λ2 ≤ · · · ≤ λn = r are the eigenval-

ues of G.

– Example: Petersen graph is a regular graph of

degree 3.

(1) Its eigenvalues are:

3, 1, 1, 1, 1, 1, −2, −2, −2, −2.

(2) Eigenvalues of its Kirchhoff matrix:

0, 2, 2, 2, 2, 2, 5, 5, 5, 5.

Number of its spanning trees: 25 × 54 = 20000.

(van Lint and Wilson, 1992)

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Figure 4: The Petersen graph.

Why Still Working On?

Even though there are known methods for counting

the number, further research is still necessary.

– Generally, the methods are not feasible for count-

ing the number of spanning trees, especially for

large graphs.

– For comparing the numbers of spanning trees in

different classes of graphs, exact formulas are

needed.

– In applications, many graphs are special. And

for these graphs it is possible to derive exact

recurrence formulas for the number of spanning

trees. Recurrence formulas usually give the exact

values in computation!

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Known Results:

Cayley’s tree formula for complete graph Kn: T (Kn) =

nn−2.

There are formulas for the numbers of spanning trees

in the following special classes of graphs. Some of

them will be mentioned later.

– Complete prism

– Fan

– Point wheel

– Ladder

– Moebius Ladder

– Cocktail party graph

– k-dimensional lattice

– Complete multipartite graph

– Multi-star related graph

– Some circulant graphs

–

–

–

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Figure 5: Two examples of circulant graphs.

2.1 Spanning trees in circulant graphs

Let 1 ≤ s1 < s2 < · · · < sk be integers.

– Undirected circulant graph Cs1,s2,···,skn : there are

n vertices labeled 0, 1, 2, · · ·, n− 1, with each

vertex i (0 ≤ i ≤ n− 1) adjacent to 2k vertices

i± s1, i± s2, · · · , i± sk mod n.

Examples of two circulant graphs: C1,25 and C2,3

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– Directed circulant graph ~Cs1,s2,···,skn : digraph on

n vertices 0, 1, 2, · · ·, n − 1; for each vertex

i (0 ≤ i ≤ n − 1), there are k arcs from i to

vertices i + s1, i + s2, · · · , i + sk mod n.

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Some Recurrence Formulas

– T (C1,2n ) = nF 2

n , Fn are Fibonacci numbers.

(Conjectured by Bedrosian (1973), and by Boesch

and Wang (1982), proven by Kleitman and Golden

(1975), and Prodinger (1986). A much simpler

proof given by Yong and Zhang (1994).

– T (C1,3n ) = na2

n,

an =√

2an−1 +√

2an−3 − an−4,

a1 = 1, a2 = 2√

2, a3 = 5, a4 = 5√

2.

(Yong, Talip and Acenjian, 1997)

– T (C2,3n ) = na2

n,

an = an−1 + an−2 + an−3 − an−4,

a1 = 1, a2 = 1, a3 = 1, a4 = 3.

(Zhang, Yong and Golin, 2000)

– T (~C2,3n ) = nan,

an = an−1 + 2an−2 + 2an−3 − 4an−4,

a1 = 1, a2 = 1, a3 = 1, a4 = 5.

(Zhang, Yong and Golin, 2000)

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General Results for Circulant Graphs

– T (Cs1,s2,···,skn ) = na2

n, (Golin, Yong and Zhang,

2000)

where an satisfies a recurrence relation of the

form ∀n ≥ 2sk−1, an =∑2sk−1

i=1 bian−i.

The initial values of an and coefficients bi (1 ≤i ≤ 2sk−1) can be found using Matrix Tree The-

orem.

– T (~Cs1,s2,···,skn ) = nan, (Golin, Yong and Zhang,

2000)

where an satisfies a linear recurrence relation of

order 2sk−1 with initial conditions and coefficients.

– When n tends to infinity (Zhang, Yong, 1999)

T (~Cs1,s2,···,skn+1 )/T (~Cs1,s2,···,sk

n ) → k.

Similar result was obtained recently for the undi-

rected case (Golin, Yong, Zhang, 2006).

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An Example of Combinatorial Consideration

Consider the graph L1,2n with n vertices {0, 1, · · · , n − 1}

and with each vertex i (0 ≤ i ≤ n − 2) adjacent to i + 1and each vertex i (0 ≤ i ≤ n − 3) adjacent to i + 2. L

′1,2n

is the multigraph derived from L1,2n+1 by combining the first

two vertices into one. Then

T (L1,2n ) = T (L1,2

n−1) + T (L′1,2n−1),

T (L′1,2n ) = T (L1,2

n−1) + 2T (L′1,2n−1).

with the initial conditions T (L1,22 ) = 1 and T (L

′1,22 ) = 2.

Sovling these equations

T (L1,2n ) =

1

2√

5((3−

√5)(

3 +√

5

2)n − (3 +

√5)(

3−√

5

2)n)

= F2n−2,

where Fn are Fibonacci numbers.

(Kleitman and Golden, 1975)

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2.2 Spanning Trees in Some Composite Graphs

Let G1 = (V1, E1) and G2 = (V2, E2) be two graphs withdisjoint vertex sets.

– The join G = G1 + G2: V = V1 ∪ V2, E = E1 ∪ E2 ∪{uv|u ∈ V1, v ∈ V2}.

– The lexicographic product G = G1[G2]: V = V1 × V2,E = {(u1, v1)(u2, v2)|u1u2 ∈ E1 or u1 = u2 and v1v2 ∈E2}.

– The Cartesian product G = G1 × G2: V = V1 × V2,E = {(u1, v1)(u2, v2)|u1 = u2 and v1v2 ∈ E2 or u1u2 ∈E1 and v1 = v2}.

– The categorical product G = G1 · G2: V = V1 × V2,E = {(u1, v1)(u2, v2)|u1u2 ∈ E1 and v1v2 ∈ E2}.

– The strong product G = G1 ⊗ G2: V = V1 × V2, E isthe union of the edge sets of the Cartesian product andcategorical product of G1 and G2.

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Figure 6: Graphs G1 and G2 and their composite graphs.

G1 + G2 : V = V1 ∪ V2,

E = E1 ∪ E2 ∪ {uv|u ∈ V1, v ∈ V2};G1[G2] : V = V1 × V2,

E = {(u1, v1)(u2, v2)|u1u2 ∈ E1 or u1 = u2 and v1v2 ∈ E2};G1 ×G2 : V = V1 × V2,

E = {(u1, v1)(u2, v2)|u1 = u2 and v1v2 ∈ E2

or u1u2 ∈ E1 and v1 = v2};G1 ·G2 : V = V1 × V2,

E = {(u1, v1)(u2, v2)|u1u2 ∈ E1 and v1v2 ∈ E2};G1 ⊗G2 : V = V1 × V2,

E = Union of edges in G1 ×G2 and G1 ·G2.

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L-eigenvalues of Composite Graphs:

L-eigenvalues of G are the eigenvalues of its Lapla-

cian matrix of G. Let

L-eigenvalues of G1: λ1(= 0), λ2, · · ·, λn;

L-eigenvalues of G2: µ1(= 0), µ2, · · ·, µm.

L-igenvalues of G1 + G2: (join)

0, m+n, λ2+m, · · ·, λn+m and µ2+n, · · ·, µm+n.

L-eigenvalues of Km[G]: (lexicographic p.)

0, mn of order m− 1, and λi + (m− 1)n of order m

for each i = 2, 3, · · · , n.

L-eigenvalues of Kp,q[G]: (lexicographic p.)

0, np of order q−1, nq of order p−1, λi+np of order

q and λi + nq of order p for each i = 2, 3, · · · , n.

L-eigenvalues of G[Km]: (lexicographic p.)

mλi, (di+1)m of order m−1 for each i = 1, 2, · · · , n.

L-eigenvalues of G1 × G2: (Cartesian p.) λi + µj

(i = 1, 2, · · · , n; j = 1, 2, · · · , m).

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Figure 7: The complete prism R3(3) = K3 × C3.

The numbers of spanning trees in compos-

ite graphs

– Complete prism graph Rn(m) = Km×Cn: (Carte-

sian p.)

T (Rn(m)) =nm[(m+2+

√m2+4m

2 )n + (m+2−√

m2+4m2 )n − 2]m−1.

(Bedrosian and Prodinger, 1986)

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Figure 8: The wheel W7 = K1 + C6.

Con’t

∗ Point wheel graph Wn = K1 + Cn−1: (join)

T (Wn) = (3+√

52 )n−1 + (3−

√5

2 )n−1 − 2.

(Bedrosian and Prodinger, 1986)

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Figure 9: The ladder L5 = K2 × P5.

Con’t

∗ Ladder graph Ln = K2 × Pn: (Cartesian p.)

T (Ln) = 12√

3[(2 +

√3)n − (2−

√3)n].

(Bedrosian and Prodinger, 1986)

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Figure 10: The Moebius ladder M5.

Con’t

∗ Moebius ladder graph Mn: (Cartesian p.)

formed from K2 × Pn by adding edge from

the first vertex on one copy of Pn to the last

vertex on the second copy of Pn and an edge

from the first vertex on the second copy to the

last vertex on the first copy.

T (Mn) = n2 [((2 +

√3)n + (2−

√3)n + 2].

(Bedrosian and Prodinger, 1986)

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Figure 11: A graph isomorphic to P5 · P5.

Con’t

Consider the graph on mn vertices {(x, y)|1 ≤ x ≤m, 1 ≤ y ≤ n}, with (x, y) adjacent to (x′, y′) if

and only if |x − x′| = |y − y′| = 1. This graph is

isomorphic to Pm ·Pn (categorical p.) which consists

of two disjoint subgraphs

ECm,n = {(x, y)|x + y is even},OCm,n = {(x, y)|x + y is odd}.

If mn is even, then ECm,n and OCm,n are isomor-

phic. Stanley (1996) conjectured that

T (OC2n+1,2n+1) = 4T (EC2n+1,2n+1)

and Knuth (1997) proved it.

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3 Graphs with maximum number of span-

ning trees

– G(n, e): class of graphs with n vertices and e

edges.

– t-optimal graph: has maximum number of span-

ning trees among G(n, e).

The general problem of finding t-optimal graph is

open.

– Partial known results:

∗ A cycle in which each edge is replaced by k

multiple edges has less trees than any other

2k-connected graph with the same numbers

of vertices and edges.

(Lomonosov and Polesskii, 1972)

∗ A graph is called almost-regular if the degrees

of any two vertices differ by no more than one.

Any t-optimal graph must be almost-regular.

(Leggett and Bedrosian, 1965)

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Figure 12: An exemplification for t-optimal graph.

– Example. Remove three edges from complete

graph K6. All possible configurations of the re-

moved edges are shown in Figure 12. Each deleted

case actually represents different equivalent cases.

The corresponding graphs have spanning trees:

384, 360, 336, 324, 300.

Case 1 generates a t-optimal graph.

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– A few more results

∗ In removal of m edges (m ≤ n/2) from the

complete graph Kn, the retained graph is t-

optimal when the removed edges are not ad-

jacent.

(Moustakides and Bedrosian, 1980)

∗ If G is a complete s-partite graph, then G is

the unique simple graph that has the maxi-

mum number of spanning trees among graphs

with same numbers of vertices and edges.

(Cheng, 1981)

∗ Most recent papers are for very special graphs:

t-optimal graph in G(n, e), e = an+ b, where

a, b are consts.

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4 Enumerating spanning trees

Let N = # of spanning trees in G.

– For graphs, Gabow and Myers (1978) gave an al-

gorithm in O(NV ) time. This algorithm is opti-

mal if all spanning trees are required to be output

explicitly. Kapoor and Ramesh designed an algo-

rithm (1991) which runs in O(N + V + E) time

and uses O(V E) space. In their algorithm, the

spanning trees are not output explicitly.

– For weighted graphs, Gabow (1977) gave an al-

gorithm that outputs sorted trees in O(NE +

N log N) time. Later, Kapoor and Ramesh (1995)

improved it to O(N log V + V E).

– For digraphs, Kapoor and Ramesh (2000) gave an

algorithm which has running time O(NV + V 3)

and needs O(V 2) space.

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No efficient algorithm developed for enu-

merating all spanning trees in a graph

– The aim of the algorithm for enumerating all

spanning trees in a graph G is to construct the

computation tree C(G). The computation tree

suffices to generate all spanning trees of G.

– Each vertex x of C(G) has a spanning tree Sx of

G associated with it. The root of C(G) can be

any spanning tree.

– Repetitions of spanning trees are avoided by main-

taining two sets, IN and OUT , at each node

of C(G) and using inclusion/exclusion principle.

For the root r of C(G), INr and OUTr are both

empty.

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Figure 13: The part of computation tree C(G).

(Con’t)

– Let x be any vertex of C(G). f is an edge of G.

f 6∈ Sx and f 6∈ OUTx. Let the fundamental

cycle of f with respect to Sx contain the follow-

ing edges of Sx − INx: (e1, e2, · · · , ek). Then x

has sons Bi, 1 ≤ i ≤ k + 1. For 1 ≤ i ≤ k,

Bi corresponds to the spanning tree obtained by

replacing ei by f . Bk+1 corresponds to the tree

same with Sx. For the sons of x, the IN and

OUT sets are defined as:

INBj= INx

⋃{e1, e2, · · · , ej−1}⋃{f}, for 1 ≤ j ≤ k

OUTBj= OUTx

⋃{ej}, for 1 ≤ j ≤ k

INBk+1 = INx

OUTBk+1 = OUTx⋃{f}

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The Techniques used mostly are:

– Combinatorial Analysis

– Graph Spectra

– Matrix Theory

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The Difficulities of the Problem are:

– Combinatorial approaches can only attack those

very special (di-)graphs.

– Algebraic approaches involve in estimating the

distribution of eigenvalues. This is usually hard

for large graphs.

– Enumurating Methods obtained so far are expo-

nential in complexity.

– New method?

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Therefore, bounding the number of spanning trees

– Lower bounding the number

– Upper bounding the number

– Asymptotic analysis

BUT THE BOUNDS OBTAINED ARE VERY ROUGH!

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5 Research trends

Research about the number of spanning trees is con-

tinuing and very active. Possible Problems would

be:

– Find more special graphs in application and count

the number of spanning trees.

– Analyze the asymptotic properties of the number

of spanning trees.

– Solve the t-optimal graph problem.

– Design efficient algorithms for enumerating the

spanning trees in directed graphs.

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