On the Number of Spanning Trees a Planar Graph Can Have

Kevin Buchin Andre SchulzESA 2010, 110-121, 2010.

Reporter ： 葉士賢 林建旻 蕭聖穎 張琮勛 賴俊鳴

Definition

• A spanning tree for a graph G is a subgraph of G that is a tree and contains all the vertices of G.

Example

(e) (2, 1)

1 2

3 4

(d) (1, 4)

1 2

3 4(c) (1, 3)

1 2

3 4(b) (1, 2)

1 2

3 4

(f) (2, 2)

1 2

3 4(g) (2, 3)

1 2

3 4(h) (2, 4)

1 2

3 4

(a) (1, 1)

1 2

3 4

(p) (4, 4)

1 2

3 4(o) (4, 3)

1 2

3 4(n) (4, 2)

1 2

3 4(m) (4, 1)

1 2

3 4

(l) (3, 4)

1 2

3 4(k) (3, 3)

1 2

3 4(j) (3, 2)

1 2

3 4(i) (3, 1)

1 2

3 4

Motivation

• Laplacian matrix

Def of Laplacian’s matrix

Kirchhoff’s theorem

• Matrix property: sum zero for column and row.

• Det[minor]• Check with Prufer sequence

Notation

Naïve thinking

Upper bound 6n

• Degree = 2*e = 2(3n-6)• Hadamard’s inequality

• Richier-Gebert 1996 positive semi-definite

Improvement

thought• Q1. why not nn-2 for T(n)?

– General graph and planar graph• Q2. why 3-connected and tri, qual…?

– 3-d grid– Increase if it has cycle(s)– A graph is 3-connected if, after the removal of any

two of its vertices, any other pair of vertices remain connected by a path

– Steinitz's theorem– the graph of every convex polyhedron is planar and 3-

connected

Notations

• t(G) = the number of spanning trees• di = the degree of vi

Triangulation?

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v1

Outdegree-One Graph?

v2

v9

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v3

v1

Outdegree-One Graph

v10

Outdegree-One Graph

• The number of outdegree-one graphs graphs contained in G exceeds t(G).

• Let S be a selection.• There are total different outdegree-one

graphs in G.• Due to Eular’s formula (e<=3n-6) the average

vertex degree is less than 6, and hence we have less than 6n outdegree-one graph of G be the geometric-arithmetic mean inequality.

S has a cycle ≡ S is disconnected

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The number of exactly spanning trees ?

• Outdegree-one graphs without cycles are exactly the (oriented) spanning trees of G.

• Let Pnc be the probability that the random graph selected by S contains no cycle.

• The exactly spanning trees for any graph G is given by

More notations

• Assume there are t cycles in G• Let Ci and Ci

c be the events that Ci = the i-th cycle occurs Ci

c = the i-th cycle does not occurs in a random outdegree-one graph.• For events Ci, Cj we denote that Ci↔Cj = they are dependent Ci↮Cj = they are independent

We say events C1, …, Cl have …

• mutually exclusive dependencies if Ci↔Cj implies Pr[Ci∩Cj] = 0.

• union-closed independencies if Ci↮Ci1, Ci↮Ci2, …, Ci↮Cik implies Ci↮(Ci1∪ …∪ Cik).

If we have these 2 properties…

• Lemma1. If events C1, …, Cl have mutually exclusive dependencies and union-closed independencies then

• We can express Pnc as .

• Instead of t(G), we bound its logarithm,

K-extension

• Assume that all cycles C are enumerated such that the first t3 cycles are the triangles in G, and the last t2 cycles are the 2-cycles of G.

• In total we consider t = t2 + t3 cycles.

• We apply Lemma 1 with k = 1 and l = t3 to bound , which is the probability that no 3-cycle occurs.

• P(A|B)P(B) = P(A∩B)=>

• For l = t and k = t3 + 1.

• Thus, we can bound log from above by

• Now rephrase log ≦

log i

log j

log i

log j(log i)/i

(log i)/i

(log i)/i

(log j)/j

(log j)/j

(log j)/j

Σlog iEach vertex

Σ(log i+log j)Each edge

log i

(log i)/i

(log i)/i

(log i)/idegree = j

degree = k

degree = h

(log i)/(i*(j-1))

(log i)/(i*(j-1))

(log i)/(i*(k-1))

(log i)/(i*(k-1))

(log i)/(i*(h-1))

(log i)/(i*(h-1))

Σ (Σlog a/(a*(i-1))+Σlog b/(b*(j-1)))Each edge

i j

a b

Each a adjacent i

Each b adjacent j

• For each 2-cycle with the same (i,j,A,B)– (i,j,A,B) supplies the enough information– has the same edge weight

• We apply the similar method by (i,j,k,A,B,C)

Σlog i = μ1 (Σlog i) + μ2 (Σlog i) + μ3 (Σlog i) + μ4 (Σlog i) = D1 + D2 + D3 + D4

i j i j

ar

ar

br

br

Constraints

• We only prove the general problem.• The restricted problems are easier to analyze

than the general problem, and can be proved by the lemma in the similar way.

Conclude

• Let G be a planar graph with n vertices.– The number of spanning trees of G is at most

– If G is 3-connected and contains no triangle, then the number of its spanning trees is bounded by

– If G is 3-connected and contains no triangle and no quadrilateral, then the number of its spanning trees is bounded by

Bound improvement

• Embedding 3-Polytopes on a Small Grid(Ribo and Rote 09)

• Let G be the graph of a 3-polytope P with n vertices. P admits a realization as combinatorial equivalent polytope with integer coordinates

Bound improvement

• Former upper bound:

• Using outgoing edge approach:– T3 :– T4 :– T5 :

Bound improvement

• The number F(n) of cycle-free graphs in a planar graph with n vertices is bounded by the number of selections of at most n−1 edges from the graphs (Aichholzer 2007)

• • We have for 0 q 1/2≦ ≦

• F(n) < 6.75 by setting m=3n and q=1/3n

Bound improvement

• We give a better bound based on the bound for the number of spanning trees.

• We bound the number F(n,k) of forests in gn with k edges.– –

• The number of cycle-free graphs is bounded by:

Bound improvement

• We use as upper bound for the binomial coefficient to obtain– –

• The computed maximal value for the minimum of f1 and f2 is realized at qn = 0.94741 n. This yields a bound of n*6.4948 for the number of cycle-free graphs

n

Future Work

• Since we consider only 2-cycles and 3-cycles from triangles, one would obtain a better bound for Pnc by taking larger cycles into account.

• Lemma 1 uses two enumerations of the events Ci to avoid the influence of the ordering. An elaborated enumeration scheme of the events Ci might give better bounds.