440 likes | 533 Views
On the Number of Spanning Trees a Planar Graph Can Have. Kevin Buchin Andre Schulz ESA 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. Motivation.
E N D
On the Number of Spanning Trees a Planar Graph Can Have Kevin Buchin Andre Schulz ESA 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.
Motivation • Laplacian matrix
Kirchhoff’s theorem • Matrix property: sum zero for column and row. • Det[minor] • Check with Prufer sequence
Upper bound 6n • Degree = 2*e = 2(3n-6) • Hadamard’s inequality • Richier-Gebert 1996 positive semi-definite
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
Outdegree-One Graph? v10 v2 v9 v8 v7 v6 v5 v4 v3 v1
Outdegree-One Graph v10 v2 v9 v8 v7 v6 v5 v4 v3 v1
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 v10 v2 v8 v7 v6 v5 v4 v3 v1 v9
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 Cic be the events that Ci = the i-th cycle occurs Cic = 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,
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.
For l = t and k = t3 + 1. • Thus, we can bound log from above by
Σ(log i+log j) Each edge log j log i Σlog i Each vertex (log j)/j log j (log i)/i (log i)/i (log j)/j log i (log j)/j (log i)/i
(log i)/(i*(h-1)) (log i)/(i*(j-1)) (log i)/(i*(h-1)) degree = h degree = j (log i)/i (log i)/i log i (log i)/(i*(j-1)) (log i)/i degree = k (log i)/(i*(k-1)) (log i)/(i*(k-1))
i j a b Σ (Σlog a/(a*(i-1))+Σlog b/(b*(j-1))) Each edge Each b adjacent j Each a adjacent i
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 br ar br
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=1/3 n
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.