1. 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

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

3. Example

4. Motivation Laplacian matrix

5. Def of Laplacian’s matrix

6. Kirchhoff’s theorem Matrix property: sum zero for column and row. Det[minor] Check with Prufer sequence

7. Notation

8. Naïve thinking

9. Upper bound 6 n Degree = 2*e = 2(3n-6) Hadamard’s inequality Richier-Gebert 1996 positive semi-definite

10. Improvement

11. thought Q1. why not n n-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 path3-connected – Steinitz's theorem Steinitz's theorem – the graph of every convex polyhedron is planar and 3- connected

12. Notations t(G) = the number of spanning trees d i = the degree of v i

13. Triangulation?

14. v2v2 v9v9 v 10 v8v8 v7v7 v6v6 v5v5 v4v4 v3v3 v1v1 Outdegree-One Graph?

15. v2v2 v9v9 v8v8 v7v7 v6v6 v5v5 v4v4 v3v3 v1v1 Outdegree-One Graph v 10

16. 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 6 n outdegree-one graph of G be the geometric-arithmetic mean inequality.

17. S has a cycle ≡ S is disconnected v2v2 v8v8 v7v7 v6v6 v5v5 v4v4 v3v3 v1v1 v9v9 v 10

18. The number of exactly spanning trees ? Outdegree-one graphs without cycles are exactly the (oriented) spanning trees of G. Let P nc be the probability that the random graph selected by S contains no cycle. The exactly spanning trees for any graph G is given by

19. More notations Assume there are t cycles in G Let C i and C i c be the events that C i = the i-th cycle occurs C i c = the i-th cycle does not occurs in a random outdegree-one graph. For events C i, C j we denote that C i ↔ C j = they are dependent C i ↮ C j = they are independent

20. We say events C 1, …, C l have … mutually exclusive dependencies if C i ↔ C j implies Pr[C i ∩C j ] = 0. union-closed independencies if C i ↮ C i1, C i ↮ C i2, …, C i ↮ C ik implies C i ↮ (C i1 ∪ … ∪ C ik ).

21. If we have these 2 properties… Lemma1. If events C 1, …, C l have mutually exclusive dependencies and union-closed independencies then

22. We can express P nc as. Instead of t(G), we bound its logarithm,

23. K-extension

24. Assume that all cycles C are enumerated such that the first t 3 cycles are the triangles in G, and the last t 2 cycles are the 2-cycles of G. In total we consider t = t 2 + t 3 cycles. We apply Lemma 1 with k = 1 and l = t 3 to bound, which is the probability that no 3-cycle occurs.

25. P(A|B)P(B) = P(A∩B) =>

26. For l = t and k = t 3 + 1. Thus, we can bound log from above by

27. Now rephrase log ≦

29. log i log j log i log j (log i)/i (log j)/j Σlog i Each vertex Σ (log i+log j) Each edge

30. log i (log i)/i degree = j degree = k degree = h (log i)/(i*(j-1)) (log i)/(i*(k-1)) (log i)/(i*(h-1))

31. Σ ( Σ log a/(a*(i-1))+ Σ log b/(b*(j-1))) Each edge ij ab Each a adjacent i Each b adjacent j

32. 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)

33. Σlog i = μ 1 (Σlog i) + μ 2 (Σlog i) + μ 3 (Σlog i) + μ 4 (Σlog i) = D1 + D2 + D3 + D4

34. ij ij arar arar brbr brbr

36. Constraints

37. 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.

38. 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

39. 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

40. Bound improvement Former upper bound: Using outgoing edge approach: – T 3 : – T 4 : – T 5 :

41. 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

42. 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:

43. 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

44. Future Work Since we consider only 2-cycles and 3-cycles from triangles, one would obtain a better bound for P nc by taking larger cycles into account. Lemma 1 uses two enumerations of the events C i to avoid the influence of the ordering. An elaborated enumeration scheme of the events C i might give better bounds.