By Jin Akiyama, Mikio Kano, Toshinori Sakai

This e-book constitutes the refereed complaints of the Thailand-Japan Joint convention on Computational Geometry and Graphs, TJJCCGG 2012, held in Bangkok, Thailand, in December 2012.

The 15 unique learn papers awarded have been chosen from between six plenary talks, one precise public speak and forty-one talks by way of individuals from approximately 20 nations world wide. TJJCCGG 2012 supplied a discussion board for researchers operating in computational geometry, graph theory/algorithms and their applications.

**Read or Download Computational Geometry and Graphs: Thailand-Japan Joint Conference, TJJCCGG 2012, Bangkok, Thailand, December 6-8, 2012, Revised Selected Papers PDF**

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**Extra info for Computational Geometry and Graphs: Thailand-Japan Joint Conference, TJJCCGG 2012, Bangkok, Thailand, December 6-8, 2012, Revised Selected Papers**

**Example text**

A. a Thus the number of the a-colorings of La containing r is j=2 |Aj |. That is, the number of the a-colorings of La which contain Bi as a subset is at most 3a−1 . Case 2. |Bi | = 2, say Bi = {r, s}. Consider an a-coloring of La containing both r and s. Without loss of generality, suppose that r ∈ A1 and s ∈ A2 . To complete an a-coloring of La , we choose the other a − 2 colors each from the remaining Aj where j = 3, 4, . . , a. Thus the number of the a-colorings of La which contain Bi as a subset is aj=3 |Aj |.

So we can bound the number of possible labeled plane straight-line drawings by |Π| < 1− 3 n−4 · 8 n n! = 1 (5n+12)(n−1)! 8 Proof (of Theorem 1). Consider an n-universal point set P ⊂ R2 with |P | = n. Being universal, in particular P has to accommodate all graphs from Tn . By Lemma 1, there are exactly 2n−4 · (n − 3)! graphs in Tn , whereas by Lemma 3 no more than 18 (5n + 12)(n − 1)! graphs from Tn admit a plane straight-line drawing on P . Combining both bounds we obtain 2n−1 ≤ (5n + 12)(n − 1)(n − 2).

N−2 Proof. By Theorem 3 at least c = 14 n2 n−1 2 2 of P are in convex position. For n odd we have c= 1 4 n−1 2 2 n−3 2 n−3 2 four element subsets 2 and for n even we have c= 1 n 4 2 n−2 2 2 n−4 2 and so 1 4 3 = 8 3 = 8 c> for all n. n−1 n−2 n−3 n−4 2 2 2 2 n − 4 n(n − 1)(n − 2)(n − 3) · · n 4·3·2 n n−4 · · , 4 n 34 J. Cardinal, M. Hoﬀmann, and V. Kusters We will use this fact to prove the following lemma. Lemma 3. On any set P ⊂ R2 of n ≥ 4 points fewer than graphs from Tn admit a plane straight-line embedding.