Combinatorial Geometry and Graph Theory: Indonesia-Japan by Jin Akiyama, Edy Tri Baskoro, Mikio Kano

By Jin Akiyama, Edy Tri Baskoro, Mikio Kano

This ebook constitutes the completely refereed post-proceedings of the Indonesia-Japan Joint convention on Combinatorial Geometry and Graph thought, IJCCGGT 2003, held in Bandung, Indonesia in September 2003.

The 23 revised papers provided have been rigorously chosen in the course of rounds of reviewing and development. one of the themes coated are coverings, convex polygons, convex polyhedra, matchings, graph colourings, crossing numbers, subdivision numbers, combinatorial optimization, combinatorics, spanning bushes, quite a few graph characteristica, convex our bodies, labelling, Ramsey quantity estimation, etc.

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9 When m is even, π(X ∪Y) ⊃ {[x, b, y], [x, d, y], [b, x, y], [d, x, y] | x, y ∈ Vm , x = y }. Proof. We will give the proof only in the case m ≡ 0 (mod 4). The proof in the case m ≡ 2 (mod 4) is similar. A 2-path [x, b, y] with 2 ≤ d(x, y) ≤ r − 1 belongs to T {Yk |1 ≤ k ≤ r − 2}. A 2-path [x, b, y] with d(x, y) = r belongs to T ∗ {Yr−1 }. A 2-path [x, b, y] with d(x, y) = 1 belongs to T ∗ {Yr , Yr+1 }. Therefore π(Y) ⊃ {[x, b, y] | x, y ∈ Vm , x =y}. A 2-path [x, d, y] with 2 ≤ d(x, y) ≤ r − 1 belongs to T {Yk |1 ≤ k ≤ r − 2}.

A 2-path [d, x, y] with y − x = 2 belongs to T ∗ {Xr , Yr−1 }. For any odd l (1 ≤ l ≤ r − 1), a 2-path [d, x, y] with y − x = l belongs to T {Xk |1 ≤ k ≤ r − 1}. For any even l (2 ≤ l ≤ r − 2), a 2-path [d, x, y] with y − x = −l belongs y}. to T {Xk |1 ≤ k ≤ r − 1}. Therefore π(X ∪ Y) ⊃ {[d, x, y] | x, y ∈ Vm , x = This completes the proof. Define r + 1 6-paths as follows: Uniform Coverings of 2-Paths with 6-Paths in the Complete Graph 31 [k + 3, e, 1, k + 1, c, 0, k + 2] (k: odd, 1 ≤ k ≤ r − 3) [−(k + 3), e, −1, −(k + 1), c, 0, −(k + 2)] (k: even, 1 ≤ k ≤ r − 3), [1, e, −1, −(r − 1), c, 0, r] (m ≡ 0 (mod 4)) Zr−2 = [−(r − 2), e, r, 0, c, r − 1, −2] (m ≡ 2 (mod 4)), Zr−1 = [r − 2, e, −2, 0, c, r, −(r − 1)], [1, 0, c, −1, r, e, −(r − 1)] (m ≡ 0 (mod 4)) Zr = [1, 0, c, −1, −(r − 1), e, r] (m ≡ 2 (mod 4)), [1, e, 0, r − 1, c, r, r − 2] (m ≡ 0 (mod 4)) Zr+1 = [0, e, 1, r − 1, c, r, r − 2] .

As the location of the point D varies, we obtain a continuum of four-sided cones. 12. We start at the lowest point with the degenerate case, which is represented by the flat-folded right triangle, and move (say counterclockwise) to the point 14 away from A on the rolling belt which is represented by the flat folded pentagon. We then move to the point 12 away from A on the rolling belt, which is represented by the regular tetrahedron and so on. 1 2 E 3 1 4 4 B A F D C Fig. 12 Fig. 13 Case 3. None of the three vertices converge at a point.

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