By Jin Akiyama, Takayasu Kuwata (auth.), Jin Akiyama, Jiang Bo, Mikio Kano, Xuehou Tan (eds.)

ISBN-10: 3642249825

ISBN-13: 9783642249822

This publication constitutes the completely refereed post-conference court cases of the China-Japan Joint convention on Computational Geometry, Graphs and functions, CGGA 2010, held in Dalian, China, in November 2010.

The 23 revised complete papers offered have been conscientiously chosen in the course of rounds of reviewing and development from a number of submissions. All features of computational and discrete geometry, graph idea, graph algorithms, and their purposes are covered.

**Read or Download Computational Geometry, Graphs and Applications: 9th International Conference, CGGA 2010, Dalian, China, November 3-6, 2010, Revised Selected Papers PDF**

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**Extra resources for Computational Geometry, Graphs and Applications: 9th International Conference, CGGA 2010, Dalian, China, November 3-6, 2010, Revised Selected Papers**

**Example text**

5), so it suﬃces to apply the projection method once to determine the foldability of the end edge closer to x. 1). 2). Then we show that the exit condition is correct, that is, prove Lemma 1 described Making Polygons by Simple Folds and One Straight Cut 33 candidate limit line limit line 1 1 2 2 2 2 1 1 (a) (b) Fig. 3. Computing limit points of end-edge foldability by (1) projecting non-end edges onto end edges and (2) further projecting portions of the end edges determined to be nonfoldable. Limit points computed by the algorithm are identiﬁed with circular markers.

A square si is a connector if its two non-shared sides are opposite, otherwise si is a turn. We assume without loss of generality that the number of left turns is larger than the number of right turns (otherwise the order of the sequence of squares is simply inverted). Consider the subpath P = {s2 , . . , sk−1 } of P . The sides of P form the left subchain L and the right subchain R with |L| + |R| = 2k−4. The length of L or R corresponds to the number of connectors on the path plus twice the number of right or left turns, respectively.

The proof of this lemma is similar in spirit to [5, Fig. 6], [6, Fig. 11], though the details diﬀer. Proof. Refer to Fig. 5. If a base angle is obtuse, say at a, then we conceptually split it into two triangles, by cutting along the segment aa orthogonal to bc. The triangle aa c, with base aa , is non-obtuse. Assuming the lemma for such triangles, we can reduce the triangle to a thin triangle against aa , at which point we can fold it into the other triangle aba . Now we are left with triangle aba with base ab, which is also non-obtuse.

### Computational Geometry, Graphs and Applications: 9th International Conference, CGGA 2010, Dalian, China, November 3-6, 2010, Revised Selected Papers by Jin Akiyama, Takayasu Kuwata (auth.), Jin Akiyama, Jiang Bo, Mikio Kano, Xuehou Tan (eds.)

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