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标题: Find minimun number of points on the plane [打印本页]

作者: yma16 时间: 2005-9-10 19:28
标题: Find minimun number of points on the plane

such that:

1 They are vertices of concave polygons.

2 All the concave polygons have more than one inner angle with degree > 180.

You can either show a picture or list the coordinates of the points.

If you are inetrested, you can change (2) to be

2' All the concave polygons have exactly two inner angles with degree > 180.

作者: husonghu 时间: 2005-9-11 01:06
标题: 我想#2是对“由这些点构成的凹多边形的全体集合”而言，对吗？

我想#2是对“由这些点构成的凹多边形的全体集合”而言，对吗？

作者: yma16 时间: 2005-9-11 07:04
标题: Yes.

Yes.

作者: yma16 时间: 2005-9-11 07:19
标题: After this, please try 3, 4...

I think we can find a pattern about the minimum number of the points and how to draw the points.

作者: husonghu 时间: 2005-9-11 08:54
标题: 你肯定此题有解吗？即便对于正好2内角>180，我都没有找到答案, .....

更不要说正好3内角、或4内角>180的情况了。如果有解。你能否给出
一个正好2内角>180的例子看看？也许我还没有理解你的题意。

作者: 乱弹 时间: 2005-9-11 17:56
标题: 没理解题意

好像是要指定一些凹多边形。否则，必然有四个点是凹多边形的顶点，它们不满足条件2

作者: fzy 时间: 2005-9-11 22:04
标题: 回复：Find minimun number of points on the plane

6个应该够了。一个大三角形里倒扣一个小三角形。

改过的题好像没有解。

作者: yma16 时间: 2005-9-11 23:54
标题: What I mean is

each polygon uses all the point as its vertices. It is not the subset.

作者: yma16 时间: 2005-9-11 23:56
标题: guys and mms, could you try

(0,2), (-1, 1), (1,1), (-4,0), and (4,0). It may work.

作者: husonghu 时间: 2005-9-12 00:52
标题: 不行，连接(-4,0)(0,2)(-1,1)(1,1)(4,0)，你只在(-1,1)得到内角>180

不行，连接(-4,0)(0,2)(-1,1)(1,1)(4,0)，你只在(-1,1)得到内角>180

作者: yma16 时间: 2005-9-12 03:21
标题: I have to try more. - eom

I have to try more. - eom

作者: 乱弹 时间: 2005-9-12 04:09
标题: 你的感觉是正确的。 [@};-][@};-][@};-]

你的感觉是正确的。

作者: yma16 时间: 2005-9-13 06:39
标题: How about 4 vertices of a square within

4 vertices of a larger square. Like

. .

You can get exactly 2 angles >180. I could not get only one such angle. Although you can get more than 2 such angles. Try to reduce the number of points until we can't get 2 angles >180.

This is for (1) and (2), not (2'). FZY said (2') is impossible. He also said 6 points (vertices of 2 triangles) are enough. Please show a picture, someone.

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