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Posted: Sun Oct 15, 2006 3:13 pm
david wrote:Anyway here's my code, in case someone wants to test it.
Your program fails on this case:

Code: Select all

``````input:
1

0 0 5 0 2 4
4 0 6 0 -4 16

output:
pair 1: yes``````

Posted: Sun Oct 15, 2006 6:37 pm
kalinov wrote:
david wrote:Anyway here's my code, in case someone wants to test it.
Your program fails on this case:

Code: Select all

``````input:
1

0 0 5 0 2 4
4 0 6 0 -4 16

output:
pair 1: yes``````
hvala

and here is one more test case that helped me:

1

0 1 2 0 3 1
1 3 2 0 3 3

output:

pair 1: yes

### Re: 11122 - Tri Tri

Posted: Mon Oct 16, 2006 6:44 am
I'm a bit confused about this problem, does interior point means a lattice point or just an a common area? I assume the first, so shouldn't this testcase output no (I see that you output yes for it)?

0 0 1 0 1 1
0 0 1 0 1 1

Posted: Mon Oct 16, 2006 8:42 am
Those two have all their points in common - I am not sure why you thought that we were looking for points with integer coordinates only.

I liked this problem - especially because I learned about this (although it might be "common sense" for some):
http://en.wikipedia.org/wiki/Separating_axis_theorem

Posted: Mon Oct 16, 2006 8:52 am
I have code to find intersection of 2 convex polygons, so I just used that, then compute the area of the resulting polygons.

Posted: Mon Oct 16, 2006 9:11 am
Darko wrote:Those two have all their points in common - I am not sure why you thought that we were looking for points with integer coordinates only.
I thought so because in the sample input we have:
0 0 2 0 0 2
1 1 3 3 2 3
and answer is no although they've a common area (infinite non integer points)! can you please explain that?

Posted: Mon Oct 16, 2006 9:17 am
cpphamza wrote:
Darko wrote:Those two have all their points in common - I am not sure why you thought that we were looking for points with integer coordinates only.
I thought so because in the sample input we have:
0 0 2 0 0 2
1 1 3 3 2 3
and answer is no although they've a common area (infinite non integer points)! can you please explain that?
No, there isn't any common point..
A point of a triangle is on the edge of the other triangle..
It is not considered as a common point by the problem statement

Posted: Mon Oct 16, 2006 11:01 am
another way to say it is that the answer is "yes" iff the intersection of the two triangles have positive area. A set of infinite number of points, even if there are uncountable number of points need not have positive area (measure)

Posted: Mon Oct 16, 2006 12:54 pm
I have code to find intersection of 2 convex polygons, so I just used that, then compute the area of the resulting polygons.
I'm interested in implementing a convex polygon intersection algorithm, can you post your algorithm? (code, resources, links ...)

Also can anyone expalin other approaches to solve this problem?
Thanks.

Posted: Mon Oct 16, 2006 3:06 pm
I projected triangles onto 6 lines (normal to each side) and looked for gaps. It works for any two convex polygons - it's from that theorem (you can separate two polygons if there is a gap in projection - separating axis could go through that gap, normal to the line).

Posted: Mon Oct 16, 2006 7:52 pm
Yeah... my previous testcase response is wrong...
I re-submit, without a fault (precision's stuffs), and I get Accepted.

Rodrigo

Posted: Tue Oct 17, 2006 12:30 am
cpphamza wrote:
I have code to find intersection of 2 convex polygons, so I just used that, then compute the area of the resulting polygons.
I'm interested in implementing a convex polygon intersection algorithm, can you post your algorithm? (code, resources, links ...)

Also can anyone expalin other approaches to solve this problem?
Thanks.
The O(n) algorithm to find intersection of 2 convex polygons is related to the idea of rotating calipers.
G.T. Toussaint. A simple linear algorithm for intersecting convex polygons. The Visual Computer. 1: 118-123. 1985.

An O(n^2) algorithm is also possible, and it is pretty much just bruteforce.
Consider polygon P,Q, then for each edge of Q, compute the intersection of the half plane defined by the edge and polygon P, and update P.

If one is only interested in whether 2 convex polygon intersects, it is possible to do it in O(log n) using Chazelle and Dobkin's algorithm.

Posted: Tue Oct 17, 2006 12:40 am
sclo wrote:The O(n) algorithm to find intersection of 2 convex polygons is related to the idea of rotating calipers.
G.T. Toussaint. A simple linear algorithm for intersecting convex polygons. The Visual Computer. 1: 118-123. 1985.
Rotating?

Just sweeping left to right is enough. Imagine a vertical line through each vertex. You get O(N) strips, and in each strip at most 4 edges => each strip can be processed in O(1) time. That's all.

Posted: Sat Oct 21, 2006 8:38 am
Just sweeping left to right is enough. Imagine a vertical line through each vertex. You get O(N) strips, and in each strip at most 4 edges => each strip can be processed in O(1) time. That's all.
Could you explain the process? step by step.. I cant figure out what I have to search in each strip. If you have a link explaing this would be helpful.

Thax

### un-precise algorithm

Posted: Thu Nov 23, 2006 4:44 pm
I got ACC with a not-always-exact solution.
I take one of the triangles and start traversing its border
by doing small steps - say 1/1000 of the length of the
triangle side we are currently on.

So in about 3000 steps I get 3000 points and for each
such point I check if that point is internal for the other
triangle. That check is easy. If the point is internal then the
answer is "yes" but if all 3000 points are not internal for the
other triangle then the answer is "probably no" which
I just assume to be "for sure no".

I know it is stupid but it says ACC for the available input.