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Edge selection on a 2D grid


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#1 ktn_vrce   Members   -  Reputation: 116

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Posted 09 May 2014 - 02:40 AM

I am working on a HTML5/javascript application which allows users to create 2d shapes on a canvas(rectangles, squares, etc).

I want to allow users to be able to select any edge of these objects using mouse clicks and modify the shapes. 

 

I am very new to HTML5/javascript or game programming. I am unable to implement the edge selection using mouse clicks. Any pointers on how to implement this?



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#2 Aardvajk   Crossbones+   -  Reputation: 5852

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Posted 09 May 2014 - 03:18 AM

You need to get the closest point on the line (segment) to the point clicked, then you can compare the distance between your click point and this point to see if it is within a certain threshold.

 

A 3D version in C++ for closest point on segment to point:

Vec3 closestPointOnSegmentToPoint(const Vec3 &segment1, const Vec3 &segment2, const Vec3 &point)
{
    Vec3 ab = segment2 - segment1;

    if(dotVectors(ab, ab) < 0.01f) return ab;

    float t = dotVectors(point - segment1, ab) / dotVectors(ab, ab);

    if(t < 0.0f) t = 0.0f;
    if(t > 1.0f) t = 1.0f;

    return segment1 + (t * ab);
}

That will return the point on the segment that is closest to the point input. You can then just form a vector from this point to the click point and check its length to see if it is close enough to be considered clicked on the line.

 

Appreciate a lot of that is over your head if you are new to this. Suggest reading up on 2D vector stuff, looking at what support Java has for vector math and so on.



#3 ktn_vrce   Members   -  Reputation: 116

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Posted 09 May 2014 - 04:14 AM

Thanks for the quick reply.

 

I do understand the logic here, however, wont it be too slow once I have many segments on the canvas? I will need to loop over all the line segments and find out which one is the closed to point?



#4 Álvaro   Crossbones+   -  Reputation: 12481

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Posted 09 May 2014 - 04:39 AM

Thanks for the quick reply.
 
I do understand the logic here, however, wont it be too slow once I have many segments on the canvas? I will need to loop over all the line segments and find out which one is the closed to point?


Well, that depends on what you mean by "many". If you really have a lot of segments, there are fancier techniques that can be used. But you can always start with the simple code and think about it harder only if it proves to be too slow.

#5 ktn_vrce   Members   -  Reputation: 116

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Posted 09 May 2014 - 04:50 AM

I will definitely start with the simpler technique like the one mentioned by Aardvajk, but it would be great if you can share some pointers on some advanced techniques. I came across R-trees while researching, is there any detail explanation on how to implement it? I am planning to create a 3D application at some point in future, and any reference material on this topic would help.



#6 Álvaro   Crossbones+   -  Reputation: 12481

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Posted 09 May 2014 - 05:35 AM

Yes, space-partitioning techniques like R-trees can be used.

You could also use something like OpenGL to render thickened segments, using a unique id for the segment as the color and a depth value that corresponds to the distance to the segment (you could use a texture for this, or perhaps a fragment shader). You would render this to a texture and then you just need to look up where the user clicked.

I am not sure the latter suggestion is particularly good in this situation, but it illustrates a general technique to be aware of.

#7 Aardvajk   Crossbones+   -  Reputation: 5852

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Posted 09 May 2014 - 06:57 AM

Also note. You'll be amazed how many times you can run a closest point to line test without it taking any significant time. And any spatial partitioning system has its own overhead in time and complexity.

 

I've dumped entire spatial partitioning systems over a broadphase AABB check because the former has been faster and far less complex.

 

What you can do here is form an AABB from the end points of the line and check the point is inside the AABB first, but I'm not even sure that will be significantly quicker in this particular case.






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