Develop Reference

Minimum distance between two polygons [duplicate]

I want to find the minimum distance between two polygons with million number of vertices(not the minimum distance between their vertices). I have to find the minimum of shortest distance between each vertex of first shape with all of the vertices of the other one. Something like the Hausdorff Distance, but I need the minimum instead of the maximum.

Perhaps you should check out (PDF warning! Also note that, for some reason, the order of the pages is reversed) "Optimal Algorithms for Computing the Minimum Distance Between Two Finite Planar Sets" by Toussaint and Bhattacharya:
It is shown in this paper that the
minimum distance between two finite
planar sets if [sic] n points can be
computed in O(n log n) worst-case
running time and that this is optimal
to within a constant factor.
Furthermore, when the sets form a
convex polygon this complexity can be
reduced to O(n).
If the two polygons are crossing convex ones, perhaps you should also check out (PDF warning! Again, the order of the pages is reversed) "An Optimal Algorithm for Computing the Minimum Vertex Distance Between Two Crossing Convex Polygons" by Toussaint:
Let P = {p1,
p2,..., pm} and Q = {q1, q2,...,
qn} be two intersecting polygons whose vertices are specified
by their cartesian coordinates in
order. An optimal O(m + n)
algorithm is presented for computing
the minimum euclidean distance between
a vertex pi in P and a
vertex qj in Q.

There is a simple algorithm that uses Minkowski Addition that allows calculating min-distance of two convex polygonal shapes and runs in O(n + m).
Links:
algoWiki, boost.org, neerc.ifmo.ru (in russian).
If Minkowski subtraction of two convex polygons covers (0, 0), then they intersect

Java: Perimeter of a convex polygon

I have an ArrayList of some Point-s. It's guaranteed that the Points are part of a convex polygon.
How can I calculate the perimeter of this convex polygon?
Update: The Points in the ArrayList are out of any order
Update 2: All the points are part of the convex polygon's edge

Are the points in order? If so, you just need to sum up the distance from each vertex to the next

Sum the distance between each two consecutive points.

If the points are not ordered, its impossible without determining the correct order. Thats because if there are more than 3 points there is more than one polygon they could form.
I'm not entirely sure the constraint that the points are forming a convex polygon is sufficient to determine a canonic shape from the point cloud.
My guess is that by taking a random point from the list, and then looking for the nearest remaining point you can build a canonic order. From there its just summing up the length of the lines formed by consecutive points.
Edit: On second thought, scratch that idea. It won't work for all cases. That leaves you with permuting the points and checking if the formed polygon is indeed convex.
The question on how to check if a polygon is convex has been asked and answered here: How do determine if a polygon is complex/convex/nonconvex?

Java: plot points based on distances

I need to plot a group of points based on distances. I have three unknown points X, Y, and Z. I then get another unknown point (A) and its distances from the originals (AX, AY, AZ). I will continue getting points and distances (B, BX, BY, BZ; C, CX, CY, CZ) etc.
My question is whether its possible to plot all of the points. If so, how many points would I need for an exact plot map? What about an approximate map?
This is similar to this question but I get a different set of distances and am not limited to the original number of points.
Also, if it would help I could add more points to the X, Y, Z group which would give me more distances for A, B, etc.What I don't know until it's been somehow calculated are any of the Distances XY, XZ, YZ, AB, AC, etc.

I am not sure exactly what you mean by exact plot map or approximate plot map. I think I might know but I am not sure. But plotting all points in this case, to me is not possible if the user can continue to "add more points to the XYZ group", which is dynamic. It would sound like you need to know what the user is going to plot before he does. Now if all this is static, it is possible

I assume you use 2D space
If it is 1D then 2 points are enough (not identical !!!).
If 2D then 3 distances is enough but the points used must not lay on the same line !!!
position/orientation of the plot
for relative plot are above conditions enough if you want also the exact orientation and position then you need to know exact position of first 3 points otherwise your plot will look the same but can be offseted,rotated and mirrored to original geometry.
knowing 1 point eliminates offset
knowing 2 point eliminates rotation
knowing 3 point eliminates mirroring
[notes]
you need n+1 points for n-D coordinate system
[edit1] equations
original question text did not contain any equations need but comments requires it so here are some:
You will need intersection point between two hyperspheres (in 2D circles, in 3D spheres,...) so look here:
circle-circle intersection
Cast circle from each point as center with radius equal to the distance from that point. Find out intersection point that is the same between all combinations of circles (0,1),(0,2),(1,2)
Yellow intersection is the same in all 3 combinations so that is the next point or for 2D just solve this system:
(x-x0)^2+(y-y0)^2=l0^2
(x-x1)^2+(y-y1)^2=l1^2
(x-x2)^2+(y-y2)^2=l2^2
where x,y is the intersection point, xi,yi are center of circle and li is distance from that point.
The first option should be simpler and more accurate if done right but need some knowledge on vector and trigonometry math. You will need to add rotation or compute on vectors and use perpendicular vector feature in 2D
V(x,y) -> V0(+y,-x),V1(-y,+x)
where V0,V1 are perpendicular to V

Collision Handling... Confused where to go

So, I'm working on a 2D physics engine, and I have an issue. I'm having a hard time conceptualizing how you would calculate this:
Take two squares:They move, collide, and at some vector based off of the velocity of the two + the shape of them.
I have two vector lists(2D double lists) that represent these two shapes, how does one get the normal vector?
The hit vector is just (s1 is the first shape, s2 the second) s2 - s1 in terms of the position of the center of mass.
Now, I know a normal vector is one perpendicular to an edge, and I know that you can get the perpendicular vector of a line by 90 degrees, but what edge?
I read in several places, it is the edge a corner collided on. How do you determine this?
It just makes no sense to me, how you would mathematically or programmatically determine what edge.
Can anyone point out what I'm doing wrong in my understanding? Sorry for providing no code to explain this, as I'm having an issue writing the code for it in the first place.

Figure1: In 2D the normal vector is perpendicular to the tangent line:
Figure2: In 3D the normal vector is perpindicular to the tangent plane
Figure3: For a square the normal vector is easy if you are not at a corner; It is just perpendicular to the side of the square (in the image above, n = 1 i + 0 j, for any point along the right side of the square).
However, at a corner it becomes a little more difficult because the tangent is not well-defined (in terms of derivatives, the tangent is discontinuous at the corner, so perpendicular is ambiguous).
Even though the normal vector is not defined at a corner, it is defined directly to the left and right of it. Therefore, you can use the average of those two normals (n1 and n2) as the normal at a corner.
To be less technical, the normal vector will be in the direction from the center of the square to the corner of the collision.
EDIT: To answer the OP's further questions in the chat below: "How do you calculate the normal vector for a generic collision between two polygons s1 and s2 by only knowing the intersecting vertices."
In general, you can calculate the norm like this (N is total verts, m is verts inside collision):
vcenter = (∑N vi) / N
vcollision = (∑m vi) / m
n = vcollision - vcenter
Fig. 1 - vcollision is only a single vertex.
Fig. 2 - vcollision is avg of two verts.
Fig. 3 - vcollision for generic polygon intersection.

Euclidean distance to calculate if something is near something

currently i have using a framework and it has a function called distance2D, and it has this description:
Calculate the Euclidean distance
between two points (considering a
point as a vector object). Disregards
the Z component of the vectors and is
thus a little faster.
and this is how i use it
if(g.getCenterPointGlobal().distance2D(target.getCenterPointGlobal()) > 1)
System.out.println("Near");
i have totally no idea what a Euclidean distance is, i am thinking that it can be used to calculate how far 2 points are? because i am trying to compare distance between 2 objects and if they are near within a certain range i want to do something. how would i be able to use this?

Euclidean distance is the distance between 2 points as if you were using a ruler. I don't know what are the dimensions of your Euclidean space, but be careful because the function you are using just takes in consideration the first two dimensions (x,y). Thus if you have a space with 3 dimensions(x,y,z) it will only use the first two(x,y of x,y,z) to calculate the distance. This may give a wrong result.
For what I understood, if you want to trigger some action when two points are within some range you should make:
<!-- language: lang-java -->
if(g.getCenterPointGlobal().distance2D(target.getCenterPointGlobal()) < RANGE)
System.out.println("Near");

The Euclidean distance is calculated tracing a straight line between two points and measuring as the hypotenuse of a imaginary isosceles triangle between the two lines and a complementary point. This measure is scalar, so it's a good metric for your calculations.

Euclidean geometry is a coordinate system in which space is flat, not curved. You don't need to care about non-Euclidean geometry unless for example you're dealing with coordinates mapped onto a sphere, such as calculating the shortest travel distance between two places on Earth.
I imagine this function will basically use Pythagoras' theorem to calculate the distance between the two objects. However, as the description says, it disregards the Z component. In otherwords, it will only give a correct answer if both points have the same Z value (aka "depth").

If you wish to compare distances and save time, use not the distance itself, but its square: (x1-x2)^2 + (y1-y2)^2. Don't take sqrt. So, your distances will work exactly as euclidian ones, but quickly.