Given an axis aligned square divided in four cells of equal size A, B, C and D.
Given a line segment going from point s1 to point s2.
What is the fastest way to find the cells traversed by the segment (if any), sorted by traversal order ?
In the example above, the correct results are :
Segment 1 : [D]
Segment 2 : [A, B]
Segment 3 : [C, D, B]
Segment 4 : []
Segment 5 : [C]
You can try "A Fast Voxel Traversal Algorithm for Ray Tracing" by Amanatides and Woo.
It is intended to deal with big grids, but the principle can be useful for your application too.
The following can be done using only a simple line intersect formula:
Observe that your grid consists of 6 line segments (3 horizontal, 3 vertical). If you were to extend each of these segments into infinity (making them lines, without start or end points), you have divided your 2D space into 16 areas.
Define a 4x4 array of areas. For each such area, store which line (if any) delimits it on the north side, on the east side, etc. This will be our traversal data structure.
Now, to find the cells traversed by a given query segment S, starting at u and ending at v, we will do a line walk from u to v using this traversal data structure to keep track of the area we're currently in and at which locations we're exiting that area.
Determine Au as the area in which u is located, and Av as the area in which v is located. Because of the axis-aligned nature of the areas, this shouldn't take more than 4 comparisons per area (2 on x-coordinate, 2 on y).
Also, define the current location as p and the current area as A; initially, these will be u and Au, respectively.
First, report A as an area that is traversed by S. Determine the first intersection point between the segment* (p, v] and the delimiting lines at each of the 4 sides of A. If such an intersection point q is found, the side that contains q determines which neighbouring area will become our new A - in that case our new p will be the intersection point q. Using this new A and p, repeat this step.
If no intersection point is found, p must lie in the same area as v and the walk is complete.
*(p, v] meaning: the segment between p and v, excluding p but including v.
Related
The problem is as it follows:
Given N (N <= 100,000) points by their Cartesian coordinates, test if every rectangle (with an area > 0) built using any two of them contains at least another point from that list either inside the rectangle or on his margin.
I know the algorithm for O(N^2) time complexity. I want to find the solution for O(N * logN). Memory is not a problem.
When I say two points define a rectangle, I mean that those two points are in two of its opposite corners. The sides of the rectangles are parallel to the Cartesian axes.
Here are two examples:
N = 4
Points:
1 1
3 5
2 4
8 8
INCORRECT: (the rectangle (1, 1) - (2, 4) doesn't have any point inside of it or on its margin).
N = 3
Points:
10 9
13 9
10 8
CORRECT.
Sort the points in order of the max of their coordinate pairs, from lowest to highest (O(n*log(n))). Starting with the lower-left point (lowest max coordinate), if the next point in the ordering does not share either the original point's x-value or its y-value (e.g. (1,2) and (5,2) share a y-coordinate of 2, but (1,2) and (2, 1) have neither coordinate in common), the set of points fails the test. Otherwise, move to the next point. If you reach the final point in this way (O(n)), the set is valid. The algorithm overall is O(n*log(n)).
Explanation
Two points that share a coordinate value lie along a line parallel to one of the axes, so there is no rectangle drawn between them (since such a rectangle would have area=0).
For a given point p1, if the next "bigger" point in the ordering, p2, is directly vertical or horizontal from p1 (i.e. it shares a coordinate value), then all points to the upper-right of p2 form rectangles with p1 that include p2, so there are no rectangles in the set that have p1 as the lower-left corner and lack an internal point.
If, however, the next-bigger point is diagonal from p2, then the p1 <-> p2 rectangle has no points from the set inside it, so the set is invalid.
For every point P = (a, b) in the set, search the nearest points of the form Y = (x, b) and X = (a, y) such that x > a and y > b.
Then, find if the rectangle defined by the two points X, Y contains any* internal point R besides P, X and Y. If that is the case, it's easy to see that the rectangle P, R does not contain any other point in the set.
If no point in the set exists matching the restrictions for X or Y, then you have to use (a, ∞) or (∞, b) respectively instead.
The computational cost of the algorithm is O(NlogN):
Looking for X or Y can be done using binary search [O(logN)] over a presorted list [O(NlogN)].
Looking for R can be done using some spatial tree structure as a quadtree or a k-d tree [O(logN)].
*) If X, Y contains more than one internal point, R should be selected as the nearest to P.
Update: the algorithm above works for rectangles defined by its botton-left and upper-right corners. In order to make it work also for rectangles defined by its botton-right and upper-left corners, a new point X' (such that it is the nearest one to P of the form X' = (a, y') where y' < b) and the corresponding rectangle defined by X', Y should also be considered for every point in the set.
I have two sets of points A and B.
I want to find all points in B that are within a certain range r to A, where a point b in B is said to be within range r to A if there is at least one point a in A whose (Euclidean) distance to b is equal or smaller to r.
Each of the both sets of points is a coherent set of points. They are generated from the voxel locations of two non overlapping objects.
In 1D this problem fairly easy: all points of B within [min(A)-r max(A)+r]
But I am in 3D.
What is the best way to do this?
I currently repetitively search for every point in A all points in B that within range using some knn algorithm (ie. matlab's rangesearch) and then unite all those sets. But I got a feeling that there should be a better way to do this. I'd prefer a high level/vectorized solution in matlab, but pseudo code is fine too :)
I also thought of writing all the points to images and using image dilation on object A with a radius of r. But that sounds like quite an overhead.
You can use a k-d tree to store all points of A.
Iterate points b of B, and for each point - find the nearest point in A (let it be a) in the k-d tree. The point b should be included in the result if and only if the distance d(a,b) is smaller then r.
Complexity will be O(|B| * log(|A|) + |A|*log(|A|))
I archived further speedup by enhancing #amit's solution by first filtering out points of B that are definitely too far away from all points in A, because they are too far away even in a single dimension (kinda following the 1D solution mentioned in the question).
Doing so limits the complexity to O(|B|+min(|B|,(2r/res)^3) * log(|A|) + |A|*log(|A|)) where res is the minimum distance between two points and thus reduces run time in the test case to 5s (from 10s, and even more in other cases).
example code in matlab:
r=5;
A=randn(10,3);
B=randn(200,3)+5;
roughframe=[min(A,[],1)-r;max(A,[],1)+r];
sortedout=any(bsxfun(#lt,B,roughframe(1,:)),2)|any(bsxfun(#gt,B,roughframe(2,:)),2);
B=B(~sortedout,:);
[~,dist]=knnsearch(A,B);
B=B(dist<=r,:);
bsxfun() is your friend here. So, say you have 10 points in set A and 3 points in set B. You want to have them arrange so that the singleton dimension is at the row / columns. I will randomly generate them for demonstration
A = rand(10, 1, 3); % 10 points in x, y, z, singleton in rows
B = rand(1, 3, 3); % 3 points in x, y, z, singleton in cols
Then, distances among all the points can be calculated in two steps
dd = bsxfun(#(x,y) (x - y).^2, A, B); % differences of x, y, z in squares
d = sqrt(sum(dd, 3)); % this completes sqrt(dx^2 + dy^2 + dz^2)
Now, you have an array of the distance among points in A and B. So, for exampl, the distance between point 3 in A and point 2 in B should be in d(3, 2). Hope this helps.
I have a set of non-intersecting lines, some of which are connected at vertices. I'm trying to find the smallest polygon, if one exists, that encloses a given point. So, in the image below, out of the list of all the line segments, given the point in red, I want to get the blue ones only. I'm using Python, but could probably adapt an algorithm from other languages; I have no idea what this problem is called.
First, remove all line segments that have at least one free endpoint, not coincident with any other segment. Do that repeatedly until no such segment remains.
Now you have a plane nicely subdivided into polygonal areas.
Find a segment closest to your point. Not the closest endpoint but the closest segment, mind you. Find out which direction along the segment you need (your point should be to the right of the directed segment). Go to the endpoint, turn right (that is, take the segment next to the one you came from, counting counterclockwise). Continue going to the next endpoint and turning right until you hit the closest segment again.
Then, check if the polygon encloses the given point. If it is not, then you have found an "island"; remove that polygon and the entire connected component it belongs to, and start over by selecting the nearest segment again. Connected components can be found with a simple DFS.
This gives you a clockwise-oriented polygon. If you want counterclockwise, which is often the default "positive" direction in both the software an the literature, start with the point to your left, and turn left at each intersection.
It surely helps if, given an endpoint, you can quickly find all segments incident with it.
This is really just an implementation of #n.m.'s answer.
This is how far I got before the bounty expired; it's completely untested.
def smallestPolygon(point,segments):
"""
:param point: the point (x,y) to surrond
:param segments: the line segments ( ( (a,b),(c,d) ) , . . . )
that will make the polygon
(assume no duplicates and no intersections)
:returns: the segments forming the smallest polygon containing the point
"""
connected = list(segments)
def endPointMatches(segment1,segment2):
"""
Which endpoints of segment1 are in segment2 (with (F,F) if they're equal)
"""
if ( segment1 == segment2 or segment1 == segment2[::-1] ):
return ( False, False )
return ( segment1[0] in segment2 , segment1[1] in segment2 )
keepPruning = True
while ( keepPruning ):
keepPruning = False
for segment in connected:
from functors import partial
endPointMatcher = partial(endPointMatches,segment1=segment)
endPointMatchings = map(endPointMatcher,connected)
if ( not and(*map(any,zip(*endPointMatchings))) ):
connected.remove(segment)
keepPruning = True
def xOfIntersection(seg,y):
"""
:param seg: a line segment ( (x0,y0), (x1,y1) )
:param y: a y-coordinate
:returns: the x coordinate so that (x,y) is on the line through the segment
"""
return seg[0][0]+(y-seg[0][1])*(seg[1][0]-seg[0][0])/(seg[1][1]-seg[0][1])
def aboveAndBelow(segment):
return ( segment[0][1] <= point[1] ) != ( segment[1][1] <= point[1] )
# look for first segment to the right
closest = None
smallestDist = float("inf")
for segment in filter(aboveAndBelow,connected):
dist = xOfIntersection(segment,point[1])-point[0]
if ( dist >= 0 and dist < smallestDist ):
smallestDist = dist
closest = segment
# From the bottom of closest:
# Go to the other end, look at the starting point, turn right until
# we hit the next segment. Take that, and repeat until we're back at closest.
# If there are an even number of segments directly to the right of point[0],
# then point is not in the polygon we just found, and we need to delete that
# connected component and start over
# If there are an odd number, then point is in the polygon we found, and we
# return the polygon
Approach.
I suggest to interpret the input as a PSLG, G, which consists of vertices and edges. Then your question reduces to finding the face of G that is hit by the point p. This is done by shooting a ray from p to some direction to find an edge of the boundary of the face and traverse the boundary of the face in some direction. However, the first edge hit could be a face that is not hit by p, but itself enclosed by the face hit by p. Hence, we may need to keep search along the ray emanated by p.
Implementational details.
In the code below I shoot a ray to east direction and run around the face in clockwise direction, i.e., at each vertex I take the next counter-clockwise edge, until I end up at the first vertex again. The resulting face is returned as a sequence of vertices of G.
If you want to return a simple polygon then you have to clean-up the input graph G by pruning of trees in G such that only the simple faces remain.
def find_smallest_enclosing_polygon(G, p, simple=False):
"""Find face of PSLG G hit by point p. If simple is True
then the face forms a simple polygon, i.e., "trees"
attached to vertices are pruned."""
if simple:
# Make G comprise simple faces only, i.e., all vertices
# have degree >= 2.
done = False
while not done:
done = True
for v in [v in vertices if degree(v) <= 1]:
# Remove vertex v and all incident edges
G.remove(v)
done = False
# Shoot a ray from p to the east direction and get all edges.
ray = Ray(p, Vector(1, 0))
for e in G.iter_edges_hit(ray):
# There is no enclosing face; p is in the outer face of G
if e is None:
return None
# Let oriented edge (u, v) point clockwise around the face enclosing p
u, v = G.get_vertices(e)
if u.y < v.y
u, v = v, u
# Construct the enclosing face in clockwise direction
face = [u, v]
# While not closed
while face[-1] != face[0]:
# Get next counter-clockwise edge at last vertex at last edge.
# If face[-1] is of degree 1 then I assume to get e again.
e = G.get_next_ccw_edge(face[-2], face[-1])
face.append(G.get_opposite_vertex(e, face[-1]))
# Check whether face encloses p.
if contains(face, p):
return face
return None
Complexity.
Let n denote the number of vertices. Note that in a PSLG the number of edges are in O(n). The pruning part may take O(n^2) time the way it is implemented above. But it could be one in O(n) time by identifying the degree-1 vertices and keep traversing from those.
The ray intersection routine can be implemented trivially in O(n) time. Constructing the face takes O(m) time, where m is the size of the polygon constructed. We may need to test multiple polygons but the sum of sizes of all polygons is still in O(n). The test contains(face, p) could be done by checking whether face contains an uneven number of edges returned by G.iter_edges_hit(ray), i.e., in O(m) time.
With some preprocessing you can build up a data structure that finds the face hit by p in O(log n) time by classical point location algorithms in computational geometry and you can store the resulting polygons upfront, for the simple and/or non-simple cases.
If you're doing this a number of times with the same lines and different points, it would be worthwhile preprocessing to figure out all the polygons. Then it's straightforward: draw a line from the point to infinity (conceptually speaking). Every time the you cross a line, increment the crossing count of each polygon the line is a part of. At the end, the first polygon with an odd crossing count is the smallest enclosing polygon. Since any arbitrary line will do just as well as any other (it doesn't even need to be straight), simplify the arithmetic by drawing a vertical or horizontal line, but watch out for crossing actual endpoints.
You could do this without preprocessing by creating the polygons as you cross each line. This basically reduces to n.m.'s algorithm but without all the special case checks.
Note that a line can belong to two polygons. Indeed, it could belong to more, but it's not clear to me how you would tell: consider the following:
+---------------------------+
| |
| +-------------------+ |
| | | |
| | +-----------+ | |
| | | | | |
| | | | | |
+---+---+-----------+---+---+
I have an array with vertices representing a triangular strip.
I need to convert it into polygon.
There are many solution to do the reverse, but I failed to find one for the above problem.
Or it could be too easy and I just cannot see it.
Please help.
OpenGL=compatible, see
http://en.wikipedia.org/wiki/Triangle_strip
Example:
for this strip http://en.wikipedia.org/wiki/File:Triangle_Strip_Small.png
I need output A B D F E C or A C E F D B
I believe the following should work:
Walk through the list of vertices. Add the first point to your polygon. Push the second point on the stack. Add the third point to the polygon. Continue alternating between pushing points on the stack and adding them to the polygon until you reach the end of the list. When you get to the end of the list, pop the points of the stack and add them to the polygon.
I'll assume your triangle strip is always connected the same way (which I believe is true for OpenGL).
The "bottom" vertices are always two
apart: A, C, E, ...
The "top"
vertices are always two apart: B, D,
F, ...
Take the "bottom" list and append the reverse of the "top" list. (ACEFDB for the example)
Or, more directly, using a zero-based index instead of letters:
// do "bottom"
for ( i = 0; i < N; i += 2 )
addVertex( i )
// do "top"
largestOddNumberLessThanN = N % 2 == 0 ? N - 1 : N - 2;
for ( i = largestOddNumberLessThanN; i >= 0; i -= 2 )
addVertex( i )
There may be a shortcut if your shape has a particularly simple structure, but in general I think you want to do the following
Create a list of vertices and edges from your list of triangles. This involves detecting shared points (hopefully they are exact matches so you can find them by hashing instead of needing some sort of fuzzy search) and shared lines (that's easy--just don't draw two edges between the same pair of shared vertices).
Find the upper-left-most point.
Find the edge that travels as upper-left-most as possible, and is in the counterclockwise direction.
Walk to the next vertex.
Find the next edge that doubles back on the previous one as much as possible.
Keep going until you hit the upper-left-most point again.
I have two line segments: X1,Y1,Z1 - X2,Y2,Z2 And X3,Y3,Z3 - X4,Y4,Z4
I am trying to find the shortest distance between the two segments.
I have been looking for a solution for hours, but all of them seem to work with lines rather than line segments.
Any ideas how to go about this, or any sources of furmulae?
I'll answer this in terms of matlab, but other programming environments can be used. I'll add that this solution is valid to solve the problem in any number of dimensions (>= 3).
Assume that we have two line segments in space, PQ and RS. Here are a few random sets of points.
> P = randn(1,3)
P =
-0.43256 -1.6656 0.12533
> Q = randn(1,3)
Q =
0.28768 -1.1465 1.1909
> R = randn(1,3)
R =
1.1892 -0.037633 0.32729
> S = randn(1,3)
S =
0.17464 -0.18671 0.72579
The infinite line PQ(t) is easily defined as
PQ(u) = P + u*(Q-P)
Likewise, we have
RS(v) = R + v*(S-R)
See that for each line, when the parameter is at 0 or 1, we get one of the original endpoints on the line returned. Thus, we know that PQ(0) == P, PQ(1) == Q, RS(0) == R, and RS(1) == S.
This way of defining a line parametrically is very useful in many contexts.
Next, imagine we were looking down along line PQ. Can we find the point of smallest distance from the line segment RS to the infinite line PQ? This is most easily done by a projection into the null space of line PQ.
> N = null(P-Q)
N =
-0.37428 -0.76828
0.9078 -0.18927
-0.18927 0.61149
Thus, null(P-Q) is a pair of basis vectors that span the two dimensional subspace orthogonal to the line PQ.
> r = (R-P)*N
r =
0.83265 -1.4306
> s = (S-P)*N
s =
1.0016 -0.37923
Essentially what we have done is to project the vector RS into the 2 dimensional subspace (plane) orthogonal to the line PQ. By subtracting off P (a point on line PQ) to get r and s, we ensure that the infinite line passes through the origin in this projection plane.
So really, we have reduced this to finding the minimum distance from the line rs(v) to the origin (0,0) in the projection plane. Recall that the line rs(v) is defined by the parameter v as:
rs(v) = r + v*(s-r)
The normal vector to the line rs(v) will give us what we need. Since we have reduced this to 2 dimensions because the original space was 3-d, we can do it simply. Otherwise, I'd just have used null again. This little trick works in 2-d:
> n = (s - r)*[0 -1;1 0];
> n = n/norm(n);
n is now a vector with unit length. The distance from the infinite line rs(v) to the origin is simple.
> d = dot(n,r)
d =
1.0491
See that I could also have used s, to get the same distance. The actual distance is abs(d), but as it turns out, d was positive here anyway.
> d = dot(n,s)
d =
1.0491
Can we determine v from this? Yes. Recall that the origin is a distance of d units from the line that connects points r and s. Therefore we can write dn = r + v(s-r), for some value of the scalar v. Form the dot product of each side of this equation with the vector (s-r), and solve for v.
> v = dot(s-r,d*n-r)/dot(s-r,s-r)
v =
1.2024
This tells us that the closest approach of the line segment rs to the origin happened outside the end points of the line segment. So really the closest point on rs to the origin was the point rs(1) = s.
Backing out from the projection, this tells us that the closest point on line segment RS to the infinite line PQ was the point S.
There is one more step in the analysis to take. What is the closest point on the line segment PQ? Does this point fall inside the line segment, or does it too fall outside the endpoints?
We project the point S onto the line PQ. (This expression for u is easily enough derived from similar logic as I did before. Note here that I've used \ to do the work.)
> u = (Q-P)'\((S - (S*N)*N') - P)'
u =
0.95903
See that u lies in the interval [0,1]. We have solved the problem. The point on line PQ is
> P + u*(Q-P)
ans =
0.25817 -1.1677 1.1473
And, the distance between closest points on the two line segments was
> norm(P + u*(Q-P) - S)
ans =
1.071
Of course, all of this can be compressed into just a few short lines of code. But it helps to expand it all out to gain understanding of how it works.
One basic approach is the same as computing the shortest distance between 2 lines, with one exception.
If you look at most algorithms for finding the shortest distance between 2 lines, you'll find that it finds the points on each line that are the closest, then computes the distance from them.
The trick to extend this to segments (or rays), is to see if that point is beyond one of the end points of the line, and if so, use the end point instead of the actual closest point on the infinite line.
For a concrete sample, see:
http://softsurfer.com/Archive/algorithm_0106/algorithm_0106.htm
More specifically:
http://softsurfer.com/Archive/algorithm_0106/algorithm_0106.htm#dist3D_Segment_to_Segment()
I would parameterize both line segments to use one parameter each, bound between 0 and 1, inclusive. Then find the difference between both line functions and use that as the objective function in a linear optimization problem with the parameters as variables.
So say you have a line from (0,0,0) to (1,0,0) and another from (0,1,0) to (0,0,0) (Yeah, I'm using easy ones). The lines can be parameterized like (1*t,0*t,0*t) where t lies in [0,1] and (0*s,1*s,0*s) where s lies in [0,1], independent of t.
Then you need to minimize ||(1*t,1*s,0)|| where t, s lie in [0,1]. That's a pretty simple problem to solve.
How about extending the line segments into infinite lines and find the shortest distance between the two lines. Then find the points on each line that are the end points of the shortest distance line segment.
If the point for each line is on the original line segment, then the you have the answer. If a point for each line is not on the original segment, then the point is one of the original line segments' end points.
Finding the distance between two finite lines based on finding between two infinite lines and then bound the infinite lines to the finite lines doesn't work always. for example try this points
Q=[5 2 0]
P=[2 2 0]
S=[3 3.25 0]
R=[0 3 0]
Based on infinite approach the algorithm select R and P for distance calculation (distance=2.2361), but somewhere in the middle of R and S has got a closer distance to the P point. Apparently, selecting P and [2 3.166] from R to S line has lower distance of 1.1666. Even this answer could get better by precise calculation and finding orthogonal line from P to R S line.
First, find the closest approach Line Segment bridging between their extended lines. Let's call this LineSeg BR.
If BR.endPt1 falls on LS1 and BR.endPt2 falls on LS2, you're done...just calculate the length of BR.
If the bridge BR intersects LS1 but not LS2, use the shorter of these two distances: smallerOf(dist(BR.endPt1, LS2.endPt1), dist(BR.endPt1, LS2.endPt2))
If the bridge BR intersects LS2 but not LS1, use the shorter of these two distances: smallerOf(dist(BR.endPt2, LS1.endPt1), dist(BR.endPt2, LS1.endPt2))
If none of these conditions hold, the closest distance is the closest pairing of endpoints on opposite Line Segs.
This question is the topic of the article On fast computation of distance between line segments by Vladimir J. Lumelksy 1985. It goes even further by finding not only the minimal Euclidean distance (MinD) but a point on each segment separated by that distance.
The general algorithm is as follows:
Compute the global MinD (global means the distance between two infinite lines containing the segments) and coordinates of both points (bases) of the line of minimum distances, see skew lines; if both bases lie inside the segments, then actual MinD is equal to the global MinD; otherwise, continue.
Compute distances between the endpoints of both segments (a total of four distances).
Compute coordinates of the base points of perpendiculars from the endpoints of one segment onto the other segment; compute the lengths of those perpendiculars whose base points lie inside the corresponding segments (up to four base point, and four distances).
Out of the remaining distances, the smallest is the sought actual MinD.
Altogether, this represents the computation of six points and of nine distances.
The article then describes and proves how to reduce the amount of tests based on the data received in initial steps of the algorithm and how to handle degenerate cases (e.g. equal endpoints of a segment).
C-language implementation by Eric Larsen can be found here, see SegPoints() function.