Recently I have started doing some research on the SAT (Separating Axis Theorem) for collision detection in a game I am making. I understand how the algorithm works and why it works, what I'm puzzled about is how it expects one to be able to so easily calculate the projection of the shape onto different axes.
I assume the projection of a polygon onto a vector is represented by line segment from point A to point B, so my best guess to find points A and B would be to find the angle of the line being projected onto and calculate the min and max x-values of the coordinates when the shape is rotated to the angle of the projection (i.e. such that it is parallel to the x-axis and the min and max values are simply the min and max values along the x-axis). But to do this for every projection would be a costly operation. Do any of you guys know a better solution, or could at least point me to a paper or document where a better solution is described?
Simple way to calculate the projection of the polygon on line is to calculate projection of all vertex onto the line and get the coordinates with min-max values like you suggested but you dont need to rotate the polygon to do so.
Here is algorithm to find projection of point on line :-
line : y = mx + c
point : (x1,y1)
projection is intersection of line perpendicular to given line and passing through (x1,y1)
perdenicular line :- y-y1 = -1/m(x-x1) slope of perpendicular line is -1/m
y = -1/m(x-x1) + y1
To find point of intersection solve the equation simultaneously :-
y = mx + c , y = -1/m(x-x1) + y1
mx + c = -1/m(x-x1) + y1
m^2*x + mc = x1-x + my1
(m^2+1)x = x1 + my1 - mc
x = (x1-my1 - mc)/(m^2+1)
y = mx + c = m(x1-my1-mc)/(m^2+1) + c
Time complexity : For each vertex it takes O(1) time so it is O(V) where V is no of vertex in the polygon
If your polygon is not convex, compute its convex hull first.
Given a convex polygon with n vertices, you can find its rotated minimum and maximum x-coordinate in n log n by binary search. You can always test whether a vertex is a minimum or a maximum by rotating an comparing it and the two adjacent vertices. Depending on the results of the comparison, you know whether to jump clockwise or counterclockwise. Jump by k vertices, each time decreasing k by half (at the start k=n/2).
This may or may not bring real speed improvement. If your typical polygon has a dozen or so vertices, it may make little sense to use binary search.
Related
Assume you have a convex polygon P(defined by an array of points p), and a set of points S(all of them outside of P), how do you choose a point s in S such that it increases the most the area of P.
Example
I have a O(|P|) formula to calculate the area of the polygon, but I can't do this for every point in S given that
3 ≤ |P|, |S| ≤ 10^5
The big dots are the points in S
No 3 points in P u S are collinear
Given fixed points p = (px, py), q = (qx, qy) and a variable point s = (sx, sy), the signed area of the triangle ∆pqs is
|px py 1|
½ |qx qy 1|
|sx sy 1| ,
which is a linear polynomial in sx, sy.
One approach is to compute cumulative sums of these polynomials where p, q are the edges in clockwise order. Use binary search to find the sublist of edges that remain in the convex hull with a given point s, add the polynomials, and evaluate for s.
You have a method to calculate the exact area that is added by a point n (and David Eisenstat posted another), but their complexity depends on the number of sides of the polygon. Ideally you'd have a method that can quickly approximate the additional area, and you'd only have to run the exact method for a limited number of points.
As Paul pointed out in a comment, such an approximation should give a result that is consistently larger than the real value; this way, if the approximation tells you that a point adds less area than the current maximum (and with randomly ordered input this will be true for a large majority of points), you can discard it without needing the exact method.
The simplest method would be one where you only measure the distance from each point to one point in the polygon; this could be done e.g. like this:
Start by calculating the area of the polygon, and then find the smallest circle that contains the whole polygon, with center point c and radius r.
Then for each point n, calculate the distance d from n to c, and approximate the additional area as:
the triangle with area r × (d - r)
plus the rectangle with area 2 × r 2 (pre-calculated)
plus the half circle with area r × π (pre-calculated)
minus the area of the polygon (pre-calculated)
This area is indicated in blue on the image below, with the real additional area slightly darker and the excess area added by the approximation slightly lighter:
So for each point, you need to calculate a distance using √ ((xn - xc)2 + (yn - yc)2) and then multiply this distance by a constant and add a constant.
Of course, the precision of this approximation depends on how irregular the shape of the polygon is; if it does not resemble a circle at all, you may be better off creating a larger simple polygon (like a triangle or rectangle) that contains the original polygon, and use the precise method on the larger polygon as an approximation.
UPDATE
In a simple test where the polygon is a 1x1 square in the middle of a 100x100 square space, with 100,000 points randomly placed around it, the method described above reduces the number of calls to the precise measuring function from 100,000 to between 150 and 200, and between 10 and 20 of these calls result in a new maximum.
While writing the precise measuring function for the square I used in the test, I realised that using an axis-aligned rectangle instead of a circle around the polygon leads to a much simpler approximation method:
Create a rectangle around the polygon, with sides A and B and center point c, and calculate the areas of the rectangle and the polygon. Then, for each point n, the approximation of the additional area is the sum of:
the triangle with base A and height abs(yn - yc) - B/2
the triangle with base B and height abs(xn - xc) - A/2
the area of the rectangle minus the area of the polygon
(If the point is above, below or next to the rectangle, then one of the triangles has a height < 0, and only the other triangle is added.)
So the steps needed for the approximation are:
abs(xn - xc) × X + abs(yn - yc) × Y + Z
where X, Y and Z are constants, i.e. 2 subtractions, 2 additions, 2 multiplications and 2 absolute values. This is even simpler than the circle method, and a rectangle is also better suited for oblong polygons. The reduction in the number of calls to the precise measuring function should be similar to the test results mentioned above.
Task
Calculate the distance d in meters between a query-point q and a polygon P.
The query-point q is defined as tuple (latitudeq, longitudeq), the polygon P as ordered list of tuples [(latitude1, longitude1), ..., (latituden, longituden)].
Problem
I can't handle latitude and longitude as if they were x- and y-coordinates of a plane as this leads to huge errors if the polygon is not small and not near (0, 0).
Tools
I know how to calculate the distance between two points given each points latitude and longitude. I do also know how to calculate the distance between a point and a great circle. But for this task I'd need to know how to calculate the distance dist(q, 1—2) between a point q and and part of a great-circle 1—2. The distance of q would be simply min(dist(q, a—b)) ∀ a—b in P.
Question
Can you provide me a formula how to calculate the distance between a query-point q defined by a tuple (latitudeq, longitudeq) and a great-circle-arc defined by pair of latitude-longitude-tuples [(latitude1, longitude1),(latitude2, longitude2)]?
Example
If you had code to compute the distance between one point x and a geodesic line
segment s, you could repeat this for each edge of your geodesic polygon.
Let s=(a,b). s is an arc of a great circle. Rotate the sphere so that
s lies on the equator, and x follows along with the sphere rotation.
Then the latitude of x essentially tells you the distance to s: It is either
the distance from x to a, or x to b, or, if x lies in the sector above/below s, it is a simple factor (2 π r) times the latitude.
I have encountered the following interesting problem while preparing for a
contest.
You have a triangle with sides of length a, b, c and a rope of length L. You need to find
the surfaced enclosed by the rope that has the maximum surface area and it has to be entirely inside the triangle.
So, if L = a + b + c, then it's the area of the triangle.
Else, we know that the circle has the biggest surface to perimeter area, so if L is smaller or equal to the perimeter of the inscribed circle of the triangle, then the area will be the area of the circle of perimeter L.
So, the remaining case is alfa < L < a + b + c, where alfa is the perimeter of the inscribed circle .
Any ideas would be great!
EDIT: I would like to know if I should focus on some kind of algorithm for solving this
or trying to figure it out a mathematical formula. The contest contains somehow a combination of both. The edges can be as long as 100 and the precision of a,b,c,L is of 4 digits after the decimal point .
After reading the answers to this question: https://math.stackexchange.com/questions/4808/why-circle-encloses-largest-area, I agree with n.m., and think the optimal curve verifies:
Curvature is either constant, or flat when it touches the triangle, meaning it is composed of segments lying on the triangle sides, and circle arcs, all sharing the same radius.
There are no angles, meaning the arcs are tangent to the triangle sides.
With these conditions, the solution is obtained by three circles of same radius R, each tangent to two sides of the triangle (see below). When R varies between 0 and the radius of the inscribed circle, we start from the triangle itself, and end to the inscribed circle, where all three circles coincide. The length of the curve is the perimeter of the circle of radius R + the perimeter (p) of the smaller triangle: L = 2*PiR + p. The area is the area (a) of the smaller triangle + one disc of radius R + the remaining rectangles: A = PiR^2 + p*R + a.
Since a circle has the largest Area/Perimeter, start with the inscribed circle. If L is less than that circumference, then shrink appropriately. If L is longer, grow whichever of the 3 arcs maximizes dA/dL. I don't know if there's a closed form, but the largest arc will be in the 3rd of the triangle with the sides most approaching parallel.
It should be trivial to solve this algorithmically. With 4 decimals of precision, increment by 0.0001 checking each arc to see which has the greatest dA/dL for that single increment.
I worked up a drawing of the geometry overnight:
The inscribed circle is constructed by bisecting each of the angles and finding the intersections of the bisectors. I've labeled the half-angle "a1" (and all related variables have '1'). The area of the non-circular portion is two trapezoids (one denoted with the red outline). We can calculate the area for a single trapezoid as L1 * (m1 + R)/2 (note that when L1, L2, L3 are all zero, these trapezoids are all zero, and we just get the inscribed circle area). The circular cap has a radius of m1 to remain tangent with the side of the triangle. For a given choice of L1, m1 = R(x1-L1)/x1.
From here you can easily calculate the perimeter and area of each of the three sectors and solve numerically.
I cannot prove that this is the largest area, just that this is how to calculate the area and perimeter of this construction.
..answering my own comment/question, it can be proved that the radii must be equal,
Here is a useful formula:
the gray area A is
A = r^2 ( alpha - Pi + 2/tan(alpha/2) ) /2
but even more useful..the arc length is simply:
s = 2 ( b - A/r )
from here it is straightforward to show the three radii must be equal to each other:
writing the rope length and enclosed area:
ropelength = trianglelength - 2 Sum[r[i] a[i] ]
ropearea = trianglearea - Sum[r[i]^2 a[i] /2 ]
where
a[i]=( alpha[i] - Pi + 2/tan(alpha[i]/2) )
after a bit of manipulation maximizing the area leads to all r[i] equal. Note the three a[i], ropelength,trainglearea,trianglelength are all constants that you do not need to work out. Pedantically solve for r[l] = f( constants, r[2],r[3] ) sub into the second expression and solve for d ropearea /d r[2] = 0 and d /d r[3] = 0 with the result:
r =(1/2) (triangle_length - rope_length) /(Sum(1/tan(alpha[i]/2)) - Pi)
(the messy expression for a[i] is substituted only at the last step ) finally..
ropearea = trianglearea - (trianglelength-ropelength)^2/(8 Sum[a[i])
= trianglearea - (1/2)(trianglelength-ropelength) r
edit -- a useful identity ..with a,b,c, the lengths of the sides.
Sum(1/tan(alpha[i]/2)) = Sqrt( S^3 / ((S-a)(S-b)(S-c)) )
S = 1/2 (a+b+c) ! S is semiperimeter not to be confused with arc length s
the above expressions then can be used to reproduce the formula for an inscribed circle,
rinscribed = Sqrt( ((S-a)(S-b)(S-c)) / S )
If the perimeter of the rope is too small or too large, the answers are trivial. The interesting case is a shape with 6 vertices that goes line-arc-line-arc-line-arc. The arc are all tangent to their neighbouring lines and their radii are equal. I don't have a rigorous proof, but imagine a 2D balloon filled with air and squeezed between the sides of the triangle.
It is easy to express the overall shape and thus the perimeter given the radius; the opposite direction (perimeter to radius) is then easily found numerically.
For 3 points in 2D :
P1(x1,y1),
P2(x2,y2),
P3(x3,y3)
I need to find a point P(x,y), such that the maximum of the manhattan distances
max(dist(P,P1),
dist(P,P2),
dist(P,P3))
will be minimal.
Any ideas about the algorithm?
I would really prefer an exact algorithm.
There is an exact, noniterative algorithm for the problem; as Knoothe pointed out, the Manhattan distance is rotationally equivalent to the Chebyshev distance, and P is trivially computable for the Chebyshev distance as the mean of the extreme coordinates.
The points reachable from P within the Manhattan distance x form a diamond around P. Therefore, we need to find the minimum diamond that encloses all points, and its center will be P.
If we rotate the coordinate system by 45 degrees, the diamond is a square. Therefore, the problem can be reduced to finding the smallest enclosing square of the points.
The center of a smallest enclosing square can be found as the center of the smallest enclosing rectangle (which is trivially computed as the max and min of the coordinates). There is an infinite number of smallest enclosing squares, since you can shift the center along the shorter edge of the minimum rectangle and still have a minimal enclosing square. For our purposes, we can simply use the one whose center coincides with the enclosing rectangle.
So, in algorithmic form:
Rotate and scale the coordinate system by assigning x' = x/sqrt(2) - y/sqrt(2), y' = x/sqrt(2) + y/sqrt(2)
Compute x'_c = (max(x'_i) + min(x'_i))/2, y'_c = (max(y'_i) + min(y'_i))/2
Rotate back with x_c = x'_c/sqrt(2) + y'_c/sqrt(2), y_c = - x'_c/sqrt(2) + y'_c/sqrt(2)
Then x_c and y_c give the coordinates of P.
If an approximate solution is okay, you could try a simple optimization algorithm. Here's an example, in Python
import random
def opt(*points):
best, dist = (0, 0), 99999999
for i in range(10000):
new = best[0] + random.gauss(0, .5), best[1] + random.gauss(0, .5)
dist_new = max(abs(new[0] - qx) + abs(new[1] - qy) for qx, qy in points)
if dist_new < dist:
best, dist = new, dist_new
print new, dist_new
return best, dist
Explanation: We start with the point (0, 0), or any other random point, and modify it a few thousand times, each time keeping the better of the new and the previously best point. Gradually, this will approximate the optimum.
Note that simply picking the mean or median of the three points, or solving for x and y independently does not work when minimizing the maximum manhattan distance. Counter-example: Consider the points (0,0), (0,20) and (10,10), or (0,0), (0,1) and (0,100). If we pick the mean of the most separated points, this would yield (10,5) for the first example, and if we take the median this would be (0,1) for the second example, which both have a higher maximum manhattan distance than the optimum.
Update: Looks like solving for x and y independently and taking the mean of the most distant points does in fact work, provided that one does some pre- and postprocessing, as pointed out by thiton.
I have a circle with two points inside it that make up a line segment. How can I calculate the distance from one endpoint to the edge of the circle where the line would intersect it?
I think the easist way is to figure out where the intersection of the line and the circle is, then just calculate the distance from the line segment point with the intersection point.
So, say your circle is described by the equation
x^2 + y^2 = 5
and your line segment is points like
(1,3), (2,4)
First, you figure out the equation for the line that is directly over the segment, which, in this case, would be
y = x + 2
You then substitute this equation into the first equation, and you get
x^2 + (x+2)^2 = 5
Simplify this into
2x^2 + 4x - 1 = 0
and solve via the quadradic formula.
You now have the x coordinates of the two intersection points. From there, plug into the line equation to get the y coordinates. Then you can just do normal point distance calculation ala Pythagoras.
sqrt ( (x1 - x2)^2 + (y1 - y2)^2 )
Use the center of the circle as a point of reference. Get the distance from the center point to your two points, then the radius of the circle. You can now draw a triangle between any three of those points (center, segment point, and circle edge.) Pythagoras can handle the rest.
Two points define a line L. Solve for an equation Cx + L = 0 where C is the equation of the circle. If I remember correctly :P Some more information here.