What is the best and faster way to resolve this problem?
Have a list of coordinates x,y => (x1,y1),(x..,y..),(xn,yn) ; x, y > 0 (intergers)
Find in all 2D space the coordinate (a,b) with the min sum manhattan distance from all point.
a, b can be different from (xn,yn).
Find median point A using only X-coordinates.
Find median point B using only Y-coordinates.
Point with components (A.X, B.Y) is what you need.
Example for 7 points:
point C is median by X - it goes 4-th in order of X-coordinates
point D is median by Y - it goes 4-th in order of Y-coordinates
min dist coordinate is (5,3) (marked by cross)
Related
I came across the below question and I am unable to find a way to find the closest pair distance using Divide and Conquer for the same, could someone please help?
L is the closest pair distance among all points with negative x co ordinate and
R is the closest pair distance among all points with positive x co ordinate.
Assume there are atleast 2 points with positive and 2 points with negative x coordinate. if L<R and that no point has x co ordinate in the interval (-L/2, R/2),
What is the Closest Pair Distance?
no point has x co ordinate in the interval (-L/2, R/2)
So, the closest distance between a point of the left (x <= -L/2) and a point on the right (x >= R/2) is if the two points are on the boundary (with the same y coordinates): Distance = L/2 + R/2
L < R
So, (L/2 + R/2) > (L/2 + L/2) = L
In other words, the shortest between a point that is on the left and one on the right is always greater than L, so L is the shortest distance.
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.
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.
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.
Picture a canvas that has a bunch of points randomly dispersed around it. Now pick one of those points. How would you find the closest 3 points to it such that if you drew a triangle connecting those points it would cover the chosen point?
Clarification: By "closest", I mean minimum sum of distances to the point.
This is mostly out of curiosity. I thought it would be a good way to estimate the "value" of a point if it is unknown, but the surrounding points are known. With 3 surrounding points you could extrapolate the value. I haven't heard of a problem like this before, doesn't seem very trivial so I thought it might be a fun exercise, even if it's not the best way to estimate something.
Your problem description is ambiguous. Which triangle are you after in this figure, the red one or the blue one?
The blue triangle is closer based on lexicographic comparison of the distances of the points, while the red triangle is closer based on the sum of the distances of the points.
Edit: you clarified it to make it clear that you want the sum of distances to be minimized (the red triangle).
So, how about this sketch algorithm?
Assume that the chosen point is at the origin (makes description of algorithm easy).
Sort the points by distance from the origin: P(1) is closest, P(n) is farthest.
Start with i = 3, s = ∞.
For each triple of points P(a), P(b), P(i) with a < b < i, if the triangle contains the origin, let s = min(s, |P(a)| + |P(b)| + |P(i)|).
If s ≤ |P(1)| + |P(2)| + |P(i)|, stop.
If i = n, stop.
Otherwise, increment i and go back to step 4.
Obviously this is O(n³) in the worst case.
Here's a sketch of another algorithm. Consider all pairs of points (A, B). For a third point to make a triangle containing the origin, it must lie in the grey shaded region in this figure:
By representing the points in polar coordinates (r, θ) and sorting them according to θ, it is straightforward to examine all these points and pick the closest one to the origin.
This is also O(n³) in the worst case, but a sensible order of visiting pairs (A, B) should yield an early exit in many problem instances.
Just a warning on the iterative method. You may find a triangle with 3 "near points" whose "length" is greater than another resulting by adding a more distant point to the set. Sorry, can't post this as a comment.
See Graph.
Red triangle has perimeter near 4 R while the black one has 3 Sqrt[3] -> 5.2 R
Like #thejh suggests, sort your points by distance from the chosen point.
Starting with the first 3 points, look for a triangle covering the chosen point.
If no triangle is found, expand you range to include the next closest point, and try all combinations.
Once a triangle is found, you don't necessarily have the final answer. However, you have now limited the final set of points to check. The furthest possible point to check would be at a distance equal to the sum of the distances of the first triangle found. Any further than this, and the sum of the distances is guaranteed to exceed the first triangle that was found.
Increase your range of points to include the last point whose distance <= the sum of the distances of the first triangle found.
Now check all combinations, and the answer is the triangle found from this set with the minimal sum of distances.
second shot
subsolution: (analytic geometry basics, skip if you are familiar with this) finding point of the opposite half-plane
Example: Let's have two points: A=[a,b]=[2,3] and B=[c,d]=[4,1]. Find vector u = A-B = (2-4,3-1) = (-2,2). This vector is parallel to AB line, so is the vector (-1,1). The equation for this line is defined by vector u and point in AB (i.e. A):
X = 2 -1*t
Y = 3 +1*t
Where t is any real number. Get rid of t:
t = 2 - X
Y = 3 + t = 3 + (2 - X) = 5 - X
X + Y - 5 = 0
Any point that fits in this equation is in the line.
Now let's have another point to define the half-plane, i.e. C=[1,1], we get:
X + Y - 5 = 1 + 1 - 5 < 0
Any point with opposite non-equation sign is in another half-plane, which are these points:
X + Y - 5 > 0
solution: finding the minimum triangle that fits the point S
Find the closest point P as min(sqrt( (Xp - Xs)^2 + (Yp - Ys)^2 ))
Find perpendicular vector to SP as u = (-Yp+Ys,Xp-Xs)
Find two closest points A, B from the opposite half-plane to sigma = pP where p = Su (see subsolution), such as A is on the different site of line q = SP (see final part of the subsolution)
Now we have triangle ABP that covers S: calculate sum of distances |SP|+|SA|+|SB|
Find the second closest point to S and continue from 1. If the sum of distances is smaller than that in previous steps, remember it. Stop if |SP| is greater than the smallest sum of distances or no more points are available.
I hope this diagram makes it clear.
This is my first shot:
split the space into quadrants
with picked point at the [0,0]
coords
find the closest point
from each quadrant (so you have 4
points)
any triangle from these
points should be small enough (but not necesarilly the smallest)
Take the closest N=3 points. Check whether the triange fits. If not, increment N by one and try out all combinations. Do that until something fits or nothing does.