I have a set of points. I want to separate them into 2 distinct sets. To do this, I choose two points (a and b) and draw an imaginary line between them. Now I want to have all points that are left from this line in one set and those that are right from this line in the other set.
How can I tell for any given point z whether it is in the left or in the right set? I tried to calculate the angle between a-z-b – angles smaller than 180 are on the right hand side, greater than 180 on the left hand side – but because of the definition of ArcCos, the calculated angles are always smaller than 180°. Is there a formula to calculate angles greater than 180° (or any other formula to chose right or left side)?
Try this code which makes use of a cross product:
public bool isLeft(Point a, Point b, Point c){
return ((b.X - a.X)*(c.Y - a.Y) - (b.Y - a.Y)*(c.X - a.X)) > 0;
}
Where a = line point 1; b = line point 2; c = point to check against.
If the formula is equal to 0, the points are colinear.
If the line is horizontal, then this returns true if the point is above the line.
Use the sign of the determinant of vectors (AB,AM), where M(X,Y) is the query point:
position = sign((Bx - Ax) * (Y - Ay) - (By - Ay) * (X - Ax))
It is 0 on the line, and +1 on one side, -1 on the other side.
You look at the sign of the determinant of
| x2-x1 x3-x1 |
| y2-y1 y3-y1 |
It will be positive for points on one side, and negative on the other (and zero for points on the line itself).
The vector (y1 - y2, x2 - x1) is perpendicular to the line, and always pointing right (or always pointing left, if you plane orientation is different from mine).
You can then compute the dot product of that vector and (x3 - x1, y3 - y1) to determine if the point lies on the same side of the line as the perpendicular vector (dot product > 0) or not.
Using the equation of the line ab, get the x-coordinate on the line at the same y-coordinate as the point to be sorted.
If point's x > line's x, the point is to the right of the line.
If point's
x < line's x, the point is to the left of the line.
If point's x == line's x, the point is on the line.
I implemented this in java and ran a unit test (source below). None of the above solutions work. This code passes the unit test. If anyone finds a unit test that does not pass, please let me know.
Code: NOTE: nearlyEqual(double,double) returns true if the two numbers are very close.
/*
* #return integer code for which side of the line ab c is on. 1 means
* left turn, -1 means right turn. Returns
* 0 if all three are on a line
*/
public static int findSide(
double ax, double ay,
double bx, double by,
double cx, double cy) {
if (nearlyEqual(bx-ax,0)) { // vertical line
if (cx < bx) {
return by > ay ? 1 : -1;
}
if (cx > bx) {
return by > ay ? -1 : 1;
}
return 0;
}
if (nearlyEqual(by-ay,0)) { // horizontal line
if (cy < by) {
return bx > ax ? -1 : 1;
}
if (cy > by) {
return bx > ax ? 1 : -1;
}
return 0;
}
double slope = (by - ay) / (bx - ax);
double yIntercept = ay - ax * slope;
double cSolution = (slope*cx) + yIntercept;
if (slope != 0) {
if (cy > cSolution) {
return bx > ax ? 1 : -1;
}
if (cy < cSolution) {
return bx > ax ? -1 : 1;
}
return 0;
}
return 0;
}
Here's the unit test:
#Test public void testFindSide() {
assertTrue("1", 1 == Utility.findSide(1, 0, 0, 0, -1, -1));
assertTrue("1.1", 1 == Utility.findSide(25, 0, 0, 0, -1, -14));
assertTrue("1.2", 1 == Utility.findSide(25, 20, 0, 20, -1, 6));
assertTrue("1.3", 1 == Utility.findSide(24, 20, -1, 20, -2, 6));
assertTrue("-1", -1 == Utility.findSide(1, 0, 0, 0, 1, 1));
assertTrue("-1.1", -1 == Utility.findSide(12, 0, 0, 0, 2, 1));
assertTrue("-1.2", -1 == Utility.findSide(-25, 0, 0, 0, -1, -14));
assertTrue("-1.3", -1 == Utility.findSide(1, 0.5, 0, 0, 1, 1));
assertTrue("2.1", -1 == Utility.findSide(0,5, 1,10, 10,20));
assertTrue("2.2", 1 == Utility.findSide(0,9.1, 1,10, 10,20));
assertTrue("2.3", -1 == Utility.findSide(0,5, 1,10, 20,10));
assertTrue("2.4", -1 == Utility.findSide(0,9.1, 1,10, 20,10));
assertTrue("vertical 1", 1 == Utility.findSide(1,1, 1,10, 0,0));
assertTrue("vertical 2", -1 == Utility.findSide(1,10, 1,1, 0,0));
assertTrue("vertical 3", -1 == Utility.findSide(1,1, 1,10, 5,0));
assertTrue("vertical 3", 1 == Utility.findSide(1,10, 1,1, 5,0));
assertTrue("horizontal 1", 1 == Utility.findSide(1,-1, 10,-1, 0,0));
assertTrue("horizontal 2", -1 == Utility.findSide(10,-1, 1,-1, 0,0));
assertTrue("horizontal 3", -1 == Utility.findSide(1,-1, 10,-1, 0,-9));
assertTrue("horizontal 4", 1 == Utility.findSide(10,-1, 1,-1, 0,-9));
assertTrue("positive slope 1", 1 == Utility.findSide(0,0, 10,10, 1,2));
assertTrue("positive slope 2", -1 == Utility.findSide(10,10, 0,0, 1,2));
assertTrue("positive slope 3", -1 == Utility.findSide(0,0, 10,10, 1,0));
assertTrue("positive slope 4", 1 == Utility.findSide(10,10, 0,0, 1,0));
assertTrue("negative slope 1", -1 == Utility.findSide(0,0, -10,10, 1,2));
assertTrue("negative slope 2", -1 == Utility.findSide(0,0, -10,10, 1,2));
assertTrue("negative slope 3", 1 == Utility.findSide(0,0, -10,10, -1,-2));
assertTrue("negative slope 4", -1 == Utility.findSide(-10,10, 0,0, -1,-2));
assertTrue("0", 0 == Utility.findSide(1, 0, 0, 0, -1, 0));
assertTrue("1", 0 == Utility.findSide(0,0, 0, 0, 0, 0));
assertTrue("2", 0 == Utility.findSide(0,0, 0,1, 0,2));
assertTrue("3", 0 == Utility.findSide(0,0, 2,0, 1,0));
assertTrue("4", 0 == Utility.findSide(1, -2, 0, 0, -1, 2));
}
First check if you have a vertical line:
if (x2-x1) == 0
if x3 < x2
it's on the left
if x3 > x2
it's on the right
else
it's on the line
Then, calculate the slope: m = (y2-y1)/(x2-x1)
Then, create an equation of the line using point slope form: y - y1 = m*(x-x1) + y1. For the sake of my explanation, simplify it to slope-intercept form (not necessary in your algorithm): y = mx+b.
Now plug in (x3, y3) for x and y. Here is some pseudocode detailing what should happen:
if m > 0
if y3 > m*x3 + b
it's on the left
else if y3 < m*x3 + b
it's on the right
else
it's on the line
else if m < 0
if y3 < m*x3 + b
it's on the left
if y3 > m*x3+b
it's on the right
else
it's on the line
else
horizontal line; up to you what you do
I wanted to provide with a solution inspired by physics.
Imagine a force applied along the line and you are measuring the torque of the force about the point. If the torque is positive (counterclockwise) then the point is to the "left" of the line, but if the torque is negative the point is the "right" of the line.
So if the force vector equals the span of the two points defining the line
fx = x_2 - x_1
fy = y_2 - y_1
you test for the side of a point (px,py) based on the sign of the following test
var torque = fx*(py-y_1)-fy*(px-x_1)
if torque>0 then
"point on left side"
else if torque <0 then
"point on right side"
else
"point on line"
end if
Assuming the points are (Ax,Ay) (Bx,By) and (Cx,Cy), you need to compute:
(Bx - Ax) * (Cy - Ay) - (By - Ay) * (Cx - Ax)
This will equal zero if the point C is on the line formed by points A and B, and will have a different sign depending on the side. Which side this is depends on the orientation of your (x,y) coordinates, but you can plug test values for A,B and C into this formula to determine whether negative values are to the left or to the right.
basically, I think that there is a solution which is much easier and straight forward, for any given polygon, lets say consist of four vertices(p1,p2,p3,p4), find the two extreme opposite vertices in the polygon, in another words, find the for example the most top left vertex (lets say p1) and the opposite vertex which is located at most bottom right (lets say ). Hence, given your testing point C(x,y), now you have to make double check between C and p1 and C and p4:
if cx > p1x AND cy > p1y ==> means that C is lower and to right of p1
next
if cx < p2x AND cy < p2y ==> means that C is upper and to left of p4
conclusion, C is inside the rectangle.
Thanks :)
#AVB's answer in ruby
det = Matrix[
[(x2 - x1), (x3 - x1)],
[(y2 - y1), (y3 - y1)]
].determinant
If det is positive its above, if negative its below. If 0, its on the line.
Here's a version, again using the cross product logic, written in Clojure.
(defn is-left? [line point]
(let [[[x1 y1] [x2 y2]] (sort line)
[x-pt y-pt] point]
(> (* (- x2 x1) (- y-pt y1)) (* (- y2 y1) (- x-pt x1)))))
Example usage:
(is-left? [[-3 -1] [3 1]] [0 10])
true
Which is to say that the point (0, 10) is to the left of the line determined by (-3, -1) and (3, 1).
NOTE: This implementation solves a problem that none of the others (so far) does! Order matters when giving the points that determine the line. I.e., it's a "directed line", in a certain sense. So with the above code, this invocation also produces the result of true:
(is-left? [[3 1] [-3 -1]] [0 10])
true
That's because of this snippet of code:
(sort line)
Finally, as with the other cross product based solutions, this solution returns a boolean, and does not give a third result for collinearity. But it will give a result that makes sense, e.g.:
(is-left? [[1 1] [3 1]] [10 1])
false
Issues with the existing solution:
While I found Eric Bainville's answer to be correct, I found it entirely inadequate to comprehend:
How can two vectors have a determinant? I thought that applied to matrices?
What is sign?
How do I convert two vectors into a matrix?
position = sign((Bx - Ax) * (Y - Ay) - (By - Ay) * (X - Ax))
What is Bx?
What is Y? Isn't Y meant to be a Vector, rather than a scalar?
Why is the solution correct - what is the reasoning behind it?
Moreover, my use case involved complex curves rather than a simple line, hence it requires a little re-jigging:
Reconstituted Answer
Point a = new Point3d(ax, ay, az); // point on line
Point b = new Point3d(bx, by, bz); // point on line
If you want to see whether your points are above/below a curve, then you would need to get the first derivative of the particular curve you are interested in - also known as the tangent to the point on the curve. If you can do so, then you can highlight your points of interest. Of course, if your curve is a line, then you just need the point of interest without the tangent. The tangent IS the line.
Vector3d lineVector = curve.GetFirstDerivative(a); // where "a" is a point on the curve. You may derive point b with a simple displacement calculation:
Point3d b = new Point3d(a.X, a.Y, a.Z).TransformBy(
Matrix3d.Displacement(curve.GetFirstDerivative(a))
);
Point m = new Point3d(mx, my, mz) // the point you are interested in.
The Solution:
return (b.X - a.X) * (m.Y - a.Y) - (b.Y - a.Y) * (m.X - a.X) < 0; // the answer
Works for me! See the proof in the photo above. Green bricks satisfy the condition, but the bricks outside were filtered out! In my use case - I only want the bricks that are touching the circle.
Theory behind the answer
I will return to explain this. Someday. Somehow...
An alternative way of getting a feel of solutions provided by netters is to understand a little geometry implications.
Let pqr=[P,Q,R] are points that forms a plane that is divided into 2 sides by line [P,R]. We are to find out if two points on pqr plane, A,B, are on the same side.
Any point T on pqr plane can be represented with 2 vectors: v = P-Q and u = R-Q, as:
T' = T-Q = i * v + j * u
Now the geometry implications:
i+j =1: T on pr line
i+j <1: T on Sq
i+j >1: T on Snq
i+j =0: T = Q
i+j <0: T on Sq and beyond Q.
i+j: <0 0 <1 =1 >1
---------Q------[PR]--------- <== this is PQR plane
^
pr line
In general,
i+j is a measure of how far T is away from Q or line [P,R], and
the sign of i+j-1 implicates T's sideness.
The other geometry significances of i and j (not related to this solution) are:
i,j are the scalars for T in a new coordinate system where v,u are the new axes and Q is the new origin;
i, j can be seen as pulling force for P,R, respectively. The larger i, the farther T is away from R (larger pull from P).
The value of i,j can be obtained by solving the equations:
i*vx + j*ux = T'x
i*vy + j*uy = T'y
i*vz + j*uz = T'z
So we are given 2 points, A,B on the plane:
A = a1 * v + a2 * u
B = b1 * v + b2 * u
If A,B are on the same side, this will be true:
sign(a1+a2-1) = sign(b1+b2-1)
Note that this applies also to the question: Are A,B in the same side of plane [P,Q,R], in which:
T = i * P + j * Q + k * R
and i+j+k=1 implies that T is on the plane [P,Q,R] and the sign of i+j+k-1 implies its sideness. From this we have:
A = a1 * P + a2 * Q + a3 * R
B = b1 * P + b2 * Q + b3 * R
and A,B are on the same side of plane [P,Q,R] if
sign(a1+a2+a3-1) = sign(b1+b2+b3-1)
equation of line is y-y1 = m(x-x1)
here m is y2-y1 / x2-x1
now put m in equation and put condition on y < m(x-x1) + y1 then it is left side point
eg.
for i in rows:
for j in cols:
if j>m(i-a)+b:
image[i][j]=0
A(x1,y1) B(x2,y2) a line segment with length L=sqrt( (y2-y1)^2 + (x2-x1)^2 )
and a point M(x,y)
making a transformation of coordinates in order to be the point A the new start and B a point of the new X axis
we have the new coordinates of the point M
which are
newX = ((x-x1)(x2-x1)+(y-y1)(y2-y1)) / L
from (x-x1)*cos(t)+(y-y1)*sin(t) where cos(t)=(x2-x1)/L, sin(t)=(y2-y1)/L
newY = ((y-y1)(x2-x1)-(x-x1)(y2-y1)) / L
from (y-y1)*cos(t)-(x-x1)*sin(t)
because "left" is the side of axis X where the Y is positive, if the newY (which is the distance of M from AB) is positive, then it is on the left side of AB (the new X axis)
You may omit the division by L (allways positive), if you only want the sign
I'm looking for ideas/code (preferably C#, but other languages work too) to create Ulam's Spiral infinitely large (limited by length of time the program is running, or until stopped).
Now the numbers are all primes so the code for those is rather irrelevant. The interesting part is how to code the arrangement in the evergrowing (infinite) spiral, what kind of data structure is good to support it, and maybe ideas for output (graphics file, text file?).
How would you go about this?
Consider the lengths of each side:
1, 1, 2, 2, 3, 3, 4, 4, ...
The straightforward thing is to iterate over each side, rendering that side.
You can use LOGO style rendering primitives:
Angle = 0;
x=0; y = 0;
int number = 1;
int sideLength = 1;
StartLine();
for (int side = 1; side < maxSize; side++) {
for (int k = 0; k < sideLength; k++) {
Forward(1);
number++;
if (isPrime(number)) {
StopLine();
Ouput(number);
StartLine();
}
}
TurnLeft();
if (side % 2 == 0) sideLength++;
}
You might improve this by only iterating over primes on a side:
The following program works by directly calculating the coordinates of a number. The method NumberToPoint() performs the following mapping.
0 => (x0 , y0 )
1 => (x0 + 1, y0 )
2 => (x0 + 1, y0 - 1)
3 => (x0 , y0 - 1)
4 => (x0 - 1, y0 - 1)
5 => (x0 - 1, y0 )
6 => ...
The rest is a very simple prime number test and a small console application.
In order to save an image I would consider two solutions. If you can create a buffer for the whole image, you can just use the program below to fill the buffer.
If the buffer would be to large, I would create a method PointToNumber() and invert the calculation - the method takes two coordinates and returns the number at this point. With this method you can iterate from top to bottom and left to right and calculate the number at this point, check if it is prime, and output the pixel as you go without a buffer. But for both solutions the image size should be be known before you start, because adding pixels at the top and left is quite expensive (but of cause possible).
Questions
Any good ideas for converting the coefficient lookup in NumberToPoint() into rock solid math without using modulo, integer division, and sign a thousand times?
Any good ideas to shorten or speed up the prime number test?
Code
using System;
using System.Drawing;
using System.Linq;
using System.Threading;
namespace UlamsSpiral
{
public static class Program
{
public static void Main()
{
Int32 width = 60;
Int32 height = 60;
Console.SetWindowSize(Math.Min(width, 120), Math.Min(height, 60));
Console.SetBufferSize(width, height);
Console.CursorVisible = false;
Int32 limit = (Int32)Math.Pow(Math.Min(width, height) - 2, 2);
for (Int32 n = 1; n <= limit; n++)
{
Point point = NumberToPoint(n - 1, width / 2 - 1, height / 2);
Console.ForegroundColor = n.IsPrime() ? ConsoleColor.DarkBlue : ConsoleColor.DarkGray;
Console.SetCursorPosition(point.X, point.Y);
Console.Write('\u25A0');
Console.SetCursorPosition(0, 0);
Console.Write(n);
Thread.Sleep(10);
}
Console.ReadLine();
}
private static Point NumberToPoint(Int32 n, Int32 x0, Int32 y0)
{
Int32[,] c = { { -1, 0, 0, -1, 1, 0 }, { -1, 1, 1, 1, 0, 0 }, { 1, 0, 1, 1, -1, -1 }, { 1, -1, 0, -1, 0, -1 } };
Int32 square = (Int32)Math.Floor(Math.Sqrt(n / 4));
Int32 index;
Int32 side = (Int32)Math.DivRem(n - 4 * square * square, 2 * square + 1, out index);
Int32 x = c[side, 0] * square + c[side, 1] * index + c[side, 2];
Int32 y = c[side, 3] * square + c[side, 4] * index + c[side, 5];
return new Point(x + x0, y + y0);
}
private static Boolean IsPrime(this Int32 n)
{
if (n < 3) return (n == 2);
return Enumerable.Range(2, (Int32)Math.Sqrt(n)).All(m => n % m != 0);
}
}
}
One possible way to do it is to create a linear array or a List to store the numbers and use a formula to determine when the direction needs to change.
As for output, I liked the example on wikipedia of drawing a black pixel for a prime and a white pixel for all other numbers.
Why not have a "generator" process/thread that creates the numbers and a "reader/display" process/thread that displays them, then you can separate the creation from the display and then the program will only really be limited by how much data the "reader/display" consumes. Since i would assume the "generator" needs a fairly constant sized set of data to work with.