Hello, I've got a question which I cannot solve so I need a bit help.
In the picture above you can see an Oriented Bounding Box specified by 4 points (A, B, C, D). There is also a point in space called P. If I cast a ray from P against the OBB the ray is going to intersect the OBB at some point. This point of intersection is called Q in the picture. By the way the ray is always going to be x-axis aligned which means its directional vector is either (1, 0) or (-1,0) if normalized. My goal is to find the point of intersection - Q. Is there a way (if possible computationaly inexpensive) to do so?
Thanks in advance.
One way to do this is to consider each side of the bounding box to be a linear equation of the form y = ax + b, where a is the slope and b is the y-intercept. Then consider the ray from P to be an equation of the form y = c, where c is a constant. Then compare this equation to each of the four other equations to see where it intersects each one. One of these intersections will be our Q, if a Q exists; it's possible that the ray will miss the bounding box entirely. We will need to do a few checks:
Firstly, eliminate all potential Q's that are on the wrong side of P.
Secondly, check each of the four intersections to make sure they are within the bounds of the lines that they represent, and eliminate the ones that are not.
Finally, if any potential Q's remain, the one closest to P will be our Q. If no potential Q's remain, this means that the ray from P misses the bounding box entirely.
For example...
The line drawn from D to B would have a slope equal to (B.y - D.y) / (B.x - D.x) and a y-intercept equal to B.y - B.x * slope. Then the entire equation is y = (B.y - D.y) / (B.x - D.x) * x + B.y - B.x * (B.y - D.y) / (B.x - D.x). Set this equation equal to y = P.y and solve for x:
x = (P.y - B.y + B.x*(B.y - D.y)/(B.x - D.x))*(B.x - D.x)/(B.y - D.y)
The result of this equation will give you the x-value of the intersection. The y-value is P.y. Do this for each of the other 3 lines as well: A-B, A-C, C-D. I will refer to these intersections as Q(D-B), Q(A-B), Q(A-C), and Q(C-D) respectively.
Next, eliminate candidate-Q's that are on the wrong side of P. In our example, this eliminates Q(A-B), since it is way off to the right side of the screen. Mathematically, Q(A-B).x > P.x.
Then eliminate all candidate-Q's that are not on the line they represent. We can do this check by finding the lowest and highest x-values and y-values given by the two points that represent the line. For example, to check that Q(A-C) is on the line A-C, check that C.x <= Q(A-C).x <= A.x and C.y <= Q(A-C).y <= A.y. Q(A-C) passes the test, as well as Q(D-B). However, Q(C-D) does not pass, as it is way off to the left side of the screen, far from the box. Therefore, Q(C-D) is eliminated from candidacy.
Finally, of the two points that remain, Q(A-C) and Q(D-B), we choose Q(D-B) to be our winner, because it is closest to P.
We now can say that the ray from P hits the bounding box at Q(D-B).
Of course, when you implement this in code, you will need to account for divisions by zero. If a line is perfectly vertical, there does not exist a point-slope equation of the line, so you will need to create a separate formula for this case. If a line is perfectly horizontal, it's respective candidate-Q should be automatically eliminated from candidacy, as the ray from P will never touch it.
Edit:
It would be more efficient to only do this process with lines whose two points are on vertically opposite sides of point P. If both points of a line are above P, or they are both below P, they would be eliminated from candidacy from the beginning.
Find the two sides that straddle p on Y. (Test of the form (Ya < Yp) != (Yb < Yp)).
Then compute the intersection points of the horizontal by p with these two sides, and keep the first to the left of p.
If the ray points to the left(right) then it must intersect an edge that connects to the point in the OOB with max(min) x-value. We can determine which edge by simply comparing the y-value of the ray with the y value of the max(min) point and its neighbors. We also need to consider OBBs that are actually axis-aligned, and thus have two points with equal max(min) x-value. Once we have the edge it's simple to confirm that the ray does in fact intersect the OBB and calculate its x-value.
Here's some Java code to illustrate (ideone):
static double nearestX(Point[] obb, int y, int dir)
{
// Find min(max) point
int n = 0;
for(int i=1; i<4; i++)
if((obb[n].x < obb[i].x) == (dir == -1)) n = i;
// Determine next or prev edge
int next = (n+1) % 4;
int prev = (n+3) % 4;
int nn;
if((obb[n].x == obb[next].x) || (obb[n].y < y) == (obb[n].y < obb[next].y))
nn = next;
else
nn = prev;
// Check that the ray intersects the OBB
if(Math.abs(y) > Math.abs(obb[nn].y)) return Double.NaN;
// Standard calculation of x from y for line segment
return obb[n].x + (y-obb[n].y)*(obb[nn].x-obb[n].x)/(obb[nn].y-obb[n].y);
}
Test:
public static void main(String[] args)
{
test("Diamond", new Point[]{p(0, -2), p(2, 0), p(0, 2), p(-2,0)});
test("Square", new Point[]{p(-2, -2), p(2, -2), p(2, 2), p(-2,2)});
}
static void test(String label, Point[] obb)
{
System.out.println(label + ": " + Arrays.toString(obb));
for(int dir : new int[] {-1, 1})
{
for(int y : new int[] {-3, -2, -1, 0, 1, 2, 3})
System.out.printf("(% d, % d) = %.0f\n", y , dir, nearestX(obb, y, dir));
System.out.println();
}
}
Output:
Diamond: [(0,-2), (2,0), (0,2), (-2,0)]
(-3, -1) = NaN
(-2, -1) = 0
(-1, -1) = 1
( 0, -1) = 2
( 1, -1) = 1
( 2, -1) = 0
( 3, -1) = NaN
(-3, 1) = NaN
(-2, 1) = 0
(-1, 1) = -1
( 0, 1) = -2
( 1, 1) = -1
( 2, 1) = 0
( 3, 1) = NaN
Square: [(-2,-2), (2,-2), (2,2), (-2,2)]
(-3, -1) = NaN
(-2, -1) = 2
(-1, -1) = 2
( 0, -1) = 2
( 1, -1) = 2
( 2, -1) = 2
( 3, -1) = NaN
(-3, 1) = NaN
(-2, 1) = -2
(-1, 1) = -2
( 0, 1) = -2
( 1, 1) = -2
( 2, 1) = -2
( 3, 1) = NaN
I'm developing an application that involves getting the camera angle in a game. The angle can be anywhere from 0-359. 0 is North, 90 is East, 180 is South, etc. I'm using an API, which has a getAngle() method in Camera class.
How would I find the average between different camera angles. The real average of 0 and 359 is 179.5. As a camera angle, that would be South, but obviously 0 and 359 are both very close to North.
You can think of it in terms of vectors. Let θ1 and θ2 be your two angles expressed in radians. Then we can determine the x and y components of the unit vectors that are at these angles:
x1 = sin(θ1)
y1 = cos(θ1)
x2 = sin(θ2)
y2 = cos(θ2)
You can then add these two vectors, and determine the x and y components of the result:
x* = x1 + x2
y* = y1 + y2
Finally, you can determine the angle of this resulting vector:
θavg = tan-1(y*/x*)
or, even better, use atan2 (a function supported by many languages):
θavg = atan2(y*, x*)
You will probably have to separately handle the cases where y* = 0 and x* = 0, since this means the two vectors are pointing in exactly opposite directions (so what should the 'average' be?).
It depends what you mean by "average". But the normal definition is the bisector of the included acute angle. You must put both within 180 degrees of each other. There are many ways to do this, but a simple one is to increment or decrement one of the angles. If the angles are a and b, then this will do it:
if (a < b)
while (abs(a - b) > 180) a = a + 360
else
while (abs(a - b) > 180) a = a - 360
Now you can compute the simple average:
avg = (a + b) / 2
Of course you may want to normalize one more time:
while (avg < 0) avg = avg + 360
while (avg >= 360) avg = avg - 360
On your example, you'd have a=0, b=359. The first loop would increment a to 360. The average would be 359.5. Of course you could round that to an integer if you like. If you round up to 360, then the final set of loops will decrement to 0.
Note that if your angles are always normalized to [0..360) none of these loops ever execute more than once. But they're probably good practice so that a wild argument doesn't cause your code to fail.
You want to bisect the angles not average them. First get the distance between them, taking the shortest way around, then divide that in half and add to one of the angles. Eg:
A = 355
B = 5
if (abs(A - B) < 180) {
Distance = abs(A - B)
if (A < B) {
Bisect = A + Distance / 2
}
else {
Bisect = B + Distance / 2
}
}
else {
Distance = 360 - abs(A - B)
if (A < B) {
Bisect = A - Distance / 2
}
else {
Bisect = B - Distance / 2
}
}
Or something like that -- "Bisect" should come out to zero for the given inputs. There are probably clever ways to make the arithmetic come out with fewer if and abs operations.
In a comment, you mentioned that all "angles" to be averaged are within 90 degrees to each other. I am guessing that there is really only one camera, but it moves around a lot, and you are creating some sort of picture stability mechanism for the camera POV.
In any case, there is only the special case where the camera may be in the 270-359 quadrant and the 0-89 quadrant. For all other cases, you can just take a simple average. So, you just need to detect that special case, and when it happens, treat the angles in the 270-359 quadrant as -90 to -1 instead. Then, after computing the simple average, adjust it back into the 270-359 quadrant if necessary.
In C code:
int quadrant (int a) {
assert(0 <= a && a < 360);
return a/90;
}
double avg_rays (int rays[], int num) {
int i;
int quads[4] = { 0, 0, 0, 0 };
double sum = 0;
/* trivial case */
if (num == 1) return rays[0];
for (i = 0; i < num; ++i) ++quads[quadrant(rays[i])];
if (quads[0] == 0 || quads[3] == 0) {
/* simple case */
for (i = 0; i < num; ++i) sum += rays[i];
return sum/num;
}
/* special case */
for (i = 0; i < num; ++i) {
if (quadrant(rays[i]) == 3) rays[i] -= 360;
sum += rays[i];
}
return sum/num + (sum < 0) * 360;
}
This code can be optimized at the expense of clarity of purpose. When you detect the special case condition, you can fix up the sum after the fact. So, you can compute sum and figure out the special case and do the fix up in a single pass.
double avg_rays_opt (int rays[], int num) {
int i;
int quads[4] = { 0, 0, 0, 0 };
double sum = 0;
/* trivial case */
if (num == 1) return rays[0];
for (i = 0; i < num; ++i) {
++quads[quadrant(rays[i])];
sum += rays[i];
}
if (quads[0] == 0 || quads[3] == 0) {
/* simple case */
return sum/num;
}
/* special case */
sum -= quads[3]*360;
return sum/num + (sum < 0) * 360;
}
I am sure it can be further optimized, but it should give you a start.
I've gotten stuck getting my euler angles out my rotation matrix.
My conventions are:
Left-handed (x right, z back, y up)
YZX
Left handed angle rotation
My rotation matrix is built up from Euler angles like (from my code):
var xRotationMatrix = $M([
[1, 0, 0, 0],
[0, cx, -sx, 0],
[0, sx, cx, 0],
[0, 0, 0, 1]
]);
var yRotationMatrix = $M([
[ cy, 0, sy, 0],
[ 0, 1, 0, 0],
[-sy, 0, cy, 0],
[ 0, 0, 0, 1]
]);
var zRotationMatrix = $M([
[cz, -sz, 0, 0],
[sz, cz, 0, 0],
[ 0, 0, 1, 0],
[ 0, 0, 0, 1]
]);
Which results in a final rotation matrix as:
R(YZX) = | cy.cz, -cy.sz.cx + sy.sx, cy.sz.sx + sy.cx, 0|
| sz, cz.cx, -cz.sx, 0|
|-sy.cz, sy.sz.cx + cy.sx, -sy.sz.sx + cy.cx, 0|
| 0, 0, 0, 1|
I'm calculating my euler angles back from this matrix using this code:
this.anglesFromMatrix = function(m) {
var y = 0, x = 0, z = 0;
if (m.e(2, 1) > 0.999) {
y = Math.atan2(m.e(1, 3), m.e(3, 3));
z = Math.PI / 2;
x = 0;
} else if (m.e(2, 1) < -0.999) {
y = Math.atan2(m.e(1, 3), m.e(3, 3));
z = -Math.PI / 2;
x = 0;
} else {
y = Math.atan2(-m.e(3, 1), -m.e(1, 1));
x = Math.atan2(-m.e(2, 3), m.e(2, 2));
z = Math.asin(m.e(2, 1));
}
return {theta: this.deg(x), phi: this.deg(y), psi: this.deg(z)};
};
I've done the maths backwards and forwards a few times, but I can't see what's wrong. Any help would hugely appreciated.
Your matrix and euler angles aren't consistent. It looks like you should be using
y = Math.atan2(-m.e(3, 1), m.e(1, 1));
instead of
y = Math.atan2(-m.e(3, 1), -m.e(1, 1));
for the general case (the else branch).
I said "looks like" because -- what language is this? I'm assuming you have the indexing correct for this language. Are you sure about atan2? There is no single convention for atan2. In some programming languages the sine term is the first argument, in others, the cosine term is the first argument.
The last and most important branch of the anglesFromMatrix function has a small sign error but otherwise works correctly. Use
y = Math.atan2(-m.e(3, 1), m.e(1, 1))
since only m.e(3, 1) of m.e(1, 1) = cy.cz and m.e(3, 1) = -sy.cz should be inverted. I haven't checked the other branches for errors.
Beware that since sz = m.e(2, 1) has two solutions, the angles (x, y, z) used to construct the matrix m might not be the same as the angles (rx, ry, rz) returned by anglesFromMatrix(m). Instead we can test that the matrix rm constructed from (rx, ry, rz) does indeed equal m.
I worked on this problem extensively to come up with the correct angles for a given matrix. The problem in the math comes from the inability to determine a precise value for the SIN since -SIN(x) = SIN(-x) and this will affect the other values of the matrix. The solution I came up with comes up with two equally valid solutions out of eight possible solutions. I used a standard Z . Y . X matrix form but it should be adaptable to any matrix. Start by findng the three angles from: X = atan(m32,m33): Y = -asin(m31) : Z = atan(m21,m11) : Then create angles X' = -sign(X)*PI+X : Y'= sign(Y)*PI-Y : Z = -sign(Z)*pi+Z . Using these angles create eight set of angle groups : XYZ : X'YZ : XYZ' : X'YZ' : X'Y'Z' : XY'Z' : X'Y'Z : XY'Z
Use these set to create the eight corresponding matrixes. Then do a sum of the difference between the unknown matrix and each matrix. This is a sum of each element of the unknown minus the same element of the test matrix. After doing this, two of the sums will be zero and those matrixes will represent the solution angles to the original matrix. This works for all possible angle combinations including 0's. As 0's are introduced, more of the eight test matrixes become valid. At 0,0,0 they all become idenity matrixes!
Hope this helps, it worked very well for my application.
Bruce
update
After finding problems with Y = -90 or 90 degrees in the solution above. I came up with this solution that seems to reproduce the matrix at all values!
X = if(or(m31=1,m31=-1),0,atan(m33+1e-24,m32))
Y = -asin(m31)
Z = if(or(m31=1,m31=-1),-atan2(m22,m12),atan2(m11+1e-24,m21))
I went the long way around to find this solution, but it wa very enlightening :o)
Hope this helps!
Bruce
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