Is there a quick way to determine if a vector is in a quadrant?

Question

I am wondering if there is a quick easy way to determine if a given vector is in a specified quadrant.

In the below image, I've define the ranges of the blue quadrants.

For example, if I am given a sample vector of -9.4,13.3 is there an easy way to figure out which of the four quadrants it is in (Q2)?

I'm looking to know if there is some vector math that can determine this relatively simply instead of doing a bunch of less than / greater than operators on a fixed range.

Considering that I could have more or less than 4 quadrants and they might not be at this nice 45 degree angle.

Answer 1

In this case, what is wrong with comparisons? It's (in this case) pretty easy, just

if(abs(x) > abs(y)
  if(x > 0) q = "q2"; else q = "q4";
else
  if(y > 0) q = "q1"; else q = "q3";

Alternatively, you could do something like (hopefully I'll get this right)

angle = atan2(y, x);
angle -= pi/4; // now (0,pi/2) is all Q1
quadrant = angle / (pi/2);
if(quadrant < 0) quadrant += 4;
quadrant++; // to make it 1 to 4. (not quite right for the numbering, but...)

This can be generalized for any evenly-spaced wedges. (They can be called "quadrants" when there's four of them.)

If the wedges aren't evenly spaced, you could still have a list of their angles, calculated when they change, also with atan2. Then it becomes a search problem, given the angle from the origin for a point of interest versus the wedge list. For a small number of wedges, a linear search is probably fine. If you have dozens, or thousands, a binary search might be nicer.

If it must be inside the circle, check the distance to origin first... Is there a reason it must be done "in vector math"?

Answer 2

This is inspired by how OpenGL handles Cubemaps:

(Pseudocode)

vector = normalize(vector);
if(abs(vector.x) > abs(vector.y)){
  if(vector.x < 0)
    //Q1
  else
    //Q3
}else{
  if(vector.y < 0)
    //Q4
  else
    //Q2
}

(i hope you don't count this as "a bunch") Another way would be to calculate the angle which would involve a arctan and that should be a lot slower than my approach.

Answer 3

Yes.

First calculate the matrix for the transformation that maps your custom quadrants to the standard quadrants numbering CCW from +ve X and +ve Y.

Second apply this matrix to the point of interest (call it (x,y)) to get a mapped point in standard quadrant space (x',y'):.

Now apply this formula to the resulting vector (x',y'):

Quadrant = 1 + (1 - sign(y')) - ( sign(y') (1 - sign(x') ) / 2

Answer 4

Yes, you can perform simple binary search on region divisor vectors.
If you pre-process and sort your divisors ccw(and you probably should), it is sublinear log(n) complexity for n regions (oposed to other answers with linear complexity), moreover it doesnt rely on inaccurate and slow functions like atan, in addition your regions does not need to be quadrants or evenly spaced, it works for any number of any regions.
All you need is to determine if queried vecor lies to left or to right to the tested divisor(that is one vector operation) and apply this greater/lesser logic on same binary search like you already know. Code example(not tested):

Divisor findRegion(vec2 query_vec) {
return *(std::lower_bound(divisors.begin(), divisors.end(), query_vec, isLeft));
}

Please note you might need to implement additional logic if you allow regions greater than pi, but I assume you need use this for more smaller regions

Answer 5

If the sectors are of equal size then one approach would be to calculate the angle, divide by a full rotation and then multiply by the number or sectors:

rad = atan2(x, y)
normalised_angle = rad / (2 * pi)
sector = (int)(normalised_angle * sector_count)

For the example given in the question we would also require a π/4 offset to account for a quadrant straddling the zero angle. This offset also requires wrapping the angle value to ensure that it remains between 0 and 2π

sector_count = 4
offset =  pi / 4

rad = atan2(x, y) + offset

while (rad < 0):
    rad += 2 * pi
while (rad > 2 * pi):
    rad -= 2 * pi

normalised_angle = rad / (2 * pi)
sector = (int)(normalised_angle * sector_count)

Answer 6

Quadrant(1,2,3,4)=4-(1+sign(x))* (1-sign(y^2))-(2-sign(y))* (1-sign(x^2))

       -sign(x^2*y^2)*((1+sign(abs(x)-abs(y)))*(1+sign(x))/2

                      +(1-sign(abs(x)-abs(y)))*(1-sign(y)/2))