Fastest way to sort vectors by angle without actually computing that angle

Question

Many algorithms (e.g. Graham scan) require points or vectors to be sorted by their angle (perhaps as seen from some other point, i.e. using difference vectors). This order is inherently cyclic, and where this cycle is broken to compute linear values often doesn't matter that much. But the real angle value doesn't matter much either, as long as cyclic order is maintained. So doing an atan2 call for every point might be wasteful. What faster methods are there to compute a value which is strictly monotonic in the angle, the way atan2 is? Such functions apparently have been called “pseudoangle” by some.

Answer 1

I started to play around with this and realised that the spec is kind of incomplete. atan2 has a discontinuity, because as dx and dy are varied, there's a point where atan2 will jump between -pi and +pi. The graph below shows the two formulas suggested by @MvG, and in fact they both have the discontinuity in a different place compared to atan2. (NB: I added 3 to the first formula and 4 to the alternative so that the lines don't overlap on the graph). If I added atan2 to that graph then it would be the straight line y=x. So it seems to me that there could be various answers, depending on where one wants to put the discontinuity. If one really wants to replicate atan2, the answer (in this genre) would be

# Input:  dx, dy: coordinates of a (difference) vector.
# Output: a number from the range [-2 .. 2] which is monotonic
#         in the angle this vector makes against the x axis.
#         and with the same discontinuity as atan2
def pseudoangle(dx, dy):
    p = dx/(abs(dx)+abs(dy)) # -1 .. 1 increasing with x
    if dy < 0: return p - 1  # -2 .. 0 increasing with x
    else:      return 1 - p  #  0 .. 2 decreasing with x

This means that if the language that you're using has a sign function, you could avoid branching by returning sign(dy)(1-p), which has the effect of putting an answer of 0 at the discontinuity between returning -2 and +2. And the same trick would work with @MvG's original methodology, one could return sign(dx)(p-1).

Update In a comment below, @MvG suggests a one-line C implementation of this, namely

pseudoangle = copysign(1. - dx/(fabs(dx)+fabs(dy)),dy)

@MvG says it works well, and it looks good to me :-).

Answer 2

I know one possible such function, which I will describe here.

# Input:  dx, dy: coordinates of a (difference) vector.
# Output: a number from the range [-1 .. 3] (or [0 .. 4] with the comment enabled)
#         which is monotonic in the angle this vector makes against the x axis.
def pseudoangle(dx, dy):
    ax = abs(dx)
    ay = abs(dy)
    p = dy/(ax+ay)
    if dx < 0: p = 2 - p
    # elif dy < 0: p = 4 + p
    return p

So why does this work? One thing to note is that scaling all input lengths will not affect the ouput. So the length of the vector (dx, dy) is irrelevant, only its direction matters. Concentrating on the first quadrant, we may for the moment assume dx == 1. Then dy/(1+dy) grows monotonically from zero for dy == 0 to one for infinite dy (i.e. for dx == 0). Now the other quadrants have to be handled as well. If dy is negative, then so is the initial p. So for positive dx we already have a range -1 <= p <= 1 monotonic in the angle. For dx < 0 we change the sign and add two. That gives a range 1 <= p <= 3 for dx < 0, and a range of -1 <= p <= 3 on the whole. If negative numbers are for some reason undesirable, the elif comment line can be included, which will shift the 4th quadrant from -1…0 to 3…4.

I don't know if the above function has an established name, and who might have published it first. I've gotten it quite a while ago and copied it from one project to the next. I have however found occurrences of this on the web, so I'd consider this snipped public enough for re-use.

There is a way to obtain the range [0 … 4] (for real angles [0 … 2π]) without introducing a further case distinction:

# Input:  dx, dy: coordinates of a (difference) vector.
# Output: a number from the range [0 .. 4] which is monotonic
#         in the angle this vector makes against the x axis.
def pseudoangle(dx, dy):
    p = dx/(abs(dx)+abs(dy)) # -1 .. 1 increasing with x
    if dy < 0: return 3 + p  #  2 .. 4 increasing with x
    else:      return 1 - p  #  0 .. 2 decreasing with x