I'm looking for an enumeration algorithm to search through a 3D array "sphering" around a given starting point.
Given an array a of size NxNxN where each N is 2^k for some k, and a point p in that array. The algorithm I'm looking for should do the following: If a[p] satisfies a certain predicate, the algorithm stops and p is returned. Otherwise the next point q is checked, where q is another point in the array that is the closest to p and hasn't been visited yet. If that doesn't match either, the next q'is checked an so on until in the worst case the whole array has been searched.
By "closest" here the perfect solution would be the point q that has the smallest Euclidean distance to p. As only discrete points have to be considered, perhaps some clever enumeration algorithm woukd make that possible. However, if this gets too complicated, the smallest Manhattan distance would be fine too. If there are several nearest points, it doesn't matter which one should be considered next.
Is there already an algorithm that can be used for this task?
You can search for increasing squared distances, so you won't miss a point. This python code should make it clear:
import math
import itertools
# Calculates all points at a certain distance.
# Coordinate constraint: z <= y <= x
def get_points_at_squared_euclidean_distance(d):
result = []
x = int(math.floor(math.sqrt(d)))
while 0 <= x:
y = x
while 0 <= y:
target = d - x*x - y*y
lower = 0
upper = y + 1
while lower < upper:
middle = (lower + upper) / 2
current = middle * middle
if current == target:
result.append((x, y, middle))
break
if current < target:
lower = middle + 1
else:
upper = middle
y -= 1
x -= 1
return result
# Creates all possible reflections of a point
def get_point_reflections(point):
result = set()
for p in itertools.permutations(point):
for n in range(8):
result.add((
p[0] * (1 if n % 8 < 4 else -1),
p[1] * (1 if n % 4 < 2 else -1),
p[2] * (1 if n % 2 < 1 else -1),
))
return sorted(result)
# Enumerates all points around a center, in increasing distance
def get_next_point_near(center):
d = 0
points_at_d = []
while True:
while not points_at_d:
d += 1
points_at_d = get_points_at_squared_euclidean_distance(d)
point = points_at_d.pop()
for reflection in get_point_reflections(point):
yield (
center[0] + reflection[0],
center[1] + reflection[1],
center[2] + reflection[2],
)
# The function you asked for
def get_nearest_point(center, predicate):
for point in get_next_point_near(center):
if predicate(point):
return point
# Example usage
print get_nearest_point((1,2,3), lambda p: sum(p) == 10)
Basically you consume points from the generator until one of them fulfills your predicate.
This is pseudocode for a simple algorithm that will search in increasing-radius spherical husks until it either finds a point or it runs out of array. Let us assume that condition returns either true or false and has access to the x, y, z coordinates being tested and the array itself, returning false (instead of exploding) for out-of-bounds coordinates:
def find_from_center(center, max_radius, condition) returns a point
let radius = 0
while radius < max_radius,
let point = find_in_spherical_husk(center, radius, condition)
if (point != null) return point
radius ++
return null
the hard part is inside find_in_spherical_husk. We are interested in checking out points such that
dist(center, p) >= radius AND dist(center, p) < radius+1
which will be our operating definition of husk. We could iterate over the whole 3D array in O(n^3) looking for those, but that would be really expensive in terms of time. A better pseudocode is the following:
def find_in_spherical_husk(center, radius, condition)
let z = center.z - radius // current slice height
let r = 0 // current circle radius; maxes at equator, then decreases
while z <= center + radius,
let z_center = (z, center.x, point.y)
let point = find_in_z_circle(z_center, r)
if (point != null) return point
// prepare for next z-sliced cirle
z ++
r = sqrt(radius*radius - (z-center.z)*(z-center.z))
the idea here is to slice each husk into circles along the z-axis (any axis will do), and then look at each slice separately. If you were looking at the earth, and the poles were the z axis, you would be slicing from north to south. Finally, you would implement find_in_z_circle(z_center, r, condition) to look at the circumference of each of those circles. You can avoid some math there by using the Bresenham circle-drawing algorithm; but I assume that the savings are negligible compared with the cost of checking condition.
Suppose more than two (nearly) line segments are given. How can I calculate the position order of them?
Example:
The result I expect from the input
(x_0a,y_0a), (x_0b,y_0b), (x_1a,y_1a), (x_1b,y_1b), ... , (x_4a,y_4a), (x_4b,y_4b)
is
L_0 -> L_4 -> L_2 -> L_1 -> L_3
Choose arbitrary segment end as base point. For example, (x0a, y0a).
Make vector for this segment (normalization is not needed)
v = (x0b,y0b) - (x0a,y0a)
Make perpendicular vector
n = (-v.y, v.x)
Calculate projection value of arbitrary end of every segment on the line, perpendicular to v (this value is the same for both ends of segment)
d(i) = (xia-x0a, yia-y0a)
p(i) = Dot(d(i), n)
Sort p(i)
Build linear equation for every segment:
Ai*x + Bi*y + Ci = 0, where
Ai = y_ib - y_ia
Bi = x_ia - x_ib
Ci = -A * x_ia - B * y_ia
For every equation substitute x with 0, and find y
Bi*y + Ci = 0
y = - Ci / Bi
Then sort segments by the resultant y.
This works when segments are not parallel to the y-axis. If they are parallel to the y-axis, then substitute y with 0, find x, and sort by x.
Is there a way to find the lines perpendicular to a given lines given all the lines are of the form Ax+By+C=0?
I came up with a solution which takes quadratic running time. Is there a better way?
this is my code:
public class Perpendicular {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n=in.nextInt();
ArrayList<Line> list_of_lines=new ArrayList<Line>();
for(long i=0;i<n;i++){
long a=in.nextLong();
long b=in.nextLong();
long c=in.nextLong();
list_of_lines.add(new Line(a,b,c));
}
long p[]=new long[n];
Arrays.fill(p,0);
for(int i=0;i<n;i++){
for(int j=1;j<n;j++){
if(list_of_lines.get(i).slope()*list_of_lines.get(j).slope()== -1){
p[i]++;//num of perpendicular lines to i
p[j]++;//num of perpendicular lines to j
}
}
}
}
}
class Line{
public long a,b,c;
public Line(long a,long b,long c){
this.a=a;
this.b=b;
this.c=c;
}
public double slope(){
return a/b;
}
}
Any line which is guided by the equation -Bx + Ay + D = 0 would be the set of lines perpendicular to the given family of lines(Ax + By + C = 0).
You just need to check and make sure that the product of their slopes are equal to -1.
So, a family of lines of the nature -Bx + Ay + D = 0 would satisfy this criteria.
To check all family of lines now is quite easy and is much easier than your expected quadratic solution.
EDIT :-
You don't need to run your loop in this way. Just ensure that when you've checked for perpendicularity between 2 lines, you don't perform the commutative check again. Ex; You checked perpendicularity between 1 and 2, then you don't need to duplicate your effort checking perpendicularity between 2 and 1. This should be avoided in your code by improving the second-loop initialisation by j = i+1;.
Using the suggested edit, you will also not need to check same lines for perpendicularity as that is impossible to be true. So, line k shouldn't be tested with itself for perpendicularity check. This has been easily resolved using the hint 1 itself.
Convert the equation to the form y = m*x + c, m is the slope and c is the y-intercept. Now, all lines perpendicular to this set of lines will have slope -1/m (perpendicular lines have slope product -1), which should give you the equation.
If you have a line Ax + By + C = 0, then perpendicular line A1x + B1y + C1 = 0 should conform the dot product v * v1 = 0 (* is dot product here). So (A, B) * (A1 * B1) = A1 * A + B1 * B = 0. Where one of the solutions is A = -B1, B = A1.
So your line has the form Bx - Ay + C1 = 0, where C1 is any point. Which means that it is O(1).
Two lines would be perpendicular iff their slopes multiply to -1. (Assuming none of them are horizontal/vertical)
You can group the lines according to their slope in O(n log n).
For each pair of groups with perpendicular slopes each pair of lines in them would be an answer which you can iterate over. O(n lg n + num_of_answers)
Thus the algorithm would be O(n lg n + num_of_answers).
Note that there could be O(n^2) such pairs; but if you just need to find the number of such pairs, it could be found in O(n lg n).
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.
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 :-).
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
I kinda like trigonometry, so I know the best way of mapping an angle to some values we usually have is a tangent. Of course, if we want a finite number in order to not have the hassle of comparing {sign(x),y/x}, it gets a bit more confusing.
But there is a function that maps [1,+inf[ to [1,0[ known as inverse, that will allow us to have a finite range to which we will map angles. The inverse of the tangent is the well known cotangent, thus x/y (yes, it's as simple as that).
A little illustration, showing the values of tangent and cotangent on a unit circle :
You see the values are the same when |x| = |y|, and you see also that if we color the parts that output a value between [-1,1] on both circles, we manage to color a full circle. To have this mapping of values be continuous and monotonous, we can do two this :
use the opposite of the cotangent to have the same monotony as tangent
add 2 to -cotan, to have the values coincide where tan=1
add 4 to one half of the circle (say, below the x=-y diagonal) to have values fit on the one of the discontinuities.
That gives the following piecewise function, which is a continuous and monotonous function of the angles, with only one discontinuity (which is the minimum) :
double pseudoangle(double dx, double dy)
{
// 1 for above, 0 for below the diagonal/anti-diagonal
int diag = dx > dy;
int adiag = dx > -dy;
double r = !adiag ? 4 : 0;
if (dy == 0)
return r;
if (diag ^ adiag)
r += 2 - dx / dy;
else
r += dy / dx;
return r;
}
Note that this is very close to Fowler angles, with the same properties. Formally, pseudoangle(dx,dy) + 1 % 8 == Fowler(dx,dy)
To talk performance, it's much less branchy than Fowler's code (and generally less complicated imo). Compiled with -O3 on gcc 6.1.1, the above function generates an assembly code with 4 branches, where two of them come from dy == 0 (one checking if the both operands are "unordered", thus if dy was NaN, and the other checking if they are equal).
I would argue this version is more precise than others, since it only uses mantissa preserving operations, until shifting the result to the right interval. This should be especially visible when |x| << |y| or |y| >> |x|, then the operation |x| + |y| looses quite some precision.
As you can see on the graph the angle-pseudoangle relation is also nicely close to linear.
Looking where branches come from, we can make the following remarks:
My code doesn't rely on abs nor copysign, which makes it look more self-contained. However playing with sign bits on floating point values is actually rather trivial, since it's just flipping a separate bit (no branch!), so this is more of a disadvantage.
Furthermore other solutions proposed here do not check whether abs(dx) + abs(dy) == 0 before dividing by it, but this version would fail as soon as only one component (dy) is 0 -- so that throws in a branch (or 2 in my case).
If we choose to get roughly the same result (up to rounding errors) but without branches, we could abuse copsign and write:
double pseudoangle(double dx, double dy)
{
double s = dx + dy;
double d = dx - dy;
double r = 2 * (1.0 - copysign(1.0, s));
double xor_sign = copysign(1.0, d) * copysign(1.0, s);
r += (1.0 - xor_sign);
r += (s - xor_sign * d) / (d + xor_sign * s);
return r;
}
Bigger errors may happen than with the previous implementation, due to cancellation in either d or s if dx and dy are close in absolute value. There is no check for division by zero to be comparable with the other implementations presented, and because this only happens when both dx and dy are 0.
If you can feed the original vectors instead of angles into a comparison function when sorting, you can make it work with:
Just a single branch.
Only floating point comparisons and multiplications.
Avoiding addition and subtraction makes it numerically much more robust. A double can actually always exactly represent the product of two floats, but not necessarily their sum. This means for single precision input you can guarantee a perfect flawless result with little effort.
This is basically Cimbali's solution repeated for both vectors, with branches eliminated and divisions multiplied away. It returns an integer, with sign matching the comparison result (positive, negative or zero):
signed int compare(double x1, double y1, double x2, double y2) {
unsigned int d1 = x1 > y1;
unsigned int d2 = x2 > y2;
unsigned int a1 = x1 > -y1;
unsigned int a2 = x2 > -y2;
// Quotients of both angles.
unsigned int qa = d1 * 2 + a1;
unsigned int qb = d2 * 2 + a2;
if(qa != qb) return((0x6c >> qa * 2 & 6) - (0x6c >> qb * 2 & 6));
d1 ^= a1;
double p = x1 * y2;
double q = x2 * y1;
// Numerator of each remainder, multiplied by denominator of the other.
double na = q * (1 - d1) - p * d1;
double nb = p * (1 - d1) - q * d1;
// Return signum(na - nb)
return((na > nb) - (na < nb));
}
The simpliest thing I came up with is making normalized copies of the points and splitting the circle around them in half along the x or y axis. Then use the opposite axis as a linear value between the beginning and end of the top or bottom buffer (one buffer will need to be in reverse linear order when putting it in.) Then you can read the first then second buffer linearly and it will be clockwise, or second and first in reverse for counter clockwise.
That might not be a good explanation so I put some code up on GitHub that uses this method to sort points with an epsilion value to size the arrays.
https://github.com/Phobos001/SpatialSort2D
This might not be good for your use case because it's built for performance in graphics effects rendering, but it's fast and simple (O(N) Complexity). If your working with really small changes in points or very large (hundreds of thousands) data sets then this won't work because the memory usage might outweigh the performance benefits.
nice.. here is a varient that returns -Pi , Pi like many arctan2 functions.
edit note: changed my pseudoscode to proper python.. arg order changed for compatibility with pythons math module atan2(). Edit2 bother more code to catch the case dx=0.
def pseudoangle( dy , dx ):
""" returns approximation to math.atan2(dy,dx)*2/pi"""
if dx == 0 :
s = cmp(dy,0)
else::
s = cmp(dx*dy,0) # cmp == "sign" in many other languages.
if s == 0 : return 0 # doesnt hurt performance much.but can omit if 0,0 never happens
p = dy/(dx+s*dy)
if dx < 0: return p-2*s
return p
In this form the max error is only ~0.07 radian for all angles.
(of course leave out the Pi/2 if you don't care about the magnitude.)
Now for the bad news -- on my system using python math.atan2 is about 25% faster
Obviously replacing a simple interpreted code doesnt beat a compiled intrisic.
If angles are not needed by themselves, but only for sorting, then #jjrv approach is the best one. Here is a comparison in Julia
using StableRNGs
using BenchmarkTools
# Definitions
struct V{T}
x::T
y::T
end
function pseudoangle(v)
copysign(1. - v.x/(abs(v.x)+abs(v.y)), v.y)
end
function isangleless(v1, v2)
a1 = abs(v1.x) + abs(v1.y)
a2 = abs(v2.x) + abs(v2.y)
a2*copysign(a1 - v1.x, v1.y) < a1*copysign(a2 - v2.x, v2.y)
end
# Data
rng = StableRNG(2021)
vectors = map(x -> V(x...), zip(rand(rng, 1000), rand(rng, 1000)))
# Comparison
res1 = sort(vectors, by = x -> pseudoangle(x));
res2 = sort(vectors, lt = (x, y) -> isangleless(x, y));
#assert res1 == res2
#btime sort($vectors, by = x -> pseudoangle(x));
# 110.437 μs (3 allocations: 23.70 KiB)
#btime sort($vectors, lt = (x, y) -> isangleless(x, y));
# 65.703 μs (3 allocations: 23.70 KiB)
So, by avoiding division, time is almost halved without losing result quality. Of course, for more precise calculations, isangleless should be equipped with bigfloat from time to time, but the same can be told about pseudoangle.
Just use a cross-product function. The direction you rotate one segment relative to the other will give either a positive or negative number. No trig functions and no division. Fast and simple. Just Google it.
I already googled for the problem but only found either 2D solutions or formulas that didn't work for me (found this formula that looks nice: http://www.ogre3d.org/forums/viewtopic.php?f=10&t=55796 but seems not to be correct).
I have given:
Vec3 cannonPos;
Vec3 targetPos;
Vec3 targetVelocityVec;
float bulletSpeed;
what i'm looking for is time t such that
targetPos+t*targetVelocityVec
is the intersectionpoint where to aim the cannon to and shoot.
I'm looking for a simple, inexpensive formula for t (by simple i just mean not making many unnecessary vectorspace transformations and the like)
thanks!
The real problem is finding out where in space that the bullet can intersect the targets path. The bullet speed is constant, so in a certain amount of time it will travel the same distance regardless of the direction in which we fire it. This means that it's position after time t will always lie on a sphere. Here's an ugly illustration in 2d:
This sphere can be expressed mathematically as:
(x-x_b0)^2 + (y-y_b0)^2 + (z-z_b0)^2 = (bulletSpeed * t)^2 (eq 1)
x_b0, y_b0 and z_b0 denote the position of the cannon. You can find the time t by solving this equation for t using the equation provided in your question:
targetPos+t*targetVelocityVec (eq 2)
(eq 2) is a vector equation and can be decomposed into three separate equations:
x = x_t0 + t * v_x
y = y_t0 + t * v_y
z = z_t0 + t * v_z
These three equations can be inserted into (eq 1):
(x_t0 + t * v_x - x_b0)^2 + (y_t0 + t * v_y - y_b0)^2 + (z_t0 + t * v_z - z_b0)^2 = (bulletSpeed * t)^2
This equation contains only known variables and can be solved for t. By assigning the constant part of the quadratic subexpressions to constants we can simplify the calculation:
c_1 = x_t0 - x_b0
c_2 = y_t0 - y_b0
c_3 = z_t0 - z_b0
(v_b = bulletSpeed)
(t * v_x + c_1)^2 + (t * v_y + c_2)^2 + (t * v_z + c_3)^2 = (v_b * t)^2
Rearrange it as a standard quadratic equation:
(v_x^2+v_y^2+v_z^2-v_b^2)t^2 + 2*(v_x*c_1+v_y*c_2+v_z*c_3)t + (c_1^2+c_2^2+c_3^2) = 0
This is easily solvable using the standard formula. It can result in zero, one or two solutions. Zero solutions (not counting complex solutions) means that there's no possible way for the bullet to reach the target. One solution will probably happen very rarely, when the target trajectory intersects with the very edge of the sphere. Two solutions will be the most common scenario. A negative solution means that you can't hit the target, since you would need to fire the bullet into the past. These are all conditions you'll have to check for.
When you've solved the equation you can find the position of t by putting it back into (eq 2). In pseudo code:
# setup all needed variables
c_1 = x_t0 - x_b0
c_2 = y_t0 - y_b0
c_3 = z_t0 - z_b0
v_b = bulletSpeed
# ... and so on
a = v_x^2+v_y^2+v_z^2-v_b^2
b = 2*(v_x*c_1+v_y*c_2+v_z*c_3)
c = c_1^2+c_2^2+c_3^2
if b^2 < 4*a*c:
# no real solutions
raise error
p = -b/(2*a)
q = sqrt(b^2 - 4*a*c)/(2*a)
t1 = p-q
t2 = p+q
if t1 < 0 and t2 < 0:
# no positive solutions, all possible trajectories are in the past
raise error
# we want to hit it at the earliest possible time
if t1 > t2: t = t2
else: t = t1
# calculate point of collision
x = x_t0 + t * v_x
y = y_t0 + t * v_y
z = z_t0 + t * v_z