Very fast 3D distance check? - c++

Question

Is there a way to do a quick and dirty 3D distance check where the results are rough, but it is very very fast? I need to do depth sorting. I use STL sort like this:
bool sortfunc(CBox* a, CBox* b)
{
return a->Get3dDistance(Player.center,a->center) <
b->Get3dDistance(Player.center,b->center);
}
float CBox::Get3dDistance( Vec3 c1, Vec3 c2 )
{
//(Dx*Dx+Dy*Dy+Dz*Dz)^.5
float dx = c2.x - c1.x;
float dy = c2.y - c1.y;
float dz = c2.z - c1.z;
return sqrt((float)(dx * dx + dy * dy + dz * dz));
}
Is there possibly a way to do it without a square root or possibly without multiplication?

Answer 1

You can leave out the square root because for all positive (or really, non-negative) numbers x and y, if sqrt(x) < sqrt(y) then x < y. Since you're summing squares of real numbers, the square of every real number is non-negative, and the sum of any positive numbers is positive, the square root condition holds.
You cannot eliminate the multiplication, however, without changing the algorithm. Here's a counterexample: if x is (3, 1, 1) and y is (4, 0, 0), |x| < |y| because sqrt(1*1+1*1+3*3) < sqrt(4*4+0*0+0*0) and 1*1+1*1+3*3 < 4*4+0*0+0*0, but 1+1+3 > 4+0+0.
Since modern CPUs can compute a dot product faster than they can actually load the operands from memory, it's unlikely that you would have anything to gain by eliminating the multiply anyway (I think the newest CPUs have a special instruction that can compute a dot product every 3 cycles!).
I would not consider changing the algorithm without doing some profiling first. Your choice of algorithm will heavily depend on the size of your dataset (does it fit in cache?), how often you have to run it, and what you do with the results (collision detection? proximity? occlusion?).

Answer 2

What I usually do is first filter by Manhattan distance
float CBox::Within3DManhattanDistance( Vec3 c1, Vec3 c2, float distance )
{
float dx = abs(c2.x - c1.x);
float dy = abs(c2.y - c1.y);
float dz = abs(c2.z - c1.z);
if (dx > distance) return 0; // too far in x direction
if (dy > distance) return 0; // too far in y direction
if (dz > distance) return 0; // too far in z direction
return 1; // we're within the cube
}
Actually you can optimize this further if you know more about your environment. For example, in an environment where there is a ground like a flight simulator or a first person shooter game, the horizontal axis is very much larger than the vertical axis. In such an environment, if two objects are far apart they are very likely separated more by the x and y axis rather than the z axis (in a first person shooter most objects share the same z axis). So if you first compare x and y you can return early from the function and avoid doing extra calculations:
float CBox::Within3DManhattanDistance( Vec3 c1, Vec3 c2, float distance )
{
float dx = abs(c2.x - c1.x);
if (dx > distance) return 0; // too far in x direction
float dy = abs(c2.y - c1.y);
if (dy > distance) return 0; // too far in y direction
// since x and y distance are likely to be larger than
// z distance most of the time we don't need to execute
// the code below:
float dz = abs(c2.z - c1.z);
if (dz > distance) return 0; // too far in z direction
return 1; // we're within the cube
}
Sorry, I didn't realize the function is used for sorting. You can still use Manhattan distance to get a very rough first sort:
float CBox::ManhattanDistance( Vec3 c1, Vec3 c2 )
{
float dx = abs(c2.x - c1.x);
float dy = abs(c2.y - c1.y);
float dz = abs(c2.z - c1.z);
return dx+dy+dz;
}
After the rough first sort you can then take the topmost results, say the top 10 closest players, and re-sort using proper distance calculations.

Answer 3

Here's an equation that might help you get rid of both sqrt and multiply:
max(|dx|, |dy|, |dz|) <= distance(dx,dy,dz) <= |dx| + |dy| + |dz|
This gets you a range estimate for the distance which pins it down to within a factor of 3 (the upper and lower bounds can differ by at most 3x). You can then sort on, say, the lower number. You then need to process the array until you reach an object which is 3x farther away than the first obscuring object. You are then guaranteed to not find any object that is closer later in the array.
By the way, sorting is overkill here. A more efficient way would be to make a series of buckets with different distance estimates, say [1-3], [3-9], [9-27], .... Then put each element in a bucket. Process the buckets from smallest to largest until you reach an obscuring object. Process 1 additional bucket just to be sure.
By the way, floating point multiply is pretty fast nowadays. I'm not sure you gain much by converting it to absolute value.

Answer 4

I'm disappointed that the great old mathematical tricks seem to be getting lost. Here is the answer you're asking for. Source is Paul Hsieh's excellent web site: http://www.azillionmonkeys.com/qed/sqroot.html . Note that you don't care about distance; you will do fine for your sort with square of distance, which will be much faster.
In 2D, we can get a crude approximation of the distance metric without a square root with the formula:
distanceapprox (x, y) =
which will deviate from the true answer by at most about 8%. A similar derivation for 3 dimensions leads to:
distanceapprox (x, y, z) =
with a maximum error of about 16%.
However, something that should be pointed out, is that often the distance is only required for comparison purposes. For example, in the classical mandelbrot set (z←z2+c) calculation, the magnitude of a complex number is typically compared to a boundary radius length of 2. In these cases, one can simply drop the square root, by essentially squaring both sides of the comparison (since distances are always non-negative). That is to say:
√(Δx2+Δy2) < d is equivalent to Δx2+Δy2 < d2, if d ≥ 0
I should also mention that Chapter 13.2 of Richard G. Lyons's "Understanding Digital Signal Processing" has an incredible collection of 2D distance algorithms (a.k.a complex number magnitude approximations). As one example:
Max = x > y ? x : y;
Min = x < y ? x : y;
if ( Min < 0.04142135Max )
|V| = 0.99 * Max + 0.197 * Min;
else
|V| = 0.84 * Max + 0.561 * Min;
which has a maximum error of 1.0% from the actual distance. The penalty of course is that you're doing a couple branches; but even the "most accepted" answer to this question has at least three branches in it.
If you're serious about doing a super fast distance estimate to a specific precision, you could do so by writing your own simplified fsqrt() estimate using the same basic method as the compiler vendors do, but at a lower precision, by doing a fixed number of iterations. For example, you can eliminate the special case handling for extremely small or large numbers, and/or also reduce the number of Newton-Rapheson iterations. This was the key strategy underlying the so-called "Quake 3" fast inverse square root implementation -- it's the classic Newton algorithm with exactly one iteration.
Do not assume that your fsqrt() implementation is slow without benchmarking it and/or reading the sources. Most modern fsqrt() library implementations are branchless and really damned fast. Here for example is an old IBM floating point fsqrt implementation. Premature optimization is, and always will be, the root of all evil.

Answer 5

Note that for 2 (non-negative) distances A and B, if sqrt(A) < sqrt(B), then A < B. Create a specialized version of Get3DDistance() (GetSqrOf3DDistance()) that does not call sqrt() that would be used only for the sortfunc().

Answer 6

If you worry about performance, you should also take care of the way you send your arguments:
float Get3dDistance( Vec3 c1, Vec3 c2 );
implies two copies of Vec3 structure. Use references instead:
float Get3dDistance( Vec3 const & c1, Vec3 const & c2 );

Answer 7

You could compare squares of distances instead of the actual distances, since d2 = (x1-x2)2 + (y1-y2)2+ (z1-z2)2. It doesn't get rid of the multiplication, but it does eliminate the square root operation.

Answer 8

How often are the input vectors updated and how often are they sorted? Depending on your design, it might be quite efficient to extend the "Vec3" class with a pre-calculated distance and sort on that instead. Especially relevant if your implementation allows you to use vectorized operations.
Other than that, see the flipcode.com article on approximating distance functions for a discussion on yet another approach.

Answer 9

Depending slightly on the number of points that you are being used to compare with, what is below is pretty much guaranteed to be the get the list of points in approximate order assuming all points change at all iteration.
1) Rewrite the array into a single list of Manhattan distances with
out[ i ] = abs( posn[ i ].x - player.x ) + abs( posn[ i ].y - player.y ) + abs( posn[ i ].z - player.z );
2) Now you can use radix sort on floating point numbers to order them.
Note that in practice this is going to be a lot faster than sorting the list of 3d positions because it significantly reduces the memory bandwidth requirements in the sort operation which all of the time is going to be spend and in which unpredictable accesses and writes are going to occur. This will run on O(N) time.
If many of the points are stationary at each direction there are far faster algorithms like using KD-Trees, although implementation is quite a bit more complex and it is much harder to get good memory access patterns.

Answer 10

If this is simply a value for sorting, then you can swap the sqrt() for a abs(). If you need to compare distances against set values, get the square of that value.
E.g. instead of checking sqrt(...) against a, you can compare abs(...) against a*a.

Answer 11

You may want to consider caching the distance between the player and the object as you calculate it, and then use that in your sortfunc. This would depend upon how many times your sort function looks at each object, so you might have to profile to be sure.
What I'm getting at is that your sort function might do something like this:
compare(a,b);
compare(a,c);
compare(a,d);
and you would calculate the distance between the player and 'a' every time.
As others have mentioned, you can leave out the sqrt in this case.

Answer 12

If you could center your coordinates around the player, use spherical coordinates? Then you could sort by the radius.
That's a big if, though.

Answer 13

If your operation happens a lot, it might be worth to put it into some 3D data structure. You probably need the distance sorting to decide which object is visible, or some similar task. In order of complexity you can use:
Uniform (cubic) subdivision
Divide the used space into cells, and assign the objects to the cells. Fast access to element, neighbours are trivial, but empty cells take up a lot of space.
Quadtree
Given a threshold, divide used space recursively into four quads until less then threshold number of object is inside. Logarithmic access element if objects don't stack upon each other, neighbours are not hard to find, space efficient solution.
Octree
Same as Quadtree, but divides into 8, optimal even if objects are above each other.
Kd tree
Given some heuristic cost function, and a threshold, split space into two halves with a plane where the cost function is minimal. (Eg.: same amount of objects at each side.) Repeat recursively until threshold reached. Always logarithmic, neighbours are harder to get, space efficient (and works in all dimensions).
Using any of the above data structures, you can start from a position, and go from neighbour to neighbour to list the objects in increasing distance. You can stop at desired cut distance. You can also skip cells that cannot be seen from the camera.
For the distance check, you can do one of the above mentioned routines, but ultimately they wont scale well with increasing number of objects. These can be used to display data that takes hundreds of gigabytes of hard disc space.