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Home/ Questions/Q 1060721
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T18:22:11+00:00 2026-05-16T18:22:11+00:00

Is there a way to do a quick and dirty 3D distance check where

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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?

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  1. Editorial Team
    Editorial Team
    2026-05-16T18:22:12+00:00Added an answer on May 16, 2026 at 6:22 pm

    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?).

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