It's been a while since my last blog entry, but the problem I posed on my earlier blog entry still persists - how to efficiently choose a good split plane for an n-vector data structure. To summarize the structure, geographic points are stored as n-vectors (unit vectors) in a binary tree. Branch nodes of this tree define a plane that bisects/halves the unit sphere - one child of the branch contains points that are below the plane, the other child contains points that are above. Like all trees, leaf nodes are turned into branch nodes as they are filled beyond a threshold. That is the basic idea of it.

Split plane determination must occur when a leaf becomes a branch or when a branch becomes unbalanced. In either case, the method called to determine the split plane is agnostic as to why it was called - it simply receives a bunch of points that it must determine a good split for. My initial implementation was naive:

bestsplit = ...
bestscore = infinite;
for (i=0; i<numpoints; i++) {
  for (j=i+1; j<numpoints; j++) {
    split = (points[i]-points[j]).normalize();
    score = 0;
    for (k=0; k<numpoints; k++) {
      dot = split.dot(points[k]);
      // accumulate some scoring heuristics, such as is each point above/below the split, how far above/below, etc.
      score += ...;
    }
    if ( bestscore > score ) {
      bestscore = score;
      bestsplit = split;
    }
  }
}

So basically - for all points, determine the point between it and every other point, normalize it, and consider this difference as the normal to the split plane, then test it to see how it "scores" using some heuristics. Probably you can see how this will perform miserably for any significant number of points. The big-O notation is something like n^3 which doesn't include the expensive normalization that is called n^2 times. Several other methods were tested, such as fixing the normal of the split plane to be perpendicular to the x, y, or z axis, but these also proved too expensive and/or also had test cases where the split determination was unsatisfactory.

Heuristics

Enter calculus. If we can represent each of the heuristics as a mathematical function, we can determine when the function reaches what is called a "critical point". Specifically we are interested in the critical point that is the global maximum or minimum, depending on which heuristic it is for. So far we have three of these.

1. Similar Distance

We don't want a split plane where, for example, all points above are nearby and all points below are far away. The points should be as evenly distributed as possible on either side. Given that the dot product of a plane normal and any other point is negative when the point is below the plane and positive when the point is above, and the absolute value of the dot product increases as distance to the plane increases, the sum of all dot products for a good split will be at or close to zero. If we let \(P\) be the array of points, \(N\) be the number of points in the array, and \(S\) the split plane, the following function will add all dot products:

\(SumOfDots = \displaystyle\sum_{i=1}^{N} P_i \cdot S\)

The summation here is not really part of a mathematical function, at least not one we can perform meaningful calculus on, since the calculation must be done in the code. We don't know ahead of time how many points there will be or what their values are, so the function should be agnostic in this regard. As written we cannot use it without inordinate complication, but consider that it is really doing this:

\(SumOfDots = \displaystyle\sum_{i=1}^{N} P_{ix} * S_x + P_{iy} * S_y + P_{iz} * S_z\)

This summation expanded will look something like:

\(SumOfDots = P_{1x} * S_x + P_{1y} * S_y + P_{1z} * S_z + P_{2x} * S_x + P_{2y} * S_y + P_{2z} * S_z + P_{3x} * S_x + P_{3y} * S_y + P_{3z} * S_z + \cdots\)

We can then rewrite this as:

\(SumOfDots = S_x*(P_{1x} + P_{2x} + P_{3x} + \cdots) + S_y*(P_{1y} + P_{2y} + P_{3y} + \cdots) + S_z*(P_{1z} + P_{2z} + P_{3z} + \cdots)\)

Or similarly:

\(SumOfDots = S_x*(\displaystyle\sum_{i=1}^{N} P_{ix}) + S_y*(\displaystyle\sum_{i=1}^{N} P_{iy}) + S_z*(\displaystyle\sum_{i=1}^{N} P_{iz})\)

As far as the mathematical function is concerned, the sums are constants and we can replace them with single characters to appear concise:

\(A = \displaystyle\sum_{i=1}^{N} P_{ix}\)

\(B = \displaystyle\sum_{i=1}^{N} P_{iy}\)

\(C = \displaystyle\sum_{i=1}^{N} P_{iz}\)

We can pre-calculate these in the code as such:

double A = 0, B = 0, C = 0;
for (i=0; i<N; i++) {
  A += points[i].x;
  B += points[i].y;
  C += points[i].z;
}

We will rewrite the function with these constants:

\(SumOfDots = S_x*A + S_y*B + S_z*C\)

This is great so far. We are interested in when this function reaches zero. To make it simpler, we can square it which makes the negative values positive, and then we become interested in when this function reaches a global minimum:

\(SquaredSumOfDots = (S_x*A + S_y*B + S_z*C)^2\)

So again, when this function reaches zero it means that the points on either side of the split plane - as denoted by \(S\) - are spread apart evenly. This does not mean that \(S\) is a good split plane overall, as the points could all lie on the plane, or some other undesirable condition occurs. For that we have other heuristics.

As a final note, since the vector \(S\) represents a unit normal to a plane, any determination of it must be constrained to space of the unit sphere:

\(S_x^2 + S_y^2 + S_z^2 = 1\)

2. Large Distance

Practically speaking, the points will not be random and will originate from a grid of points or combination of grids. But random or not, if the points form a shape that is not equilateral - in other words if they form a rectangular shape instead of a square or an ellipse instead of a circle - the larger axis of the shape should be split so that child areas are not even less equilateral. To achieve this we can emphasize that the sum of the absolute value of all the dot products is large, meaning that the points are farther away from the split plane. To do this, we want to find the global maximum of a function that calculates the sum:

\(SumOfAbsoluteDots=\displaystyle\sum_{i=1}^{N} |P_i \cdot S|\)

Unfortunately there is no way, that I know of, to handle absolute values and still determine critical points, so we need to rewrite this function without the absolute value operator. Really all we are interested in is when this function reaches a maximum, so we can replace it with a square:

\(SumOfSquaredDots=\displaystyle\sum_{i=1}^{N} (P_i \cdot S)^2\)

Not unlike the previous heuristic function, we need to rewrite this so that \(S\) is not contained in the summation, and we can extract the constant values. If we skip some of the expanding and reducing the squared dot product we will arrive at this step:

\(SumOfSquaredDots=(S_x^2*\displaystyle\sum_{i=1}^{N} P_{ix}^2) + (S_y^2*\displaystyle\sum_{i=1}^{N} P_{iy}^2) + (S_z^2*\displaystyle\sum_{i=1}^{N} P_{iz}^2) + (S_x * S_y * 2 * \displaystyle\sum_{i=1}^{N} P_{ix}*P_{iy})+ (S_x * S_z * 2 * \displaystyle\sum_{i=1}^{N} P_{ix}*P_{iz})+ (S_y * S_z * 2 * \displaystyle\sum_{i=1}^{N} P_{iy}*P_{iz})\)

Again we will create some named constants to be concise:

\(D = \displaystyle\sum_{i=1}^{N} P_{ix}^2\)

\(E = \displaystyle\sum_{i=1}^{N} P_{iy}^2\)

\(F = \displaystyle\sum_{i=1}^{N} P_{iz}^2\)

\(G = 2 * \displaystyle\sum_{i=1}^{N} P_{ix}*P_{iy}\)

\(H = 2 * \displaystyle\sum_{i=1}^{N} P_{ix}*P_{iz}\)

\(I = 2 * \displaystyle\sum_{i=1}^{N} P_{iy}*P_{iz}\)

As before, these can be pre-calculated:

double D = 0, E = 0, F = 0, G = 0, H = 0, I = 0;
for (i=0; i<N; i++) {
  D += points[i].x * points[i].x;
  E += points[i].y * points[i].y;
  F += points[i].z * points[i].z;
  G += points[i].x * points[i].y;
  H += points[i].x * points[i].z;
  I += points[i].y * points[i].z;
}
G *= 2.0;
H *= 2.0;
I *= 2.0;

And then the function becomes:

\(SumOfSquaredDots=(S_x^2*D) + (S_y^2*E) + (S_z^2*F) + (S_x * S_y * G)+ (S_x * S_z * H) + (S_y * S_z * I)\)

Again when this function maximizes, the points are farthest away from the split plane, which is what we want.

3. Similar number of points

A good split plane will also have the same number of points on either side. We can again use the dot product since it is negative for points below the plane and positive for points above. But we cannot simply sum the dot products themselves, since a large difference for a point on one side will cancel out several smaller differences on the other. To account for this we normalize the distance to be either +1 or -1:

\(SumOfNormalizedDots=\displaystyle\sum_{i=1}^{N} \frac{P_i \cdot S}{\sqrt{(P_i \cdot S)^2}}\)

Expanding this becomes:

\(SumOfNormalizedDots=\displaystyle\sum_{i=1}^{N} \frac{P_{ix} * S_x + P_{iy} * S_y + P_{iz} * S_z}{\sqrt{(S_x^2*P_{ix}^2) + (S_y^2*P_{iy}^2) + (S_z^2*P_{iz}^2) + (S_x*S_y*2*P_{ix}*P_{iy})+ (S_x*S_z*2*P_{ix}*P_{iz})+ (S_y*S_z*2*P_{iy}*P_{iz})}}\)

Unfortunately there is no way to reduce this function so that we can extract all \(S\) references out of the summation, and as with the previous heuristics, put all \(P\) references into constants that we pre-calculate and use to simplify the function. So at present, this heuristic is not able to be used. I am still working on it. The problem is that each iteration of the sum is dependent on the values of \(S\) in such a way that it cannot be extracted.

Putting it all together

What we ultimately want to do is combine the heuristic functions into one function, then use this one function find the critical point - either the global minimum or global maximum depending on how we combine them. The issue is that for \(SquaredSumOfDots\) we want the global minimum and for \(SumOfSquaredDots\) we want the global maximum. We can account for this by inverting the former so that we instead want the global maximum for it. The combined function then becomes:

\(Combined = SumOfSquaredDots - SquaredSumOfDots\)

Applying the terms from each function, we get:

\(Combined = (S_x^2*D) + (S_y^2*E) + (S_z^2*F) + (S_x * S_y * G)+ (S_x * S_z * H) + (S_y * S_z * I) - (S_x*A + S_y*B + S_z*C)^2\)

Expanding the square on the right side and combining some terms will give us:

\(Combined = (S_x^2*(D-A^2)) + (S_y^2*(E - B^2)) + (S_z^2*(F - C^2)) + (S_x * S_y * (G - (2 * A * B)))+ (S_x * S_z * (H - (2 * A * C))) + (S_y * S_z * (I - (2 * B * C)))\)

Again we can combine/pre-calculate some constants to simplify:

\(J = D-A^2\)

\(K = E-B^2\)

\(L = F-C^2\)

\(M = G - (2 * A * B)\)

\(N = H - (2 * A * C)\)

\(O = I - (2 * B * C)\)

double J = D - (A * A);
double K = E - (B * B);
double L = F - (C * C);
double M = G - (2 * A * B);
double N = H - (2 * A * C);
double O = I - (2 * B * C);

And then apply these to the function:

\(Combined = (S_x^2*J) + (S_y^2*K) + (S_z^2*L) + (S_x * S_y * M)+ (S_x * S_z * N) + (S_y * S_z * O)\)

Because we are lazy, we then plug this into a tool that does the calculus for us. And this is where I am currently blocked on this issue, as I have found no program that can perform this computation. I've actually purchased Wolfram Mathematica and attempted it with the following command:

Maximize[{((x^2)*j) + ((y^2)*k) + ((z^2)*l) + (x*y*m) + (x*z*n) + (y*z*o), ((x^2) + (y^2) + (z^2)) == 1}, {x, y, z}]

After 5 or 6 days this had not finished and I had to restart the computer it was running on. I assumed that if it took that long, it would not complete in a reasonable amount of time. I will update this blog entry if I make any progress on this problem as well as the 3rd heuristic function.

Further Optimizations

While I have not gotten this far yet, it may be ultimately necessary to emphasize (or de-emphasize) one heuristic over the others in order to further optimize the split determination. This could be done by simply multiplying each heuristic function by a scalar value to either increase or decrease it's emphases on the final result. The reason I haven't researched this yet is because if I cannot find the global maximum without these extra variables, I certainly cannot do it with them.