Support vectors in three dimensions

[Yellin]

10 events are randomly placed in a cube according to a probability density that is non-zero and finite everywhere in the cube. The events are classified into two classes in a linearly separable manner with at least one event in each class. What is the maximum possible number of support vectors needed with non-zero probability to maximize the margin?

a) 2
b) 3
c) 4
d) 5
e) none of the above

[kkkkk]

I would think that the maximum no. of support vectors for any type of scenario (with 2 classes linearly separable) is N.

[markland]

If I understand the question, it's asking what's the maximum number S such that there is a non-zero probability that S support vectors will be needed. I can see how N SVs might be needed, but the probability of that event would be 0.

BTW I don't know the answer. I'm guessing it's 4?

[IamMrBB]

Good question. I, as always, dislike 'non of the above option'. maybe add 10 and/ or 9 as options. Think answer is c. 4.

[Yellin]

ANSWER (read no further if you want to solve the problem yourself):

There must be at least one event on each of the parallel planes maximizing the margin. Given two fixed points, one on each of two parallel planes, the maximum distance between the planes while keeping them parallel is achieved by rotating them until the normals of the planes are parallel to the vector joining the two points. But that can be done without putting any events between the two planes if and only if there are exactly two support vectors. So two is the minimum number of support vectors. If, at the other extreme, there is no possible orientation change of the parallel planes without a support vector separating from one of them, there must be two more events on the planes, one for each degree of rotational freedom to be prevented. That gives the maximum number of support vectors, four. Any additional event on a plane could only be there with zero probability, because the volume within planes is zero fraction of the cube's volume.

[mic00]

I like this problem, though I'm a little confused about this part of the proof:

Originally Posted by Yellin

If, at the other extreme, there is no possible orientation change of the parallel planes without a support vector separating from one of them, there must be two more events on the planes, one for each degree of rotational freedom to be prevented. That gives the maximum number of support vectors, four.

I see it like this: for P>0 we should expect 30 degrees of freedom. If there are 4 SVs, then we have the following degrees of freedom: rotational orientation of the planes (2 degrees), distance between the planes (1 degree), offset from the origin (1 degree), and 2 degrees for each of the 4 SVs. That's 12 degrees for the SVs, plus 6*3 for the other points -- 30 total.

Originally Posted by Yellin

Any additional event on a plane could only be there with zero probability, because the volume within planes is zero fraction of the cube's volume.

That's clear - restricting any other event to be an SV reduces its degrees of freedom by one.

Incidentally, now I'm a little bit curious about the number of SVs in general. I think I'd previously assumed that #SV < d+1 also had P=0, but that doesn't seem to be the case, and I wonder about problems that could be written around testing for that (the relationship between #SVs, number of dimensions, and size of training set)...

[Yellin]

The degrees of freedom I was writing about are not the 3N ones of the N events. I was mentally adding support vectors until the parallel planes defining the margin are fixed by them. Perhaps I shouldn't have mentioned degrees of freedom at all, and just asked you to visualize how many support vectors there would have to be in order to freeze the planes defining the margin. But when I wrote about there being two degrees of freedom I meant the degrees of freedom of the orientation of the pair of parallel planes once they have been partly locked in position by one support vector on each. There are two coordinates needed to define the direction of the common normal of the pair of planes. To fix the orientation, continue adding support vectors. If a third support vector is added, there must be two on one plane. That plane will still not be fixed in orientation; it can rotate about an axis joining the two support vectors, and the parallel plane can be reoriented to keep it parallel, without separating any support vector from its plane. So only one degree of freedom of the normal is eliminated by having a second support vector on one plane. One more support vector is needed on either of the planes to completely fix the orientation of the two planes. So four support vectors is the maximum (with non-zero probability). There can be fewer than four support vectors, in which case the orientation of the two planes must have adjusted itself to maximize the margin given whatever number of support vectors there is.

The problem can be generalized to d dimensions and N events, and the answer is that the number of support vectors can be anything from 2 to min(d+1,N), depending on where the events are and how they're classified. Here's why:

Call "" the unit normal to the planes and call "" and "" two support vectors, one on one plane and one on the other. Then the remaining support vectors must be different from both and , and they must satisfy either or . To completely specify the d dimensional we need d equations, one of which is that is a unit vector. That leaves d-1 additional support vectors required, for a total of d+1. There can be more support vectors if they accidentally happen to land on one of the planes, but that has zero probability. There must be fewer if the number of points, N, is less than d+1. There can be fewer if the plane orientations are not completely fixed by the support vectors, but have adjusted themselves as best they can to maximize the margin.

The problem could have been made harder by asking about the more general case of d dimensions, but then it would have been more of a mathematical exercise than of thinking about what support vectors really are. And I hadn't thought of the above explanation at the time I wrote the problem.