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)...