Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The regression depth of a line is the minimum number of points it has to cross to take it from its initial position to vertical. The undirected depth of a point is the minimum number of lines a ray originating at the point will cross before escaping. If we use projective duality, then regression depth is same as undirected depth. I cannot see how that would be.

My attempt is as follows: By duality, points will map to lines, and lines will map to points. Suppose I am trying to compute the undirected depth of a point. So there is a point such that all rays emanating from it will meet n/3 points or more in each direction. In the dual plane, the point will coincide with a line. But what will rays correspond to?

EDIT: Quoting from paper by Nina Amenta, Marshall Bern et al, "Regression Depth and Center Points": Geometrically, the regression depth of a hyperplane is the minimum number of points intersected by the hyperplane as it undergoes any continuous motion taking it from its initial position to vertical. In the dual setting of hyperplane arrangements, the undirected depth of a point in an arrangement is the minimum number of hyperplanes touched by or parallel to a ray originating at the point. Standard techniques of projective duality transform any statement about regression depth to a mathematically equivalent statement about undirected depth and vice versa.

I was trying to come up with this duality but havent so far succeeded. In R_2, when we consider lines and points, I was considering a point that has high undirected depth, but am unable to make a transformation that would show exactly how the undirected depth of the point will map to the regression depth of a line in the dual plane.

share|improve this question
1  
There is presumably some background arrangement of lines and points going on that makes this question make sense, but you really need to be more precise about this. –  Qiaochu Yuan Apr 30 '10 at 19:19
    
I have added the exact definition to be clearer. I am trying to come up with the transformation that would show how regression depth will map to undirected depth and vice versa. –  Umar Sheikh Apr 30 '10 at 19:38
add comment

2 Answers

up vote 0 down vote accepted

(For some context: the question here is written in terms of points in a plane, and my answer is in the same terms, but the same duality works in higher dimensions as well. The paper in question is arxiv:cs.CG/9809037.)

Let P be the plane in which you are measuring regression depth; regression depth is defined in a Euclidean plane, not a projective one, but we view P as being a Euclidean plane embedded as a subset of its projective completion P*, so that a line is vertical iff it passes through the point at vertical infinity (the intersection point in P* of two vertical lines).

Now take a dual projective plane Q*, and again view Q* as containing a Euclidean plane Q and a line at infinity; choose Q and Q* in such a way that the line at infinity in Q* is dual to the point at vertical infinity in P*. (It's easy enough to do this with coordinates, if you prefer them to this sort of conceptual explanation.)

Then rotating a line in P until it is vertical is the same thing as moving the projectively dual point in Q until it reaches infinity, and the number of points crossed by the rotating line is the same as the number of lines crossed by the moving point.

share|improve this answer
add comment

The construction probably goes like this.

Let $l$ be the line whose regression depth you are evaluating, and le $l'$ be its final (vertical) position after some continuous motion. Now assuming $l \ne l'$, there is some unique point $p$ that they intersect in.

Without loss of generality now, we can assume that the continuous motion taking $l$ to $l'$ is a rotation centered at $p$, since it will sweep the same area. Let $S$ be the set of lines in the double wedge between $l$ and $l'$ with apex at $p$.

The dual of a double wedge is a line segment. Construct the dual of $l$ and $l'$. these are two points that mark the endpoints of the line segment. By duality, this line segment intersects the duals of all points in the wedge and is the desired ray originating at a point (which is the dual point of $l$)

share|improve this answer
    
Your answer is very easy to follow without defining or worrying about projective vs euclidean plane. It also resolves what a ray is. it is a line segment whose dual is a wedge. –  Umar Sheikh May 1 '10 at 0:14
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.