Linear mapping of sliced hyperbox

[Yaron Berman]

Thanks,
But I think the assumption is not true. Look at the following counter
example in R^3 -> R^2:
The hyperbox is [-1,1]*[-1,1]*[-1,1]
The (closed) halfspace is H = { (x1,x2,x3) in R^3 | x3 <= 0 }
The transformation matrix is (1 0 0.5;
					0 1 0.5)
The transformation projects the plane x3=0 to the entire plane in R^2,
because the first two column vectors are linearly independent.
So the intersection of the halfspace projection with the original
zonogon leaves the zonogon the same, while the zonogon defined by the
transformation on [-1,0]*[-1,0]*[-1,1] is definitely different.

Regards,
Yaron

> Isn't it true that p(hypercube ^ halfspace) = p(hypercube) ^
p(halfspace)
> ?
> where p(x) -> y projects an n-dimensional volume x into a
2-dimensional
> region y
> 
> If so, you could compute the zonogon the usual way (project the edge
> vectors
> and sort them by angle, and then string them together in sorted order
to
> create the zonogon) and after that slice with the projected halfspace.
> 


> ----- Original Message -----
> From: "Yaron Berman" <yaronber at cs.huji.ac.il>
> To: <compgeom-discuss at research.bell-labs.com>
> Sent: Tuesday, October 05, 2004 3:43 AM
> Subject: Linear mapping of sliced hyperbox
> 
> 
> > Hello,
> >
> > It is well known that a linear mapping of an n-dimensional hyperbox
is a
> > zonotope, and when the mapping is to R^2, it is simply a convex
> > centrally symmetric polygon with at most 2n vertices, which is easy
to
> > compute (in O(nlogn) time).
> > My question is about the linear map of a hyperbox intersected with a
> > halfspace whose boundary contains the origin (which is also the
center
> > of the box), that is a box sliced through its center.
> > Is there some efficient way to compute the map (still a convex
polygon),
> > other than enumerating all the 2^n vertices of the sliced box and
> > computing the convex hull of their map? If there is an efficient
method,
> > can it be extended to the case of a box intersected with a cone?
> >
> > Thanks very much for the assistance,
> > Yaron


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