Subject: Re: CCL:Smallest rectangular box

From: [email-masked] (Jack Smith)
Date: Thu, 13 Jun 1996 09:16:48 -0400

OK, here's my intuitive answer (i.e. no proof):

  Generate a distance matrix and find the two most distal points.  Let
these two points define the Z direction (i.e. parallel to Z-axis) of the
rectangular box.  Now generate another distance matrix where the distance
is the projected distance perpendicular to the Z direction. Find the two
most distal points and define the Y direction from the projection of their
connecting vector in a plane perpendicular to Z.  The X direction is also
now fixed - perpendicular to the Y and Z directions.  The dimensions of the
rectangular box are now determined by the most distal points in each of the
axial directions, and its center (origin) is defined by the midpoints.

  There's probably a way to prove that the six most distal points defining
the faces of the enclosing rectangular box are unique and that all the
interior points are in some sense "bound" by these six points, independent
of the rotational frame or origin.

- Jack

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From: Pat Harvey <[email-masked]>
Date: Thu, 13 Jun 1996 15:40:48 +0100 (bst)

> of the rotational frame or origin.

Surely this is not correct. Consider the trivial case of two points. The
smallest bounding rectangle will have the two points at opposite corners.

Pat Harvey                  Tel:   +44-71-594-5847                           
New building Room 164 
Department of Chemistry     email: [email-masked]                         
Imperial College                                                           
London. SW7 2AY             Grad Student

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From: Pat Harvey <[email-masked]>
Date: Thu, 13 Jun 1996 16:08:50 +0100 (bst)

> of the rotational frame or origin.
Consider 4 point defining a rectangle. The smallest bounding rectangle is
the rectangle defined by the points. The most distal points are the
diagonals.

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From: Ryszard Czerminski <[email-masked]>
Date: Thu, 13 Jun 1996 12:10:22 -0400 (EDT)

> smallest bounding rectangle will have the two points at opposite corners.

but if you consider an example of square with
side lenght a, then procedure below will give you as a bounding box
square with side lenght = a*sqrt(2)

Best regards,

Ryszard

 ..................................
Ryszard Czerminski
Molecular Simulations Incorporated
16 New England Executive Park
Burlington, MA 01803-5297
tel: (617)229-8875 x 217
fax: (617)229-9899
e-mail: [email-masked]

On Thu, 13 Jun 1996, Pat Harvey wrote:

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