Re: Motion 11
Dan Zuras Intervals wrote:
Date: Tue, 16 Feb 2010 10:11:46 +0100
From: =?ISO-8859-1?Q?Fr=E9d=E9ric_Goualard?=
<Frederic.Goualard@xxxxxxxxxxxxxx>
To: Dan Zuras Intervals <intervals08@xxxxxxxxxxxxxx>
CC: stds-1788@xxxxxxxxxxxxxxxxx
Subject: Re: Motion 11
Dear Dan,
Dan Zuras Intervals wrote:
I believe your observation about Corollary 1 & arguments
for the need for reverse operations both contribute to an
approach along the lines George has suggested.
For example, Nate has pointed out the the lack of inverse
operators within ordinary intervals can be repaired by the
use of the Dual() operator. Where
Dual([a,b]) = [b,a].
Thus, you can solve for yy in xx + yy = zz by
xx + yy = zz
(xx - Dual(xx)) + yy = zz - Dual(xx)
[0,0] + yy = zz - Dual(xx)
yy = zz - Dual(xx)
All 4 basic operations may be inverted in this way. (I
will leave it to Nate to discuss the details on grounds
of incompetence on my part. :-)
I do not know enough about modal interval arithmetic to have any
definitive argument about that. However, I doubt you can easily
obtain
the same results as with reverse division. Once again, consider my
example:
A=[0,2]
B=[-1,1]
C=[1,1]
with \circ=\times
and compute \times_1^-(B,C,A) = hull({x\in[0,2]\mid\exists
b\in[-1,1], x\times b\in[1,1]})
Can you obtain [1, 2] as a result? The n+1-ary reverse division will.
Ah, yes. We have discussed this before.
Still, I believe that the answer is that such a result
CAN be obtained with the proper inverse of B. But I'll
let Nate describe the details.
Nate?
In this example, my observation is reverse mode appears to be the same as
the normal interval expression
([1,1]/[-1,1]) \intersect [0,2].
Kaucher/modal arithmetic is an unrelated topic, I believe.
IMO, the only tricky issue of Frederic's example appears to be how to deal
with the fact division by [-1,1] bifurcates the problem into two disjoint
interval results.
On the one hand, reverse mode gives [1,2] in a single operation. On the
other hand, bisecting B=[-1,1] into B_1=[-1,0] and B_2=[0,1] at the
branch-cut and then processing each side with normal interval operations
yields
branch1: ([1,1]/[-1,0]) \intersect [0,2] = empty
branch2: ([1,1]/[0,1]) \intersect [0,2] = [1,2].
This gives the same result [1,2] obtained by the reverse mode operator.
So the question in my mind is: does a dedicated reverse-mode operation in
hardware provide benefit over using the normal interval operations? My
feeling is a deeply pipelined interval processor will compute the normal
interval operations very quickly. A more scientific analysis might be
helpful. :)
The same applies if A=[-2,2], however in this case it seems clear to me the
normal interval operations are more desireable, i.e.,
branch1: ([1,1]/[-1,0]) \intersect [-2,2] = [-2,-1]
branch2: ([1,1]/[0,1]) \intersect [-2,2] = [1,2]
In interval Newton, processing intervals [-2,-1] and [1,2] as two separate
branches is advantageous, plus it eliminates the need to design hardware
that must return lists of disjoint interval results.
Nate