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Re: Vectors summing to zero
Posted:
Sep 2, 2010 7:09 AM
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On Sep 1, 6:36 pm, Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email>
wrote:
> In article
> <13059615-753f-4932-9a69-729a4047a...@z28g2000yqh.googlegroups.com>,
> "Tim Golden BandTech.com" <tttppp...@yahoo.com> wrote:
>
>
>
> > On Aug 31, 10:58 pm, Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email>
> > wrote:
> > > In article
> > > <4945f2c3-43da-4fbf-8748-2edc6c81c...@s9g2000yqd.googlegroups.com>,
> > > "Tim Golden BandTech.com" <tttppp...@yahoo.com> wrote:
>
> > > > On Aug 31, 1:25 pm, TXB <txb1...@gmail.com> wrote:
> > > > > I have a collection of vectors that span some n-dimensional space.
> > > > > These vectors are typically unit vectors of the dimensional space or
> > > > > the
> > > > > difference of two unit vectors. I want to find all the three-vector
> > > > > combinations that sum to zero. Currently I do this by selecting every
> > > > > possible 3 vector combination to see if a non-trivial linear
> > > > > combination
> > > > > exists that sums to zero. Specifically, I select a vector (v1) and
> > > > > perform
> > > > > a least-squares solution to find a and b so that a*v2 + b*v3 = v1. If
> > > > > the
> > > > > sum is zero (and a, b = +1 or -1), I record the grouping of (v1, v2,
> > > > > and
> > > > > v3) and continue. This is working, however, I feel like there is a
> > > > > simpler
> > > > > methodology that I am missing. Eventually I use this information to
> > > > > create
> > > > > an adjacency matrix for these vectors and display it as a graph. I
> > > > > know
> > > > > very little graph theory (I am an engineer, not a mathematician) so I
> > > > > am
> > > > > not sure if I could use any graph theory to help.
>
> > > > > Thanks.
>
> > > > > txb
>
> > > > The vectors will always be coplanar. This means that the cross product
> > > > of each combination will be colinear; the same value when normalized.
>
> > > > U( v1 x v2 ) = U( v1 x v3 ) .
>
> > > And how, pray tell, do you propose to define/compute the cross product
> > > in n-dimensional space, when n > 3?
>
> > Yeah, I overlooked that point. Still, the coplanar comment can stand.
> > Is it true that the condition is merely
>
> > v1 + v2 + v3 = 0 ?
>
> > Or is TXB looking for scaled versions as well? e.g. is
> > (U( 5, 1, 0, 0 )), ( 1, 0, 0, 0 ), ( 0, 1, 0, 0 )
> > where U is a unit vector scalar conversion in the criterion?
>
> > If the dimension is large then the chance of hitting a coplanar triple
> > is slight. There may be a dot product solution lurking here that is so
> > simple. But then, the sum itself seems so simple that I don't see much
> > puzzle anyway. It's a need for speed, eh? Use partial data seems a
> > good method of elimination, but depends on the qualities of the data.
>
> > The sum of the angles will always be 2 pi and so by computing these
> > angles another process of elimination can be had. This is simple data
> > without the dimensional complexities. Integerizing that angle could
> > get pretty fast results as type char or short since the resulting
> > angle can be coded to overflow or some such hardware condition. Also
> > no actual trig operation is necessary since a coarse LUT would be
> > adequate. All other angular sums will be smaller, but the condition
> > queried above will also be included within this criterion. Depending
> > on how many candidates are eliminated this method could be fast if the
> > list for final precision verification is short.
>
> You know, if you actually knew some mathematics, you'd be dangerous.
>
> --
> Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
Well, Gerry, you should check out polysign numbers.
http://bandtechnology.com/PolySigned
That is how I come to have these strong opinions. What about
Frederick's claim of a 7D cross product? I have no idea what he is
talking about but it sure sounds interesting.
It is interesting that angle in high dimension is still well behaved.
Anytime you have an angle of pi you do have an inverse of your
original vector, regardless of the dimensional qualities of the
vector.
As I go over it again I will stand by the 2 pi claim. If you have a
falsification then please present it. I don't necessarily believe all
of mainstream mathematics so I find it acceptable that you do not
believe mine. Simply stated: the sum of the angles between three
coplanar vectors will always be 2 pi, in any dimension. The converse
holds: If the sum of the angles of three vectors is 2 pi, then the
vectors are coplanar. This is not a proof, but is an observation. I'm
open to falsification, but doubt that you will find one.
- Tim
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