Orientation consistency of a group

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The discussion centers on determining the orientation consistency of a set of vectors, specifically how to efficiently verify if a given set of directional relationships, represented as d(Ax, Ay), is consistent. The user Mike outlines that if two vectors point in the same direction, d(Ax, Ay) equals 1, while opposite directions yield -1. The conversation suggests that a matrix representation of these relationships can be permuted into a specific form to simplify the consistency check, moving away from brute force methods. Stephen confirms this approach and offers insights into efficient algorithms for testing matrix forms.

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SanDiegoMike
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Hello,

I apologise if this has been asked before. I did a search, but quite frankly, I'm not even sure what to search for.

My gut feeling is that this problem has already been well defined and solved, so if someone could point me in the proper direction, that would be great.


Consider a set of n vectors which can point in the positive or negative direction: {A1, A2, ... An }

if Ax and Ay point in the same direction, then d(Ax,Ay) = 1.

Similarly, if they point in the opposite direction, then d(Ax,Ay) = -1.

Now if d(A1,A2) = 1, and d(A2,A3) = 1, then d(A1,A3) must equal 1 to be consistent.

Similarly, if d(A1,A4) = 1 and d(A4,A5) = -1, then d(A1,A5) must equal -1 to be consistent.

[I'm guessing for the trivial case of d(Ax,Ax), the answer would be 1]

the problem:

Assume that I'm given all d(Ax,Ay) for x,y = 1...n. What is an efficient way of determining if this set is consistent?

[note: I can do this brute force, but as n gets big, it becomes a mess.]

Thanks!
-Mike.
 
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You also want d(Ax,Ay) = d(Ay,Ax), I assume.

One thought is that if you look at d(X_i,Y_j) as an entry in a matrix, you would expect the matrix to be permutable to one that has two square blocks of the 1's along the diagonal and -1 elsewhere. For example:

\begin{pmatrix} 1 &amp; 1 &amp; 1 &amp; 1 &amp; -1 &amp; -1 \\<br /> 1 &amp; 1 &amp; 1 &amp; 1 &amp; -1 &amp; -1\\<br /> 1 &amp; 1 &amp; 1 &amp; 1 &amp; -1 &amp; -1\\<br /> 1 &amp; 1 &amp; 1 &amp; 1 &amp; -1 &amp; -1\\<br /> -1&amp;-1&amp;-1&amp;-1&amp;1&amp;1\\<br /> -1&amp;-1&amp;-1&amp;-1&amp;1&amp;1 \end{pmatrix}

If you could prove the matrix had to have that form then you could test for this condition.
I'm assuming permuting a given matrix to that form is less work than "brute force".
 
Thanks Stephen, I do believe you're correct. That seems to be a very clever way forward.

It's been a long time since my undergraduate course work in linear algebra. Can you point me in the direction of an efficient algorithm for testing for such a matrix?
 
no worries, I figured out how to do the test. Very simple indeed. Thanks!
 

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