> On 06/04/2008, Alan G Isaac <[EMAIL PROTECTED]> wrote:
>> Just checking:
>> it's important to me that this won't change
>> the behavior of boolean matrices, but I don't
>> see a test for this. E.g., ::
>> >>> import numpy as N
>> >>> A = N.mat('1 0;1 1',dtype='bool')
>> >>> A**2
>> matrix([[ True, False],
>> [ True, True]], dtype=bool)
On Sun, 6 Apr 2008, Anne Archibald apparently wrote:
> I have no desire to change the behaviour of boolean matrices, and I'll
> write a test, but what is it supposed to do with them? Just produce
> reduce(dot,[A]*n)?
That's the part I care about.
> For zero it will give the identity,
Yes.
> and for negative powers some sort of floating-point
> inverse.
That deserves discussion.
Not all "invertible" boolean matrices have an inverse in the algebra.
Just the orthogonal ones do.
I guess I would special case inverses for Boolean matrices.
Just test if the matrix B is orthogonal (under Boolean
multiplication) and if so return B's transpose.
> Currently for positive powers it should produce the right
> answer provided multiplication is associative (which
> I think it is).
Yes; N×N boolean matrices are I believe a semi-group under multiplication.
> The inverse actually poses a problem: should it return
> (A**(-1))**n or (A**n)**(-1)? (No, they're not the same
> for boolean matrices.)
I think it must be the latter ... ?
By associativity, if B has an inverse A,
then B**n must have inverse A**n.
So you are observing that with boolean matrices
we might find B**n is invertible even though B is not.
Right? So the latter will work in more cases.
So again: form B**n, test for orthogonality,
and return the transpose if the test passes.
Cheers,
Alan Isaac
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