Talk:Permanent

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Do permanents have anything to do with Bose statistics? I mean is there something analogous to the Slater determinant using permanents?

[edit] how to compute

I'm unfamiliar with the concept of groups, and the more with symmetry-groups. Would it be possible for me to compute the "permanent" anyway?

Say, I'm already familiar with the implementation of an algorithm to compute the determinant, with a recursive method using pivot elements; and stepping from the size 2x2-definition to a 3x3-definition and higher recursively. Also I know that there are shortcuts for the 3x3-matrix (but say this would be a special cases only applicable for computing determinants). Then how could I compute the permanent of a 3x3 after I have the (simple) definition of a permanent of a 2x2 matrix? With a 3x3-determinant of

a0 b0 c0
a1 b1 c1
a2 b2 c2 

I can compute the determinant via the subdeterminants like

 a0 * det(b1,c2,c1,b2) - b0*det(c1,a2,c2,a1) + c0*det(a1,b2,b1,a2)

(don't know exactly about the signs at the moment)

Is there any way to compute the permanent similar without refering to the "symmetric group"-concept? (if I started learning about groups and symmetric groups I assume I would have a lot of errors in the beginning and would not be able to determine the permanent correctly). Thanks -

--Gotti 10:31, 13 January 2007 (UTC)

[edit] Complexity

"The permanent is also more difficult to compute than the determinant."

Is this known to be true or is it based on an unproven assumption? —Preceding unsigned comment added by Deepmath (talk • contribs) 01:52, 12 March 2008 (UTC)