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Re: Calculating matrix permanent
Posted:
Jan 26, 2013 11:09 AM


"Patrick D. Rockwell" <prockwell@thegrid.net> wrote in message news:3cd635d38c004c8fb5b3af9c45759521@v9g2000pbi.googlegroups.com... > I've read how to calculate the permanent of a matrix, > but I found the notation hard to understand so I'd > Like a practical demonstration. For 2 by 2 matrices > It's easy. For example, the permanent of > > > 2 4 > 9 6 > > Is 2*6+4*9=36+12=48 > > But what about > > 5 3 1 > 8 2 4 > 7 9 6 > > What is48 the permanent of the above matrix > and how do you calculate it?
I'd never heard of the permanent before, but checking Wickipedia it seems it's defined just like the determinant, but with all plus signs. So in your example:
Permanent ( 5 3 1 8 2 4 7 9 6 ) = 5*2*6 + 5*4*9 + 3*8*6 + 3*4*7 + 1*8*9 + 1*2*7
Each of the products contains exactly one matrix entry from each line, and exactly one from each column. There are six ways to select such entries, corresponding to the six permutations of a set of three elements, which is why we have the 6 products appearing in the sum. (Also, that's why the definition in Wikipedia makes use of permutations in it's notation.)
For calculation, it is probably easier to group the products based on, say, the top row:
... = 5*(2*6 + 4*9) + 3*(8*6 + 4*7) + 1*(8*9 + 2*7)
noting that the expressions in brackets are then just the permanents of the appropriate submatrices
... = 5*Permanent(2 4 6 9) + 3*Permanent(8 4 7 6) + 1*Permanent(8 2 7 9)
I.e. all this is just like with determinants, but without the "confusion" of the negative signs appearing.
HTH Mike.



