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| Title: | Mathematics at DEC |
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| Moderator: | RUSURE::EDP |
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| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
387.0. "Permanents" by TOOLS::STAN () Thu Nov 21 1985 17:06
If M is a square matrix, the permanent of M, denoted per(M),
is just like the determinant, except that all terms are positive,
whereas for a determinant, some products are positive and some
are negative.
For example, the permanent of
1 2 3
4 5 6
7 8 9
is (1)(5)(9)+(2)(6)(7)+(3)(4)(8)+(1)(6)(8)+(2)(4)(9)+(3)(5)(7).
Lots of neat formulas are known for determinants of matrices of various
patterns, but literally nothing is known for permanents.
For example, suppose M[n] is the n X n matrix whose (i,j) entry
is n(i-1)+j-1. For example, M[4] is
0 1 2 3
4 5 6 7
8 9 10 11 .
12 13 14 15
Is there some neat formula for per(M[n]) in terms of n?
I have a lot of data, but I see no pattern.
Similarly, what about the matrix whose (i,j) entry is i+j?
or i+j-2? or (-1)**(i+j) * (i+j) ?
| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 387.1 | | ADVAX::J_ROTH | | Thu Nov 21 1985 17:28 | 6 |
| I recall seeing some information on permanents in "Advanced Combinatorics"
by Comtet. I think they also arise in the study of certain stochastic
processes, so some results are known, for example, I believe permanents
are known to be very difficult to evaluate, unlike determinants.
- Jim
|
| 387.2 | | HITECH::JENSEN | | Thu Nov 21 1985 19:06 | 8 |
| I seem to recall studying this problem in an automata theory course
(too) many years ago. I believe someone proved that there
is not only no closed form solution, there aren't even any shortcuts.
I'm holed up in the basement of ZK2 this week, but when I get back
to CA I'll try to dig up my notes & post anything interesting.
/P Jensen
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