T.R | Title | User | Personal Name | Date | Lines |
---|
1304.1 | | GUESS::DERAMO | Dan D'Eramo | Thu Oct 04 1990 18:32 | 3 |
| This problem is also discussed in SIEVAX::ALGORITHMS topic 88.
Dan
|
1304.2 | Mathematical <-> algorithm | ELIS::BUREMA | In the middle of life is if... | Fri Oct 05 1990 04:00 | 8 |
| That is correct. I posted the original note (88) there. However
_generating_ all the possibilities is an algorithm to me. The question
of the maximum number of combinations is (al least to me) a
mathematical problem. Agree?
0^0
Wildrik |
\-/
|
1304.3 | | GUESS::DERAMO | Dan D'Eramo | Fri Oct 05 1990 14:11 | 5 |
| I agree it is a mathematical problem. Reply .1 was giving
a pointer to more information on the problem to other readers
here, it wasn't saying that posting .0 here was inappropriate.
Dan
|
1304.4 | I stand corrected, sorry | ELIS::BUREMA | In the middle of life is if... | Wed Oct 10 1990 04:19 | 1 |
|
|
1304.5 | Rook polynomials?
| VIVIAN::MILTON | I'm thinking about it! | Wed Oct 10 1990 08:47 | 3 |
| Can rook polynomials be used for this sort of problem?
Tony.
|
1304.6 | The answer my friend ... | ELIS::BUREMA | In the middle of life is if... | Thu Oct 11 1990 04:59 | 6 |
| If I knew what "rook polynomials" are, I could answer. As it is I don't
know. Could you explain?
0^0
Wildrik |
\~/
|
1304.7 | | GUESS::DERAMO | Dan D'Eramo | Thu Oct 11 1990 10:32 | 4 |
| They have something to do with how many different ways a
rook could move from A to B on a chess board. I think.
Dan
|
1304.8 | Please expand, I'm curious | ELIS::BUREMA | In the middle of life is if... | Tue Oct 16 1990 04:11 | 4 |
| Could you give an example. And do they also apply to non-square chess
boards (e.g. 4x6, 7x4, 9x8 ...)?
Wildrik (-8
|
1304.9 | | GUESS::DERAMO | Dan D'Eramo | Tue Oct 16 1990 11:15 | 4 |
| I don't really know all that much about them, I just recognized
the name.
Dan
|
1304.10 | Example wanted | ELIS::BUREMA | In the middle of life is if... | Wed Oct 17 1990 11:43 | 8 |
| Ok, then. Is there anyone who can give an example, or provide pointers
to books etc. so that I can get at least an idea of what rook
polynomials are; what they are used for, and if they are applicable to
my problem?
0^0
Wildrik, | (Confused).
\~/
|
1304.11 | Books on Rooks | VIVIAN::MILTON | I'm thinking about it! | Wed Oct 17 1990 17:11 | 16 |
| The following books contain discussions of rook polynomials:-
C.L. Liu, Introduction to Combinatorial Mathematics, McGraw-Hill, New
York, 1968.
I. Anderson, A First Course in Combinatorial Mathematics, Oxford
University Press, New York, 1974.
I'll try to give an example in a following reply but the basic
technique is about the number of ways of placing non-taking rooks on a
board of not only any size but any shape and including prohibited
squares (this may be of use in your problem but will need quite a lot
of work).
Tony.
|
1304.12 | Simple example | VIVIAN::MILTON | I'm thinking about it! | Wed Oct 17 1990 17:55 | 52 |
|
Consider a job allocation problem in which a number of people are skilled at
various different jobs. The problem is to determine the number of ways of
assigning an appropriately skilled person to each job. For example, we may have
the following situation, in which x in a square indicates those people skilled
at particular jobs.
People
A B C D
+---+---+---+---+
1 | x | x | x | x |
+---+---+---+---+
2 | x | | | x |
Jobs +---+---+---+---+
3 | x | x | | x |
+---+---+---+---+
4 | x | x | x | x |
+---+---+---+---+
To solve this problem using rook polynomials, we find the rook polynomial of
the board formed by the white squares (prohibited), if this rook polynomial
is:-
2 n
1 + r x + r x + ... + r x
1 2 n
then the number of ways of assigning people to jobs is:-
n
n! - r (n-1)! + r (n-2)! - ... + (-1) r
1 2 n
For example, the rook polynomial of the white squares above is:-
2 3 4
1 + 3x + 2x + 0x + 0x
(1 way to place no rooks, 3 ways to place 1 rook, 2 ways to place 2 rooks, no
ways to place 3 rooks and no ways to place 4 rooks)
and so the number of ways of assigning people to jobs is:-
4! - 3x3! + 2x2! - 0x1! + 0 = 10.
This has been a simple case as more complex boards require and understanding of
the various operations on board shapes (product, inclusion-exclusion) and the
resultant algebra looks like chopped up bits of chess board!
Hope this helps.
Tony.
|
1304.13 | 8+8 --> 4 rounds | ELIS::BUREMA | In the middle of life is if... | Thu Oct 18 1990 10:44 | 8 |
| > is n/2. However for 8 + 8 people, I can never come up with more than
> 3 *unique* rounds.
I *have* come up with 4 unique rounds. However it was by trial and
error, and does not lend itself to a general solution. But is it the
maximum.... ?
Wildrik 8-)
|
1304.14 | | TRACE::GILBERT | Ownership Obligates | Thu Oct 18 1990 11:17 | 8 |
| .12> (1 way to place no rooks, 3 ways to place 1 rook, 2 ways to place 2 rooks,
.12> no ways to place 3 rooks and no ways to place 4 rooks)
This should say "1 way to place 2 rooks".
.13> I *have* come up with 4 unique rounds.
Great! Would you post the solution?
|
1304.15 | Here it is ... | ELIS::BUREMA | In the middle of life is if... | Thu Oct 18 1990 12:53 | 55 |
| Re: .14 This is my solution:
First of all I created a matrix and assigned the entries as follows:
+--+--+--+--+
|M1|F3|M5|F7| M for one gender
+--+--+--+--+ F for the *other*
|F1|M3|F5|M7|
+--+--+--+--+
|M2|F4|M6|F8|
+--+--+--+--+
|F2|M4|F6|M8|
+--+--+--+--+
(Notice the simularity to a 4x4 chessbboard, where M's are the black
squares (No pun intended!)).
I then assigned "courts" to the players:
+--+--+--+--+ +--+--+--+--+ +--+--+--+--+ +--+--+--+--+
| 1| 2| 3| 4| | 1| 3| 3| 1| | 1| 2| 1| 2| | 1| 1| 4| 4|
+--+--+--+--+ +--+--+--+--+ +--+--+--+--+ +--+--+--+--+
| 1| 2| 3| 4| | 3| 1| 1| 3| | 4| 3| 4| 3| | 3| 3| 2| 2|
+--+--+--+--+ +--+--+--+--+ +--+--+--+--+ +--+--+--+--+
| 1| 2| 3| 4| | 2| 4| 4| 2| | 2| 1| 2| 1| | 2| 2| 3| 3|
+--+--+--+--+ +--+--+--+--+ +--+--+--+--+ +--+--+--+--+
| 1| 2| 3| 4| | 4| 2| 2| 4| | 3| 4| 3| 4| | 4| 4| 1| 1|
+--+--+--+--+ +--+--+--+--+ +--+--+--+--+ +--+--+--+--+
Writing it out I obtain:
Round 1 M1+F1 M3+F3 M5+F5 M7+F7
M2+F2 M4+F4 M6+F6 M8+F8
Round 2 M1+F5 M2+F6 M5+F1 M6+F4
M3+F7 M4+F8 M7+F3 M8+F2
Round 3 M1+F4 M2+F7 M3+F6 M4+F5
M5+F8 M6+F3 M7+F2 M8+F1
Round 4 M1+F3 M2+F4 M3+F1 M4+F2
M8+F6 M7+F5 M6+F8 M5+F7
(Barring typo's).
Again, this was done by hand, so I have no code to do it. Neither
do I have generalization(sp?) for Nx4 or MxN.
N.B. This is partly due to the algorithm of Nasser in note 88 in the ALGORITHMS
conference (credit where credit's due).
0^0
Wildrik |
\ /
|