QueenAlice.com


Username:

Password:

Remember me



Forgot Password?
Registration FREE!





Topic: How many?
Back to Forum Index
Back to Forums List


Author

Message
richerbyUnited Kingdom flag
kingdave wrote:
Here is a another strategy to try. Define B(n) to be the Blutigeroo count, where his criteria for exclusion would have to be generalized to the n case. I suspect that finding a closed form for B(n) is doable.

I doubt it, since both the number and nature of the constraints depends on n. Combinatorial counting problems tend to be computationally intractable (very often #P-complete, which is at least as bad as being NP-hard and probably much worse), often even when the problem of finding a single solution is easy (polynomial time). A closed form, depending on how exactly you define the term, would suggest a polynomial-time algorithm.

Dave.

Blutigeroo
I'm with Richerby on this one. I'm not suggesting that anyone give up on it but it is looking more and more like the kind of complexity that does not allow a formulaic approach. For small numbers of n players, an algorithm seems the only solution. For large n, that too becomes problematic ... at least until quantum computing takes hold!

Richerby, I recycled a constraint that I had remmed out but it was not the best one to use which is simply:
if ((b > 12) or (c > 12) or (d > 12)) continue;

... that shaves the time down a tiny bit further (down to 32% of the first algorithm). Anyhow, the point is taken.

Kingdave, your approach seems to be the best one although the original error is a concern. Were you able to eliminate any possibility of that in further runs? If so, I think I can say after a quick test that B(n) fails for n larger than 4. This is probably because further constraints are needed (n-2 constraints?). For B(5) I get 43451 after confirming that my generalized algorithm agrees with Kingdave for n < 5.

kingdaveUnited States flag
Blutigeroo, for n = 5 you would need to add another constraint: the sum of any 3 numbers has to be <= 36. Can you try adding that and see what you get? Your other constraint would be the sum of any 2 numbers is <= 28, right?


Blutigeroo
Thanks ... that got it. I had tried the constraint but I had it as a formula {i.e. (n-1)*12-8}. I didn't notice that it was incorrect and therefore failed ... the last constant should have been a 12. I've confirmed your result: 39801

Previous 1 2 3 4 5

©2004-2024 Queen Alice Internet Chess Club
All rights reserved.