The new forums will be named Coin Return (based on the most recent vote)! You can check on the status and timeline of the transition to the new forums here.
The Guiding Principles and New Rules document is now in effect.
[solved?] I'm a noob.
Legionnaired
Registered User regular
Registered User regular
I had a flash of insight last night, and I think I may have figured out how to do the TSP in O(log(2^n)^4.7). (Current best might be something like (2^n)(n^2)...)
I'm writing out a test program now, but if it works where do I go from here?
So... what? Do I talk to my professor/advisor? Do I write out the algorithm, get it signed and notarized and mail it to myself first?
I'm writing out a test program now, but if it works where do I go from here?
So... what? Do I talk to my professor/advisor? Do I write out the algorithm, get it signed and notarized and mail it to myself first?
Legionnaired on
0
This discussion has been closed.
Posts
Basically, buy one of those cardboard backed composition books, write down your ideas and algorithms and then in the upper hand corner, write down the date and sign it.
I doubt your advisor/professor would take the credit for it, but I'd also be surprised if the approach holds up. If it does, your professor should be able to help with the paper and publication that you'll need to do to get this properly dealt with.
No, I think (offhand) it's O((log(2^n))^C).
And yeah Kakodaimonos, I'm pretty skeptical myself. It may turn into just another pretty good heuristic attempt, but it might 'accidentally' find the optimal solution more often than not.
What he means is that lg( 2^n ) = n. It's a property of logarithms.
You don't have the goods, I'm certain, but hypothetically someone with the algorithm would also need to be careful with their well being. I certainly wouldn't post openly on a forum about it, because that's embarrassing if wrong and dangerous if right. You might end up on the side of a milk cartoon. Luckily you fall into the former category, but next time, try to be a bit more secretive. :rotate:
Where b is the base of the original logarithm. 1 / log2(b) is a constant, so it can be pulled out of the O-notation.
So O(log(2^n)) = O(n).
*sigh*