bio | website | |
---|---|---|
location | ||
age | 26 | |
visits | member for | 2 years, 5 months |
seen | 7 hours ago | |
stats | profile views | 310 |
Jun 24 |
revised |
Unique Nash equilibrium games
Links to corresponding articles added |
Jun 21 |
revised |
Unique Nash equilibrium games
added 84 characters in body; edited tags |
Jun 19 |
asked | Unique Nash equilibrium games |
Jul 3 |
comment |
Number of Configurations in the optimal Hanoi tower
"It seems forcing the number of disks on all 3 towers specifies a unique position" _ this seems to imply that $2^n \in O(n^3)$ :) |
Jun 25 |
awarded | Informed |
May 2 |
revised |
Number of Configurations in the optimal Hanoi tower
added 450 characters in body; edited tags; edited title |
May 2 |
comment |
Number of Configurations in the optimal Hanoi tower
Well, I see these connections can be arbitrary and there might be no nice way to show the equivalence. I just wanted to have a closed formula for Hanoi tower configurations, from the computational perspective, formulas I see with the equal sequences can be calculated in linear time with respect to $n$, whilst recurrent formula for Hanoi configurations takes cubic time. |
Apr 30 |
awarded | Critic |
Apr 30 |
asked | Number of Configurations in the optimal Hanoi tower |
Jan 11 |
awarded | Editor |
Jan 11 |
revised |
computational complexity
edited tags; edited title |
Jan 9 |
awarded | Scholar |
Jan 9 |
accepted | computational complexity |
Jan 8 |
awarded | Supporter |
Jan 8 |
comment |
computational complexity
I agree with connected components and retraction of boundary, but after these operations we are still dealing with quite arbitrary structure. We can even make things "easier" and assume that every edge is incident with at most 3 triangles by simple topological transformation. If we consider dual problem, take 1 vertex in each triangle and connect two vertices if their respective triangles share edges, we get: in somewhat special graph we need to find cubic planar 3-connected subgraph(corresponding to triangulation). Finding cubic subgraph in planar graph is NP-complete. I see some connection. |
Jan 8 |
awarded | Autobiographer |
Jan 8 |
revised |
computational complexity
edited tags |
Jan 8 |
revised |
computational complexity
edited tags |
Jan 8 |
comment |
computational complexity
If we are given set of triangles as a certificate, first we will check connectivity, afterwards we check that the link of each vertex is a cycle. If these conditions are satisfied, we know for sure that we have a closed surface. Then calculate Euler characteristic, we are dealing with sphere if and only if Euler characteristic is 2. en.wikipedia.org/wiki/Surface#Classification_of_closed_surfaces here you can check what are the Euler characteristics of closed surfaces. |
Jan 8 |
awarded | Student |