bio | website | |
---|---|---|
location | ||
age | 26 | |
visits | member for | 2 years, 7 months |
seen | Aug 26 at 20:42 | |
stats | profile views | 319 |
Jul
19 |
revised |
Unique Nash equilibrium games
one more reference given |
Jul
7 |
awarded | Enthusiast |
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 |