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visits | member for | 2 years, 3 months |
seen | Sep 17 at 20:12 | |
stats | profile views | 235 |
Jul 19 |
comment |
Vertex transitive and edge transitive and line graph
$G=K_3+K_{1,3}$ is a counterexample. The line-graph of $G$ is vertex transitive but $G$ is not edge transitive. |
Jul 18 |
awarded | Civic Duty |
Jul 12 |
awarded | Critic |
Jun 23 |
revised |
Estimate for the travelling salesman problem for balls inside a grid
added 83 characters in body |
Jun 23 |
revised |
Estimate for the travelling salesman problem for balls inside a grid
added 83 characters in body |
Jun 23 |
answered | Estimate for the travelling salesman problem for balls inside a grid |
Jun 23 |
comment |
Estimate for the travelling salesman problem for balls inside a grid
I see... so then it might be a good idea to edit the question and replace "path" by "walk", as a "path" in a graph usually means that it visits no vertex more than once. |
Jun 20 |
comment |
Estimate for the travelling salesman problem for balls inside a grid
Regarding question 2), which paths do you consider? What is a "path", actually, by your definition? |
Jun 20 |
comment |
Estimate for the travelling salesman problem for balls inside a grid
So, by "ball", you mean the ball in the $\ell_1$-norm, that is, a square with sides of angles $\pi/4$ with respect to the axes? |
Jun 19 |
answered | Angle of a regular simplex |
Jun 17 |
answered | A bound on a set |
Jun 17 |
comment |
A bound on a set
@Joybangla As GH from MO correctly noticed, I misread the question and assumed that $\alpha_i \in \{-1,1\}$. |
Jun 16 |
comment |
A bound on a set
Yes, Sperner Theorem gives the upper bound ${n \choose \lfloor n/2\rfloor}$. For $\alpha>2$, one can use the generalization to families of subsets with no chains of length $r$, where $r=\lceil\alpha/2\rceil+1$. See en.wikipedia.org/wiki/Sperner%27s_theorem#No_long_chains |
Jun 13 |
comment |
Hiding $k$ disks inside a larger disk
And for four discs, there are a continuum many configurations where the integral is $\pi +4$. |
Jun 12 |
comment |
Hiding $k$ disks inside a larger disk
Now for two such touching discs, the integral is $\pi + 2$, and for three discs like on the picture the integral is $\pi + 3$. Asymptotically, and approximately, it seems you want to pack the discs into another disc of smallest diameter (and thus perimeter). |
Jun 11 |
answered | Hiding $k$ disks inside a larger disk |
May 29 |
awarded | Yearling |
May 20 |
revised |
Wait time to grid network disconnection with failing edges
now considering also separators formed by two noncontractible curves |
May 20 |
comment |
Wait time to grid network disconnection with failing edges
You are absolutely right, I missed those disconnected disconnectors; they should be included in the calculations. Fortunately, we have to consider only pairs of boundary curves. Since we are on the torus, at least one of the components of every disconnected subgraph has at most two boundary components (either one contractible curve of a pair of parallel noncontractible curves). |
May 19 |
revised |
Wait time to grid network disconnection with failing edges
added the second sentence, corrected the upper bound |