169 reputation
6
bio website cynic.cc/blog
location Sao Paulo, Brazil
age 38
visits member for 3 years, 3 months
seen 2 days ago

Computer Scientist, Free Software Advocate, Wannabe Mathematician, attracted by beautiful Typography.

I am the current maintainer of the algorithms bundle for LaTeX, which includes the packages algorithm and algorithmic. It is openly developed at github.

By the way, my github account also features many of my projects (and I am also collaborator on other repositories).

I am also a proud Debian Maintainer


Dec
18
awarded  Caucus
Mar
2
revised How can I solve equations like $A(z)=1+z+(z+z^2)A(z)$?
Removed thanks, reworded for introduction of the context before the question, instead of after the question.
Mar
2
suggested approved edit on How can I solve equations like $A(z)=1+z+(z+z^2)A(z)$?
Dec
22
accepted Graph on the cover of Bollobás's “Combinatorics”
Dec
21
comment Graph on the cover of Bollobás's “Combinatorics”
BTW, all of the answers are superb and I don't know what to choose as "the" answer to this question...
Dec
21
comment Graph on the cover of Bollobás's “Combinatorics”
@ThomasAndrews, Great. Translating from your description to wikipedia's notation: for any two adjacent vertices, there is exactly 1 path of length 1 between then, which means that any 2 adjacent vertices have exactly 1 neighbor in common (in wikipedia's notation, $\lambda = 1$). For any 2 non-adjacent vertices, there is exactly 2 (vertex disjoint) paths of length 2 between them, which means that 2 non-adjacent vertices have 2 neighbors in common (in wikipedia's notation, $\mu = 2$). The graph is regular of degree $k = 4$ with $v = 9$ vertices and is, therefore, $sgr(9, 4, 1, 2)$.
Dec
20
asked Graph on the cover of Bollobás's “Combinatorics”
Dec
15
accepted Showing that the infinite grid is Eulerian
Dec
15
comment Showing that the infinite grid is Eulerian
Your explanation of what you meant with the ellipsis is helpful in understanding what you meant. But there is one problem, though, if I were to transform this into a proof by induction: after I traversed the whole plane, how do you define "the last 3 or 4 or 5" last steps? Perhaps I misunderstood you?
Dec
15
comment Showing that the infinite grid is Eulerian
The definition of Eulerian given in the book for infinite graphs is that you simply have a path that extends from its two end vertices indefinitely, is allowed to pass through any vertex any number of times, but each edge only a finite number of times.
Dec
15
comment Showing that the infinite grid is Eulerian
@GerryMyerson, I didn't get what Eppstein said about an even sided diamond shaped region. That would include the particular case of squares, but I can't even see the case for squares! :) OTOH, I am following (again) his hint of walking the edges of the graph such that the endpoints of a visited region of the graph always lie on its border and it has been working so far for a drawing... I am distilling some observations, but I don't know if I will be able to transform them into an algorithm to cover the whole graph. Let's see how things progress now. :)
Dec
15
comment Showing that the infinite grid is Eulerian
How do you get from the 5th step to the 6th step? In the 5th step, let's call the two extremes $u$ (at the left) and $v$ (at the right). The only way that I see that jump is to get $v$, visit 3 sides of the upper square and then finish there at the top. Is this right? You are basically just walking with one of the endpoints, right? But in your last figure, how do you get of the dead-end?
Dec
15
awarded  Scholar
Dec
15
awarded  Student
Dec
15
asked Showing that the infinite grid is Eulerian
Nov
16
suggested rejected edit on Burnside's Lemma, applications
Nov
14
awarded  Editor
Nov
14
comment How to show that $1 \over \sqrt{1 - 4x} $ generate $\sum_{n=0}^\infty \binom{2n}{n}x^n $
It would be a good thing to note that, while we are using the same variable names (namely $n$) both on the left and on the right side of the equations, for one to derive that $na = 2$ and $n(n-1)a^2 = 12$, as you did, you have to consider that, on the RHS, when setting $x = 0$, you need the coefficients to be already known to be constant---not involving the variable $n$ of the LHS.
Nov
14
revised Too old to start math
Fixed leftover typo for consistency. (As edits can't be this small, I made some other cosmetic changes.)
Nov
14
suggested approved edit on Too old to start math