bio | website | |
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visits | member for | 4 years, 4 months |
seen | 2 days ago | |
stats | profile views | 97 |
Sep 22 |
comment |
Characterization of a subset of [0,1] $II$
Given the possibility of excluding arbitrary countable subsets, I guess the restriction has to be that for $t\in T, t\neq 1$, the intersection $T\cap[t,t+\delta]$ must be uncountable for all $\delta>0$. |
Jan 29 |
answered | Is there a dense rational sequence of positive separation? |
Jan 16 |
comment |
An extension of the real semiring with multiple degrees of infinity
I still don't quite get the "dominance" property of divisibility; assuming $(a/b)*b$ should be $a$, your second example gives $2^*=1^*+2^*$, so that either $1^*=0$, or your semiring is not cancellable - are you ok with the latter? |
Oct 11 |
awarded | Caucus |
Oct 11 |
awarded | Constituent |
Jun 25 |
awarded | Yearling |
May 14 |
comment |
two boy scouts problems
Your original problem does sound solvable to me (essentially, you need a $(2n-1)$-edge coloring of $K_{2n}$). Is there an additional constraint (e.g. only one match of each sport per round)? |
Feb 14 |
comment |
Who knows this convex polytope?
No real answer, but I think you want six parallelepipeds, not four. |
Oct 26 |
answered | fixed points of system of quadratic equations |
Jun 18 |
comment |
Resources-Aware Combinatorial Game Theory
Just to clarify: the resources spent on moves are removed from the game? I'm asking because there is also the related version in which they go to the opponent, so that you can actually regain resources (and they don't a priori limit they number of moves). |
Feb 27 |
comment |
Affine space partition of a general set
Do you have any restrictions on the kind of covering you want (minimum number of spaces, say?) After all, every single point of $F_q^n$ is an affine subspace. |
Feb 6 |
comment |
Linear Algebra - Find an N-Dimensional Vector Orthogonal to A Given Vector
Are there any other constraints? Otherwise, let $a$ be the given vector. If $a_1=0$, then $(1,0,0,\dots)$ works; if $a_2=0$, then $(0,1,0,0,\dots)$ works; otherwise, $(a_2,-a_1,0,0,\dots)$ works. |
Jan 11 |
comment |
Optimization of a Specific Polynomial
Following up on Pietro's comment: Suppose wlog that $c_1,c_2,c_3\neq 0$. Consider the behaviour of $f$ along the line $\{(x,x,x,0,\dots,0)|x\in\mathbb{R}\}$: It reduces to $g(x) = c_1c_2c_3x^3+h(x)$, where $h$ is quadratic, so $g$, and therefore also $f$, is unbounded in both directions. |
Oct 10 |
comment |
Is there an upper bound to strongly connected components
Sorry, cannot edit my comment - that should of course be $n(n-1) + m(m+1)\geq 2e$. |
Oct 10 |
awarded | Commentator |
Oct 10 |
comment |
Is there an upper bound to strongly connected components
If $e$ is greater than that, you can still get up to $n-m$ SCCs, where $m$ is the least integer such that $n(n-1)+m(m+1)\leq 2e$, by starting with the maximal acyclic graph and adding all back edges into a single SCC of minimal size. |
Jun 23 |
comment |
What the heck is the Continuum Hypothesis doing in Weibel's Homological Algebra?
And even "essentially countable" structures (at least as I understand the term) are not entirely safe, see for example Friedman's paper arxiv.org/abs/math/9811187 |
May 25 |
comment |
General integer solution for $x^2+y^2-z^2=\pm 1$
oops - yes, of course. Thanks! |
May 25 |
comment |
n-partite n-clique
You would need at least some constraint on the number of isolated vertices, wouldn't you? Currently, the graph with no edges at all is a counterexample. |
May 25 |
comment |
General integer solution for $x^2+y^2-z^2=\pm 1$
Looks good. One way of getting there is setting $a:=x+z, b:=x-y$ to obtain $ab=(1+y)(1-y)$, so that there must be $r,s,t,u$ with $rs=a,tu=b,1+y=rt,1-y=su$. |