bio | website | cs.ox.ac.uk/people/… |
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location | Oxford, United Kingdom | |
age | 51 | |
visits | member for | 4 years, 1 month |
seen | 5 hours ago | |
stats | profile views | 1,865 |
Sep 22 |
revised |
nonnegativity conditions for a polynomial in two variables
more references, and an explanation |
Sep 22 |
answered | nonnegativity conditions for a polynomial in two variables |
Sep 22 |
comment |
Complexity of Untwisting Polygons
generating random polygons is a different story then; they care about the uniformity, etc... |
Sep 22 |
comment |
Complexity of Untwisting Polygons
I'm not sure it's not better to forget all the edges and just construct a polygon directly. |
Sep 22 |
awarded | Custodian |
Sep 22 |
reviewed | Close Algorithm to find the “optimal” path in a given graph |
Sep 22 |
reviewed | Approve Non-trivial bounds for polynomials at a fixed point |
Sep 22 |
answered | Complexity of Untwisting Polygons |
Sep 20 |
reviewed | Approve Changing Left Comodules into Right Comodules via the Antipode, and Comodule Tensor Products |
Sep 20 |
comment |
Fast computation of a Groebner basis - What is Possible
arguably the fastest Groebner bases implementations are www-salsa.lip6.fr/~jcf/Software |
Sep 3 |
comment |
Is this problem on weighted bipartite graph solvable in polynomial time or it is NP-Complete
Why do you talk about finding a sequence? The order you pick up your $v_i$'s does not matter. Better talk about a subset of A. Do you allow weights to be negative? If not, then you trivially will have to take the whole A. |
Sep 3 |
comment |
On the positivity of matrices
The vector space of matrices commuting with G has a basis {$I$,$\pi+\pi^4$,$\pi^2+\pi^3$}, where $\pi$ is the permutation matrix corresponding to the cyclic permutation (1,2,3,4,5). Thus $P$ will be a linear combination of these 3 matrices. |
Sep 3 |
comment |
On the positivity of matrices
$G$ is able to guarantee that $P_{ij}=p$, as the dimension of the centraliser algebra (i.e. the vector space of matrices commuting with every element in $G$) is 3, which can be e.g. checked directly, or by computing the permutation character of $G$. All these things I don't define may be found in books on permutation groups, e.g. P.Cameron's books.google.co.uk/books/about/… |
Sep 2 |
comment |
On the positivity of matrices
G depends on M only, and can be checked to be of size 10. |
Sep 2 |
comment |
On the positivity of matrices
One way to conclude that for $P$ being the average over an orbit all these $P_{ij}=p$, would be to note that $P$ commutes with all $\Pi\in G$. One can check that the algebra of all matrices commuting with $G$ coincides with the algebra of symmetric circulant matrices. Thus $P$ has the claimed form. |
Sep 2 |
revised |
On the positivity of matrices
added a link to the article I referred to |
Sep 2 |
comment |
On the positivity of matrices
@Hansen: I don't see how you would complete a proof "by hand" without being able to reduce the number of parameters significantly. |
Sep 2 |
comment |
On the positivity of matrices
here it is: journals.cambridge.org/article_S0305004100036185 |
Sep 2 |
awarded | Enlightened |
Sep 1 |
comment |
On the positivity of matrices
What I am saying is: take an offending P, and average it over G. This gives a symmetric circulant matrix which is still offending, and which only depends on one parameter, $p$. |