bio | website | cs.ox.ac.uk/people/… |
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location | Oxford, United Kingdom | |
age | 50 | |
visits | member for | 3 years, 7 months |
seen | yesterday | |
stats | profile views | 1,713 |
Jun 27 |
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dual problem of SDP
this looks like a standard exercise in the duality theory of SDPs, apart from there seems to be a typo in the objective function (unless there is an extra assumption that $x$ is real). |
Jun 24 |
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Permutation Groups Containing non-commuting $p$-cycles
@GeoffRobinson: I guess it's due to Frieder's primitivity argument really needing prime length. Anyhow, it seems that Jordan's result can often tell you more about groups in your theorem. |
Jun 19 |
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Find a path that covers as many nodes as possible
@NickS : mere existence of a walk of $\ell$ steps is NP-hard, disregarding the edge lengths. |
Jun 19 |
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Find a path that covers as many nodes as possible
@NickS : calculating powers usually allows one to solve shortest paths problems; here it's more like longest path problem... |
Jun 19 |
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Find a path that covers as many nodes as possible
right, the poster means "walk". |
Jun 19 |
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Permutation Groups Containing non-commuting $p$-cycles
@FriederLadisch : just a remark that "generated" is important here. It seems that you might have primitive action on each block this way (and this is indeed a reduction to the primitive case, more or less). |
Jun 19 |
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Permutation Groups Containing non-commuting $p$-cycles
@GeoffRobinson: I'm not saying it might be a consequence. I meant to say that perhaps some part of it might be known to him, perhaps if you omit the corner cases of degree being close to p. Also, do you know a transitive imprimitive permutation group containing two non-commuting p-cycles? |
Jun 19 |
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Find a path that covers as many nodes as possible
if you are able to solve your problem exactly for graphs with each edge of length 1, and for $d$ sufficiently big, then you can solve Hamiltonian Path exactly. That is, your problem is NP-hard. |
Jun 19 |
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Find a path that covers as many nodes as possible
just think of all your edges having length 1 each. |
Jun 19 |
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Permutation Groups Containing non-commuting $p$-cycles
It won't be too surprising if Camille Jordan knew this result already (due to en.wikipedia.org/wiki/Jordan's_theorem_(symmetric_group)) |
Jun 19 |
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Compressing a system of linear equations
do you know the rank of $A$? |
Jun 19 |
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Find a path that covers as many nodes as possible
check out en.wikipedia.org/wiki/Hamiltonian_path_problem |
Jun 18 |
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Expected number of connected components in a random graph
IIRC, there is an answer in N.Alon & J.Spenser "The Probabilistic Method". |
Jun 17 |
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Kneser graphs eigenvalues
as the eigenvalues and their multiplicities are known, this seems to be a straightforward exercise. |
Jun 7 |
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A semisimple group ring
perhaps you might want to edit the question to make it more "watertight"... |
Jun 5 |
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System of quadratic complex equations
Thank you - I actually published in computer algebra. I read the question as a request for a symbolic procedure... |
Jun 4 |
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System of quadratic complex equations
just as computer algebra systems do. E.g. by specifying Thom encoding, i.e. the signs of the derivatives at the root. |
Jun 3 |
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System of quadratic complex equations
the question explicitly says "without using a numerical method". |
Jun 1 |
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Elements of finite order of $\mathrm{PGL}(n,\mathbb{Q})$
there seems to be a lot known about finite subgroups of $GL_n(\mathbb{Q})$, cf. e.g. references in ams.org/journals/proc/1997-125-12/S0002-9939-97-04283-4/… E.g. each of them is conjugate to one in $GL_n(\mathbb{Z})$. |
May 31 |
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System of quadratic complex equations
the conditions $|x_k-1|\leq 1$ make this a system of quadratic equations and (quadratic) inequalities. |