bio | website | |
---|---|---|
location | Scotland | |
age | 32 | |
visits | member for | 4 years |
seen | Jan 13 at 13:40 | |
stats | profile views | 867 |
Research Fellow at the University of St Andrews.
I'm currently working on various things to do with interactions between semigroups and automata.
Jan 16 |
awarded | Scholar |
Jan 16 |
accepted | Status of an open problem about semilinear sets |
Jan 16 |
comment |
Status of an open problem about semilinear sets
Thanks very much, this is exactly the sort of thing I was looking for. This not only confirms that the problem is (at least as far as the authors of the paper are aware) still open, but gives a potential alternative approach to determining whether or not it is decidable. |
Jan 2 |
comment |
Status of an open problem about semilinear sets
Thanks, I haven't had time to look at it in detail yet, but since it doesn't seem to say anything about stratification, I'm not sure how helpful it will be. |
Jul 4 |
awarded | Yearling |
Oct 2 |
awarded | Nice Question |
Jun 15 |
comment |
An extension of Lagrange's theorem to semigroups?
(I was typing that comment quite slowly, so didn't see the two preceding comments until afterwards.) |
Jun 15 |
comment |
An extension of Lagrange's theorem to semigroups?
That's a good example. It's already clear from considering finite monogenic semigroups that the order of a subsemigroup of a 'Smarandache Semigroup' doesn't have to divide the order of the semigroup, but this example shows that nothing I would call an 'analogue of Lagrange's theorem' holds. |
Jun 10 |
comment |
A semigroup with the property that $x^n = a$ has at least one solution
Is $n$ a constant? |
Jun 8 |
comment |
Certain type of regular languages
@Dylan: Thanks for fixing it! I think I used to know about that, but I haven't written anything on mathoverflow for a long time, so I forgot. |
Jun 7 |
comment |
Certain type of regular languages
Sorry about the LaTeX failure in the second line. I don't know what that's about. |
Jun 7 |
answered | Certain type of regular languages |
Mar 25 |
awarded | Enthusiast |
Mar 19 |
comment |
Question about $\omega$-regular languages
If I could see the specific example you are looking at, I might have more to say. In complete generality I could only suggest the approach I would probably take. (I don't know much about $\omega$-regular languages, but in general I prefer working with automata to other representations of languages.) |
Mar 17 |
comment |
Question about $\omega$-regular languages
I think it's easy, given a Büchi automaton, to describe the finite prefixes of words accepted by the automaton. So perhaps you could first find a Büchi automaton accepting your language? (I think it's straightforward to do that from the $\omega$-regular expression.) |
Mar 2 |
comment |
Mapping from a finite index subgroup onto the whole group
@Ben: Ah yes, of course! Thanks. |
Mar 2 |
comment |
Mapping from a finite index subgroup onto the whole group
Nice! Do you happen to know of an example if we remove the requirement that $H$ has finite index in $G$? I rather expect it's possible then, but I haven't thought of an example yet. |
Mar 2 |
comment |
Mapping from a finite index subgroup onto the whole group
Hi Victor. I knew I must be missing something, but I can't believe it was something so obvious! I should have checked what you wrote more thoroughly. Anyway, I deleted my answer, since it contributes nothing and so it's better for the question to still have 0 answers. |
Feb 27 |
comment |
Non-isomorphic groups with the same oriented Cayley graph
Oh, you're right, sorry! I guess I thought you meant they were the only choices because you said 'the resulting Cayley graph is going to be...', which looked as if you were saying it would always be of that form. (By the way, I don't agree that there is 'not much choice' of 'mag' generating set for your $H$, or that it has to be symmetric. Consider $S = \{a,ab\}$ for example, where $a$ and $b$ are the generators of the factors.) |
Feb 27 |
comment |
Non-isomorphic groups with the same oriented Cayley graph
Yes, it is for undirected graphs. |