4,358 reputation
1031
bio website
location Scotland
age 32
visits member for 2 years, 6 months
seen 1 hour ago

I do research on groups, semigroups, and especially interactions between algebra and formal language theory. My non-research interests include combinatorics, elementary number theory and topology.


Mar
1
awarded  Yearling
Mar
1
awarded  Nice Answer
Jan
13
reviewed Approve suggested edit on Why - not how - do you solve Differential Equations?
Dec
19
revised What exactly are eigen-things?
deleted 6 characters in body
Aug
13
reviewed No Action Needed How do I find out the coordinates, interpolating across an angled line?
Jul
28
revised Structure theorem for finitely generated commutative $semi$groups.
Changed 'Tasmura' to 'Tamura' (I know one-letter edits are generally discouraged, but this seemed significant).
Jul
20
comment Mathematical writing: why should we not use the phrase “we have that”?
@ShreevatsaR: It's not just you, although I phrase things slightly differently, for example "We have $A=B$" I would read as "We have $A$ equal to $B$".
Jul
5
revised Whether $L=\{(a^m,a^n)\}^*$ is regular or not?
edited tags; edited title
Jul
4
comment Formal grammar for the language $L = \{w\in\{a,b\}^*,\,w=xx,\,x=a^nb^na^mb^m,\,n\ge0,\,m\ge0\}$
@Harald: Indeed this language isn't context-free. Probably it's easier to prove in this special case, but a generalisation is contained in my paper on poly-context-free groups (Theorem 3.12: the language $L$ in this question is essentially what I am calling $L^{(2,2)}$ - defined on p.19).
Jul
4
answered Whether $L=\{(a^m,a^n)\}^*$ is regular or not?
Jul
4
comment Whether $L=\{(a^m,a^n)\}^*$ is regular or not?
@Rudong: Your question still doesn't make sense. $L$ cannot be a regular language if you consider it as a 'language' over $A^*\times A^*$, because, as Boris has already pointed out, $A^*\times A^*$ is not a finite set.
May
26
answered Size of the image (rank) in $T_n$
May
26
revised Semigroup with exactly one left(right) identity?
specified that the functions act on the right.
May
26
comment Semigroup with exactly one left(right) identity?
@user4514: Ah, I should have specified which side the functions are acting on, as obviously it makes a difference. I'll fix this now.
May
15
comment Push Down Automata which has 2 stacks
Well, depending on how you define a PDA with two stacks, it can even be as powerful as a Turing machine!
May
11
reviewed Approve suggested edit on How does Tor and the Tensor functor interact?
May
11
reviewed Leave Open Derivative of a ratio of geometric series
May
10
comment Push Down Automata which has 2 stacks
What precisely do you mean by 'recognised by two PDA's'? And do you in fact mean by two PDA's, as in the body of the question, or by a PDA with two stacks, as in the title?
May
10
comment irreducible words in a semigroup
@Ewan: There is a lot of sense in distinguishing $0$ from $1$. In semigroups, $1$ is usually used to represent a multiplicative identity, while $0$ is used to represent a multiplicative zero, i.e. an element such that $x0 = 0x = 0$ for all $x\in S$. There is no mixing up of additive and multiplicative notation going on here.
May
7
awarded  Caucus