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location Fiji
age 37
visits member for 2 years, 5 months
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I'm a lecturer of mathematics at the University of the South Pacific. My research interests are in algebraic topology and metric geometry.


Oct
16
answered Why is Division harder than Multiplication?
Oct
15
answered Generalization of a certain riddle and ultrafilters (?)
Oct
15
answered Hausdorff Topological Space
Oct
14
comment N vs NP. Existence or Constructive.
I don't quite agree. There are plenty of important algorithms with polynomial time of degree 3 or 4. Such problems are hardly feasible for only moderately large inputs. Making computers faster usually does not improve things much since the size of problems tend to increase as well. Parallelism also doesn't help much since it does not reduce the degree of polynomials. The best justification for heuristic techniques is that they are effective for solving both polynomial time problems and other problems.
Oct
14
comment N vs NP. Existence or Constructive.
A few comments. Firstly, heuristic algorithms are justified regardless of a positive or negative answer to $P=NP$. Heuristic algorithms are used also for problems with polynomial solutions which are still too computationally expensive. Even if $P=NP$ we will certainly still use heuristics to more efficiently solve problems. Secondly, I'm not entirely sure but I don't think that P vs. NP can be undecidable. What is certain is that P vs. NP can only be undecidable if $P\ne NP$. Of course, the answer may simply be beyond human capacity. E.g., if polynomial algos for NP-complete probs are too long
Oct
14
comment N vs NP. Existence or Constructive.
@Durian The Clay institutes simply asks for a resolution of whether or not P is NP. Any proof, constructive or not, will qualify. Of course, very few people believe P=NP, so most likely the answer is $P\ne NP$, in which case, I think, your (slightly vague) question dissolves. In the odd case where it will be shown, even constructively, that P=NP it may very well be the case that no efficient polynomial solutions exist to some (most?) NP (-hard) problems. In such an event, there will be little practical value to the resolution. In any case, the P=NP problem is one of theoretical value (probably
Oct
14
answered Proving that if $p$ is a prime number then $gcd (p, (p-1)!) =1$
Oct
13
comment sequence ${1/n}$ converges in R (to 0) but fails to converge in the set of all positive real number. Why?
yes yes yes yes.
Oct
13
answered Can you have open intervals that are sequentially compact?
Oct
12
reviewed Approve suggested edit on Showing this metric space is complete
Oct
12
comment Showing this metric space is complete
what is $1/x_n$???
Oct
9
answered How do I show two things are not isomorphic
Oct
9
comment terminology: accumulation points, limit point, cluster point
Thanks @BrianM.Scott
Oct
8
comment Is Z2 is simple.
yes, of course.
Oct
8
comment Nomenclature for posets s.t. for all $x<y$, there exists $z$ with $x<z<y$.
the property you mention is named 'dense'.
Oct
8
answered Is Z2 is simple.
Oct
8
comment If $ y \in [0,1] \times [0,1]$, is $[0,1] \times [0,1]-y$ connected?
you consider the restriction since that gives you a counter example. The original unrestricted $f$ is a supposed homeomorphism between connected spaces. There is nothing a-priori wrong about that.
Oct
8
comment If $ y \in [0,1] \times [0,1]$, is $[0,1] \times [0,1]-y$ connected?
it would seem OP was asking about how to prove a particular space is (path) connected, not for the general proof that path-connectivity implies connectivity.
Oct
8
comment If $ y \in [0,1] \times [0,1]$, is $[0,1] \times [0,1]-y$ connected?
basically yes, but you try to explain (i.e., find a theorem) that makes this manifestly impossible. Intuitively, you are correct, being (path) connected is something that is preserves under homeomorphisms (i.e., it is a topological invariant).
Oct
8
answered If $ y \in [0,1] \times [0,1]$, is $[0,1] \times [0,1]-y$ connected?