
Re: Stone Cech
Posted:
Mar 22, 2013 10:43 AM


On Fri, 22 Mar 2013 05:34:37 0500, quasi <quasi@null.set> wrote:
>Butch Malahide wrote: >>Butch Malahide wrote: >>>Butch Malahide wrote: >>>>quasi wrote: >>>>>Butch Malahide wrote: >>>>>>quasi wrote: >>>>>>>quasi wrote: >>>>>>>>quasi wrote: >>>>>>>>>quasi wrote: >>>>>>>>>>Butch Malahide wrote: >>>>>>>>>>>David C. Ullrich wrote: >>>>>>>>>>>>Butch Malahide wrote >>>>>>>>>>>>>William Elliot wrote: >>>>>>>>>>>>>>David Hartley wrote: >>>>>>>>>>>>>>>William Elliot wrote: >>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>Perhaps you could illustrate with the five >>>>>>>>>>>>>>>>different one to four point point >>>>>>>>>>>>>>>>compactifications of two open end line >>>>>>>>>>>>>>>>segements. >>>>>>>>>>>>>>> >>>>>>>>>>>>>>>(There are seven.) >>>>>>>>>>>>>> >>>>>>>>>>>>>>Ok, seven nonhomeomophic finite Hausdorff >>>>>>>>>>>>>>compactications. >>>>>>>>>>>>> >>>>>>>>>>>>>How many will there be if you start with n segments >>>>>>>>>>>>>instead of 2? >>>>>>>>>>>> >>>>>>>>>>>>Surely there's no simple formula for that? >>>>>>>>>>>> >>>>>>>>>>>> ... >>>>>>>>>>> >>>>>>>>>>> ... >>>>>>>>>>> >>>>>>>>>>>I wasn't necessarily expecting a *complete* answer, >>>>>>>>>>>such as an explicit generating function. Maybe someone >>>>>>>>>>>could give a partial answer, such as an asymptotic >>>>>>>>>>>formula, or nontrivial upper and lower bounds, or a >>>>>>>>>>>reference to a table of small values, or the ID number >>>>>>>>>>>in the Encyclopedia of Integer Sequences, or just the >>>>>>>>>>>value for n = 3. (I got 21 from a hurried hand count.) >>>>>>>>>> >>>>>>>>>>For n = 3, my hand count yields 19 distinct >>>>>>>>>>compactifications, up to homeomorphism. >>>>>>>>>> >>>>>>>>>>Perhaps I missed some cases. >>>>>>>>>> >>>>>>>>>I found 1 more case. >>>>>>>>> >>>>>>>>>My count is now 20. >>>>>>>> >>>>>>>>I found still 1 more case. >>>>>>>> >>>>>>>>So 21 it is! >>>>>>>> >>>>>>>>But after that, there are no more  I'm certain. >>>>>>> >>>>>>>Oops  the last one I found was bogus. >>>>>>> >>>>>>>So my count is back to 20. >>>>>> >>>>>>Hmm. I counted them again, and I still get 21. >>>>>> >>>>>>4 3component spaces: OOO, OO, O, . >>>>>> >>>>>>7 2component spaces: OO, O, O6, O8, , 6, 8. >>>>>> >>>>>>10 connected spaces: O, , 6, 8, Y, theta, dumbbell, and >>>>>>the spaces obtained by taking a Y and gluing one, two, or >>>>>>all three of the endpoints to the central node. >>>>> >>>>>Thanks. >>>>> >>>>>It appears I missed the plain "Y", but other than that, >>>>>everything matches. >>>>> >>>>>So yes, 21 distinct types. >>>> >>>> For n = 4 I get 56 types. If I counted right (very iffy), >>> >>> Found two more. Never mind! >> >>And now I get 61. The hell with it. > >Ullrich predicted it (hopeless squared).
Hey, you guys are making progress! Now all we need is an f such that f(2) = 7, f(3) = 21 or whatever it was, and f(4) = 61. I doubt there's more than one such f...
> >For small n, say n < 10, it might be feasible to get the counts >via a computer program, but my sense is that the development of >such a program would be fairly challenging. If I get a chance, >I may give it a try.
My prediiction is that anything remotely resembling brute force will take forever to run, even in a compiled language. Just enumerating all the possible equivalence relations on 2n points is going to take a long time; now a brute force check whether _one_ pair of equivalence relations defines homeomorphic spaces will also take a long time, and there will be N(N1)/2 such checks, where N is huge even for small n.
But don't let me stop you. I'd start in Python, which would be much too slow  the point being that Python is so easy to _write_ that it would be the quickest way to get the algorithm straight. I'd test the Python code for n = 2 and 3 maybe; for n = 3 it would take forever. Once I had a correct algorithm I'd rewrite it in a compiled language. Then ask for the answer for n = 4 and maybe take a trip somewhere while waiting for the answer...
> >quasi

