alex @alexthecamel@schelling.pt
queering the randos in my dms asking for nudes/ randos in my dms asking for math advice binary
twitter: pawnofcthulhu
Joined Dec 2022
@johncarlosbaez
when all the rectangles are oriented the same way you could also think of these as networks of unit ohm resistors?
@johncarlosbaez
in this particular case the side length must be 0
but for more general configurations which you can't get from guillotine cuts it seems like a useful idea
@johncarlosbaez
actually I guess for the purposes of the "divide square into similar rectangles" problem we could just allow the side length of the interior rectangle to be negative
@johncarlosbaez
roughly the idea is to treat vertical lines as vertices and boxes as edges. thing is, this shape you had before gives the tetrahedron with this rule, but there are 2 such shapes.
so to do this in more generality we'd need to think of these as directed graphs and add some more criteria?
@johncarlosbaez
the first nontrivial graph is the tetrahedron which explains the 90-91 difference
@johncarlosbaez
(oops schoder numbers run 1,2,6,22,90; typo)
@johncarlosbaez
(i'd had in my head the problem of counting 3-connected planar graphs for a while, just learning someone else had done something like it helped me put the pieces together)
@johncarlosbaez
"build": glue 2 graphs along an edge and remove this edge
"trivial": anything you can build out of k-cycles and the graph with 2 vertices joined by k edges (k>=2)
@johncarlosbaez
oh i just accidentally ran into these the number of rooted 2 connected planar graphs with n edges (allowing multiple edges between vertices) runs as G_2=1,2,6,22,91 whereas the schroder numbers run 1,2,6,22,91.
roughly you can build 2-connected planar graphs out of 3-connected planar graphs and the schroder numbers count graphs which are "trivial" in this sense
@sgf @ppscrv @johncarlosbaez @gregeganSF @rogers
Hmm, if you know the multiplication operation you've got here is associative you've got an idempotent monoid generated by 3 elements, which must be a surjective image of the free such monad, right?
@ppscrv @johncarlosbaez @gregeganSF @rogers @sgf
I guess a useful thing is we now can work out what canonical form a term *should* reduce to
@ppscrv @johncarlosbaez @gregeganSF @rogers @sgf
I've been thinking of this as a "canonical form" of say
$f(a_1,...,a_{n-1})a_na_{n-1}g(a_1,...,a_{n-2},a_n$ or similarly with f and g both missing the same variable
and wondering if when you left-multiply or right multipy this by stuff you still get something of this form
@johncarlosbaez @gregeganSF @rogers @sgf
why this is all there is to know about a term I'm not sure.
but assuming this is true if we're only looking at elements made up of these terms there's not that much structure, we're stuck with the set of initial and set of final substrings we start with but I suspect we can recombine them more or less how we like.
@johncarlosbaez @gregeganSF @rogers @sgf
the trick is if you look at the initial substring until you've seen all n elements
so $t = f(a_1,...,a_{n-1})a_n g(a_1,...,a_n)$ say
and both what $a_n$ is and what $f(a_1,...,a_{n-1})$ is as a term containing (n-1) generators don't change with our rewrite rules (ok, f might change but it'll change to something equivalent)
and you do the same reasoning with terminal substrings
and build things up inductively, and you get the same formula as the OEIS
@robinhouston @gregeganSF @sgf @johncarlosbaez @rogers
I'm as surprised as you!
@robinhouston @rogers @gregeganSF @johncarlosbaez @sgf
still not exactly sure how to do this systematically i think i've gotten to "looking at the formula for the number of elements of the monoid i can find this many elements and i think i can prove they're different but i'm not sure why I can reduce everything to them"
@robinhouston @rogers @gregeganSF @johncarlosbaez @sgf
ok pulling the example of square free words of wikipedia I'm guessing something like
(012)1021201(210)->
01(20121021)(20121021)0->
01201210210->
01210210->01210
@johncarlosbaez @gregeganSF @rogers @sgf
F(3) seems probably computable with a lot of work, beyond that I'm not sure
@johncarlosbaez @gregeganSF @rogers @sgf
guess is 33538195 for n=3
and \((2*(2^p-1)^2+p^2+1))
where \(p=\prod_{i=1}^n i^{2^{n-i-1}}\) more generally
@johncarlosbaez @gregeganSF @rogers @sgf
if the elements of the free idempotent rig on n elements work how I *think* they do I suspect you can get an exact formula for the number of elements made up of terms containing all the generators (the equivalent of ab,ba,aba,bab)
queering the randos in my dms asking for nudes/ randos in my dms asking for math advice binary
twitter: pawnofcthulhu
Joined Dec 2022
