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Re: Potluck dish -- a game
From: |
Ścisław Dercz |
Subject: |
Re: Potluck dish -- a game |
Date: |
Wed, 19 Feb 2014 09:52:55 +0100 |
Hi Mateusz!
Nice to meet someone mathematically-oriented! I'm delighted with your care
about the proper use of algebraic language and am always eager to be
corrected. However this time there is no need to do so -- it IS a monoid,
one of those you meet almost everywhere, i.e. a monoid of X->X
transformations. The identity map is it's neutral element (identity, unit)
and composition (X->X)x(X->X)->(X->X) is its action. Notice that it is not
a group since some transformations cannot be reverted -- e.g. the ones
produced with mk-remover.
The code is a bit messy and it mostly comes from a one-night hackaton, and
I will not be able to polish it until next weekend for sure.
I will translate the Polish comments. At the moment I can only provide you
with the last one, a neat mixture of Gombrowicz and Maslowska: "koniec
bomba a kto czytal ten ssie galy eurocwelom" -- the translation does not
seem trivial, perhaps something like "The end and the bomb, who read is an
euro-pansy cocksucker" -- however it might look a bit homophobic (as many
Polish curses do), perhaps you or Panicz could propose something better?
Greeting,
d.
PS I just found your answer, so now I guess we do agree on the monoid
thing?
As of monads I do understand that the class of endofunctors with
composition and identity as unit form a monoid, no doubt, that's the
construction I mentioned at the beginning.
But that's the monad-monoid relation in the direction opposite to what I
was asking for. And technically I am not sure whether "the endofunctor
definition" applies to Kleisli triplets (which you probably meant) that
easily, as it seems either broader, or that the "all told" part hides some
important constraints, like that the category C you consider has enough
means (perhaps binary products or some other universal constructions) to
represent "the new type" T(A) for any A in C... But again, I am an ignorant
so please do correct me.