On May 29, 2006, at 9:33 AM, Dominic Steinitz wrote:
Hi Dominic - I hope it's ok for me to ask this question and I'm absolutely burning with curiosity to find out the answer... How did you know to write ((.).(.)) instead of (\f g a b -> f (g a b)) ?
Brian,
I can't remember. I certainly don't find it intuitive. I think it was discussed on the Haskell mailing list a long time ago.
It may have been this thread, where Tom Pledger points out that ($) makes a good "0-ary" case of the progression: http://www.haskell.org/pipermail/haskell/2003-July/012259.html But then it's only 3 years old, so clearly not a "long time ago" :) . -- Fritz PS: In my original posting on that thread, I said that abstracting this out as a fold of compositions over lists of compositions (foldr (.) id (replicate n (.)) was quite difficult for typing reasons. I now realize that Oleg probably does that sort of thing idly doodling on the back of a napkin while he chats on the phone with his other hand. My mistake. :) ---------- From the thread (quoting Tom Pledger quoting me): Tom Pledger wrote:
K. Fritz Ruehr writes: : | But Jerzy Karczmarczuk enlightened me as to the full generality possible | along these lines (revealing the whole truth under the influence of at | least one beer, as I recall). Namely, one can define a sequence of | functions (let's use a better notation now, with "c" for composition): | | c1 = (.) -- good old composition | | c2 = (.) . (.) -- my (.<) from above | | c3 = (.) . (.) . (.) | | c4 = (.) . (.) . (.) . (.) | | -- etc.
Nice!
There's also
c0 = ($)
which is clearer if you use 'non-pointfree' notation
... c2 f g x y = f (g x y) c1 f g x = f (g x) c0 f g = f g
- Tom