# Hudak state emulation discussion - can you give me some idea?

6 messages
Open this post in threaded view
|

## Hudak state emulation discussion - can you give me some idea?

 Hi Everyone, I was going through the classic Hudak paper "Conception, Evolution, and Application of Functional Programming". Probably the most valuable part is towards the end where he talks about how state is represented in Haskell. On Page 48 of the PDF (page 406 of the ACM Computing survey volume), he gives the following equations "For expository purposes we would like to make the state as implicit as possible, and thus we express the result as a composition of higher-order functions. To facilitate this and to make the result look as much like the original program as possible, we define the following higher-order infix operators and functions": 1 f:=g =\s -> +fs(gs) 2? ?f;g =\s -> g(fs)=\s-+fs(gs) 3? ?f;g =\s -> g(fs) 4 goto f = f 5 f+?g=\s-+fs+gs 6 f
Open this post in threaded view
|

## Re: Hudak state emulation discussion - can you give me some idea?

 On Tue, 17 Mar 2009 14:32:24 +0530, Girish Bhat <[hidden email]> wrote: >Hi Everyone, > >I was going through the classic Hudak paper "Conception, Evolution, >and Application of Functional Programming". >Probably the most valuable part is towards the end where he talks >about how state is represented in Haskell. >On Page 48 of the PDF (page 406 of the ACM Computing survey volume), >he gives the following equations >"For expository purposes we would like to make the state as implicit >as possible, and thus we express the result as a composition of >higher-order functions. To facilitate this and to make the result look >as much like the original program as possible, we define the following >higher-order infix operators and functions": > > >1 f:=g =\s -> +fs(gs) > >2? ?f;g =\s -> g(fs)=\s-+fs(gs) > >3? ?f;g =\s -> g(fs) > >4 goto f = f > >5 f+?g=\s-+fs+gs > >6 f >7 if? p c = \s + (if (p s) then (c s) else s) > > > >Unfortunately I am unable to parse/understand what he is doing here. >My closure understanding too seems to be wonky. :) >Would someone be kind enough to translate each of the above into plain >english and explain how to read such eqns in the future? (FYI, there is a copy of the above-mentioned paper that doesn't require an ACM account available at http://www.cs.berkeley.edu/~jcondit/pl-prelim/hudak89functional.pdf.) Hudak is just defining a series of higher-order infix operators and functions.  (You have made some notational errors in the above-mentioned notation, so I am revising it to match the paper as below.)  A backslash denotes a lambda symbol; i.e., whatever immediately follows the lambda is a parameter in the following equation.  Specifically:   >1 f := g =\s -> f s (g s) In other words, the infix operator ':=' works between functions f and g such that lambda s -> f s (g s). >2 f ; g = \s -> g (f s) In other words, the infix operator ';' works between functions f and g such that lambda s -> g (f s). >3? ?f;g =\s -> g(fs) I couldn't find this equation on p. 406 of the volume; where did you find it? >4 goto f = f In other words, the function "goto" applied to f is defined as simply f. >5 f +' g = \s -> f s + g s In other words, the infix operator '+'' works between functions f and g such that lambda s -> f s + g s. >6 f <' g = \s -> f s < g s In other words, the infix operator '<'' works between functions f and g such that lambda s -> f s < g s. >7 if' p c = \s -> (if (p s) then (c s) else s) In other words, the function "if'" applied to functions p and c is such that lambda p c -> (if (p s) then (c s) else s). Hope this helps.... -- Benjamin L. Russell -- Benjamin L. Russell  /   DekuDekuplex at Yahoo dot com http://dekudekuplex.wordpress.com/Translator/Interpreter / Mobile:  +011 81 80-3603-6725 "Furuike ya, kawazu tobikomu mizu no oto." -- Matsuo Basho^
Open this post in threaded view
|

## Re: Hudak state emulation discussion - can you give me some idea?

 > (FYI, there is a copy of the above-mentioned paper that doesn't > require an ACM account available at > http://www.cs.berkeley.edu/~jcondit/pl-prelim/hudak89functional.pdf.) > > Hudak is just defining a series of higher-order infix operators and > functions.  (You have made some notational errors in the > above-mentioned notation, so I am revising it to match the paper as > below.)  A backslash denotes a lambda symbol; i.e., whatever Sorry about that. > >3? ?f;g =\s -> g(fs) > > I couldn't find this equation on p. 406 of the volume; where did you > find it? > Again my transcription error. > In other words, the function "if'" applied to functions p and c is > such that lambda p c -> (if (p s) then (c s) else s). > > Hope this helps.... > Thanks! It does. I think what threw me was that while there is enough redundancy in what he states for someone more clever than me, he would have explicitly stated before hand  that he was defining the operators [:=], [+'], ['if] etc.:) thanks again.
Open this post in threaded view
|

## Re: Hudak state emulation discussion - can you give me some idea?

 On Wed, 18 Mar 2009 13:02:34 +0530, Girish Bhat <[hidden email]> wrote: >>[...] >> >> Hope this helps.... >> >Thanks! It does. I think what threw me was that while there is enough >redundancy in what he states for someone more clever than me, he would >have explicitly stated before hand  that he was defining the operators >[:=], [+'], ['if] etc.:) But he did; specifically, on pages 405 (the previous page) to 406 of the volume, Hudak writes as follows: >For expository purposes we would like to >make the state as implicit as possible, and >thus we express the result as a composition >of higher-order functions. To facilitate this >and to make the result look as much like >the original program as possible, we define >the following higher-order infix operators >and functions(22): So he did in fact explicitly state beforehand that he was defining those operators. -- Benjamin L. Russell -- Benjamin L. Russell  /   DekuDekuplex at Yahoo dot com http://dekudekuplex.wordpress.com/Translator/Interpreter / Mobile:  +011 81 80-3603-6725 "Furuike ya, kawazu tobikomu mizu no oto." -- Matsuo Basho^