[Haskell] Monad transformer question

[Chris Kuklewicz]

Quick comments below

Cyril Schmidt wrote:
> Working on a Monte-Carlo simulation where I have to
> calculate the values of a certain function on the given set of inputs,
> I noticed that some of the input variables change for every iteration,
> while others do not.
> 
> To give a simple example, let's suppose I have a function
> 
> f a1 a2 p = a1*a2 + p
> 
> and I have to get its values for
> [ (a1,a2,p) | a1 <- [0.1,0.2], a2 <- [0.1,0.2], p <- [0..9] ]
> 
> For efficiency, I want to pre-calculate (a1*a2) for each pair of a1 and a2,
> and then calculate f for each p.
> 
> (The real function is far more complicated, but the idea is the same:
> pre-calculate all that depends on seldom-changing variables, and then run
> the rest of the iterations using the pre-calculated value).
> 
> I built my function f as a Reader monad because my real f has many
> input parameters, so it is handy to have them packed in the Reader's
> environment.
> There are two Readers, in fact: the first environment contains the seldom-
> -changing variables, the second contains the variable that changes often.
> 
> My code looked as follows.
> 
>> {-# OPTIONS -fno-monomorphism-restriction #-}
>> module Main where
>> import Control.Monad.Reader
> 
> The variable that changes most often:
> 
>> data Inner = Inner { p1 :: Double }
> 
> The variables that change seldom
> 
>> data Outer = Outer { a1 :: Double,
>>                     a2 :: Double }
> 
> Function precalc pre-calculates (a1*a2)
> 
>> precalc = do
>>    a1 <- asks a1
>>    a2 <- asks a2
>>    let r = {-# SCC "r" #-} a1*a2
>>    return r

'precalc is of type Reader Outer Double

> 
> Function f in monadic form:
> 
>> f :: Reader Outer (Reader Inner Double)

Here 'f' is a monadic computation of
instance Monad (Reader Outer) ...
that returns a value of type (Reader Inner Double)

>> f = do
>>    r <- precalc

This acts as a closure to cache the 'precalc' computation value of Double

>>    return $ do { p1 <- asks p1
>>                ; let s = {-# SCC "s" #-} r+p1
>>                ; return s }
> 
> Function runf runs f over all values of p:
> 
>> runf (a1,a2) = do
>>    let reader  = runReader f $ Outer a1 a2

Here 'reader' is the cached result of running 'f' and is a (Reader Inner Double).

>>        results = map (runReader reader) [ Inner { p1 = x } | x <- [0..9] ]
>>    putStrLn $ "a1 = "++show a1++", a2= "++show a2++", results = "
>>               ++show results
> 
> The main function
> 
>> main = mapM_ runf [(a1,a2) | a1 <- [0.1,0.2], a2 <- [0.1,0.2]]
> 
> This all works fine; the profiler shows that the (a1*a2) calculation is
> performed
> exactly 4 times, while addition, just as expected, 40 times.
> 
> I noticed that f
> f :: Reader Outer (Reader Inner Double)
> can be implemented using monad transformer:
> f' :: ReaderT Outer (Reader Inner) Double
> 
> The only difference in the implementation is that f' uses lift instead of
> return:
> 
>> f' :: ReaderT Outer (Reader Inner) Double

f' is a ReaderT with is a Monad instance via
instance (...) =>  Monad (ReaderT Outer (Reader Inner)) Double ...
So f' is a Monad that computes a value of type Double.

>> f' = do
>>    r <- precalc
>>    lift   $ do { p1 <- asks p1
>>                ; let s = {-# SCC "s" #-} r+p1
>>                ; return s }
>>
>> runf' (a1,a2) = do
>>    let reader  = runReaderT f' $ Outer a1 a2

This forms the same (Reader Inner Double) type for 'reader' as before, but not
by computing any part of f'.  It just packages f' and (Outer a1 a2).  It turns
f' into a different Monad that computes a value of type Double and calls it
'reader'.

>>        results = map (runReader reader) [ Inner { p1 = x } | x <- [0..9] ]
>>    putStrLn $ "a1 = "++show a1++", a2= "++show a2++", results = "
>>               ++show results
> 
> 
> However similar they look, f and f' have very different behaviour (their
> results are the same, of course).
> 
> When I use runf' instead of runf, the profiler shows that precalc is
> invoked 40 times,
> which means that all the benefits of pre-calculating (a1*a2) are gone. (In
> the real application, I pre-calculate a much more complicated and
> expensive expression, that's why it matters).
> 
> I am curious why this happens. As far as I can see, the lift function of
> ReaderT is the same as return of Reader, and the >>= in Reader and ReaderT
> are pretty similar to each other, so why is the behaviour different?
> 
> This is a question of a purely theoretical significance for me; it does
> not hinder my work in any way. Still, I would greatly appreciate any
> ideas.
> By the way, I am using GHC 6.4.2 on Windows.
> 
> Kind regards,
> 
> Cyril
> 
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