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Today's Topics:
1. Re: Need better explanation of the 'flipThree' example in
LYAH (Olumide)
----------------------------------------------------------------------
Message: 1
Date: Wed, 19 Sep 2018 11:03:04 +0100
From: Olumide <50...@web.de>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Need better explanation of the
'flipThree' example in LYAH
Message-ID: <299b0121-ca10-25a2-a3ec-ee9aefa4d...@web.de>
Content-Type: text/plain; charset=utf-8; format=flowed
Thanks Francesco. You are right. I had one too many rbraces.
However from the definition
fmap f (Prob xs) = Prob $ map (\(x,p) -> (f x , p)) xs
and x = Head | Tails suggests to me that f ought to be a function like
any (==Heads) that applies to [x]
Therefore I've changed the desugared expression to
flatten( fmap (\a -> flatten( fmap ( \b -> return ( any (== Heads) [a,b]
) ) coin ) ) coin )
So that from the definition of coin, the inner expression
fmap ( \b -> return ( any (== Heads) [a,b] ) ) coin
expands to
fmap ( \b -> return ( any (== Heads) [a,b] ) ) Prob[ (Heads, 1%2) ,
(Tails, 1%2) ]
Detour: writing all this made me realize that (1) the lambda is applied
the lambda to all elements(?) of the Prob, and (2) the term a represents
each element of the outer Prob.
So that the above expression becomes
Prob $ [ (any (==Heads)[a,Head], 1%2) , (any (==Head)[a,Tails], 1%2) ]
writing all this made me realize that (1) the lambda is applied to all
elements(?) of the Prob, and that (2) the term a represents each element
of the outer Prob -- I think I am starting to see the
similarity/connection between the do-notation and list comprehensions.
Returning to the above expression, when the term a is Heads, we get
Prob [ (return any (==Heads)[Heads,Head], 1%2) , (return any
(==Head)[Heads,Tails], 1%2) ]
and because
return any (==Heads)[Heads,Head] = Prob[True,1%1] , and
return any (==Heads)[Heads,Tails] = Prob[True,1%1]
The above expression becomes
Prob [ (Prob[True,1%1] , 1%2 ) , ( Prob[True,1%1] , 1%2 ) ]
The double Prob makes it clear why flatten is needed.
I'll stop here and wait for your feedback.
BTW, I'm considering using GHC.Proof to assert that this sequence of
reductions is equivalent.
Regards,
- Oumide
On 17/09/18 01:09, Francesco Ariis wrote:
> Hello Olumide,
>
> On Mon, Sep 17, 2018 at 12:15:58AM +0100, Sola Aina wrote:
>> flatten( fmap (\a -> ( flatten( fmap ( \b -> return [a,b] ) coin ) ) coin )
>>
>> Is this how to proceed and how am I to simplify this expression?
>
> I get a slightly different final expression:
>
> flatten (fmap (\a -> flatten (fmap (\b -> return [a,b]) coin)) coin)
> -- notice the lack of `(` before the second `flatten`
>
> This could be rewritten as:
>
> flatMap (\a -> flatMap (\b -> return [a,b]) coin) coin
>
> -- flatMap :: (a1 -> Prob a2) -> Prob a1 -> Prob a2
> -- flatMap f x = flatten (fmap f x)
>
> which is a bit easier on the eyes. This expression cannot be simplified
> any further if we don't bring to the table the definition of `coin`!
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