Re: [Haskell-beginners] Need better explanation of the 'flipThree' example in LYAH
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`!
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`! _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
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Francesco Ariis -
Olumide