[Haskell-beginners] Need better explanation of the 'flipThree' example in LYAH

[Olumide]

Hi Francesco,

Thanks as always for your reply. (I note that you've replied many of my 
questions.)

I have not replied because I've been thinking very hard about this 
problem and in particular about the way the do notation is desugared in 
this case and something doesn't sit right with me. (Obviously there's 
something I've failed to understand.)

Take for example

do
     a <- coin
     b <- coin
     return ([a,b])

This would desugar into

do a <- coin
     do b <- coin
         return [a,b]

and ultimately into

coin >>= ( \a ->
coin >>= ( \b ->
return [a,b] ))

Noting that m >>= f = flatten (fmap f m) , and recursively applying 
expanding >>= I get

flatten( fmap (\a -> coin >>= ( \b -> return [a,b] ) ) coin )

and

flatten( fmap (\a -> ( flatten( fmap ( \b -> return [a,b] ) coin ) ) coin )

Is this how to proceed and how am I to simplify this expression?

Regards,

- Olumide




On 21/08/18 02:00, Francesco Ariis wrote:
> Hello Olumide,
> 
> On Tue, Aug 21, 2018 at 01:04:01AM +0100, Olumide wrote:
>> My understanding of what's going on here is sketchy at best. One of several
>> explanations that I am considering is that all combination of a, b and c are
>> evaluated in (==Tails) [a,b,c] but I cannot explain how the all function
>> creates 'fuses' the list [f a, f b, f c]. I know that all f xs = and . map f
>> xs (the definition on hackage is a lot more complicated) but, again, I
>> cannot explain how the and function 'fuses' the list [f a, f b, f c].
> 
> Let's copy the relevant monad instance:
> 
>      instance Monad Prob where
>          return x = Prob [(x,1%1)]
>          m >>= f = flatten (fmap f m)
> 
> and desugar `flipThree:
> 
>      flipThree = coin       >>= \a ->
>                  coin       >>= \b ->
>                  loadedCoin >>= \c ->
>                  return (all (==Tails) [a,b,c])
> 
> 
> Now it should be clearer: `coin >>= \a -> ...something...` takes `coin`
> (Prob [(Heads,1%2),(Tails,1%2)]), applies a function (\a -> ...) to all
> of its elements, flattens (probability wise) the result.
> So approximately we have:
> 
>      1. some list ([a, b])
>      2. nested lists after applying `\a -> ...` [[a1, a2], [b1, b2]]
>      3. some more flattened list [a1, a2, b1, b2]
> 
> `\a -> ...` itself contains `\b ->` which cointains `\c ->`, those are
> nested rounds of the same (>>=) trick we saw above.
> At each time the intermediate result is bound to a variable (\a, \b
> and \c), so for each triplet we can use `all`.
> 
>> If I'm on the right track I realize that I'm going to have to study the list
>> the between list comprehensions and the do-notation in order how all the
>> return function create one Prob.
> 
> Indeed I recall working the example on paper the first time I read it:
> once you do it, it should stick!
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