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Today's Topics:
1. Need better explanation of the 'flipThree' example in LYAH
(Olumide)
2. Re: Need better explanation of the 'flipThree' example in
LYAH (Francesco Ariis)
----------------------------------------------------------------------
Message: 1
Date: Tue, 21 Aug 2018 01:04:01 +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: [Haskell-beginners] Need better explanation of the
'flipThree' example in LYAH
Message-ID: <83c66b35-f934-4a09-4274-24585cb0d...@web.de>
Content-Type: text/plain; charset=utf-8; format=flowed
Dear List,
I'm trying to understand the following example from LYAH
import Data.List (all)
flipThree :: Prob Bool
flipThree = do
a <- coin
b <- coin
c <- loadedCoin
return (all (==Tails) [a,b,c])
Where
import Data.Ratio
newtype Prob a = Prob { getProb :: [(a,Rational)] } deriving Show
and
data Coin = Heads | Tails deriving (Show, Eq)
coin :: Prob Coin
coin = Prob [(Heads,1%2),(Tails,1%2)]
loadedCoin :: Prob Coin
loadedCoin = Prob [(Heads,1%10),(Tails,9%10)]
The result:
ghci> getProb flipThree
[(False,1 % 40),(False,9 % 40),(False,1 % 40),(False,9 % 40),
(False,1 % 40),(False,9 % 40),(False,1 % 40),(True,9 % 40)]
See http://learnyouahaskell.com/for-a-few-monads-more#making-monads.
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].
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.
Regards,
- Olumide
------------------------------
Message: 2
Date: Tue, 21 Aug 2018 03:00:57 +0200
From: Francesco Ariis <fa...@ariis.it>
To: beginners@haskell.org
Subject: Re: [Haskell-beginners] Need better explanation of the
'flipThree' example in LYAH
Message-ID: <20180821010057.cyu6pg6mdb5fq...@x60s.casa>
Content-Type: text/plain; charset=us-ascii
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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