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!