Lets say I have the following:
f :: b -> a -> b
x :: b
l :: [a]
and
foldl' f x l
runs in constant space. That is f is suitably strict.
Now consider if I have:
f2 :: b -> a -> Maybe b
f2 x y = if (pred x y) then Just $! (f x y) else Nothing
will
foldM f2 x l
reliably run in constant space? Or is there something else I need to do to ensure I have both constant space but still the short circuiting behaviour of Maybe?
(Note whilst I've asked this question about Maybe, I actually want to do this with Either, but I suspect the approach is similar)
In the library source code foldM is defined as foldlM, which in turn is defined as
foldlM :: (Foldable t, Monad m) => (b -> a -> m b) -> b -> t a -> m b
foldlM f z0 xs = foldr c return xs z0
where c x k z = f z x >>= k
Assuming, c x k z = f2 z x >>= k, let's see what happens when we call it. To see if it's constant space or not, we will only reduce the expressions by applying the topmost function without reducing the subexpressions.
foldlM f2 z0 (x:xs)
=
foldr c return (x:xs) z0
=
c x (foldr c return xs) z0
=
f2 z0 x >>= foldr c return xs
Since >>= is strict on the first arg, we evaluate f2 z0 x first. If that returns Nothing, we ignore the rest (short-circuiting, as you mentioned). If that returns Just y, we have
Just y >>= foldr c return xs
=
foldr c return xs y
and we are ready for the next loop.
This did not cause our term to grow, so it looks like it runs in constant space (provided f2 keeps the size of y constant, of course).
So I'm playing with the Cantor pairing function, and trying to follow the wikipedia formulas as close as possible.
type N = Int
toCantor :: (N, N) -> N
fromCantor :: N -> (N, N)
toCantor (x, y) = (x + y) * (x + y + 1) `div` 2 + y
type N so I can easily change to Integer later (some of the intermediate calcs will get big).
uncurried form, partly to follow wp, partly so (fromCantor . toCantor) === id and (toCantor . fromCantor) === id.
Again following wp:
fromCantor z = (x, y) where
x = w - y
y = z - t
t = (w * w + w) `div` 2
w = floor $ (sqrt (fromIntegral (z * 8 + 1)) - 1.0) / 2.0
This works and everything but gee that formula for w is ugly!
It needs all those parens because I've got a formula nested inside a function call and a loose-binding (-) nested inside a tight-binding (/).
(And both those operators are non-commutative, so I must be careful.)
Q 1. Is there a way to make that formula prettier/pointfree?
I see the formula starts from z and builds outwards. So I can pipeline the calculation:
(.|) :: a -> (a -> b) -> b -- pipelining
infixl 0 .|
x .| f = f x -- aka flip ($)
wP :: N -> N -- w with Pipelining
wP z = z
.| (* 8)
.| (+ 1)
.| fromIntegral
.| sqrt
.| subtract 1.0
.| (/ 2.0)
.| floor
Is this style prior art? Is (.|) a good way to spell that operation -- I think I've seen it as a Lens operator(?)
Q 2. I've (deliberately) laid that out in pseudo-monad style. Could it actually be a do block?
First I need a monad. I could use Maybe or (Either e) -- which would be a Good Thing because several of those functions are partial, and I ought to be using a safe version.
Then instead of z I'd put return z.
But the binding goes the wrong way round. Instead of Monad m => m a -> (a -> m b) -> m b, I want Monad m => m a -> (a -> b) -> m b. That looks like an fmap, but flipped.
I could apply some sort of lifting to the functions/operators, but that then obscures the arithmetic with monad plumbing.
Rebindable syntax?
Your operator .| already exists in Data.Function as &. To make it pointfree, you can either use >>> from Control.Arrow, or invert the order of everything and just use .. For Monad m => m a -> (a -> b) -> m b, you want <&>, from Data.Functor.
This could be a way of constructing
Is there a (ideally standard) way of accomplishing
f :: Int -> Int
f x = 2*x
g :: Int -> String
g x = show x
h = (f, g)
fmap h 5 -- results in: (10, "5")
In general, for functions going from A->T_i for some variable types T_i and a fixed type A, I think this would just be a simplification of a BiFunctor, at lease for a 2-tuple of 1-argument functions - it would be great to see a generalization going beyond 2-tuples.
You could use uncurry (&&&), as follows:
> import Control.Arrow
> f :: Int->Int ; f x = 2*x
> g :: Int->String ; g x = show x
> h = (f, g)
> uncurry (&&&) h 5
(10,"5")
Mind the pure function below, in an imperative language:
def foo(x,y):
x = f(x) if a(x)
if c(x):
x = g(x)
else:
x = h(x)
x = f(x)
y = f(y) if a(y)
x = g(x) if b(y)
return [x,y]
That function represents a style where you have to incrementally update variables. It can be avoided in most cases, but there are situations where that pattern is unavoidable - for example, writing a cooking procedure for a robot, which inherently requires a series of steps and decisions. Now, imagine we were trying to represent foo in Haskell.
foo x0 y0 =
let x1 = if a x0 then f x0 else x0 in
let x2 = if c x1 then g x1 else h x1 in
let x3 = f x2 in
let y1 = if a y0 then f y0 else y0 in
let x4 = if b y1 then g x3 else x3 in
[x4,y1]
That code works, but it is too complicated and error prone due to the need for manually managing the numeric tags. Notice that, after x1 is set, x0's value should never be used again, but it still can. If you accidentally use it, that will be an undetected error.
I've managed to solve this problem using the State monad:
fooSt x y = execState (do
(x,y) <- get
when (a x) (put (f x, y))
(x,y) <- get
if c x
then put (g x, y)
else put (h x, y)
(x,y) <- get
put (f x, y)
(x,y) <- get
when (a y) (put (x, f y))
(x,y) <- get
when (b y) (put (g x, x))) (x,y)
This way, need for tag-tracking goes away, as well as the risk of accidentally using an outdated variable. But now the code is verbose and much harder to understand, mainly due to the repetition of (x,y) <- get.
So: what is a more readable, elegant and safe way to express this pattern?
Full code for testing.
Your goals
While the direct transformation of imperative code would usually lead to the ST monad and STRef, lets think about what you actually want to do:
You want to manipulate values conditionally.
You want to return that value.
You want to sequence the steps of your manipulation.
Requirements
Now this indeed looks first like the ST monad. However, if we follow the simple monad laws, together with do notation, we see that
do
x <- return $ if somePredicate x then g x
else h x
x <- return $ if someOtherPredicate x then a x
else b x
is exactly what you want. Since you need only the most basic functions of a monad (return and >>=), you can use the simplest:
The Identity monad
foo x y = runIdentity $ do
x <- return $ if a x then f x
else x
x <- return $ if c x then g x
else h x
x <- return $ f x
y <- return $ if a x then f y
else y
x <- return $ if b y then g x
else y
return (x,y)
Note that you cannot use let x = if a x then f x else x, because in this case the x would be the same on both sides, whereas
x <- return $ if a x then f x
else x
is the same as
(return $ if a x then (f x) else x) >>= \x -> ...
and the x in the if expression is clearly not the same as the resulting one, which is going to be used in the lambda on the right hand side.
Helpers
In order to make this more clear, you can add helpers like
condM :: Monad m => Bool -> a -> a -> m a
condM p a b = return $ if p then a else b
to get an even more concise version:
foo x y = runIdentity $ do
x <- condM (a x) (f x) x
x <- fmap f $ condM (c x) (g x) (h x)
y <- condM (a y) (f y) y
x <- condM (b y) (g x) x
return (x , y)
Ternary craziness
And while we're up to it, lets crank up the craziness and introduce a ternary operator:
(?) :: Bool -> (a, a) -> a
b ? ie = if b then fst ie else snd ie
(??) :: Monad m => Bool -> (a, a) -> m a
(??) p = return . (?) p
(#) :: a -> a -> (a, a)
(#) = (,)
infixr 2 ??
infixr 2 #
infixr 2 ?
foo x y = runIdentity $ do
x <- a x ?? f x # x
x <- fmap f $ c x ?? g x # h x
y <- a y ?? f y # y
x <- b y ?? g x # x
return (x , y)
But the bottomline is, that the Identity monad has everything you need for this task.
Imperative or non-imperative
One might argue whether this style is imperative. It's definitely a sequence of actions. But there's no state, unless you count the bound variables. However, then a pack of let … in … declarations also gives an implicit sequence: you expect the first let to bind first.
Using Identity is purely functional
Either way, the code above doesn't introduce mutability. x doesn't get modified, instead you have a new x or y shadowing the last one. This gets clear if you desugar the do expression as noted above:
foo x y = runIdentity $
a x ?? f x # x >>= \x ->
c x ?? g x # h x >>= \x ->
return (f x) >>= \x ->
a y ?? f y # y >>= \y ->
b y ?? g x # x >>= \x ->
return (x , y)
Getting rid of the simplest monad
However, if we would use (?) on the left hand side and remove the returns, we could replace (>>=) :: m a -> (a -> m b) -> m b) by something with type a -> (a -> b) -> b. This just happens to be flip ($). We end up with:
($>) :: a -> (a -> b) -> b
($>) = flip ($)
infixr 0 $> -- same infix as ($)
foo x y = a x ? f x # x $> \x ->
c x ? g x # h x $> \x ->
f x $> \x ->
a y ? f y # y $> \y ->
b y ? g x # x $> \x ->
(x, y)
This is very similar to the desugared do expression above. Note that any usage of Identity can be transformed into this style, and vice-versa.
The problem you state looks like a nice application for arrows:
import Control.Arrow
if' :: (a -> Bool) -> (a -> a) -> (a -> a) -> a -> a
if' p f g x = if p x then f x else g x
foo2 :: (Int,Int) -> (Int,Int)
foo2 = first (if' c g h . if' a f id) >>>
first f >>>
second (if' a f id) >>>
(\(x,y) -> (if b y then g x else x , y))
in particular, first lifts a function a -> b to (a,c) -> (b,c), which is more idiomatic.
Edit: if' allows a lift
import Control.Applicative (liftA3)
-- a functional if for lifting
if'' b x y = if b then x else y
if' :: (a -> Bool) -> (a -> a) -> (a -> a) -> a -> a
if' = liftA3 if''
I'd probably do something like this:
foo x y = ( x', y' )
where x' = bgf y' . cgh . af $ x
y' = af y
af z = (if a z then f else id) z
cgh z = (if c z then g else h) z
bg y x = (if b y then g else id) x
For something more complicated, you may want to consider using lens:
whenM :: Monad m => m Bool -> m () -> m ()
whenM c a = c >>= \res -> when res a
ifM :: Monad m => m Bool -> m a -> m a -> m a
ifM mb ml mr = mb >>= \b -> if b then ml else mr
foo :: Int -> Int -> (Int, Int)
foo = curry . execState $ do
whenM (uses _1 a) $
_1 %= f
ifM (uses _1 c)
(_1 %= g)
(_1 %= h)
_1 %= f
whenM (uses _2 a) $
_2 %= f
whenM (uses _2 b) $ do
_1 %= g
And there's nothing stopping you from using more descriptive variable names:
foo :: Int -> Int -> (Int, Int)
foo = curry . execState $ do
let x :: Lens (a, c) (b, c) a b
x = _1
y :: Lens (c, a) (c, b) a b
y = _2
whenM (uses x a) $
x %= f
ifM (uses x c)
(x %= g)
(x %= h)
x %= f
whenM (uses y a) $
y %= f
whenM (uses y b) $ do
x %= g
This is a job for the ST (state transformer) library.
ST provides:
Stateful computations in the form of the ST type. These look like ST s a for a computation that results in a value of type a, and may be run with runST to obtain a pure a value.
First-class mutable references in the form of the STRef type. The newSTRef a action creates a new STRef s a reference with an initial value of a, and which can be read with readSTRef ref and written with writeSTRef ref a. A single ST computation can use any number of STRef references internally.
Together, these let you express the same mutable variable functionality as in your imperative example.
To use ST and STRef, we need to import:
{-# LANGUAGE NoMonomorphismRestriction #-}
import Control.Monad.ST.Safe
import Data.STRef
Instead of using the low-level readSTRef and writeSTRef all over the place, we can define the following helpers to match the imperative operations that the Python-style foo example uses:
-- STRef assignment.
(=:) :: STRef s a -> ST s a -> ST s ()
ref =: x = writeSTRef ref =<< x
-- STRef function application.
($:) :: (a -> b) -> STRef s a -> ST s b
f $: ref = f `fmap` readSTRef ref
-- Postfix guard syntax.
if_ :: Monad m => m () -> m Bool -> m ()
action `if_` guard = act' =<< guard
where act' b = if b then action
else return ()
This lets us write:
ref =: x to assign the value of ST computation x to the STRef ref.
(f $: ref) to apply a pure function f to the STRef ref.
action `if_` guard to execute action only if guard results in True.
With these helpers in place, we can faithfully translate the original imperative definition of foo into Haskell:
a = (< 10)
b = even
c = odd
f x = x + 3
g x = x * 2
h x = x - 1
f3 x = x + 2
-- A stateful computation that takes two integer STRefs and result in a final [x,y].
fooST :: Integral n => STRef s n -> STRef s n -> ST s [n]
fooST x y = do
x =: (f $: x) `if_` (a $: x)
x' <- readSTRef x
if c x' then
x =: (g $: x)
else
x =: (h $: x)
x =: (f $: x)
y =: (f $: y) `if_` (a $: y)
x =: (g $: x) `if_` (b $: y)
sequence [readSTRef x, readSTRef y]
-- Pure wrapper: simply call fooST with two fresh references, and run it.
foo :: Integral n => n -> n -> [n]
foo x y = runST $ do
x' <- newSTRef x
y' <- newSTRef y
fooST x' y'
-- This will print "[9,3]".
main = print (foo 0 0)
Points to note:
Although we first had to define some syntactical helpers (=:, $:, if_) before translating foo, this demonstrates how you can use ST and STRef as a foundation to grow your own little imperative language that's directly suited to the problem at hand.
Syntax aside, this matches the structure of the original imperative definition exactly, without any error-prone restructuring. Any minor changes to the original example can be mirrored directly to Haskell. (The addition of the temporary x' <- readSTRef x binding in the Haskell code is only in order to use it with the native if/else syntax: if desired, this can be replaced with an appropriate ST-based if/else construct.)
The above code demonstrates giving both pure and stateful interfaces to the same computation: pure callers can use foo without knowing that it uses mutable state internally, while ST callers can directly use fooST (and for example provide it with existing STRefs to modify).
#Sibi said it best in his comment:
I would suggest you to stop thinking imperatively and rather think in a functional way. I agree that it will take some time to getting used to the new pattern, but try to translate imperative ideas to functional languages isn't a great approach.
Practically speaking, your chain of let can be a good starting point:
foo x0 y0 =
let x1 = if a x0 then f x0 else x0 in
let x2 = if c x1 then g x1 else h x1 in
let x3 = f x2 in
let y1 = if a y0 then f y0 else y0 in
let x4 = if b y1 then g x3 else x3 in
[x4,y1]
But I would suggest using a single let and giving descriptive names to the intermediate stages.
In this example unfortunately I don't have a clue what the various x's and y's do, so I cannot suggest meaningful names. In real code you would use names such as x_normalized, x_translated, or such, instead of x1 and x2, to describe what those values really are.
In fact, in a let or where you don't really have variables: they're just shorthand names you give to intermediate results, to make it easy to compose the final expression (the one after in or before the where.)
This is the spirit behind the x_bar and x_baz below. Try to come up with names that are reasonably descriptive, given the context of your code.
foo x y =
let x_bar = if a x then f x else x
x_baz = f if c x_bar then g x_bar else h x_bar
y_bar = if a y then f y else y
x_there = if b y_bar then g x_baz else x_baz
in [x_there, y_bar]
Then you can start recognizing patterns that were hidden in the imperative code. For example, x_bar and y_bar are basically the same transformation, applied respectively to x and y: that's why they have the same suffix "_bar" in this nonsensical example; then your x2 probably doesn't need an intermediate name , since you can just apply f to the result of the entire "if c then g else h".
Going on with the pattern recognition, you should factor out the transformations that you are applying to variables into sub-lambdas (or whatever you call the auxiliary functions defined in a where clause.)
Again, I don't have a clue what the original code did, so I cannot suggest meaningful names for the auxiliary functions. In a real application, f_if_a would be called normalize_if_needed or thaw_if_frozen or mow_if_overgrown... you get the idea:
foo x y =
let x_bar = f_if_a x
y_bar = f_if_a y
x_baz = f (g_if_c_else_h x_bar)
x_there = g_if_b x_baz y_bar
in [x_there, y_bar]
where
f_if_a x
| a x = f x
| otherwise = x
g_if_c_else_h x
| c x = g x
| otherwise = h x
g_if_b x y
| b y = g x
| otherwise = x
Don't disregard this naming business.
The whole point of Haskell and other pure functional languages is to express algorithms without the assignment operator, meaning the tool that can modify the value of an existing variable.
The names you give to things inside a function definition, whether introduced as arguments, let, or where, can only refer to one value (or auxiliary function) throughout the entire definition, so that your code can be more easily reasoned about and proven correct.
If you don't give them meaningful names (and conversely giving your code a meaningful structure) then you're missing out on the entire purpose of Haskell.
(IMHO the other answers so far, citing monads and other shenanigans, are barking up the wrong tree.)
I always prefer layering state transformers to using a single state over a tuple: it definitely declutters things by letting you "focus" on a specific layer (representations of the x and y variables in our case):
import Control.Monad.Trans.Class
import Control.Monad.Trans.State
foo :: x -> y -> (x, y)
foo x y =
(flip runState) y $ (flip execStateT) x $ do
get >>= \v -> when (a v) (put (f v))
get >>= \v -> put ((if c v then g else h) v)
modify f
lift $ get >>= \v -> when (a v) (put (f v))
lift get >>= \v -> when (b v) (modify g)
The lift function allows us to focus on the inner state layer, which is y.
Is it possible to write the Y Combinator in Haskell?
It seems like it would have an infinitely recursive type.
Y :: f -> b -> c
where f :: (f -> b -> c)
or something. Even a simple slightly factored factorial
factMaker _ 0 = 1
factMaker fn n = n * ((fn fn) (n -1)
{- to be called as
(factMaker factMaker) 5
-}
fails with "Occurs check: cannot construct the infinite type: t = t -> t2 -> t1"
(The Y combinator looks like this
(define Y
(lambda (X)
((lambda (procedure)
(X (lambda (arg) ((procedure procedure) arg))))
(lambda (procedure)
(X (lambda (arg) ((procedure procedure) arg)))))))
in scheme)
Or, more succinctly as
(λ (f) ((λ (x) (f (λ (a) ((x x) a))))
(λ (x) (f (λ (a) ((x x) a))))))
For the applicative order
And
(λ (f) ((λ (x) (f (x x)))
(λ (x) (f (x x)))))
Which is just a eta contraction away for the lazy version.
If you prefer short variable names.
Here's a non-recursive definition of the y-combinator in haskell:
newtype Mu a = Mu (Mu a -> a)
y f = (\h -> h $ Mu h) (\x -> f . (\(Mu g) -> g) x $ x)
hat tip
The Y combinator can't be typed using Hindley-Milner types, the polymorphic lambda calculus on which Haskell's type system is based. You can prove this by appeal to the rules of the type system.
I don't know if it's possible to type the Y combinator by giving it a higher-rank type. It would surprise me, but I don't have a proof that it's not possible. (The key would be to identify a suitably polymorphic type for the lambda-bound x.)
If you want a fixed-point operator in Haskell, you can define one very easily because in Haskell, let-binding has fixed-point semantics:
fix :: (a -> a) -> a
fix f = f (fix f)
You can use this in the usual way to define functions and even some finite or infinite data structures.
It is also possible to use functions on recursive types to implement fixed points.
If you're interested in programming with fixed points, you want to read Bruce McAdam's technical report That About Wraps it Up.
The canonical definition of the Y combinator is as follows:
y = \f -> (\x -> f (x x)) (\x -> f (x x))
But it doesn't type check in Haskell because of the x x, since it would require an infinite type:
x :: a -> b -- x is a function
x :: a -- x is applied to x
--------------------------------
a = a -> b -- infinite type
If the type system were to allow such recursive types, it would make type checking undecidable (prone to infinite loops).
But the Y combinator will work if you force it to typecheck, e.g. by using unsafeCoerce :: a -> b:
import Unsafe.Coerce
y :: (a -> a) -> a
y = \f -> (\x -> f (unsafeCoerce x x)) (\x -> f (unsafeCoerce x x))
main = putStrLn $ y ("circular reasoning works because " ++)
This is unsafe (obviously). rampion's answer demonstrates a safer way to write a fixpoint combinator in Haskell without using recursion.
Oh
this wiki page and
This Stack Overflow answer seem to answer my question.
I will write up more of an explanation later.
Now, I've found something interesting about that Mu type. Consider S = Mu Bool.
data S = S (S -> Bool)
If one treats S as a set and that equals sign as isomorphism, then the equation becomes
S ⇋ S -> Bool ⇋ Powerset(S)
So S is the set of sets that are isomorphic to their powerset!
But we know from Cantor's diagonal argument that the cardinality of Powerset(S) is always strictly greater than the cardinality of S, so they are never isomorphic.
I think this is why you can now define a fixed point operator, even though you can't without one.
Just to make rampion's code more readable:
-- Mu :: (Mu a -> a) -> Mu a
newtype Mu a = Mu (Mu a -> a)
w :: (Mu a -> a) -> a
w h = h (Mu h)
y :: (a -> a) -> a
y f = w (\(Mu x) -> f (w x))
-- y f = f . y f
in which w stands for the omega combinator w = \x -> x x, and y stands for the y combinator y = \f -> w . (f w).