Lifting a value in the State monad in Haskell - haskell

Question

I am writing a Sudoku generator/solver in Haskell as a learning exercise.
My solve function takes in a UArray but returns a State Int (UArray ...) so that it can also return the maximum difficulty level that it found while solving.
This is my function so far (still in the very experimental early stage):
import Control.Monad.State (State, put)
import Control.Monad.Trans.Class (lift)
import Data.Array.MArray (thaw)
import Data.Array.ST (runSTUArray)
import Data.Array.Unboxed (UArray)
-- ...
type Cell = Word16
solve :: UArray (Int, Int) Cell -> State Int (UArray (Int, Int) Cell)
solve grid = do
return $ runSTUArray $ do
arr <- thaw grid
lift $ put 42
return arr
It does not really do anything with the mutable array yet. I am simply trying to get it to type check with the put 42, but currently get the following error:
• Couldn't match kind ‘*’ with ‘* -> *’
When matching the kind of ‘ST’
• In a stmt of a 'do' block: lift $ put 42
In the second argument of ‘($)’, namely
‘do arr <- thaw grid
lift $ put 42
return arr’
In the second argument of ‘($)’, namely
‘runSTUArray
$ do arr <- thaw grid
lift $ put 42
return arr’
|
128 | lift $ put 42
| ^^^^^^^^^^^^^

Answer 1

runSTUArray ... is a pure value, it does not know anything about "outer monad". And State cares about how you use it, you cannot pass it opaquely into ST.
What you could do:
Option1: change the whole program to move more logic to ST side. Instead of State you'd use STRef then:
solve :: ST s (STRef Int) -> ST s (UArray (Int, Int) Cell) -> ST s ()
...
Option2: manually extract it and pass it to ST, then get back and put explicitly. But there is complication. runSTUArray does not allow getting another value together with the array. I don't know how it can be done safely with current array functions. Unsafely you could re-implement better runSTUArray which can pass another value. You could also add fake cells and encode the new state there.
The way to export another value exists in the vector package, there is (in new versions) createT function which can take not bare vector but a structure containing it (or even several vectors). So, overall, your example would be like:
import Control.Monad.State (State, put, get)
import Data.Word (Word16)
import qualified Data.Vector.Unboxed as DVU
type Cell = Word16
solve :: DVU.Vector Cell -> State Int (DVU.Vector Cell)
solve grid = do
oldState <- get
let (newState, newGrid) = DVU.createT (do
arr <- DVU.thaw grid
pure (oldState + 42, arr))
put newState
pure newGrid
vectors are one-dimensional only, unfortunately

Answer 2

solve grid has form return $ .... This means that State Int (UArray (Int, Int) Cell) is just specialized Monad m => m (UArray (Int, Int) Cell) - the ... does not have access to the features of this specific monad, it's just a UArray (Int, Int) Cell value that you return.

Answer 3

I was able to get a slight variation to compile and run after changing the State monad to a tuple (Int, Grid):
import Control.Monad.ST (ST, runST)
import Data.Array.MArray (freeze, thaw, writeArray)
import Data.Array.ST (STUArray)
import Data.Array.Unboxed (UArray)
import Data.Word (Word16)
type Cell = Word16
type Grid = UArray (Int, Int) Cell
solve :: Grid -> (Int, Grid)
solve grid =
runST $ do
mut <- thaw grid :: ST s (STUArray s (Int, Int) Cell)
writeArray mut (0, 0) 0 -- test that I can actually write
frozen <- freeze mut
return (42, frozen)
This works fine for my application.