I have to write a program in Haskell that will solve some nondeterministic problem. I think i understand List Monad in 75% so it is oblivious choice but...
(My problem is filling n x m board with ships and water i am given sums of rows and colums every part of ship has its value etd its not important right now).
I want to guard as early as possible to make algoritm effective the problem is that possibility of insertion of ship is dependant from what i am given / what i have inserted in previus moves lets call it board state and i have no idea how to pass it cuz i can't generate a new state from board alone)
My Algoritm is: 1. Initialize First Board 2. Generate First Row trying applying every possible insertion (i can insert sheep verticaly so i need to remember to insert other parts of sheep in lower rows) 3. Solve the problem for smaller board (ofc after generating each 2 rows i check is everything ok)
But i have no idea how can I pass new states cuz as far as i have read about State Monad it generates new state from old state alone and this is impossible for me to do i would want to generate new state while doing operations on value).
I am sorry for my hatred towards Haskell but after few years of programing in imperative languages being forced to fight with those Monads to do things which in other languages i could write almost instantly makes me mad. (well other things in Haskell are fine for me and some of them are actually quite nice).
Combine StateT with the list monad to get your desired behavior.
Here's a simple example of using the non-determinism of the list monad while still keeping a history of previous choices made:
import Control.Monad
import Control.Monad.Trans.Class
import Control.Monad.Trans.State
fill :: StateT [Int] [] [Int]
fill = do
history <- get
if (length history == 3)
then return history
else do
choice <- lift [0, 1, 2]
guard (choice `notElem` history)
put (choice:history)
fill
fill maintains a separate history for each path that it tries out. If it fills up the board it returns successfully, but if the current choice overlaps with a previous choice it abandons that solution and tries a different path.
You run it using evalStateT, supplying an initial empty history:
>>> evalStateT fill []
[[2,1,0],[1,2,0],[2,0,1],[0,2,1],[1,0,2],[0,1,2]]
It returns a list of all possible solutions. In this case, that just happens to be the list of all permutations in which we could have filled up the board.
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