A few months ago, I tried my hand at Seven Languages in Seven Weeks,

and it was an incredibly enlightening experience.

There was one excersize that kept me at odds for weeks though, so I

thought I’d share my experience.

## Haskell Day 3: Creating and writing a Maze Solver

The excersize calls for accomplishing two tasks:

- Creating a data structure for storing a maze
- Write a method to solve it

I think there’s a few ways to organize the data, but I chose a format that was fairly readable:

data Exits = North | West | East | South deriving (Show, Eq) data Node = NodePath (Int, Int) [Exits] | TerminalNode (Int, Int) deriving (Show, Eq) type Maze = [[Node]] testMaze :: Maze testMaze = [ [ (NodePath (0,0) [South]), (NodePath (1, 0) []), (NodePath (2, 0) []) ], [ (NodePath (0,1) [East]), (NodePath (1, 1) [East]), (NodePath (2, 1) [North, South]) ], [ (NodePath (0,2) []), (NodePath (1, 2) []), (TerminalNode (2, 2)) ] ]

Basically, this creates a few types:

- Exits, to deal with the directions from which one can move from the current position
- Node, which consists of nodes with paths (a NodePath), and a Final node (TerminalNode).
- Maze, which is simply a two-dimensional array of nodes.

I think there’s liberties here as well, but I wanted to note my choice

a NodePath and TerminalNode type: it seemed like creating completely

different types altogether allowed me to rely on the strict type of

Haskell better to solve my problems, instead of emebedding logic. But YRMV.

One big downside: putting data in structures like this made it hard to

get the data I want, and also rely on the typing. Ultimately I had to

create several utility methods to help me:

-- getNode: Returns a node object given a maze and a position getNode :: Maze -> (Int, Int)-> Node getNode maze (x,y) = maze !! y !! x -- getExits: return all the exits for a node getExits :: Node -> [Exits] getExits (NodePath _ exits) = exits getExits (TerminalNode _) = [] -- getPosition: return a (x,y) of the position of a node getPosition :: Node -> (Int, Int) getPosition (NodePath (x,y) _) = (x, y) getPosition (TerminalNode (x,y)) = (x, y)

I definitely must be doing something wrong here. The rigid typing of

Haskell should allow me to take advantage of the inner data without

creating accessors like this. But I’m not a Haskell expert, so I made

do with what I understood.

Finally, I have all the tools I need to write my solver. Here it is:

-- getNextNode: given a node, maze, and an exit, return the next node from the maze getNextNode :: Node -> Maze -> Exits -> Node getNextNode node maze exit = let (x, y) = getPosition node in case exit of North -> getNode maze (x, y - 1) West -> getNode maze (x - 1, y) East -> getNode maze (x + 1, y) South -> getNode maze (x, y + 1) -- If the element already exists in the path, we're at a dead end. -- solveRoute: returns a list of the valid routes to the exit solveRoute :: Maze -> Node -> [Node] -> Exits -> Maybe [Node] solveRoute maze node path exit = let nextNode = getNextNode node maze exit in if (nextNode `elem` path) then Nothing else solveMaze maze nextNode (node:path) -- solveMaze: solve the maze by taking solveRoute, filtering the successful results, and taking the first one. solveMaze :: Maze -> Node -> [Node] -> Maybe [Node] solveMaze maze node path = case node of TerminalNode _ -> Just (node:path) NodePath _ _-> let nodes = (filter (\x -> x /= Nothing) (map (solveRoute maze node path) (getExits node))) in if (length nodes > 0) then head nodes else Nothing

This code probably seems a bit contrived. Basically here’s what solveMaze does:

- delegates the logic to solveRoute if the node isn’t a terminalNode
- solveRoute gets the exits for the node. it loops through them, takes

the valid ones (the ones where the same position isn’t in there

twice), and passes them back into solveMaze

So solveMaze and solveRoute call each other until they find a valid

solution. I could have added them into the same method, but this

seemed like a logical split that made the code a little easier to understand.

This works. Give it a try:

mazeStart = getNode testMaze (0, 0) mazeSolution = solveMaze testMaze mazeStart []

One of the big issues I have with solution, however, is the fact that

it doesn’t use a list monad in any way. And I’m still a bit confused

as to how it comes in handy here. From my understanding, a list monad

flattens a list of lists into a single list. So ultimately, my

solution might not be taking advantage of the real power of

Haskell. It is purely functional though, so maybe it is.

Here’s the code in full:

module Day3 where import Data.List -- data Node = NodePath ((Int, Int), [Node]) | TerminalNode (Int, Int) data Exits = North | West | East | South deriving (Show, Eq) data Node = NodePath (Int, Int) [Exits] | TerminalNode (Int, Int) deriving (Show, Eq) type Maze = [[Node]] testMaze :: Maze testMaze = [ [ (NodePath (0,0) [South]), (NodePath (1, 0) []), (NodePath (2, 0) []) ], [ (NodePath (0,1) [East]), (NodePath (1, 1) [East]), (NodePath (2, 1) [North, South]) ], [ (NodePath (0,2) []), (NodePath (1, 2) []), (TerminalNode (2, 2)) ] ] -- getNode getNode :: Maze -> (Int, Int)-> Node getNode maze (x,y) = maze !! y !! x -- getExists getExits :: Node -> [Exits] getExits (NodePath _ exits) = exits getExits (TerminalNode _) = [] -- getPosition getPosition :: Node -> (Int, Int) getPosition (NodePath (x,y) _) = (x, y) getPosition (TerminalNode (x,y)) = (x, y) -- getNextNode getNextNode :: Node -> Maze -> Exits -> Node getNextNode node maze exit = let (x, y) = getPosition node in case exit of North -> getNode maze (x, y - 1) West -> getNode maze (x - 1, y) East -> getNode maze (x + 1, y) South -> getNode maze (x, y + 1) -- If the element already exists in the path, we're at a dead end. solveRoute :: Maze -> Node -> [Node] -> Exits -> Maybe [Node] solveRoute maze node path exit = let nextNode = getNextNode node maze exit in if (nextNode `elem` path) then Nothing else solveMaze maze nextNode (node:path) -- solveMaze 2 solveMaze :: Maze -> Node -> [Node] -> Maybe [Node] solveMaze maze node path = case node of TerminalNode _ -> Just (node:path) NodePath _ _-> let nodes = (filter (\x -> x /= Nothing) (map (solveRoute maze node path) (getExits node))) in if (length nodes > 0) then head nodes else Nothing mazeStart = getNode testMaze (0, 0) mazeSolution = solveMaze testMaze mazeStart []