CODEWITHSUNDEEP
algorithmhaskellrecursionpattern-matchingtowers-of-hanoi
basti12354
basti12354

Reputation: 2520

Recursive function in haskell

I am trying to solve some issues with programming in haskell.

I am having this three types and the Hanoi-algorithm: 

type Position = Int
type Move = (Position,Position)
type Towers = ([Int],[Int],[Int])

hanoi 1 i j = [(i,j)]
hanoi n i j = hanoi n' i otherT ++ [(i,j)] ++ hanoi n' otherT j
    where      
            n' = n-1
            otherT = 1+2+3-i-j -- other tower

Now I wrote a function with makes one single MOVE.

move ::([Move],Towers) -> ([Move],Towers) 
move ::([Move],Towers) -> ([Move],Towers)
move ([],(xs,ys,zs) ) = ((error "Error"),(xs,ys,zs) )
move (((a,b): tail), (xs,ys,zs) )
              | a > 3 =    (tail, ((error "Error"),ys,zs) )                
              | b > 3 =    (tail, ((error "Error"),ys,zs ) )                
              | otherwise = hilfsfunktion (((a,b): tail), (xs,ys,zs) )



hilfsfunktion (((1,2): tail), ((x:xs),(y:ys),zs) ) 
            | x < y = (tail, (xs, (x:y:ys),zs) )
            | x > y = (tail, (xs, (error "too big"),(error "too big")))
hilfsfunktion (((1,2): tail), ((x:xs), [],zs) ) = (tail, (xs, x:[],zs) )

The function is much longer, but you can see that I solved this with Pattern Matching.

Now I try to write a function, that called this "move"-function as long as the list is not empty. So at first I used Pattern Matching for the case, that the list is empty.

all_moves:: ([Move], Towers) -> Towers
all_moves ([],(xs,ys,zs) ) = (xs,ys,zs)
all_moves (((a,b): tail), (xs,ys,zs) ) = ????

Now I need some help, in Java I would use a loop to fix this. I think I have to call the function "move" recursive, but I don't know how to to this. How can I solve this function?

Upvotes: 0

Views: 308

Answers (2)

Kai-Chun Lin
Kai-Chun Lin

Reputation: 363

I would like to do like this way even though my method is nor recursive as well.

allllmove :: Towers -> Towers

allmove ([0],[0],[0]) = ([0],[0],[0])

allmove ((x:xs),y,z)

length(x:xs) > length z && length z > length y = (xs,(x:y),z)

length(x:xs) >= length z && length (x:xs) > length y = (xs,(x:y),z) ... .. .

Upvotes: 0

AlexJ136
AlexJ136

Reputation: 1282

I'm not sure exactly how your hanoi solution works, but I think this answers your question:

all_moves :: ([Move], Towers) -> Towers
all_moves ([], (xs, ys, zs)) = (xs, ys, zs)
all_moves movetowers         = all_moves (move movetowers)

Hopefully you can see why this works - we keep doing a move and then a recursive call until there are no moves left.

Upvotes: 1

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