Today we’re going to learn to solve the classic and ageless logic problems without any data structures besides List’s monadic properties as a MonadPlus!

We are going to be solving this old-as-time logic puzzle, which Wikipedia claims dates back to the 9th century:

A farmer has a wolf, a goat, and a cabbage that he wishes to transport across a river. Unfortunately, his boat can carry only one thing at a time with him. He can’t leave the wolf alone with the goat, or the wolf will eat the goat. He can’t leave the goat alone with the cabbage, or the goat will eat the cabbage. How can he properly transport his belongings to the other side one at a time, without any disasters?

We’re going to assume a somewhat basic familiarity with functional programming concepts and a basic understanding of monads (if you don’t know that much, check out adit’s great concise guide). If you aren’t familiar with MonadPlus/Alternative (and how they work as monads) check out Part 1 and Part 2, which should provide all the background and most of the syntax. Most Haskell syntax is either be explained here as we get to it or in the previous parts. Still, if you have any questions, feel free to leave a comment, give Learn You A Haskell a quick read, or stop by freenode’s friendly #haskell!

A MonadPlus Review

The usefulness of a monad depends on how you define the characteristic “bind” or “chaining” behavior. For this article, MonadPlus refers to the design pattern (and Haskell typeclass) where you model this “chaining” as a “success/fail” process .

There is a common language with to talk about this process: mzero means “fail here” and return x means “succeed with a result of the value x here”. So chaining is implemented such that chaining anything to a failure will propagate that failure forward. That is, mzero >> return x = mzero .

Our Approach

So, armed with what we learned in Part 2, let’s formulate a general plan for finding all solutions in n moves.

Now, in the List monad, we can think of things as “journeys” or stories: subject your value to a long and arduous journey, specifying at every step of the way what choices it has to continue. Then specify where journeys fail and end. At the end of it all, the result is a list of the finishing values of all trails that have completed the journey.

With the List monad, we say “Here is the description of a (single) journey. What journeys following this description succeed?”

So what could this journey be for us? Well, we think of a journey in this situation as the accumulation of moves to a plan. We start out with a blank plan (“Do nothing”). The next step, we add one move to our plan (“Just move the fox”, for example). Then the next step, we add another move (“First move the fox, then move the farmer”).

Start with a blank plan; a tabula rasa. Add a legal and safe move to it. Repeat Step 2 n times Fail if you aren’t a solution; succeed if you are.

Simple, right? We just laid out the path of a single plan, from its birth to its eventual death or ascension.

This is the most significant thing about this approach: it allows you to describe one journey, in general terms, and List will “automatically” find out all successful journeys that fit your mold. You don’t ever have to worry about the ensemble or manually deal with explicit branching or filtering. Cognitively, all you have to do is write one story. Just one. That is the power of the List Monad abstraction.

Our Types

The first thing we do when writing any Haskell program: define our types!

data Character = Farmer | Wolf | Goat | Cabbage -- 1 deriving ( Show , Eq , Enum ) newtype Move = MoveThe Character -- 2 deriving ( Eq ) instance Show Move where -- 3 show ( MoveThe Farmer ) = "F" show ( MoveThe Wolf ) = "W" show ( MoveThe Goat ) = "G" show ( MoveThe Cabbage ) = "C" type Plan = [ Move ] -- 4 data Position = West | East -- 5 deriving ( Show , Eq )

First, we define the enumerated type Character all the characters we will be working with: the farmer, the wolf, the goat, and the cabbage. Next, we define a simple Move container, which just contains a character. A MoveThe Farmer will represent a movement of only the farmer, a MoveThe Wolf will represent the movement of both the farmer and the wolf, etc. For the purposes of easy debugging, we’re going to define our own instance of Show for moves so that we can use print on them. A simple type synonym; a Plan is just a list of Move s. Note that we are not using this list as a MonadPlus — it’s just a plain dumb list of moves in our plan. A Position type: either on the west bank or on the east bank of the river. Everyone starts out on the west bank, and we want them all to end up on the east bank.

Welcome to Haskell! Hi! These “Welcome to Haskell” asides are for people unfamiliar with Haskell, mostly for Haskell syntax stuff. If you already feel comfortable, feel free to skip them. There’s a lot of Haskell syntax and concepts here; mostly, all we are doing is declaring new types. We declare that Character is “either” a Farmer , Wolf , Goat , or Cabbage . This is like saying that a Bool is either a False or a True : in fact, you could define your own Bool with something like this: (or even your own Int ) data Bool = False | True data Int = - 536870912 ... | - 1 | 0 | 1 | 2 | ... 536870911 The deriving syntax tells the compiler to automatically derive functions for printing the type (Show), testing for equality (Eq), and enumerating through them (Enum) We declare a new type Move which is just a wrapper around a Character . We can create a new Move by using MoveThe : > : t MoveThe ghci MoveThe :: Character -> Move > : t MoveThe Wolf ghci MoveThe Wolf :: Move ( ghci> represents a command at the interactive prompt ghci, and :t asks for the type of whatever comes after it) Here we define custom functions for printing out a Move Here is a type synonym Plan . Every time we use Plan as a type, we really mean [Move] , and the compiler treats the two things as the same. Position : same deal as Character .

Implementation

The Final Step

We’re going to skip to the end and write our final step and what it is supposed to be, and then fill in the functions that are necessary to make it happen.

The last stage of our journey is after we have made all n moves, we end the journey if it is not a solution.

makeNMoves :: Int -> [ Plan ] -- 1 isSolution :: Plan -> Bool findSolutions :: Int -> [ Plan ] -- 2 = do findSolutions n p <- makeNMoves n -- 3 makeNMoves n $ isSolution p -- 4 guardisSolution p return p -- 5

The type signatures of the helper functions we will be using. findSolutions n is going to be the all successful plans after n moves. Let p be a plan after n moves have been added to it. Note that makeNMoves is itself a journey — a sub-journey. So p is a single plan that has already gone through the makeNMoves journey. We are continuing that journey. End the journey unless p is a solution (all characters are on the east side) Succeed with p if the journey has not yet ended.

Hm. Sounds good! We’re done!

So now we only need to implement makeNMoves and isSolution !

Welcome to Haskell! Haskell is a functional language…but that “do” block sure looks very imperative to me. What gives? As explained in Part 1, all do blocks are just syntactical sugar for repeated applications of >>= : findSolutions :: Int -> [ Plan ] = findSolutions >>= (\p -> guard (isSolution p) >> return p) makeNMoves n(\pguard (isSolution p)p) And >>= is just the (hopefully) familiar bind. Again, look at Part 1 or adit’s tutorial for a fuller explanation.

makeNMoves

makeNMoves is going to be the main logic of our program. We want it to be a journey, itself — a journey of a single plan going through n additions of moves.

That means we want something like:

makeMove :: Plan -> [ Plan ] startingPlan :: Plan = [] startingPlan[] makeNMoves :: Int -> [ Plan ] = do makeNMoves n <- makeMove startingPlan m1makeMove startingPlan <- makeMove m1 m2makeMove m1 <- makeMove m2 m3makeMove m2 -- ... (n times) <- makeMove mx mnmakeMove mx return mn mn

Which says “The journey of makeNMoves is repeatedly making a move n times.”

Of course we have seen that particular type of do block before, it is simply:

makeNMoves :: Int -> [ Plan ] = makeNMoves n >>= makeMove makeMove startingPlanmakeMove >>= makeMove >>= makeMove -- ... makeMovemakeMove >>= makeMove -- (n times) makeMove

Luckily there is a function in the standard library that allows us to repeatedly apply a function n times: iterate :: (a -> a) -> a -> [a] . iterate f x takes a function f :: a -> a and repeatedly applies it to a starting value x :: a and yields the results as a list:

iterate f x = [ x, f x, f (f x), f (f (f x)) ... ] f x[ x, f x, f (f x), f (f (f x))

And so to get the n th application, we use iterate f x !! n ( !! being the indexing function, getting the n th element of the list)

So now we can define makeNMoves :

makeNMoves :: Int -> [ Plan ] = iterate ( >>= makeMove) ( return startingPlan) !! n makeNMoves nmakeMove) (startingPlan)

We say “apply (>>= makeMove) n times, starting the single starting plan”.

Welcome to Haskell! Remember that return x >>= f is the same as f x . You can see this here: = do foo1 y <- return x f y -- identical = f x foo2f x Where return x says “succeed with the value x ”, and y <- says “set y to the value of that success”. Of course, y is just going to be x , because we had just said "succeed with the value of x . That means that f y is the same as f x .

Even though the syntax is not the cleanest, it is important to remember here that what we are doing is simply defining the journey makeNMoves as the result of taking n makeMove journeys one after the other. The same as that first do block.

isSolution

Let’s define our helper function isSolution :: Plan -> Bool . Basically, we want to check if the positions of all of the characters are East .

First, we need a way to get the position of a farmer/animal after a given plan has been executed.

positionOf

Our function positionOf :: Plan -> Character -> Position is going to take a Plan and a Character , and report what side of the river the character is on.

Because every single move swaps the position of the farmer, the final position of the farmer depends only on the even-/odd-ness of the number of total moves. If it is even, then the farmer is on the west bank still (consider 0 moves, two moves, etc.). If it is odd, then the farmer is on the east bank.

positionOf :: Plan -> Character -> Position = case c of positionOf p c Farmer -> positionFromCount $ length p positionFromCount _ -> undefined where | even n = West positionFromCount n | othherwise = East othherwise

Now, what if we want to know about non-farmers?

Instead of finding the total number of moves, we only need to find the number of moves involving that given animal.

Let’s first filter the Plan p by moves involving the character c :

filter ( == MoveThe c) p c) p

This will return a new Plan, but with only the moves involving the character c . We can then use the length of that.

Welcome to Haskell! filter :: (a -> Bool) -> [a] -> [a] is a common function that takes a predicate a -> Bool and a list, and returns a new list with only the items for which the predicate returns true. (== MoveThe c) is a function that returns true if the move is equal to MoveThe c .

Putting it all together:

positionOf :: Plan -> Character -> Position = case c of positionOf p c Farmer -> positionFromCount . length $ p positionFromCount c -> positionFromCount . length $ filter ( == MoveThe c) p positionFromCountc) p where | even n = West positionFromCount n | othherwise = East othherwise

Welcome to Haskell! What is positionFromCount . length $ p ? In Haskell, the (.) operator represents function composition. (f . g) x is equivalent to f (g x) . “Apply g first, then apply f ”. Also recall that you can think of $ as adding an implicit parentheses around both sides of it. You visualize it like the spine of a butterfly — the “wings” are wrapped parentheses around either side of it. In that sense, f . g $ x is the same as (f . g) (x) (A rather lopsided butterfly). So, altogether, positionFromCount . length $ p is the same as (positionFromCount . length) p , which says “first, find the length of p , then turn that length into a position.” In the same way, positionFromCount . length $ filter (== MoveThe c) p is (positionFromCount . length) (filter (== MoveThe c) p) — find the length of the filtered list, then turn that length into a position. We use $ mostly because we don’t like writing parentheses everywhere when we don’t have to.

Does this actually work? Let’s try out some examples.

> let p = [ MoveThe Goat , MoveThe Farmer , MoveThe Wolf , MoveThe Goat ] ghci > positionOf p Goat ghcipositionOf p West > positionOf p Wolf ghcipositionOf p East > positionOf p Farmer ghcipositionOf p West

It works! By the way, as an unrelated note, isn’t it cool that our Plan literal reads a lot like English? MoveThe Goat, MoveThe Farmer, MoveThe Wolf…

Checking the Path

Now we have to check that the plan is a solution.

Simple — that means that all Characters are on the east side.

We can check this manually:

isSolution :: Plan -> Bool = isSolution p Farmer == East positionOf p && positionOf p Wolf == East positionOf p && positionOf p Goat == East positionOf p && positionOf p Cabbage == East positionOf p

Hm. Rather ugly.

We see a common pattern that we need positionOf p c for all c s. That looks like a map!

We also compare all of them to East . That sounds like a job for the prelude function all :: (a -> Bool) -> [a] -> Bool , which takes a predicate and a list and returns true if all items in the list satisfy the predicate.

Let’s piece it all together:

= all ( == East ) positions isSolution p) positions where = map (positionOf p) [ Farmer .. ] positions(positionOf p) [

Welcome to Haskell! map is probably the most ubiquitous concept in functional programming — it takes a function and a list and returns a new list with the function applied to every item. For example, map f [x,y,z] = [f x, f y, f z] . If we wanted to find the lengths of a list of strings, we’d do: map length [ "alice" , "bob" ] = [ length "alice" , length "bob" ] = [ 5 , 3 ] So in our case: map (positionOf p) [ Farmer , Wolf , Goat , Cabbage ] (positionOf p) [ = [ positionOf p Farmer -- Position of the farmer [ positionOf p Wolf -- Position of the wolf , positionOf p Goat -- Position of the goat , positionOf p Cabbage -- Position of the cabbage , positionOf p ]

We use [Farmer ..] as shorthand for [Farmer, Wolf, Goat, Cabbage] — this is because Character is an Enum, so it can be enumerated using enumeration syntax. It basically means “ Farmer , etc.”

makeMove

So let’s get down to the meat of our journey. How do we make a move?

makeMove :: Plan -> [ Plan ]

makeMove will be a function that takes a plan and returns all the successful ways you can add a move to that plan. It takes a plan and takes it through a journey of adding a move, and returns the results of all of the successful ways it can fulfill this journey. This is similar to our old halveOrDouble :: Int -> [Int] , which takes an int and returns the successful paths our int could have taken (it could have been halved…or doubled).

What does a plan have to “go through” in its journey in adding a move?

First, we get the move we want to add. We could pick a MoveThe Farmer , a MoveThe Goat , or anything! Then, we fail/end the journey if we pick a move that isn’t legal. For example, we can’t move the goat if the farmer is not on the same side of the river that the goat is on. Now, we add that move that we got to the plan. Then, we fail/end the journey if that new plan is “unsafe” — if it leaves either the Wolf and Goat alone on a riverbank or the Goat and Cabbage. At the end of it all, we succeed with the new plan.

Let’s try this out:

moveLegal :: Plan -> Move -> Bool -- 1 safePlan :: Plan -> Bool makeMove :: Plan -> [ Plan ] = do makeMove p <- MoveThe <$> [ Farmer .. Cabbage ] -- 2 next $ moveLegal p next -- 3 guardmoveLegal p next let = p ++ [next] -- 4 p'[next] $ safePlan p' -- 5 guardsafePlan p' return p' -- 6 p'

Here are the types of the helper functions we will be using. In this context, MoveThe <$> means to apply MoveThe to whatever we choose out of [Farmer .. Cabbage] . Kind of an “intercept it on the way out, and turn it into a Move”. So next is MoveThe Farmer or MoveThe Wolf , etc.; next is one of those. For every journey, we pick one of the possible moves. We insta-fail if the move is not legal with the given plan. By this, we mean that we can’t possibly move an animal unless the farmer is on the same side as the animal. Let’s let p' be next appended to the original plan p . We insta-fail unless the new plan is safe. If we haven’t failed yet, then we succeed with the new plan as the result.

Welcome to Haskell! Okay, so I was slightly hand-wavey with <$> . But it is true that something like: x <- ( * 2 ) <$> Just 3 will put 6 ( 3 * 2 ) into x — it’ll take out the 3 and then apply (*2) to it before storing it in x . What’s going on under the hood is actually less magical. <$> basically says “apply inside”. It is like $ , but “inside”. Remember how we can do: > ( * 2 ) $ 3 ghci 6 to apply (*2) to 3? We can then also do: > ( * 2 ) $ 3 ghci 6 > ( * 2 ) <$> Just 3 ghci Just 6 > ( * 2 ) <$> [ 3 ] ghci [ 6 ] Now, if we think of a List like a list of possible successes, then applying a function “inside” means applying the function to all of the possible successes: > ( * 2 ) <$> [ 3 , 4 , 5 ] ghci [ 6 , 8 , 10 ] > MoveThe $ Farmer ghci MoveThe Farmer > MoveThe <$> [ Farmer , Wolf , Goat , Cabbage ] ghci [ MoveThe Farmer , MoveThe Wolf , MoveThe Goat , MoveThe Cabbage ] So when I say <- MoveThe <$> [ Farmer , Wolf , Goat , Cabbage ] next I really mean <- [ MoveThe Farmer , MoveThe Wolf , MoveThe Goat , MoveThe Cabbage ] next But still, it sometimes is cool to think of it as “Get the item inside, and then apply this function to it before you bind it to your variable”, if only for funsies.

Thought experiment

So let’s say our plan is, currently, [MoveThe Goat, MoveThe Farmer, MoveThe Wolf] . At the end of it all, our goat, wolf, and farmer are on the east bank, and the cabbage is on the west bank.

What happens on a typical journey of makeMove ?

First, we pick something to move. Let’s say next is MoveThe Farmer . This move is legal (moving the farmer is always legal). Our new plan is [MoveThe Goat, MoveThe Farmer, MoveThe Wolf, MoveThe Farmer] This plan is not safe. If we move the farmer, the goat and the wolf will be alone, and that is bad news for the goat. We fail at the second guard. We don’t return anything, because this journey is a total and utter failure.

Huh. How unfortunate. Let’s try again with another pick for next :

Let’s pick MoveThe Cabbage this time for next . This move isn’t even legal! The cabbage is on the west bank but the farmer is on the east. Failure!

Well, that’s kind of depressing. Let’s try another:

We pick MoveThe Goat for next . This move is legal; both the goat and the farmer are on the east bank. Our new plan is [MoveThe Goat, MoveThe Farmer, MoveThe Wolf, MoveThe Goat] . This plan is indeed safe. The goat and the cabbage are now on the west bank, but so is the farmer. Because all is well, we return our new plan!

Hooray!

As an exercise, see how the journey fares if we had picked MoveThe Wolf for next .

Anyways, at the end of it all, makeMove will return all new plans from the successful journeys. So it won’t be returning the plans with MoveThe Farmer and MoveThe Cabbage added to it, but will likely be retuning the plans with MoveThe Goat and MoveThe Wolf added to it. And it’ll return those two together in a List structure.

We’re almost there! Now to just define our helper predicates moveLegal and safePlan .

moveLegal

What makes a move legal? Well, the farmer has to be on the same side as whatever is being moved.

We can re-use our positionOf :: Plan -> Character -> Position function here.

moveLegal :: Plan -> Move -> Bool MoveThe Farmer ) = True moveLegal p ( MoveThe c) = positionOf p c == positionOf p Farmer moveLegal p (c)positionOf p cpositionOf p

safePlan

One last piece. How can we tell if a plan is safe or not?

The plan is safe if nothing can eat anything else. That means if the wolf and goat or goat and cabbage sit on the same bank, so too must the farmer. Some boolean arithmetic will show that this is the same as if either the farmer is on the same side as the goat or the goat and cabbage are both “safe” (not on the side of their predators).

safePlan :: Plan -> Bool = goatPos == farmerPos || safeGoat && safeCabbage safePlan pgoatPosfarmerPossafeGoatsafeCabbage where = positionOf p Goat goatPospositionOf p = positionOf p Farmer farmerPospositionOf p = goatPos /= positionOf p Wolf safeGoatgoatPospositionOf p = positionOf p Cabbage /= goatPos safeCabbagepositionOf pgoatPos

And…that’s it! We finished!

Exercise

Notice that sometimes we are going to make “redundant moves”. For example, we could move the farmer or goat twice in a row. How can we add another guard to check if the move isn’t redundant? That is, that the move we are adding isn’t identical to the last move of the plan?

The implementation is in the final solution later on, but think about how you would do it and compare the final solution to yours!

Wrapping Up

The final code for this project is available on Github so you can follow along yourself. You can also load it interactively online on FPComplete, a great online Haskell IDE where you can test your code right there in the browser.

So…let’s test it!

Tests

First, let’s load it up on ghci:

> : l WolfGoatCabbage.hs ghcil WolfGoatCabbage.hs Ok , modules loaded : Main . , modules loaded

Let’s try a few plan lengths and see when we get one that has a valid solution:

> findSolutions 5 ghcifindSolutions [] > findSolutions 6 ghcifindSolutions [] > findSolutions 7 ghcifindSolutions G , F , W , G , C , F , G ],[ G , F , C , G , W , F , G ]] [[],[]]

Great, we have two solutions of length 7. If we try them out, it seems like they both work! Notice that, interestingly enough, the two solutions are their own reverses. This makes sense, because any solution of getting from the west bank to the east bank must also be, backwards, a valid solution of getting from the east bank to the west bank.

It turns out that the solutions of length 9 and 11 are both identical to the solutions for length 7, just with some redundant moves thrown in (moving the farmer twice in a row, moving the goat twice in a row, etc.). Also, note that all possible solutions are of odd lengths, because for even lengths, the farmer ends up on the west bank.

If we add the filter on redundant moves mentioned earlier, the next valid solutions with no direct redundancies come at length 13, and then at 19:

> findSolutions 13 ghcifindSolutions G , F , W , G , C , W , G , C , W , G , C , F , G ] [[ G , F , C , G , W , C , G , W , C , G , W , F , G ]] ,[]] > findSolutions 19 ghcifindSolutions G , F , W , G , C , W , G , C , W , G , C , W , G , C , W , G , C , F , G ] [[ G , F , C , G , W , C , G , W , C , G , W , C , G , W , C , G , W , F , G ]] ,[]]

Again note that both of these solutions come in pairs, with one being the reverse of the other. Also curious is the fact that they are actually identical to the length 7 solutions, just with cycles of W,G,C (or C,G,W ) over and over again in the middle.

Reflections

We have solved the classic logic puzzle without using any control flow other than the List’s MonadPlus instance. The solution isn’t necessarily optimal, but it is interesting that we can model something like this simply as saying: “Here is the description of a journey. What journeys following this description succeed?”

With the List MonadPlus, you can solve any problem that can be described as the result of a nondeterministic journey with choices and pitfalls along the way.

In this particular puzzle, you could have done something similar from the start using only maps and filters. However, sometimes it is more useful or more insightful to, instead of using maps and filters, use abstractions that help you frame the problem in a more meaningful way.

Hopefully as a result of this three part series and through playing around with the source code, you can appreciate the wonders of Succeed/Fail and MonadPlus!

The future