Functionally Solving Problems

In this chapter, we’ll take a look at a few interesting problems and how to think functionally to solve them as elegantly as possible. We probably won’t be introducing any new concepts, we’ll just be flexing our newly acquired Elm muscles and practicing our coding skills. Each section will present a different problem. First we’ll describe the problem, then we’ll try and find out what the best (or least worst) way of solving it is.

Reverse Polish notation calculator

Usually when we write mathematical expressions in school, we write them in an infix manner. For instance, we write 10 - (4 + 3) * 2. +, * and - are infix operators, just like the infix functions we met in Elm (+, <|, <<, etc.). This makes it handy because we, as humans, can parse it easily in our minds by looking at such an expression. The downside to it is that we have to use parentheses to denote precedence.

Reverse Polish notation is another way of writing down mathematical expressions. Initially it looks a bit weird, but it’s actually pretty easy to understand and use because there’s no need for parentheses and it’s very easy to punch into a calculator. While most modern calculators use infix notation, some people still swear by RPN calculators. This is what the previous infix expression looks like in RPN: 10 4 3 + 2 * -. How do we calculate what the result of that is? Well, think of a stack. You go over the expression from left to right. Every time a number is encountered, push it on to the stack. When we encounter an operator, take the two numbers that are on top of the stack (we also say that we pop them), use the operator and those two and then push the resulting number back onto the stack. When you reach the end of the expression, you should be left with a single number if the expression was well-formed and that number represents the result.

this expression

Let’s go over the expression 10 4 3 + 2 * - together! First we push 10 on to the stack and the stack is now 10. The next item is 4, so we push it to the stack as well. The stack is now 10, 4. We do the same with 3 and the stack is now 10, 4, 3. And now, we encounter an operator, namely +! We pop the two top numbers from the stack (so now the stack is just 10), add those numbers together and push that result to the stack. The stack is now 10, 7. We push 2 to the stack, the stack for now is 10, 7, 2. We’ve encountered an operator again, so let’s pop 7 and 2 off the stack, multiply them and push that result to the stack. Multiplying 7 and 2 produces a 14, so the stack we have now is 10, 14. Finally, there’s a -. We pop 10 and 14 from the stack, subtract 14 from 10 and push that back. The number on the stack is now -4 and because there are no more numbers or operators in our expression, that’s our result!

Now that we know how we’d calculate any RPN expression by hand, let’s think about how we could make an Elm function that takes as its parameter a string that contains a RPN expression, like "10 4 3 + 2 * -" and gives us back its result.

What would the type of that function be? We want it to take a string as a parameter and produce a number as its result. So it will probably be something like solveRPN : String -> number.

Protip: it really helps to first think what the type declaration of a function should be before concerning ourselves with the implementation and then write it down. In Elm, a function’s type declaration tells us a whole lot about the function, due to the very strong type system.


Cool. When implementing a solution to a problem in Elm, it’s also good to think back on how you did it by hand and maybe try to see if you can gain any insight from that. Here we see that we treated every number or operator that was separated by a space as a single item. So it might help us if we start by breaking a string like "10 4 3 + 2 * -" into a list of items like ["10","4","3","+","2","*","-"].

Next up, what did we do with that list of items in our head? We went over it from left to right and kept a stack as we did that. Does the previous sentence remind you of anything? Remember, in the section about folds, we said that pretty much any function where you traverse a list from left to right or right to left one element by element and build up (accumulate) some result (whether it’s a number, a list, a stack, whatever) can be implemented with a fold.

In this case, we’re going to use a left fold, because we go over the list from left to right. The accumulator value will be our stack and hence, the result from the fold will also be a stack, only as we’ve seen, it will only have one item.

One more thing to think about is, well, how are we going to represent the stack? I propose we use a list. Also I propose that we keep the top of our stack at the head of the list. That’s because adding to the head (beginning) of a list is much faster than adding to the end of it. So if we have a stack of, say, 10, 4, 3, we’ll represent that as the list [3,4,10].

Now we have enough information to roughly sketch our function. It’s going to take a string, like, "10 4 3 + 2 * -" and break it down into a list of items by using words to get ["10","4","3","+","2","*","-"]. Next, we’ll do a left fold over that list and end up with a stack that has a single item, so [-4]. We take that single item out of the list and that’s our final result!

So here’s a sketch of that function:

import List exposing (..)
import String exposing (words)

solveRPN : String -> Maybe number
solveRPN expression =
    let foldingFunction stack item = ...
    head (foldl foldingFunction [] (words expression))

We take the expression and turn it into a list of items. Then we fold over that list of items with the folding function. Mind the [], which represents the starting accumulator. The accumulator is our stack, so [] represents an empty stack, which is what we start with. After getting the final stack with a single item, we then call head on that list to get the item out… maybe.

Something interesting happened here. Originally we were thinking the type signature for this function might look something like String -> number. However, as we started to sketch out an implementation, our function had to adapt to meet real world requirements. As we discussed earlier, the head function returns a Maybe type, as otherwise it would not have any kind of meaningful value to return when passed an empty list. We could have eliminated this type from our function, e.g. by using the Maybe.withDefault function to return a default value (0 seems like it would be a natural fit) when passed an empty expression. However, as we’ll see later, due to the possibility of failure when parsing the expression, this type signature may not be such a bad fit after all.

So all that’s left now is to implement a folding function that will take a stack, like [4,10], and an item, like "3" and return a new stack [3,4,10]. If the stack is [4,10] and the item "*", then it will have to return [40]. But before that, let’s turn our function into point-free style because it has a lot of parentheses that are kind of freaking me out:

import List exposing (..)
import String exposing (words)

solveRPN : String -> number
solveRPN = 
        foldingFunction stack item = ...
    head << foldl foldingFunction [] << words

Ah, there we go. Much better. So, the folding function will take a stack and an item and return a new stack. We’ll use pattern matching to get the top items of a stack and to pattern match against operators like "*" and "-". Also, as mentioned earlier, our fold finction will be performing an operation String.toFloat which can potentially fail (e.g. if it’s passed a string which can’t be converted to a number). We could allow this possibility for failure to be reflected in our function’s type (that’s what the Result type represents), but since we’re already dealing with Maybe’s, we can also convert this Result to a Maybe, just to keep things easy for ourselves. As such, we’ll introduce the flatten function, which has a signature like Maybe (Maybe a) -> Maybe a, to help wrangle nested Maybe’s.

solveRPN : String -> Maybe number
solveRPN =
        flatten maybe = case maybe of
            Just x -> x 
            Nothing -> Nothing
        foldingFunction item stack = case (item, stack) of
            ("*", (x::y::ys)) ->
                (Maybe.map2 (\a b -> (a * b)) x y) :: ys

            ("+", (x::y::ys)) ->
                (Maybe.map2 (\a b -> (a + b)) x y) :: ys

            ("-", (x::y::ys)) ->
                (Maybe.map2 (\a b -> (b - a)) x y) :: ys

            (numberString, xs) -> 
                (Result.toMaybe <| String.toFloat numberString) :: xs
    flatten << head << foldl foldingFunction [] << words

With a concrete implementation in hand, you’ll notice that our type signature has changed again slightly. Our String.toFloat conversion function returns a Result which is a type that represents the possibility of failure. It can either be OK value (meaning, in this case, that it was able to convert the String to a Float) or it can be Err error (as would happen here if, e.g. you tried to convert the string "foobar"), with value and error representing the value of a successfully completed computation and an error respectively. We’re converting this to a Maybe, so nothing really new there. However, the conversion functions available to us all have concrete numeric types (either String.toInt : String -> Result Int, String.toFloat : String -> Result Float), so our hand is forced a bit, and we need to choose one or the other.

So, we laid our foldingFunction out as four patterns. The patterns will be tried from top to bottom. First the folding function will see if the current item is "*". If it is, then it will take a list like [Just 3,Just 4,Just 9,Just 3] and Maybe.map2 its first two elements x and y respectively. So in this case, x would be Just 3 and y would be Just 4. ys would be [Just 9,Just 3]. It will return a list that’s just like ys, only it has Just (x * y) as its head. So with this we pop the two topmost numbers off the stack, multiply them and push the result back on to the stack. If the item is not "*", the pattern matching will fall through and "+" will be checked, and so on.

If the item is none of the operators, then we assume it’s a string that represents a number. We call String.toFloat on that string to get a number from it and return the previous stack but with Just that number pushed to the top. If it wasn’t a number, we return Nothing, which is what will ultimately bubble up as the value returned from solveRPN.

And that’s it!

For the list of items ["2","3","+"], our function will start folding from the left. The intial stack will be []. It will call the folding function with [] as the stack (accumulator) and "2" as the item. Because that item is not an operator, it will be read and then added to the beginning of []. So the new stack is now [Just 2] and the folding function will be called with [Just 2] as the stack and "3" as the item, producing a new stack of [Just 3, Just 2]. Then, it’s called for the third time with [Just 3,Just 2] as the stack and "+" as the item. This causes these two numbers to be popped off the stack, added together and pushed back. The final stack is [Just 5], which is the number that we return.

Let’s play around with our function:

> solveRPN "10 4 3 + 2 * -"
Just -4 : Maybe.Maybe Float
> solveRPN "2 3 +"
Just 5 : Maybe.Maybe Float
> solveRPN "90 34 12 33 55 66 + * - +"
Just -3947 : Maybe.Maybe Float
> solveRPN "90 34 12 33 55 66 + * - + -"
Just 4037 : Maybe.Maybe Float
> solveRPN "90 34 12 33 55 66 + * - + -"
Just 4037 : Maybe.Maybe Float
> solveRPN "90 3 -"
Just 87 : Maybe.Maybe Float
> solveRPN "90 34 12 foo 55 66 bar * - baz "
Nothing : Maybe.Maybe Float

Cool, it works! One nice thing about this function is that it can be easily modified to support various other operators. They don’t even have to be binary operators. For instance, we can make an operator "log" that just pops one number off the stack and pushes back its logarithm. We can also make a ternary operators that pop three numbers off the stack and push back a result or operators like "sum" which pop off all the numbers and push back their sum.

Let’s modify our function to take a few more operators. For simplicity’s sake, we’ll change its type declaration so that it returns a number of type Float.

solveRPN : String -> Maybe Float
solveRPN =
        flatten maybe = case maybe of
            Just x -> x 
            Nothing -> Nothing
        foldingFunction item stack = case (item, stack) of
            ("*", (x::y::ys)) ->
                (Maybe.map2 (\a b -> (a * b)) x y) :: ys

            ("+", (x::y::ys)) ->
                (Maybe.map2 (\a b -> (a + b)) x y) :: ys

            ("-", (x::y::ys)) ->
                (Maybe.map2 (\a b -> (b - a)) x y) :: ys

            ("/", (x::y::ys)) ->
                (Maybe.map2 (\a b -> (b / a)) x y) :: ys

            ("^", (x::y::ys)) ->
                (Maybe.map2 (\a b -> (b ^ a)) x y) :: ys

            ("ln", (x::ys)) ->
                ( (\a -> (logBase e a)) x) :: ys

            ("sum", xs) ->
                [List.foldl (Maybe.map2 (+)) (Just 0) xs]

            (numberString, xs) -> 
                (Result.toMaybe <| String.toFloat numberString) :: xs
    flatten << head << foldl foldingFunction [] << words

Wow, great! / is division of course and ^ is floating point exponentiation. With the logarithm operator, we just pattern match against a single element and the rest of the stack because we only need one element to perform its natural logarithm. With the sum operator, we just return a stack that has only one element, which is the sum of the stack so far.

> solveRPN "2.7 ln"
Just 0.9932517730102834 : Maybe.Maybe Float
> solveRPN "10 10 10 10 sum 4 /"
Just 10 : Maybe.Maybe Float
> solveRPN "10 10 10 10 10 sum 4 /"
Just 12.5 : Maybe.Maybe Float
> solveRPN "10 2 ^"
Just 100 : Maybe.Maybe Float

Heathrow to London

Our next problem is this: your plane has just landed in England and you rent a car. You have a meeting really soon and you have to get from Heathrow Airport to London as fast as you can (but safely!).

There are two main roads going from Heathrow to London and there’s a number of regional roads crossing them. It takes you a fixed amount of time to travel from one crossroads to another. It’s up to you to find the optimal path to take so that you get to London as fast as you can! You start on the left side and can either cross to the other main road or go forward.

Heathrow - London

As you can see in the picture, the shortest path from Heathrow to London in this case is to start on main road B, cross over, go forward on A, cross over again and then go forward twice on B. If we take this path, it takes us 75 minutes. Had we chosen any other path, it would take more than that.

Our job is to make a program that takes input that represents a road system and print out what the shortest path across it is. Here’s what the input would look like for this case:


To mentally parse the input file, read it in threes and mentally split the road system into sections. Each section is comprised of a road A, road B and a crossing road. To have it neatly fit into threes, we say that there’s a last crossing section that takes 0 minutes to drive over. That’s because we don’t care where we arrive in London, as long as we’re in London.

Just like we did when solving the RPN calculator problem, we’re going to solve this problem in three steps:

In the RPN calculator section, we first figured out that when calculating an expression by hand, we’d keep a sort of stack in our minds and then go over the expression one item at a time. We decided to use a list of strings to represent our expression. Finally, we used a left fold to walk oevr the list of strings while keeping a stack to produce a solution.

Okay, so how would we figure out the shortest path from Heathrow to London by hand? Well, we can just sort of look at the whole picture and try to guess what the shortest path is and hopefully we’ll make a guess that’s right. That solution works for very small inputs, but what if we have a road that has 10,000 sections? Yikes! We also won’t be able to say for certain that our solution is the optimal one, we can just sort of say that we’re pretty sure.

That’s not a good solution then. Here’s a simplified picture of our road system:


Alright, can you figure out what the shortest path to the first crossroads (the first blue dot on A, marked A1) on road A is? That’s pretty trivial. We just see if it’s shorter to go directly forward on A or if it’s shorter to go forward on B and then cross over. Obviously, it’s cheaper to go forward via B and then cross over because that takes 40 minutes, whereas going directly via A takes 50 minutes. What about crossroads B1? Same thing. We see that it’s a lot cheaper to just go directly via B (incurring a cost of 10 minutes), because going via A and then crossing over would take us a whole 80 minutes!

Now we know what the cheapest path to A1 is (go via B and then cross over, so we’ll say that’s B, C with a cost of 40) and we know what the cheapest path to B1 is (go directly via B, so that’s just B, going at 10). Does this knowledge help us at all if we want to know the cheapest path to the next crossroads on both main roads? Gee golly, it sure does!

Let’s see what the shortest path to A2 would be. To get to A2, we’ll either go directly to A2 from A1 or we’ll go forward from B1 and then cross over (remember, we can only move forward or cross to the other side). And because we know the cost to A1 and B1, we can easily figure out what the best path to A2 is. It costs 40 to get to A1 and then 5 to get from A1 to A2, so that’s B, C, A for a cost of 45. It costs only 10 to get to B1, but then it would take an additional 110 minutes to go to B2 and then cross over! So obviously, the cheapest path to A2 is B, C, A. In the same way, the cheapest way to B2 is to go forward from A1 and then cross over.

Maybe you’re asking yourself: but what about getting to A2 by first crossing over at B1 and then going on forward? Well, we already covered crossing from B1 to A1 when we were looking for the best way to A1, so we don’t have to take that into account in the next step as well.

Now that we have the best path to A2 and B2, we can repeat this indefinitely until we reach the end. Once we’ve gotten the best paths for A4 and B4, the one that’s cheaper is the optimal path!

So in essence, for the second section, we just repeat the step we did at first, only we take into account what the previous best paths on A and B. We could say that we also took into account the best paths on A and on B in the first step, only they were both empty paths with a cost of 0.

Here’s a summary. To get the best path from Heathrow to London, we do this: first we see what the best path to the next crossroads on main road A is. The two options are going directly forward or starting at the opposite road, going forward and then crossing over. We remember the cost and the path. We use the same method to see what the best path to the next crossroads on main road B is and remember that. Then, we see if the path to the next crossroads on A is cheaper if we go from the previous A crossroads or if we go from the previous B crossroads and then cross over. We remember the cheaper path and then we do the same for the crossroads opposite of it. We do this for every section until we reach the end. Once we’ve reached the end, the cheapest of the two paths that we have is our optimal path!

So in essence, we keep one shortest path on the A road and one shortest path on the B road and when we reach the end, the shorter of those two is our path. We now know how to figure out the shortest path by hand. If you had enough time, paper and pencils, you could figure out the shortest path through a road system with any number of sections.

Next step! How do we represent this road system with Elm’s data types? One way is to think of the starting points and crossroads as nodes of a graph that point to other crossroads. If we imagine that the starting points actually point to each other with a road that has a length of one, we see that every crossroads (or node) points to the node on the other side and also to the next one on its side. Except for the last nodes, they just point to the other side.

type Node = Node Road Road | EndNode Road
type Road = Road Int Node

A node is either a normal node and has information about the road that leads to the other main road and the road that leads to the next node or an end node, which only has information about the road to the other main road. A road keeps information about how long it is and which node it points to. For instance, the first part of the road on the A main road would be Road 50 a1 where a1 would be a node Node x y, where x and y are roads that point to B1 and A2.

Another way would be to use Maybe for the road parts that point forward. Each node has a road part that point to the opposite road, but only those nodes that aren’t the end ones have road parts that point forward.

type Node = Node Road (Maybe Road)
type Road = Road Int Node

This is an alright way to represent the road system in Elm and we could certainly solve this problem with it, but maybe we could come up with something simpler? If we think back to our solution by hand, we always just checked the lengths of three road parts at once: the road part on the A road, its opposite part on the B road and part C, which touches those two parts and connects them. When we were looking for the shortest path to A1 and B1, we only had to deal with the lengths of the first three parts, which have lengths of 50, 10 and 30. We’ll call that one section. So the road system that we use for this example can be easily represented as four sections: 50, 10, 30, 5, 90, 20, 40, 2, 25, and 10, 8, 0.

It’s always good to keep our data types as simple as possible, although not any simpler!

type alias Section = { getA : Int, getB : Int, getC : Int }
type RoadSystem = List Section

This is pretty much perfect! It’s as simple as it goes and I have a feeling it’ll work perfectly for implementing our solution. Section is a simple record type that holds three integers for the lengths of its three road parts. We introduce abother type alias as well, saying that RoadSystem is a list of sections.

We could also use a triple of (Int, Int, Int) to represent a road section. Using tuples instead of making your own data types is good for some small localized stuff, but it’s usually better to make a new type for things like this. It gives the type system more information about what’s what. We can use (Int, Int, Int) to represent a road section or a vector in 3D space and we can operate on those two, but that allows us to mix them up. If we use Section and Vector data types, then we can’t accidentally add a vector to a section of a road system.

Our road system from Heathrow to London can now be represented like this:

heathrowToLondon : RoadSystem
heathrowToLondon = [Section 50 10 30, Section 5 90 20, Section 40 2 25, Section 10 8 0]

All we need to do now is to implement the solution that we came up with previously in Elm. What should the type declaration for a function that calculates a shortest path for any given road system be? It should take a road system as a parameter and return a path. We’ll represent a path as a list as well. Let’s introduce a Label type that’s just an enumeration of either A, B or C. We’ll also make a type alias: Path.

type Label = A | B | C
type alias Path = List (Label, Int)

Our function, we’ll call it optimalPath should thus have a type declaration of optimalPath : RoadSystem -> Path. If called with the road system heathrowToLondon, it should return the following path:


We’re going to have to walk over the list with the sections from left to right and keep the optimal path on A and optimal path on B as we go along. We’ll accumulate the best path as we walk over the list, left to right. What does that sound like? Ding, ding, ding! That’s right, A LEFT FOLD!

When doing the solution by hand, there was a step that we repeated over and over again. It involed checking the optimal paths on A and B so far and the current section to produce the new optimal paths on A and B. For instance, at the beginning the optimal paths were [] and [] for A and B respectively. We examined the section Section 50 10 30 and concluded that the new optimal path to A1 is [(B,10),(C,30)] and the optimal path to B1 is [(B,10)]. If you look at this step as a function, it takes a section and a pair of paths and produces a new pair of paths. The type is Section -> (Path, Path) -> (Path, Path). Let’s go ahead and implement this function, because it’s bound to be useful.

Hint: it will be useful because Section -> (Path, Path) -> (Path, Path) can be used as the binary function for a left fold, which has to have a type of b -> a -> a

roadStep : Section -> (Path, Path) -> (Path, Path)
roadStep {getA, getB, getC} (pathA, pathB) =
    let priceA = sum <| Tuple.second pathA
        priceB = sum <| Tuple.second pathB
        forwardPriceToA = priceA + getA
        crossPriceToA = priceB + getB + getC
        forwardPriceToB = priceB + getB
        crossPriceToB = priceA + getA + getC
        newPathToA = if forwardPriceToA <= crossPriceToA
                        then (A,getA)::pathA
                        else (C,getC)::(B,getB)::pathB
        newPathToB = if forwardPriceToB <= crossPriceToB
                        then (B,getB)::pathB
                        else (C,getC)::(A,getA)::pathA
    in (newPathToA, newPathToB)

this is you

What’s going on here? First, calculate the optimal price on road A based on the best so far on A and we do the same for B. We do sum <| Tuple.second pathA, so if pathA is something like [(A,100),(C,20)], priceA becomes 120. forwardPriceToA is the price that we would pay if we went to the next crossroads on A if we went there directly from the previous crossroads on A. It equals the best price to our previous A, plus the length of the A part of the current section. crossPriceToA is the price that we would pay if we went to the next A by going forward from the previous B and then crossing over. It’s the best price to the previous B so far plus the B length of the section plus the C length of the section. We determine forwardPriceToB and crossPriceToB in the same manner.

Now that we know what the best way to A and B is, we just need to make the new paths to A and B based on that. If it’s cheaper to go to A by just going forwards, we set newPathToA to be (A,getA):pathA. Basically we prepend the Label A and the section length getA to the optimal path on A so far. Basically, we say that the best path to the next A crossroads is the path to the previous A crossroads and then one section forward via A. Remember, A is just a label, whereas getA has a type of Int. Why do we prepend instead of doing pathA ++ [(A,getA)]? Well, adding an element to the beginning of a list (also known as consing) is much faster than adding it to the end. This means that the path will be the wrong way around once we fold over a list with this function, but it’s easy to reverse the list later. If it’s cheaper to get to the next A crossroads by going forward from road B and then crossing over, then newPathToA is the old path to B that then goes forward and crosses to A. We do the same thing for newPathToB, only everything’s mirrored.

Finally, we return newPathToA and newPathToB in a pair.

Let’s run this function on the first section of heathrowToLondon. Because it’s the first section, the best paths on A and B parameter will be a pair of empty lists.

> (roadStep ([], [])) <| (List.head heathrowToLondon)
Just ([(C,30),(B,10)],[(B,10)]) : Maybe.Maybe (Path, Path)

Remember, the paths are reversed, so read them from right to left. From this we can read that the best path to the next A is to start on B and then cross over to A and that the best path to the next B is to just go directly forward from the starting point at B.

Optimization tip: when we do priceA = sum <| Tuple.second pathA, we’re calculating the price from the path on every step. We wouldn’t have to do that if we implemented roadStep as a Section -> (Path, Path, Int, Int) -> (Path, Path, Int, Int) function where the integers represent the best price on A and B.

Now that we have a function that takes a pair of paths and a section and produces a new optimal path, we can just easily do a left fold over a list of sections. roadStep is called with ([],[]) and the first section and returns a pair of optimal paths to that section. Then, it’s called with that pair of paths and the next section and so on. When we’ve walked over all the sections, we’re left with a pair of optimal paths and the shorter of them is our answer. With this in mind, we can implement optimalPath.

optimalPath : RoadSystem -> Path
optimalPath roadSystem =
      (bestAPath, bestBPath) = List.foldl roadStep ([],[]) roadSystem
    if List.sum ( Tuple.second bestAPath) <=
        List.sum ( Tuple.second bestBPath)
            then List.reverse bestAPath
            else List.reverse bestBPath

We left fold over roadSystem (remember, it’s a list of sections) with the starting accumulator being a pair of empty paths. The result of that fold is a pair of paths, so we pattern match on the pair to get the paths themselves. Then, we check which one of these was cheaper and return it. Before returning it, we also reverse it, because the optimal paths so far were reversed due to us choosing consing over appending.

Let’s test this!

> optimalPath heathrowToLondon
[(B,10),(C,30),(A,5),(C,20),(B,2),(B,8),(C,0)] : Path

This is the result that we were supposed to get! Awesome! It differs from our expected result a bit because there’s a step (C,0) at the end, which means that we cross over to the other road once we’re in London, but because that crossing doesn’t cost anything, this is still the correct result.

We have the function that finds an optimal path, now we just have to read a textual representation of a road system from the standard input, convert it into a type of RoadSystem, run that through our optimalPath function and print the path.

First off, let’s make a function that takes a list and splits it into groups of the same size. We’ll call it groupsOf. For a parameter of List.range 1 10, groupsOf 3 should return Just [[1,2,3],[4,5,6],[7,8,9],[10]].

groupsOf : Int -> List a -> Maybe (List (List a))
groupsOf n list =
    if n <= 0 then
        case (n, list) of
        (_, []) ->
            Just []

        (n, xs) ->
   (\rest -> List.take n xs :: rest)
                (groupsOf n (List.drop n xs))

A standard recursive function. For an xs of List.range 1 10 and an n of 3, this equals Just [1,2,3] :: groupsOf 3 [4,5,6,7,8,9,10]. When the recursion is done, we get our list in groups of three. As you can see, our function isn’t well defined for n <= 0 (how would you form groups of -2?). This is reflected in the type signature. If we receive a grouping value that doesn’t make sense, we’ll return Nothing.

And here’s our logPath function, which takes a String, makes a RoadSystem out of it and returns a String message with the shortest path:

import List exposing (..)
import String exposing (lines)

combine : List (Maybe a) -> Maybe (List a)
combine =
    step e acc =
      case e of
        Nothing -> Nothing
        Just x -> ((::)x) acc
    List.foldr step (Just [])

flatten : Maybe (Maybe a) -> Maybe a
flatten maybe =
    case maybe of
        Just x -> x 
        Nothing -> Nothing

convertPath : Maybe (List (List String)) -> Maybe (List (List Int))
convertPath p =
  case p of
    Nothing -> Nothing
    Just xs ->
      <| map (combine << map (Result.toMaybe << String.toInt))
      <| xs

logPath pathInput =
        threes = convertPath <| groupsOf 3 (lines pathInput)
        roadSystem = flatten <| (combine << map (\ls -> 
          case ls of
            (a::b::c::xs) -> Just (Section a b c)
            _ -> Nothing
        )) threes
        path = optimalPath roadSystem
        pathString = (concat <| map Tuple.first) path
        pathPrice = (sum <| map Tuple.second) path
      Maybe.map2 (\path price ->
        "The best path to take is: " ++ (toString path) ++
        "The price is: " ++ (toString price)) pathString pathPrice  

First, we we call lines with our input string to convert something like "50\n10\n30\n..." to ["50","10","30"... and then we group the results by 3. Then we convert the list of strings to a list of numbers. Then we convert the list of lists to a list of sections. So roadSystem is now our system of roads and it even has the correct type, namely Maybe RoadSystem (or Maybe (List Section)). We call optimalPath with that and then get the path and the price in a nice textual representation.

We pass the following text

input = """50

to our program.

> case logPath input of \
|   Just result -> result \
|   Nothing -> ""
"The best path to take is: [B,C,A,C,B,B,C]\nThe price is: [10,30,5,20,2,8,0]" : String

Works like a charm! You can use your knowledge of the Random module to generate a much longer system of roads, which you can then feed to what we just wrote.