learn-you-a-haskell-remastered
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 Haskell 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 Haskell
(+, `elem`, 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.
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 a Haskell 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 :: (Num a) => String -> a.
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 Haskell, 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 Haskell, 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 Data.List
solveRPN :: (Num a) => String -> a
solveRPN expression = head (foldl foldingFunction [] (words expression))
where foldingFunction stack item = ...
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 call head on that list to get the
item out and then we apply read.
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 Data.List
solveRPN :: (Num a) => String -> a
solveRPN = head . foldl foldingFunction [] . words
where foldingFunction stack item = ...
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 "-".
solveRPN :: (Num a, Read a) => String -> a
solveRPN = head . foldl foldingFunction [] . words
where foldingFunction (x:y:ys) "*" = (x * y):ys
foldingFunction (x:y:ys) "+" = (x + y):ys
foldingFunction (x:y:ys) "-" = (y - x):ys
foldingFunction xs numberString = read numberString:xs
We laid this 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 [3,4,9,3] and call its
first two elements x and y respectively. So in this case, x would be 3
and y would be 4. ys would be [9,3]. It will return a list that’s just
like ys, only it has x and y multiplied 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. If it’s a number, we just call read on that string
to get a number from it and return the previous stack but with that
number pushed to the top.
And that’s it! Also noticed that we added an extra class constraint of
Read a to the function declaration, because we call read on our string
to get the number. So this declaration means that the result can be of
any type that’s part of the Num and Read typeclasses (like Int, Float,
etc.).
For the list of items ["2","3","+"], our function will start folding
from the left. The initial 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 the added to the
beginning of []. So the new stack is now [2] and the folding function
will be called with [2] as the stack and ["3"] as the item, producing a
new stack of [3,2]. Then, it’s called for the third time with [3,2] 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
[5], which is the number that we return.
Let’s play around with our function:
ghci> solveRPN "10 4 3 + 2 * -"
-4
ghci> solveRPN "2 3 +"
5
ghci> solveRPN "90 34 12 33 55 66 + * - +"
-3947
ghci> solveRPN "90 34 12 33 55 66 + * - + -"
4037
ghci> solveRPN "90 34 12 33 55 66 + * - + -"
4037
ghci> solveRPN "90 3 -"
87
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.
import Data.List
solveRPN :: String -> Float
solveRPN = head . foldl foldingFunction [] . words
where foldingFunction (x:y:ys) "*" = (x * y):ys
foldingFunction (x:y:ys) "+" = (x + y):ys
foldingFunction (x:y:ys) "-" = (y - x):ys
foldingFunction (x:y:ys) "/" = (y / x):ys
foldingFunction (x:y:ys) "^" = (y ** x):ys
foldingFunction (x:xs) "ln" = log x:xs
foldingFunction xs "sum" = [sum xs]
foldingFunction xs numberString = read numberString:xs
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.
ghci> solveRPN "2.7 ln"
0.9932518
ghci> solveRPN "10 10 10 10 sum 4 /"
10.0
ghci> solveRPN "10 10 10 10 10 sum 4 /"
12.5
ghci> solveRPN "10 2 ^"
100.0
Notice that we can include floating point numbers in our expression
because read knows how to read them.
ghci> solveRPN "43.2425 0.5 ^"
6.575903
I think that making a function that can calculate arbitrary floating point RPN expressions and has the option to be easily extended in 10 lines is pretty awesome.
One thing to note about this function is that it’s not really fault
tolerant. When given input that doesn’t make sense, it will just crash
everything. We’ll make a fault tolerant version of this with a type
declaration of solveRPN :: String -> Maybe Float once we get to know
monads (they’re not scary, trust me!). We could make one right now, but
it would be a bit tedious because it would involve a lot of checking for
Nothing on every step. If you’re feeling up to the challenge though, you
can go ahead and try it! Hint: you can use reads to see if a read was
successful or not.
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.
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:
50
10
30
5
90
20
40
2
25
10
8
0
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:
- Forget Haskell for a minute and think about how we’d solve the problem by hand
- Think about how we’re going to represent our data in Haskell
- Figure out how to operate on that data in Haskell so that we produce at a solution
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 over 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 Haskell’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.
data Node = Node Road Road | EndNode Road
data 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.
data Node = Node Road (Maybe Road)
data Road = Road Int Node
This is an alright way to represent the road system in Haskell 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!
data Section = Section { getA :: Int, getB :: Int, getC :: Int } deriving (Show)
type RoadSystem = [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 algebraic data type that holds three integers for the lengths of
its three road parts. We introduce a type synonym 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 algebraic 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 useSectionandVectordata 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 Haskell. 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 synonym: Path.
data Label = A | B | C deriving (Show)
type Path = [(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:
[(B,10),(C,30),(A,5),(C,20),(B,2),(B,8)]
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 involved 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 pair of paths and a section and produces a new pair of paths.
The type is (Path, Path) -> Section -> (Path, Path). Let’s go ahead
and implement this function, because it’s bound to be useful.
Hint: it will be useful because
(Path, Path) -> Section -> (Path, Path)can be used as the binary function for a left fold, which has to have a type ofa -> b -> a
roadStep :: (Path, Path) -> Section -> (Path, Path)
roadStep (pathA, pathB) (Section a b c) =
let priceA = sum $ map snd pathA
priceB = sum $ map snd pathB
forwardPriceToA = priceA + a
crossPriceToA = priceB + b + c
forwardPriceToB = priceB + b
crossPriceToB = priceA + a + c
newPathToA = if forwardPriceToA <= crossPriceToA
then (A,a):pathA
else (C,c):(B,b):pathB
newPathToB = if forwardPriceToB <= crossPriceToB
then (B,b):pathB
else (C,c):(A,a):pathA
in (newPathToA, newPathToB)
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 $ map snd 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,a):pathA. Basically we
prepend the Label A and the section length a to the optimal path 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 a has a type of Int. Why do
we prepend instead of doing pathA ++ [(A,a)]? 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.
ghci> roadStep ([], []) (head heathrowToLondon)
([(C,30),(B,10)],[(B,10)])
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 $ map snd pathA, we’re calculating the price from the path on every step. We wouldn’t have to do that if we implementedroadStepas a(Path, Path, Int, Int) -> Section -> (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 =
let (bestAPath, bestBPath) = foldl roadStep ([],[]) roadSystem
in if sum (map snd bestAPath) <= sum (map snd bestBPath)
then reverse bestAPath
else 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!
ghci> optimalPath heathrowToLondon
[(B,10),(C,30),(A,5),(C,20),(B,2),(B,8),(C,0)]
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 based on, 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
[1..10], groupsOf 3 should return [[1,2,3],[4,5,6],[7,8,9],[10]].
groupsOf :: Int -> [a] -> [[a]]
groupsOf 0 _ = undefined
groupsOf _ [] = []
groupsOf n xs = take n xs : groupsOf n (drop n xs)
A standard recursive function. For an xs of [1..10] and an n of 3, this
equals [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. And here’s our main function,
which reads from the standard input, makes a RoadSystem out of it and
prints out the shortest path:
import Data.List
main = do
contents <- getContents
let threes = groupsOf 3 (map read $ lines contents)
roadSystem = map (\[a,b,c] -> Section a b c) threes
path = optimalPath roadSystem
pathString = concat $ map (show . fst) path
pathPrice = sum $ map snd path
putStrLn $ "The best path to take is: " ++ pathString
putStrLn $ "The price is: " ++ show pathPrice
First, we get all the contents from the standard input. Then, we call
lines with our contents to convert something like "50\n10\n30\n... to
["50","10","30".. and then we map read to that to convert it to a list
of numbers. We call groupsOf 3 on it so that we turn it to a list of
lists of length 3. We map the lambda (\[a,b,c] -> Section a b c) over
that list of lists. As you can see, the lambda just takes a list of
length 3 and turns it into a section. So roadSystem is now our system of
roads and it even has the correct type, namely RoadSystem (or
[Section]). We call optimalPath with that and then get the path and the
price in a nice textual representation and print it out.
We save the following text
50
10
30
5
90
20
40
2
25
10
8
0
in a file called paths.txt and then feed it to our program.
$ cat paths.txt | runhaskell heathrow.hs
The best path to take is: BCACBBC
The price is: 75
Works like a charm! You can use your knowledge of the Data.Random module
to generate a much longer system of roads, which you can then feed to
what we just wrote. If you get stack overflows, try using foldl' instead
of foldl, because foldl' is strict.