Hell Difficulty Part
Congratulations for getting so far! Now, some of the really hardcore stuff can start.
If you are like me, you should get the functional style. You should also understand a bit more the advantages of laziness by default. But you also don't really understand where to start in order to make a real program. And in particular:
- How do you deal with effects?
- Why is there a strange imperative-like notation for dealing with IO?
Be prepared, the answers might be complex. But they all be very rewarding.
Deal With IO
Too long; didn't read:
A typical function doing IO looks a lot like an imperative program:
f :: IO a
f = do
x <- action1
action2 x
y <- action3
action4 x y- To set a value to an object we use
<-. The type of each line is
IO *; in this example:action1 :: IO baction2 x :: IO ()action3 :: IO caction4 x y :: IO ax :: b,y :: c
- Few objects have the type
IO a, this should help you choose. In particular you cannot use pure functions directly here. To use pure functions you could doaction2 (purefunction x)for example.
In this section, I will explain how to use IO, not how it works. You'll see how Haskell separates the pure from the impure parts of the program.
Don't stop because you're trying to understand the details of the syntax. Answers will come in the next section.
What to achieve?
Ask a user to enter a list of numbers. Print the sum of the numbers
toList :: String -> [Integer]
toList input = read ("[" ++ input ++ "]")
main = do
putStrLn "Enter a list of numbers (separated by comma):"
input <- getLine
print $ sum (toList input)It should be straightforward to understand the behavior of this program. Let's analyze the types in more detail.
putStrLn :: String -> IO ()
getLine :: IO String
print :: Show a => a -> IO ()Or more interestingly, we note that each expression in the do block has a type of IO a.
main = do putStrLn "Enter ... " :: IO () getLine :: IO String print Something :: IO ()
We should also pay attention to the effect of the <- symbol.
do
x <- somethingIf something :: IO a then x :: a.
Another important note about using IO.
All lines in a do block must be of one of the two forms:
action1 :: IO a
-- in this case, generally a = ()or
value <- action2 -- where
-- bar z t :: IO b
-- value :: bThese two kinds of line will correspond to two different ways of sequencing actions. The meaning of this sentence should be clearer by the end of the next section.
Now let's see how this program behaves.
For example, what occur if the user enter something strange?
Try to write foo instead of a list of integer:
toList :: String -> [Integer]
toList input = read ("[" ++ input ++ "]")
main = do
putStrLn "Enter a list of numbers (separated by comma):"
input <- getLine
print $ sum (toList input)Argh! An evil error message and a crash! The first evolution will be to answer with a more friendly message.
In order to do this, we must detect that something went wrong.
Here is one way to do this.
Use the type Maybe.
It is a very common type in Haskell.
import Data.MaybeWhat is this thing? Maybe is a type which takes one parameter.
Its definition is:
data Maybe a = Nothing | Just aThis is a nice way to tell there was an error while trying to create/compute
a value.
The maybeRead function is a great example of this.
This is a function similar to the function read[^1],
but if something goes wrong the returned value is Nothing.
If the value is right, it returns Just <the value>.
Don't try to understand too much of this function.
I use a lower level function than read; reads.
maybeRead :: Read a => String -> Maybe a
maybeRead s = case reads s of
[(x,"")] -> Just x
_ -> NothingNow to be a bit more readable, we define a function which goes like this:
If the string has the wrong format, it will return Nothing.
Otherwise, for example for "1,2,3", it will return Just [1,2,3].
getListFromString :: String -> Maybe [Integer]
getListFromString str = maybeRead $ "[" ++ str ++ "]"We simply have to test the value in our main function.
import Data.Maybe
maybeRead :: Read a => String -> Maybe a
maybeRead s = case reads s of
[(x,"")] -> Just x
_ -> Nothing
getListFromString :: String -> Maybe [Integer]
getListFromString str = maybeRead $ "[" ++ str ++ "]"
main :: IO ()
-- show
main = do
putStrLn "Enter a list of numbers (separated by comma):"
input <- getLine
let maybeList = getListFromString input in
case maybeList of
Just l -> print (sum l)
Nothing -> error "Bad format. Good Bye."
-- /showIn case of error, we display a nice error message.
Note that the type of each expression in the main's do block remains of the form IO a.
The only strange construction is error.
I'll say error msg will simply take the needed type (here IO ()).
One very important thing to note is the type of all the functions defined so far.
There is only one function which contains IO in its type: main.
This means main is impure.
But main uses getListFromString which is pure.
It is then clear just by looking at declared types which functions are pure and
which are impure.
Why does purity matter? I certainly forget many advantages, but the three main reasons are:
- It is far easier to think about pure code than impure one.
- Purity protects you from all the hard to reproduce bugs due to side effects.
- You can evaluate pure functions in any order or in parallel without risk.
This is why you should generally put as most code as possible inside pure functions.
Our next evolution will be to prompt the user again and again until she enters a valid answer. We create a function which will ask the user for an list of integers until the input is right.
askUser :: IO [Integer]
askUser = do
putStrLn "Enter a list of numbers (separated by comma):"
input <- getLine
let maybeList = getListFromString input in
case maybeList of
Just l -> return l
Nothing -> askUserThis function is of type IO [Integer].
Such a type means that we retrieved a value of type [Integer] through some IO actions.
Some people might explain while waving their hands:
«This is an
[Integer]inside anIO»
If you want to understand the details behind all of this, you'll have to read the next section. But sincerely, if you just want to use IO. Just practice a little and remember to think about the type.
Finally our main function is quite simpler:
import Data.Maybe
maybeRead :: Read a => String -> Maybe a
maybeRead s = case reads s of
[(x,"")] -> Just x
_ -> Nothing
getListFromString :: String -> Maybe [Integer]
getListFromString str = maybeRead $ "[" ++ str ++ "]"
askUser :: IO [Integer]
askUser = do
putStrLn "Enter a list of numbers (separated by comma):"
input <- getLine
let maybeList = getListFromString input in
case maybeList of
Just l -> return l
Nothing -> askUser
-- show
main :: IO ()
main = do
list <- askUser
print $ sum list
-- /showWe have finished with our introduction to IO.
This was quite fast. Here are the main things to remember:
- in the
dobloc, each expression must have the typeIO a. You are then limited in the number of expressions available. For example,getLine,print,putStrLn, etc... - Try to externalize the pure functions as much as possible.
- the
IO atype means: an IO action which returns an element of typea.IOrepresents actions; under the hood,IO ais the type of a function. Read the next section if you are curious.
If you practice a bit, you should be able to use IO.
Exercises:
- Make a program that sums all of its arguments. Hint: use the function
getArgs.
IO trick explained
Here is a tldr for this section.
To separate pure and impure parts,
main is defined as a function
which modifies the state of the world
main :: World -> WorldA function is guaranteed to have side effects only if it has this type. But look at a typical main function:
main w0 =
let (v1,w1) = action1 w0 in
let (v2,w2) = action2 v1 w1 in
let (v3,w3) = action3 v2 w2 in
action4 v3 w3We have a lot of temporary elements (here w1, w2 and w3)
which must be passed on to the next action.
We create a function bind or (>>=).
With bind we don't need temporary names anymore.
main =
action1 >>= action2 >>= action3 >>= action4Bonus: Haskell has syntactical sugar for us:
main = do
v1 <- action1
v2 <- action2 v1
v3 <- action3 v2
action4 v3Why did we use this strange syntax, and what exactly is this IO type?
It looks a bit like magic.
For now let's just forget all about the pure parts of our program, and focus on the impure parts:
askUser :: IO [Integer]
askUser = do
putStrLn "Enter a list of numbers (separated by commas):"
input <- getLine
let maybeList = getListFromString input in
case maybeList of
Just l -> return l
Nothing -> askUser
main :: IO ()
main = do
list <- askUser
print $ sum listFirst remark; it looks like an imperative structure.
Haskell is powerful enough to make impure code look imperative.
For example, if you wish you could create a while in Haskell.
In fact, for dealing with IO, imperative style is generally more appropriate.
But you should had noticed the notation is a bit unusual. Here is why, in detail.
In an impure language, the state of the world can be seen as a huge hidden global variable. This hidden variable is accessible by all functions of your language. For example, you can read and write a file in any function. The fact that a file exists or not can be seen as different states of the world.
For Haskell this state is not hidden.
It is explicitly said main is a function that potentially changes the state of the world.
Its type is then something like:
main :: World -> WorldNot all functions may have access to this variable. Those which have access to this variable are impure. Functions to which the world variable isn't provided are pure[2].
Haskell considers the state of the world as an input variable to main.
But the real type of main is closer to this one[3]:
main :: World -> ((),World)The () type is the null type.
Nothing to see here.
Now let's rewrite our main function with this in mind:
main w0 =
let (list,w1) = askUser w0 in
let (x,w2) = print (sum list,w1) in
xFirst, we note that all functions which have side effects must have the type:
World -> (a,World)Where a is the type of the result.
For example, a getChar function should have the type World -> (Char,World).
Another thing to note is the trick to fix the order of evaluation.
In Haskell, in order to evaluate f a b, you have many choices:
- first eval
athenbthenf a b - first eval
bthenathenf a b. - eval
aandbin parallel thenf a b
This is true, because we should work in a pure language.
Now, if you look at the main function, it is clear you must eval the first line before the second one since, to evaluate the second line you have to get a parameter given by the evaluation of the first line.
Such trick works nicely.
The compiler will at each step provide a pointer to a new real world id.
Under the hood, print will evaluate as:
- print something on the screen
- modify the id of the world
- evaluate as
((),new world id).
Now, if you look at the style of the main function, it is clearly awkward. Let's try to do the same to the askUser function:
askUser :: World -> ([Integer],World)Before:
askUser :: IO [Integer]
askUser = do
putStrLn "Enter a list of numbers:"
input <- getLine
let maybeList = getListFromString input in
case maybeList of
Just l -> return l
Nothing -> askUserAfter:
askUser w0 =
let (_,w1) = putStrLn "Enter a list of numbers:" in
let (input,w2) = getLine w1 in
let (l,w3) = case getListFromString input of
Just l -> (l,w2)
Nothing -> askUser w2
in
(l,w3)This is similar, but awkward.
Look at all these temporary w? names.
The lesson, is, naive IO implementation in Pure functional languages is awkward!
Fortunately, there is a better way to handle this problem. We see a pattern. Each line is of the form:
let (y,w') = action x w inEven if for some line the first x argument isn't needed.
The output type is a couple, (answer, newWorldValue).
Each function f must have a type similar to:
f :: World -> (a,World)Not only this, but we can also note that we always follow the same usage pattern:
let (y,w1) = action1 w0 in
let (z,w2) = action2 w1 in
let (t,w3) = action3 w2 in
...Each action can take from 0 to n parameters. And in particular, each action can take a parameter from the result of a line above.
For example, we could also have:
let (_,w1) = action1 x w0 in
let (z,w2) = action2 w1 in
let (_,w3) = action3 x z w2 in
...And of course actionN w :: (World) -> (a,World).
IMPORTANT, there are only two important patterns to consider:
let (x,w1) = action1 w0 in let (y,w2) = action2 x w1 inand
let (_,w1) = action1 w0 in let (y,w2) = action2 w1 in
Now, we will do a magic trick.
We will make the temporary world symbol "disappear".
We will bind the two lines.
Let's define the bind function.
Its type is quite intimidating at first:
bind :: (World -> (a,World))
-> (a -> (World -> (b,World)))
-> (World -> (b,World))But remember that (World -> (a,World)) is the type for an IO action.
Now let's rename it for clarity:
type IO a = World -> (a, World)Some example of functions:
getLine :: IO String
print :: Show a => a -> IO ()getLine is an IO action which takes a world as parameter and returns a couple (String,World).
Which can be summarized as: getLine is of type IO String.
Which we also see as, an IO action which will return a String "embeded inside an IO".
The function print is also interesting.
It takes one argument which can be shown.
In fact it takes two arguments.
The first is the value to print and the other is the state of world.
It then returns a couple of type ((),World).
This means it changes the state of the world, but doesn't yield anymore data.
This type helps us simplify the type of bind:
bind :: IO a
-> (a -> IO b)
-> IO bIt says that bind takes two IO actions as parameter and return another IO action.
Now, remember the important patterns. The first was:
let (x,w1) = action1 w0 in
let (y,w2) = action2 x w1 in
(y,w2)Look at the types:
action1 :: IO a
action2 :: a -> IO b
(y,w2) :: IO bDoesn't it seem familiar?
(bind action1 action2) w0 =
let (x, w1) = action1 w0
(y, w2) = action2 x w1
in (y, w2)The idea is to hide the World argument with this function. Let's go: As an example imagine if we wanted to simulate:
let (line1,w1) = getLine w0 in
let ((),w2) = print line1 in
((),w2)Now, using the bind function:
(res,w2) = (bind getLine (\l -> print l)) w0As print is of type (World -> ((),World)), we know res = () (null type).
If you didn't see what was magic here, let's try with three lines this time.
let (line1,w1) = getLine w0 in
let (line2,w2) = getLine w1 in
let ((),w3) = print (line1 ++ line2) in
((),w3)Which is equivalent to:
(res,w3) = bind getLine (\line1 ->
bind getLine (\line2 ->
print (line1 ++ line2)))Didn't you notice something? Yes, no temporary World variables are used anywhere! This is MA. GIC.
We can use a better notation.
Let's use (>>=) instead of bind.
(>>=) is an infix function like
(+); reminder 3 + 4 ⇔ (+) 3 4
(res,w3) = getLine >>=
\line1 -> getLine >>=
\line2 -> print (line1 ++ line2)Ho Ho Ho! Happy Christmas Everyone! Haskell has made syntactical sugar for us:
do
x <- action1
y <- action2
z <- action3
...Is replaced by:
action1 >>= \x ->
action2 >>= \y ->
action3 >>= \z ->
...Note you can use x in action2 and x and y in action3.
But what about the lines not using the <-?
Easy, another function blindBind:
blindBind :: IO a -> IO b -> IO b
blindBind action1 action2 w0 =
bind action (\_ -> action2) w0I didn't simplify this definition for clarity purpose.
Of course we can use a better notation, we'll use the (>>) operator.
And
do
action1
action2
action3Is transformed into
action1 >>
action2 >>
action3Also, another function is quite useful.
putInIO :: a -> IO a
putInIO x = IO (\w -> (x,w))This is the general way to put pure values inside the "IO context".
The general name for putInIO is return.
This is quite a bad name when you learn Haskell. return is very different from what you might be used to.
To finish, let's translate our example:
import Data.Maybe
maybeRead :: Read a => String -> Maybe a
maybeRead s = case reads s of
[(x,"")] -> Just x
_ -> Nothing
getListFromString :: String -> Maybe [Integer]
getListFromString str = maybeRead $ "[" ++ str ++ "]"
-- show
askUser :: IO [Integer]
askUser = do
putStrLn "Enter a list of numbers (separated by commas):"
input <- getLine
let maybeList = getListFromString input in
case maybeList of
Just l -> return l
Nothing -> askUser
main :: IO ()
main = do
list <- askUser
print $ sum list
-- /show Is translated into:
import Data.Maybe
maybeRead :: Read a => String -> Maybe a
maybeRead s = case reads s of
[(x,"")] -> Just x
_ -> Nothing
getListFromString :: String -> Maybe [Integer]
getListFromString str = maybeRead $ "[" ++ str ++ "]"
-- show
askUser :: IO [Integer]
askUser =
putStrLn "Enter a list of numbers (sep. by commas):" >>
getLine >>= \input ->
let maybeList = getListFromString input in
case maybeList of
Just l -> return l
Nothing -> askUser
main :: IO ()
main = askUser >>=
\list -> print $ sum list
-- /show You can compile this code to verify it keeps working.
Imagine what it would look like without the (>>) and (>>=).
Monads
Now the secret can be revealed: IO is a monad.
Being a monad means you have access to some syntactical sugar with the do notation.
But mainly, you have access to a coding pattern which will ease the flow of your code.
Important remarks:
- Monad are not necessarily about effects! There are a lot of pure monads.
- Monad are more about sequencing
For the Haskell language Monad is a type class.
To be an instance of this type class, you must provide the functions (>>=) and return.
The function (>>) will be derived from (>>=).
Here is how the type class Monad is declared (mostly):
class Monad m where
(>>=) :: m a -> (a -> m b) -> m b
return :: a -> m a
(>>) :: m a -> m b -> m b
f >> g = f >>= \_ -> g
-- You should generally safely ignore this function
-- which I believe exists for historical reason
fail :: String -> m a
fail = errorRemarks:
- the keyword
classis not your friend. A Haskell class is not a class like in object model. A Haskell class has a lot of similarities with Java interfaces. A better word should have beentypeclass. That means a set of types. For a type to belong to a class, all functions of the class must be provided for this type.- In this particular example of type class, the type
mmust be a type that takes an argument. for exampleIO a, but alsoMaybe a,[a], etc...To be a useful monad, your function must obey some rules. If your construction does not obey these rules strange things might happens:
return a >>= k == k a m >>= return == m m >>= (\x -> k x >>= h) == (m >>= k) >>= h
Maybe is a monad
There are a lot of different types that are instance of Monad.
One of the easiest to describe is Maybe.
If you have a sequence of Maybe values, you can use monads to manipulate them.
It is particularly useful to remove very deep if..then..else.. constructions.
Imagine a complex bank operation. You are eligible to gain about 700€ only if you can afford to follow a list of operations without being negative.
deposit value account = account + value
withdraw value account = account - value
eligible :: (Num a,Ord a) => a -> Bool
eligible account =
let account1 = deposit 100 account in
if (account1 < 0)
then False
else
let account2 = withdraw 200 account1 in
if (account2 < 0)
then False
else
let account3 = deposit 100 account2 in
if (account3 < 0)
then False
else
let account4 = withdraw 300 account3 in
if (account4 < 0)
then False
else
let account5 = deposit 1000 account4 in
if (account5 < 0)
then False
else
True
main = do
print $ eligible 300 -- True
print $ eligible 299 -- FalseNow, let's make it better using Maybe and the fact that it is a Monad
deposit :: (Num a) => a -> a -> Maybe a
deposit value account = Just (account + value)
withdraw :: (Num a,Ord a) => a -> a -> Maybe a
withdraw value account = if (account < value)
then Nothing
else Just (account - value)
eligible :: (Num a, Ord a) => a -> Maybe Bool
eligible account = do
account1 <- deposit 100 account
account2 <- withdraw 200 account1
account3 <- deposit 100 account2
account4 <- withdraw 300 account3
account5 <- deposit 1000 account4
Just True
main = do
print $ eligible 300 -- Just True
print $ eligible 299 -- NothingNot bad, but we can make it even better:
deposit :: (Num a) => a -> a -> Maybe a
deposit value account = Just (account + value)
withdraw :: (Num a,Ord a) => a -> a -> Maybe a
withdraw value account = if (account < value)
then Nothing
else Just (account - value)
eligible :: (Num a, Ord a) => a -> Maybe Bool
eligible account =
deposit 100 account >>=
withdraw 200 >>=
deposit 100 >>=
withdraw 300 >>=
deposit 1000 >>
return True
main = do
print $ eligible 300 -- Just True
print $ eligible 299 -- NothingWe have proven that Monads are a good way to make our code more elegant.
Note this idea of code organization, in particular for Maybe can be used
in most imperative language.
In fact, this is the kind of construction we make naturally.
An important remark:
The first element in the sequence being evaluated to
Nothingwill stop the complete evaluation. This means you don't execute all lines. You have this for free, thanks to laziness.
You could also replay these example with the definition of (>>=) for Maybe
in mind:
instance Monad Maybe where
(>>=) :: Maybe a -> (a -> Maybe b) -> Maybe b
Nothing >>= _ = Nothing
(Just x) >>= f = f x
return x = Just xThe Maybe monad proved to be useful while being a very simple example.
We saw the utility of the IO monad.
But now a cooler example, lists.
The list monad
The list monad helps us to simulate non deterministic computations. Here we go:
import Control.Monad (guard)
allCases = [1..10]
resolve :: [(Int,Int,Int)]
resolve = do
x <- allCases
y <- allCases
z <- allCases
guard $ 4*x + 2*y < z
return (x,y,z)
main = do
print resolveMA. GIC.
For the list monad, there is also a syntactical sugar:
print $ [ (x,y,z) | x <- allCases,
y <- allCases,
z <- allCases,
4*x + 2*y < z ]I won't list all the monads, but there are many monads. Using monads simplifies the manipulation of several notions in pure languages. In particular, monad are very useful for:
- IO,
- non deterministic computation,
- generating pseudo random numbers,
- keeping configuration state,
- writing state,
- ...
If you have followed me until here, then you've done it! You know monads (Well, you'll certainly need to practice a bit to get used to them and to understand when you can use them and create your own. But you already made a big step in this direction.)!
Appendix
This section is not so much about learning Haskell. It is just here to discuss some details further.
More on Infinite Tree
In the section Infinite Structures we saw some simple constructions. Unfortunately we removed two properties from our tree:
- no duplicate node value
- well ordered tree
In this section we will try to keep the first property. Concerning the second one, we must relax it but we'll discuss how to keep it as much as possible.
Our first step is to create some pseudo-random number list:
shuffle = map (\x -> (x*3123) `mod` 4331) [1..]Just as a reminder, here is the definition of treeFromList
treeFromList :: (Ord a) => [a] -> BinTree a
treeFromList [] = Empty
treeFromList (x:xs) = Node x (treeFromList (filter (<x) xs))
(treeFromList (filter (>x) xs))and treeTakeDepth:
treeTakeDepth _ Empty = Empty
treeTakeDepth 0 _ = Empty
treeTakeDepth n (Node x left right) = let
nl = treeTakeDepth (n-1) left
nr = treeTakeDepth (n-1) right
in
Node x nl nrSee the result of:
import Data.List
data BinTree a = Empty
| Node a (BinTree a) (BinTree a)
deriving (Eq,Ord)
-- declare BinTree a to be an instance of Show
instance (Show a) => Show (BinTree a) where
-- will start by a '<' before the root
-- and put a : a begining of line
show t = "< " ++ replace '\n' "\n: " (treeshow "" t)
where
treeshow pref Empty = ""
treeshow pref (Node x Empty Empty) =
(pshow pref x)
treeshow pref (Node x left Empty) =
(pshow pref x) ++ "\n" ++
(showSon pref "`--" " " left)
treeshow pref (Node x Empty right) =
(pshow pref x) ++ "\n" ++
(showSon pref "`--" " " right)
treeshow pref (Node x left right) =
(pshow pref x) ++ "\n" ++
(showSon pref "|--" "| " left) ++ "\n" ++
(showSon pref "`--" " " right)
-- show a tree using some prefixes to make it nice
showSon pref before next t =
pref ++ before ++ treeshow (pref ++ next) t
-- pshow replace "\n" by "\n"++pref
pshow pref x = replace '\n' ("\n"++pref) (show x)
-- replace on char by another string
replace c new string =
concatMap (change c new) string
where
change c new x
| x == c = new
| otherwise = x:[] -- "x"
shuffle = map (\x -> (x*3123) `mod` 4331) [1..]
treeFromList :: (Ord a) => [a] -> BinTree a
treeFromList [] = Empty
treeFromList (x:xs) = Node x (treeFromList (filter (<x) xs))
(treeFromList (filter (>x) xs))
treeTakeDepth _ Empty = Empty
treeTakeDepth 0 _ = Empty
treeTakeDepth n (Node x left right) = let
nl = treeTakeDepth (n-1) left
nr = treeTakeDepth (n-1) right
in
Node x nl nr
-- show
main = do
putStrLn "take 10 shuffle"
print $ take 10 shuffle
putStrLn "\ntreeTakeDepth 4 (treeFromList shuffle)"
print $ treeTakeDepth 4 (treeFromList shuffle)
-- /showYay! It ends! Beware though, it will only work if you always have something to put into a branch.
For example
treeTakeDepth 4 (treeFromList [1..])will loop forever.
Simply because it will try to access the head of filter (<1) [2..].
But filter is not smart enought to understand that the result is the empty list.
Nonetheless, it is still a very cool example of what non strict programs have to offer.
Left as an exercise to the reader:
- Prove the existence of a number
nso thattreeTakeDepth n (treeFromList shuffle)will enter an infinite loop. - Find an upper bound for
n. - Prove there is no
shufflelist so that, for any depth, the program ends.
In order to resolve these problem we will modify slightly our
treeFromList and shuffle function.
A first problem, is the lack of infinite different number in our implementation of shuffle.
We generated only 4331 different numbers.
To resolve this we make a slightly better shuffle function.
shuffle = map rand [1..]
where
rand x = ((p x) `mod` (x+c)) - ((x+c) `div` 2)
p x = m*x^2 + n*x + o -- some polynome
m = 3123
n = 31
o = 7641
c = 1237This shuffle function has the property (hopefully) not to have an upper nor lower bound. But having a better shuffle list isn't enough not to enter an infinite loop.
Generally, we cannot decide whether filter (<x) xs is empty.
Then to resolve this problem, I'll authorize some error in the creation of our binary tree.
This new version of code can create binary tree which don't have the following property for some of its nodes:
Any element of the left (resp. right) branch must all be strictly inferior (resp. superior) to the label of the root.
Remark it will remains mostly an ordered binary tree. Furthermore, by construction, each node value is unique in the tree.
Here is our new version of treeFromList. We simply have replaced filter by safefilter.
treeFromList :: (Ord a, Show a) => [a] -> BinTree a
treeFromList [] = Empty
treeFromList (x:xs) = Node x left right
where
left = treeFromList $ safefilter (<x) xs
right = treeFromList $ safefilter (>x) xsThis new function safefilter is almost equivalent to filter but don't enter infinite loop if the result is a finite list.
If it cannot find an element for which the test is true after 10000 consecutive steps, then it considers to be the end of the search.
safefilter :: (a -> Bool) -> [a] -> [a]
safefilter f l = safefilter' f l nbTry
where
nbTry = 10000
safefilter' _ _ 0 = []
safefilter' _ [] _ = []
safefilter' f (x:xs) n =
if f x
then x : safefilter' f xs nbTry
else safefilter' f xs (n-1)Now run the program and be happy:
import Debug.Trace (trace)
import Data.List
data BinTree a = Empty
| Node a (BinTree a) (BinTree a)
deriving (Eq,Ord)
instance (Show a) => Show (BinTree a) where
-- will start by a '<' before the root
-- and put a : a begining of line
show t = "< " ++ replace '\n' "\n: " (treeshow "" t)
where
treeshow pref Empty = ""
treeshow pref (Node x Empty Empty) =
(pshow pref x)
treeshow pref (Node x left Empty) =
(pshow pref x) ++ "\n" ++
(showSon pref "`--" " " left)
treeshow pref (Node x Empty right) =
(pshow pref x) ++ "\n" ++
(showSon pref "`--" " " right)
treeshow pref (Node x left right) =
(pshow pref x) ++ "\n" ++
(showSon pref "|--" "| " left) ++ "\n" ++
(showSon pref "`--" " " right)
-- show a tree using some prefixes to make it nice
showSon pref before next t =
pref ++ before ++ treeshow (pref ++ next) t
-- pshow replace "\n" by "\n"++pref
pshow pref x = replace '\n' ("\n"++pref) (" " ++ show x)
-- replace on char by another string
replace c new string =
concatMap (change c new) string
where
change c new x
| x == c = new
| otherwise = x:[] -- "x"
treeTakeDepth _ Empty = Empty
treeTakeDepth 0 _ = Empty
treeTakeDepth n (Node x left right) = let
nl = treeTakeDepth (n-1) left
nr = treeTakeDepth (n-1) right
in
Node x nl nr
shuffle = map rand [1..]
where
rand x = ((p x) `mod` (x+c)) - ((x+c) `div` 2)
p x = m*x^2 + n*x + o -- some polynome
m = 3123
n = 31
o = 7641
c = 1237
treeFromList :: (Ord a, Show a) => [a] -> BinTree a
treeFromList [] = Empty
treeFromList (x:xs) = Node x left right
where
left = treeFromList $ safefilter (<x) xs
right = treeFromList $ safefilter (>x) xs
safefilter :: (a -> Bool) -> [a] -> [a]
safefilter f l = safefilter' f l nbTry
where
nbTry = 10000
safefilter' _ _ 0 = []
safefilter' _ [] _ = []
safefilter' f (x:xs) n =
if f x
then x : safefilter' f xs nbTry
else safefilter' f xs (n-1)
-- show
main = do
putStrLn "take 10 shuffle"
print $ take 10 shuffle
putStrLn "\ntreeTakeDepth 8 (treeFromList shuffle)"
print $ treeTakeDepth 8 (treeFromList $ shuffle)
-- /showYou should realize the time to print each value is different. This is because Haskell compute each value when it needs it. And in this case, this is when asked to print it on the screen.
Impressively enough, try to replace the depth from 8 to 100.
It will work without killing your RAM!
The flow and the memory management is done naturally by Haskell.
Left as an exercise to the reader:
- Even with large constant value for
deepandnbTry, it seems to work nicely. But in the worst case, it can be exponential. Create a worst case list to give as parameter totreeFromList. hint: think about ([0,-1,-1,....,-1,1,-1,...,-1,1,...]). I first tried to implement
safefilteras follow:safefilter' f l = if filter f (take 10000 l) == [] then [] else filter f lExplain why it doesn't work and can enter into an infinite loop.- Suppose that
shuffleis real random list with growing bounds. If you study a bit this structure, you'll discover that with probability 1, this structure is finite. Using the following code (suppose we could usesafefilter'directly as if was not in the where of safefilter) find a definition offsuch that with probability1, treeFromList' shuffle is infinite. And prove it. Disclaimer, this is only a conjecture.
treeFromList' [] n = Empty
treeFromList' (x:xs) n = Node x left right
where
left = treeFromList' (safefilter' (<x) xs (f n)
right = treeFromList' (safefilter' (>x) xs (f n)
f = ???Thanks
Thanks to /r/haskell and
/r/programming.
Your comment were most than welcome.
Particularly, I want to thank Emm a thousand times for the time he spent on correcting my English. Thank you man.
