I am trying to write a code that generates all binary trees with n nodes (so the program has to return a list in which we can find all the different binary trees with n nodes).
Here is the way I represent binary trees :
type 'a tree = Empty | Node of 'a * 'a tree * 'a tree
So I am trying to implement a function all_tree : int -> tree list such that :
all_tree 0 = [Empty]
all_tree 1 = [Node('x',Empty,Empty)]
all_tree 2 = [Node('x',Node('x',Empty,Empty),Empty); Node('x',Empty,Node('x',Empty,Empty))]
...
I tried several ideas but it didn't work out. For example we could try the following :
let rec all_tree result = function
|0 -> r
|s -> all_tree ((List.map (fun i -> Node('x',i,Empty)) result)#(List.map (fun i -> Node('x',Empty,i)) result) ) (s-1)
in all_tree [Empty] (*some number*)
This code doesn't work because it doesn't generate every possibility.
Here is one possible answer.
let rec all_trees = function
| 0 -> [Empty]
| n ->
let result = ref [] in
for i = 0 to n-1 do
let left_side = all_trees i
and right_side = all_trees (n-1-i) in
List.iter
(fun left_tree ->
List.iter
(fun right_tree ->
result := (Node('x', left_tree, right_tree)) :: (!result)
)
right_side
)
left_side
done;
!result
;;
It's pretty simple: a tree with n>0 nodes is a tree with 1 node at the top, and then n-1 nodes below split between a certain number on the left and a certain number on the right. So we loop for i from 0 to n-1 through all possible numbers of values on the left side, and n-i-1 is going to be the number of nodes on the right side. We recursively call all_trees to get the trees with i and n-i-1 nodes, and simply aggregate them.
Notice that it's a very poor implementation. It has everything a recursive function should avoid. See something like this page on recursive implementations of the Fibonacci sequence to see how to improve it (one of the first things to do would be to cache the results rather than recompute the same things many many times).
I do agree with the question's comments though that writing a printer would be step 1 in that kind of project, because it's really annoying having to read through messy things like [Node ('x', Node ('x', Empty, Node ('x', Node ('x', Empty, Empty), Empty)), Empty);. Naming variables better would also make it easier for people to read your code and will increase the chance someone will help you. And generally, listening to the comments when people give you advice on how to properly ask your questions will make it easier for you to get answers both right now and in your future questions. For instance, in my own code, I used i as the loop index. It makes sense to me while I'm coding it, but when you read the code, maybe you would have preferred to read something like left_side_nodes or something like that, which would have made it obvious what this variable was supposed to do. It's the same in your own scenario: you could call i something like subtree or maybe something even more explicit. Actually, properly naming it could make you realize what's wrong with your code. Often, if you can't properly name a variable, it's that you don't really understand what it's doing (even local variables).
Related
So, I found out that Ocaml supports the creation of circular lists using let rec.
utop # let rec ones = 1::ones;;
val ones : int list = [1; <cycle>]
That is pretty neat, and it even prints out in utop without blowing up.
But when I try to use List.map on this kind of data it does blow up:
utop # let twos = List.map ((+) 1) ones;;
Stack overflow during evaluation (looping recursion?).
Raised by primitive operation at Stdlib__List.map in file "list.ml", line 92, characters 32-39
Called from Stdlib__List.map in file "list.ml", line 92, characters 32-39
...
That is somewhat disapointing, though not totally unexpected.
Now the question, would it be possible to implement a 'better' map function that can handle this properly. I.e. you would do something like:
let twos = betterMap ((+) 1) ones;;
And instead of blowing up it would be able to detect the cycle properly and produce:
val twos : int list = [2; <cycle>]
Since the list of ones, though looping back on itself, is effectively a finite structure, it feels like this should be possible. But how?
It is only possible to create cyclic lists when the cycle is statistically known. It is thus impossible to create a map function that works on any cyclic lists without knowing in advance the topology of cycles in the list. For instance, this function works for lists that are 1-cycle:
let map_1_cycle f = function
| [] -> []
| a :: l ->
let rec answer = f a :: answer in
answer
The generic solution is to use sequences since as a form of lazy list, they have a much better support for infinite sequences of elements:
let ones = Seq.repeat 1
let twos = Seq.map ((+) 1) ones
I'm trying to implement a function to add like terms of a sorted list of tuples (first number represents polynomial's constant, the second represents the power). I'm an ocaml noob and don't really know what I'm doing wrong or how to do this correctly.
I tried to write it, but it doesn't work
https://gyazo.com/d37bb66d0e6813537c34225b6d4048d0
let rec simp list =
match list with
| (a,b)::(c,d)::remainder where b == d -> (a+c,b)::simp(remainder)
| (a,b)::(c,d)::remainder where b != d -> (a,b)::(c,d)::simp(remainder)
| _ -> list;;
This should combine all the terms with the same second value and just return one tuple with their first values added to the new list. ie: [(3,2);(4,2)] -> [(7,2)].
I am not familiar with the where keyword - there is ocaml-where which provides it, but it seems to be doing something different than what you are expecting. As such, the syntax is just wrong, and where is unexpected.
You probably meant when instead of where.
I want to find a list of nodes that currently given nodes directly or indirectly connect to.
For example, I have a list of nodes:
[1,2]
and a list of tuples, and each of the tuples represents a direct edge:
[(1,5),(2,4),(4,6)]
So, the nodes I am looking for are
[1,2,5,4,6]
Because, 1 connects to 5, 2 connects to 4. Then, 4 is connected to 6.
To achieve this, I need two a queues, and a list. Each time a new node is discovered, we append the new node to the queue and the list. Then, we remove the first node of the queue, and go to next node. If a new node is connected to the current node of the queue. Then, we add new node to both the queue and the list.
We keep doing this until the queue is empty and we return the list.
So now, I have an append function which appends a list to another list:
fun append(xs, ys) =
case ys of
[] => xs
| (y::ys') => append(xs # [y], ys')
Then, I have a function called getIndirectNodes, which intends to return the lists of nodes that the given nodes indirectly connected to, but throws "unresolved flex record". List1 and List2 have the same items supposedly. But, List1 serves the queue, and list2 servers as the list to be returned.
fun getIndirectNode(listRoleTuples, list1, list2) =
if list1 = []
then list2
else if hd(list1) = #1(hd(listRoleTuples))
then (
append(list1,#2(hd(listRoleTuples)) :: []);
append(list2,#2(hd(listRoleTuples)) :: []);
getIndirectNode(listRoleTuples,tl(list1),list2)
)
else
getIndirectNode(listRoleTuples,tl(list1),list2)
If I remove the else if statement, it works perfectly fine. But, it's not what I intended to do. The problem is in the else if statement. What can I do to fix it?
SML needs to know exactly what shape a tuple has in order to deconstruct it.
You could specify the type of the parameter - listRoleTuples : (''a * ''a) list - but using pattern matching is a better idea.
(There are many other problems with that code, but that's the answer to your question.)
It seems that one of your classmates had this exact tuple problem in a very related task.
Make sure you browse the StackOverflow Q&A's before you ask the same question again.
As for getting the indirect nodes, this can be solved by fixed-point iteration.
First you get all the direct nodes, and then you get the direct nodes of the direct nodes.
And you do this recursively until no more new nodes occur this way.
fun getDirectNodes (startNode, edges) =
List.map #2 (List.filter (fn (node, _) => node = startNode) edges)
fun toSet xs =
... sort and remove duplicates ...
fun getReachableNodes (startNodes, edges) =
let
fun f startNode = getDirectNodes (startNode, edges)
val startNodes = toSet startNodes
val endNodes = toSet (List.concat (List.map f startNodes))
in
if startNodes = endNodes
then endNodes
else getReachableNodes (startNodes # endNodes, edges)
end
This doesn't exactly find indirect end-nodes; it finds all nodes directly or indirectly reachable by startNodes, and it includes startNodes themselves even if they're not directly or indirectly reachable by themselves.
I've tried to make this exercise easier by using sets as a datatype; it would be even neater with an actual, efficient implementation of a set type, e.g. using a balanced binary search tree. It is easier to see if there are no new nodes by adding elements to a set, since if a set already contains an element, it will be equivalent to itself before and after the addition of the element.
And I've tried to use higher-order functions when this makes sense. For example, given a list of things where I want to do the same thing on each element, List.map produces a list of results. But since that thing I want to do, getDirectNodes (startNode, edges) produces a list, then List.map f produces a list of lists. So List.concat collapses that into a single list.
List.concat (List.map f xs)
is a pretty common thing to do.
I have the following erlang code:
lists:all(fun(Element) -> somefunction(TestCase -- [Element]) end, TestCase).
Where TestCase is an array. I'm trying to iterate over the list/array with one element missing.
The problem is this code takes O(N^2) time worst case because of the copies of the TestCase array everytime -- is called. There is a clear O(N) Solution in a non functional language.
saved = TestCase[0]
temp = 0
NewTestCase = TestCase[1:]
for a in range(length(NewTestCase)):
somefunction(NewTestCase)
temp = NewTestCase[a]
NewTestCase[a] = saved
saved = temp
... or something like that.
Is there an O(N) solution in erlang?
Of course there is, but it's a little bit more complicated. I am assuming that some_function/1 is indeed a boolean function and you want to test whether it returns true for every sub-list.
test_on_all_but_one([], _Acc) -> true;
test_on_all_but_one([E|Rest], Acc) ->
case somefunction(lists:reverse(Acc,Rest)) of
true -> test_on_all_but_one(Rest, [E|Acc]);
false -> false
end.
This implementation is still O(length(List)^2) as the lists:reverse/2 call will still need O(length(Acc)). If you can modify somefunction/1 to do it's calculation on a list split into two parts, then you can modify the previous call to somefunction(lists:reverse(Acc,Rest)) with somefunction(Acc, Rest) or something similar and avoid the reconstruction.
The modification depends on the inner workings of somefunction/1. If you want more help with that, give some code!
You can split the list into 2 sublists, if it's acceptable of course.
witerate(Fun, [Tail], Acc) ->
Fun([], Acc);
witerate(Fun, [Head | Tail], Acc) ->
Fun(Tail, Acc),
witerate(Fun, Tail, [Head | Acc]).
I have a question about tuples and lists in Haskell. I know how to add input into a tuple a specific number of times. Now I want to add tuples into a list an unknown number of times; it's up to the user to decide how many tuples they want to add.
How do I add tuples into a list x number of times when I don't know X beforehand?
There's a lot of things you could possibly mean. For example, if you want a few copies of a single value, you can use replicate, defined in the Prelude:
replicate :: Int -> a -> [a]
replicate 0 x = []
replicate n | n < 0 = undefined
| otherwise = x : replicate (n-1) x
In ghci:
Prelude> replicate 4 ("Haskell", 2)
[("Haskell",2),("Haskell",2),("Haskell",2),("Haskell",2)]
Alternately, perhaps you actually want to do some IO to determine the list. Then a simple loop will do:
getListFromUser = do
putStrLn "keep going?"
s <- getLine
case s of
'y':_ -> do
putStrLn "enter a value"
v <- readLn
vs <- getListFromUser
return (v:vs)
_ -> return []
In ghci:
*Main> getListFromUser :: IO [(String, Int)]
keep going?
y
enter a value
("Haskell",2)
keep going?
y
enter a value
("Prolog",4)
keep going?
n
[("Haskell",2),("Prolog",4)]
Of course, this is a particularly crappy user interface -- I'm sure you can come up with a dozen ways to improve it! But the pattern, at least, should shine through: you can use values like [] and functions like : to construct lists. There are many, many other higher-level functions for constructing and manipulating lists, as well.
P.S. There's nothing particularly special about lists of tuples (as compared to lists of other things); the above functions display that by never mentioning them. =)
Sorry, you can't1. There are fundamental differences between tuples and lists:
A tuple always have a finite amount of elements, that is known at compile time. Tuples with different amounts of elements are actually different types.
List an have as many elements as they want. The amount of elements in a list doesn't need to be known at compile time.
A tuple can have elements of arbitrary types. Since the way you can use tuples always ensures that there is no type mismatch, this is safe.
On the other hand, all elements of a list have to have the same type. Haskell is a statically-typed language; that basically means that all types are known at compile time.
Because of these reasons, you can't. If it's not known, how many elements will fit into the tuple, you can't give it a type.
I guess that the input you get from your user is actually a string like "(1,2,3)". Try to make this directly a list, whithout making it a tuple before. You can use pattern matching for this, but here is a slightly sneaky approach. I just remove the opening and closing paranthesis from the string and replace them with brackets -- and voila it becomes a list.
tuplishToList :: String -> [Int]
tuplishToList str = read ('[' : tail (init str) ++ "]")
Edit
Sorry, I did not see your latest comment. What you try to do is not that difficult. I use these simple functions for my task:
words str splits str into a list of words that where separated by whitespace before. The output is a list of Strings. Caution: This only works if the string inside your tuple contains no whitespace. Implementing a better solution is left as an excercise to the reader.
map f lst applies f to each element of lst
read is a magic function that makes a a data type from a String. It only works if you know before, what the output is supposed to be. If you really want to understand how that works, consider implementing read for your specific usecase.
And here you go:
tuplish2List :: String -> [(String,Int)]
tuplish2List str = map read (words str)
1 As some others may point out, it may be possible using templates and other hacks, but I don't consider that a real solution.
When doing functional programming, it is often better to think about composition of operations instead of individual steps. So instead of thinking about it like adding tuples one at a time to a list, we can approach it by first dividing the input into a list of strings, and then converting each string into a tuple.
Assuming the tuples are written each on one line, we can split the input using lines, and then use read to parse each tuple. To make it work on the entire list, we use map.
main = do input <- getContents
let tuples = map read (lines input) :: [(String, Integer)]
print tuples
Let's try it.
$ runghc Tuples.hs
("Hello", 2)
("Haskell", 4)
Here, I press Ctrl+D to send EOF to the program, (or Ctrl+Z on Windows) and it prints the result.
[("Hello",2),("Haskell",4)]
If you want something more interactive, you will probably have to do your own recursion. See Daniel Wagner's answer for an example of that.
One simple solution to this would be to use a list comprehension, as so (done in GHCi):
Prelude> let fstMap tuplist = [fst x | x <- tuplist]
Prelude> fstMap [("String1",1),("String2",2),("String3",3)]
["String1","String2","String3"]
Prelude> :t fstMap
fstMap :: [(t, b)] -> [t]
This will work for an arbitrary number of tuples - as many as the user wants to use.
To use this in your code, you would just write:
fstMap :: Eq a => [(a,b)] -> [a]
fstMap tuplist = [fst x | x <- tuplist]
The example I gave is just one possible solution. As the name implies, of course, you can just write:
fstMap' :: Eq a => [(a,b)] -> [a]
fstMap' = map fst
This is an even simpler solution.
I'm guessing that, since this is for a class, and you've been studying Haskell for < 1 week, you don't actually need to do any input/output. That's a bit more advanced than you probably are, yet. So:
As others have said, map fst will take a list of tuples, of arbitrary length, and return the first elements. You say you know how to do that. Fine.
But how do the tuples get into the list in the first place? Well, if you have a list of tuples and want to add another, (:) does the trick. Like so:
oldList = [("first", 1), ("second", 2)]
newList = ("third", 2) : oldList
You can do that as many times as you like. And if you don't have a list of tuples yet, your list is [].
Does that do everything that you need? If not, what specifically is it missing?
Edit: With the corrected type:
Eq a => [(a, b)]
That's not the type of a function. It's the type of a list of tuples. Just have the user type yourFunctionName followed by [ ("String1", val1), ("String2", val2), ... ("LastString", lastVal)] at the prompt.