According to the theory of ADTs (Algebraic Data Types) the concatenation of two lists has to take O(n) where n is the length of the first list. You, basically, have to recursively iterate through the first list until you find the end.
From a different point of view, one can argue that the second list can simply be linked to the last element of the first. This would take constant time, if the end of the first list is known.
What am I missing here ?
Operationally, an Haskell list is typically represented by a pointer to the first cell of a single-linked list (roughly). In this way, tail just returns the pointer to the next cell (it does not have to copy anything), and consing x : in front of the list allocates a new cell, makes it point to the old list, and returns the new pointer. The list accessed by the old pointer is unchanged, so there's no need to copy it.
If you instead append a value with ++ [x], then you can not modify the original liked list by changing its last pointer unless you know that the original list will never be accessed. More concretely, consider
x = [1..5]
n = length (x ++ [6]) + length x
If you modify x when doing x++[6], the value of n would turn up to be 12, which is wrong. The last x refer to the unchanged list which has length 5, so the result of n must be 11.
Practically, you can't expect the compiler to optimize this, even in those cases in which x is no longer used and it could, theoretically, be updated in place (a "linear" use). What happens is that the evaluation of x++[6] must be ready for the worst-case in which x is reused afterwards, and so it must copy the whole list x.
As #Ben notes, saying "the list is copied" is imprecise. What actually happens is that the cells with the pointers are copied (the so-called "spine" on the list), but the elements are not. For instance,
x = [[1,2],[2,3]]
y = x ++ [[3,4]]
requires only to allocate [1,2],[2,3],[3,4] once. The lists of lists x,y will share pointers to the lists of integers, which do not have to be duplicated.
What you're asking for is related to a question I wrote for TCS Stackexchange some time back: the data structure that supports constant-time concatenation of functional lists is a difference list.
A way of handling such lists in a functional programming language was worked out by Yasuhiko Minamide in the 90s; I effectively rediscovered it a while back. However, the good run-time guarantees require language-level support that's not available in Haskell.
It's because of immutable state. A list is an object + a pointer, so if we imagined a list as a Tuple it might look like this:
let tupleList = ("a", ("b", ("c", [])))
Now let's get the first item in this "list" with a "head" function. This head function takes O(1) time because we can use fst:
> fst tupleList
If we want to swap out the first item in the list with a different one we could do this:
let tupleList2 = ("x",snd tupleList)
Which can also be done in O(1). Why? Because absolutely no other element in the list stores a reference to the first entry. Because of immutable state, we now have two lists, tupleList and tupleList2. When we made tupleList2 we didn't copy the whole list. Because the original pointers are immutable we can continue to reference them but use something else at the start of our list.
Now let's try to get the last element of our 3 item list:
> snd . snd $ fst tupleList
That happened in O(3), which is equal to the length of our list i.e. O(n).
But couldn't we store a pointer to the last element in the list and access that in O(1)? To do that we would need an array, not a list. An array allows O(1) lookup time of any element as it is a primitive data structure implemented on a register level.
(ASIDE: If you're unsure of why we would use a Linked List instead of an Array then you should do some more reading about data structures, algorithms on data structures and Big-O time complexity of various operations like get, poll, insert, delete, sort, etc).
Now that we've established that, let's look at concatenation. Let's concat tupleList with a new list, ("e", ("f", [])). To do this we have to traverse the whole list just like getting the last element:
tupleList3 = (fst tupleList, (snd $ fst tupleList, (snd . snd $ fst tupleList, ("e", ("f", [])))
The above operation is actually worse than O(n) time, because for each element in the list we have to re-read the list up to that index. But if we ignore that for a moment and focus on the key aspect: in order to get to the last element in the list, we must traverse the entire structure.
You may be asking, why don't we just store in memory what the last list item is? That way appending to the end of the list would be done in O(1). But not so fast, we can't change the last list item without changing the entire list. Why?
Let's take a stab at how that might look:
data Queue a = Queue { last :: Queue a, head :: a, next :: Queue a} | Empty
appendEnd :: a -> Queue a -> Queue a
appendEnd a2 (Queue l, h, n) = ????
IF I modify "last", which is an immutable variable, I won't actually be modifying the pointer for the last item in the queue. I will be creating a copy of the last item. Everything else that referenced that original item, will continue referencing the original item.
So in order to update the last item in the queue, I have to update everything that has a reference to it. Which can only be done in optimally O(n) time.
So in our traditional list, we have our final item:
List a []
But if we want to change it, we make a copy of it. Now the second last item has a reference to an old version. So we need to update that item.
List a (List a [])
But if we update the second last item we make a copy of it. Now the third last item has an old reference. So we need to update that. Repeat until we get to the head of the list. And we come full circle. Nothing keeps a reference to the head of the list so editing that takes O(1).
This is the reason that Haskell doesn't have Doubly Linked Lists. It's also why a "Queue" (or at least a FIFO queue) can't be implemented in a traditional way. Making a Queue in Haskell involves some serious re-thinking of traditional data structures.
If you become even more curious about how all of this works, consider getting the book Purely Funtional Data Structures.
EDIT: If you've ever seen this: http://visualgo.net/list.html you might notice that in the visualization "Insert Tail" happens in O(1). But in order to do that we need to modify the final entry in the list to give it a new pointer. Updating a pointer mutates state which is not allowed in a purely functional language. Hopefully that was made clear with the rest of my post.
In order to concatenate two lists (call them xs and ys), we need to modify the final node in xs in order to link it to (i.e. point at) the first node of ys.
But Haskell lists are immutable, so we have to create a copy of xs first. This operation is O(n) (where n is the length of xs).
Example:
xs
|
v
1 -> 2 -> 3
1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7
^ ^
| |
xs ++ ys ys
Related
I was trying to implement permutation to cycles in Haskell without using Monad. The problem is as follow: given a permutation of numbers [1..n], output the correspondence disjoint cycles. The function is defined like
permToCycles :: [Int] -> [[Int]]
For the input:
permToCycles [3,5,4,1,2]
The output should be
[[3,4,1],[5,2]]
By the definition of cyclic permutation, the algorithm itself is straightforward. Since [3,5,4,1,2] is a permutation of [1,2,3,4,5], we start from the first element 3 and follow the orbit until we get back to 3. In this example, we have two cycles 3 -> 4 -> 1 -> 3. Continue to do so until we traverse all elements. Thus the output is [[3,4,1],[5,2]].
Using this idea, it is fairly easy to implement in any imperative language, but I have trouble with doing it in Haskell. I find something similar in the module Math.Combinat.Permutations, but the implementation of function permutationToDisjointCycles uses Monad, which is not easy to understand as I'm a beginner.
I was wondering if I could implement it without Monad. Any help is appreciated.
UPDATE: Here is the function implemented in Python.
def permToCycles(perm):
pi_dict = {i+1: perm[i]
for i in range(len(perm))} # permutation as a dictionary
cycles = []
while pi_dict:
first_index = next(iter(pi_dict)) # take the first key
this_elem = pi_dict[first_index] # the first element in perm
next_elem = pi_dict[this_elem] # next element according to the orbit
cycle = []
while True:
cycle.append(this_elem)
# delete the item in the dict when adding to cycle
del pi_dict[this_elem]
this_elem = next_elem
if next_elem in pi_dict:
# continue the cycle
next_elem = pi_dict[next_elem]
else:
# end the cycle
break
cycles.append(cycle)
return cycles
print(permToCycles([3, 5, 4, 1, 2]))
The output is
[[3,4,1],[5,2]]
I think the main obstacle when implementing it in Haskell is how to trace the marked (or unmarked) elements. In Python, it can easily be done using a dictionary as I showed above. Also in functional programming, we tend to use recursion to replace loops, but here I have trouble with thinking the recursive structure of this problem.
Let's start with the basics. You hopefully started with something like this:
permutationToDisjointCycles :: [Int] -> [[Int]]
permutationToDisjointCycles perm = ...
We don't actually want to recur on the input list so much as we want to use an index counter. In this case, we'll want a recursive helper function, and the next step is to just go ahead and call it, providing whatever arguments you think you'll need. How about something like this:
permutationToDisjointCycles perm = cycles [] 0
where
cycles :: [Int] -> Int -> [[Int]]
cycles seen ix = ...
Instead of declaring a pi_dict variable like in Python, we'll start with a seen list as an argument (I flipped it around to keeping track of what's been seen because that ends up being a little easier). We do the same with the counting index, which I here called ix. Let's consider the cases:
cycles seen ix
| ix >= length perm = -- we've reached the end of the list
| ix `elem` seen = -- we've already seen this index
| otherwise = -- we need to generate a cycle.
That last case is the interesting one and corresponds to the inner while loop of the Python code. Another while loop means, you guessed it, more recursion! Let's make up another function that we think will be useful, passing along as arguments what would have been variables in Python:
| otherwise = let c = makeCycle ix ix in c : cycles (c ++ seen) (ix+1)
makeCycle :: Int -> Int -> [Int]
makeCycle startIx currentIx = ...
Because it's recursive, we'll need a base case and recursive case (which corresponds to the if statement in the Python code which either breaks the loop or continues it). Rather than use the seen list, it's a little simpler to just check if the next element equals the starting index:
makeCycle startIx currentIx =
if next == start
then -- base case
else -- recursive call, where we attach an index onto the cycle and recur
where next = perm !! i
I left a couple holes that need to be filled in as an exercise, and this version works on 0-indexed lists rather than 1-indexed ones like your example, but the general shape of the algorithm is there.
As a side note, the above algorithm is not super efficient. It uses lists for both the input list and the "seen" list, and lookups in lists are always O(n) time. One very simple performance improvement is to immediately convert the input list perm into an array/vector, which has constant time lookups, and then use that instead of perm !! i at the end.
The next improvement is to change the "seen" list into something more efficient. To match the idea of your Python code, you could change it to a Set (or even a HashSet), which has logarithmic time lookups (or constant with a hashset).
The code you found Math.Combinat.Permutations actually uses an array of Booleans for the "seen" list, and then uses the ST monad to do imperative-like mutation on that array. This is probably even faster than using Set or HashSet, but as you yourself could tell, readability of the code suffers a bit.
positions :: Eq a => a -> [a] -> [Int]
positions x xs = [i | (x',i) <- zip xs [0..], x == x']
i need to create a Test function for the positions function, which passes the quickcheck.
Does someone has an idea?
A possible test could perform the following operations:
randomly generate xs, ys :: [Int] and y :: Int
define list = xs ++ y : ys
test length xs `elem` positions y list
You might also want to write tests for missing elements.
That being said, it's weird to craft tests from the code. One should design test using the specification used to write the code instead. Otherwise, if the code has some quirks, they end up in tests as well: instead of checking for what the code should do, we check for what the code does, which can be pointless.
Good question! I had a bit of trouble with this too when I started using QuickCheck, but as I used it more I started to get a feel for what sort of properties you should test. Often, a good place to start is to think about what sort of relationships should hold between the input(s) and output. Your function finds the indices of an element in a list; now what properties should hold in this case? Here's two which I can think of:
The list of indices should have the same number of elements as there are occurrences of the value in the list.
Each index should correspond to the correct value.
And here's an outline of how these properties could be tested:
Generate a random list, apply positions to it, then count the number of indices returned and check that it matches with the number of occurrences of the searched value.
Generate a random list, then apply positions to it and check that the element at each of the returned indices is the value which was searched for.
I also think that #chi's answer makes a good point in saying that you should design tests from the specification rather than the code; this ties in with what I was saying above, in that a specification can help you find relationships between input(s) and output which may not be immediately apparent from the code.
So i am currently working with Haskell at my college, but kinda struggling with pattern-matching and to be more specific i'll give the program, that i am to solve:
My function is awaiting a list of lists ( with each list containing at least 3 elements ) and in each list the second element is replaced with the number of the third element.
Example:
[[1,2,3],[4,5,6,7],[8,9,10]] should become [[1,3,3],[4,6,6,7],[8,10,10]
So far I've made the following:
f [] = []
f [[x]] = []
f [[x,y]] = []
f [[x,y,z]] = [[x,z,z]]
f ([x,y,z]:xs) = [x,z,z]:f(xs)
My questions are:
How can i identify, that some lists may contain more than 3 elements and that the list must remain the same, only the 2nd element changes.
How can i make the recursion, so that it handles exceptions (for example the first list has 4 elements).
Thank you in advance!
It may help to first write the function that swaps the second value in a list with the third.
swap (x:y:z:rest) = x:z:z:rest
swap xs = xs
In the above, x:y:z:rest matches a list of at least length 3. The value of rest, since it is at the end of the pattern, will match any kind of list, both empty and full. The second pattern match of xs is a catch-all for any other type of list, which will just return the same list if it is less than 3 items long.
From there, you can write f in terms of mapping over the outer list and applying swap:
f = map swap
We're in the process of converting our imperative brains to a mostly-functional paradigm. This function is giving me trouble. I want to construct an array that EITHER contains two pairs or three pairs, depending on a condition (whether refreshToken is null). How can I do this cleanly using a FP paradigm? Of course with imperative code and mutation, I would just conditionally .push() the extra value onto the end which looks quite clean.
Is this an example of the "local mutation is ok" FP caveat?
(We're using ReadonlyArray in TypeScript to enforce immutability, which makes this somewhat more ugly.)
const itemsToSet = [
[JWT_KEY, jwt],
[JWT_EXPIRES_KEY, tokenExpireDate.toString()],
[REFRESH_TOKEN_KEY, refreshToken /*could be null*/]]
.filter(item => item[1] != null) as ReadonlyArray<ReadonlyArray<string>>;
AsyncStorage.multiSet(itemsToSet.map(roArray => [...roArray]));
What's wrong with itemsToSet as given in the OP? It looks functional to me, but it may be because of my lack of knowledge of TypeScript.
In Haskell, there's no null, but if we use Maybe for the second element, I think that itemsToSet could be translated to this:
itemsToSet :: [(String, String)]
itemsToSet = foldr folder [] values
where
values = [
(jwt_key, jwt),
(jwt_expires_key, tokenExpireDate),
(refresh_token_key, refreshToken)]
folder (key, Just value) acc = (key, value) : acc
folder _ acc = acc
Here, jwt, tokenExpireDate, and refreshToken are all of the type Maybe String.
itemsToSet performs a right fold over values, pattern-matching the Maye String elements against Just and (implicitly) Nothing. If it's a Just value, it cons the (key, value) pair to the accumulator acc. If not, folder just returns acc.
foldr traverses the values list from right to left, building up the accumulator as it visits each element. The initial accumulator value is the empty list [].
You don't need 'local mutation' in functional programming. In general, you can refactor from 'local mutation' to proper functional style by using recursion and introducing an accumulator value.
While foldr is a built-in function, you could implement it yourself using recursion.
In Haskell, I'd just create an array with three elements and, depending on the condition, pass it on either as-is or pass on just a slice of two elements. Thanks to laziness, no computation effort will be spent on the third element unless it's actually needed. In TypeScript, you probably will get the cost of computing the third element even if it's not needed, but perhaps that doesn't matter.
Alternatively, if you don't need the structure to be an actual array (for String elements, performance probably isn't that critical, and the O (n) direct-access cost isn't an issue if the length is limited to three elements), I'd use a singly-linked list instead. Create the list with two elements and, depending on the condition, append the third. This does not require any mutation: the 3-element list simply contains the unchanged 2-element list as a substructure.
Based on the description, I don't think arrays are the best solution simply because you know ahead of time that they contain either 2 values or 3 values depending on some condition. As such, I would model the problem as follows:
type alias Pair = (String, String)
type TokenState
= WithoutRefresh (Pair, Pair)
| WithRefresh (Pair, Pair, Pair)
itemsToTokenState: String -> Date -> Maybe String -> TokenState
itemsToTokenState jwtKey jwtExpiry maybeRefreshToken =
case maybeRefreshToken of
Some refreshToken ->
WithRefresh (("JWT_KEY", jwtKey), ("JWT_EXPIRES_KEY", toString jwtExpiry), ("REFRESH_TOKEN_KEY", refreshToken))
None ->
WithoutRefresh (("JWT_KEY", jwtKey), ("JWT_EXPIRES_KEY", toString jwtExpiry))
This way you are leveraging the type system more effectively, and could be improved on further by doing something more ergonomic than returning tuples.
I was trying to write a function to get all subsequences of a list that are of size n, but I'm not sure how to go about it.
I was thinking that I could probably use the built-in Data.List.subsequences and just filter out the lists that are not of size n, but it seems like a rather roundabout and inefficient way of doing it, and I'd rather not do that if I can avoid it, so I'm wondering if you have any ideas?
I want it to be something like this type
subseqofsize :: Int -> [a] -> [[a]]
For further clarification, here's an example of what I'm looking for:
subseqofsize 2 [1,2,3,3]
[[1,2],[1,3],[2,3],[1,3],[2,3],[3,3]]
Also, I don't care about the order of anything.
I'm assuming that this is homework, or that you are otherwise doing this as an exercise to learn, so I'll give you an outline of what the solution looks like instead of spoon-feeding you the correct answer.
Anyway, this is a recursion question.
There are two base cases:
sublistofsize 0 _ = ...
sublistofsize _ [] = ...
Then there are two recursive steps:
sublistofsize n (x : xs) = sublistsThatStartWithX ++ sublistsThatDon'tStartWithX
where sublistsThatStartWithX = ...
sublistsThatDon'tStartWithX = ...
Remember that the definitions of the base cases need to work appropriately with the definitions in the recursive cases. Think carefully: don't just assume that the base cases both result in an empty list.
Does this help?
You can think about this mathematically: to compute the sublists of size k, we can look at one element x of the list; either the sublists contain x, or they don't. In the former case, the sublist consists of x and then k-1 elements chosen from the remaining elements. In the latter case, the sublists consist of k elements chosen from the elements that aren't x. This lends itself to a (fairly) simple recursive definition.
(There are very very strong similarities to the recursive formula for binomial coefficients, which is expected.)
(Edit: removed code, per dave4420's reasons :) )