I am having two lists in SML, lets say list A [(a,b,c),(d,e,f)] and list B [b,e]. I want to count how many occurrence of each item in B that matches the second element of each triple in A. The output should be 2. Because b and e each occurs once in A.
This is my code so far but my counter is always set to 0 when I move from one element to another in B. I know in Java this will just be a simple double for loop.
fun number_in_months (d : (int * int * int ) list, m : (int) list) =
if null m then 0
else if null d then number_in_months(d, tl m)
else if (#2(hd d)) = (hd m) then 1 + number_in_months (tl d, m)
else number_in_months(tl d, m)
The code is not accumulating a value between recursive calls. There may be other logic errors too.
Accumulating a value using recursion and functions is a common pattern which you can read more about here. It's essence is to deconstruct a list using head and tail until the list is empty and accumulate some value at each call. The sum function below is a simple example to show this. This could be adapted to your example to accumulate acc when b or e are found in list A.
fun sum(numbers: (int) list) =
let fun sumR(numbers: (int) list, acc: int) =
if null numbers
then acc
else
sumR(tl numbers, hd numbers + acc)
in
sumR(numbers, 0)
end
Running on [1,2,3] gives:
val sum = fn : int list -> int
- sum([1,2,3]);
val it = 6 : int
Note I am intentionally vague with this answer since this is a question regarding Coursera homework for the Programming Languages class.
As you mention, it would be a nested/double loop in any imperative programming language. What you are actually missing is the second loop.
Your "inner" loop goes through all elements of d, and when this is done, your "outer" loop tries to pop the top element of m and start all over, as seen from this line of your code:
else if null d then number_in_months(d, tl m)
However as you can see, you have just tested the list d to be empty and you supply this (exact same list) to your recursive call on the tail of m, which will then fall in this same case for each successive call until m is also empty and you return 0.
Thus what you are missing is to "keep a copy" of the original input list m. This can be done in various ways, but an inner (helper) function is properly the most used one and it even "looks" like a nested loop
fun number_in_months (d, m) =
let
fun nim' ([], y::ys) = nim (d, ys) (* 1 *)
| nim' (_, []) = 0 (* 2 *)
| nim' ((_, x2, _) :: xs, yss as (y::ys)) = ... (* 3 *)
in
nim'(d, m)
end
Using pattern matching the above code gets much simpler and less error prone. In case 1, the "inner" loop has gone through all elements in d, thus the recursive call using d from the outer function which is not changed at any time. In case 2, the "outer" loop has gone through all elements of m and we return 0 (the neutral element of addition). In case 3 we do the actual work. Here pattern matching is used such that we don't need to enforce the type of the argument and we don't need to pull out the 2nd element of the triple, we already have it in the variable x2. All that is needed is to do the computation and make a recursive call with xs and yss.
When doing it this way, the inner (helper) function is using a "copy" of the original input list d and stepping through its elements (potentially modifying it), but we always got a reference to the original input list, which we can use if/when needed.
Related
I want to create a non-recursive function for my minimum
but I have some troubles with it
Can you help me please.
`let min_list lst=
let n=list.length lst ;;
let a=list.nth lst ;;
for i = 1 to n-1 ;;
let b=list.nth lst i;;
if a >b then a=b lst done ;;`
Honesly,It's difficult with non recursive fonction.So this is just for learning.I still have erreur in ligne 6
let min_list lst=
let a=List.hd lst in
let n=List.length lst in
for j =1 to n-1 do
let b=List.nth lst j in
if a > b then (let a=b) done ;;
Thank you it's useful It help me a lot .I have one other question what the difference between this
let min_array a =
let min =ref (List.hd a) in
for i = 1 to List.length a -1 do
if List.nth a i < !min then min := List.nth a i
done;
!min;;
print_int (min_array [ 10 ; 5 ; 7 ; 8 ; 12 ]);;
and
let min_array a =
let min =ref (List.hd a) in
for i = 1 to List.length a -1 do
if ref (List.nth a i) < min then min := List.nth a i
done;
!min;;
print_int (min_array [ 10 ; 5 ; 7 ; 8 ; 12 ]);;
It's the same ?I think
Why don't you want to use a recursive function ?
Liste are made to be crossed by recursive function. Everytime you use List.nth l n Ocaml has to cross n values until he found the nth element.
In Ocaml you can't change variable value as you do in other languages. You want a to be a ref.
Also your function won't return anything you'll have to put a !a between the done and the ;;. There will be a ! Because a will be a ref.
But if you want to practice use arrays instead because what you do here is in complexity O(n²) instead of O(n).
As said in the answer from Butanium, this kind of non-recursive function might be more relevant with arrays. And to work with mutable values, you need to use a reference.
A solution might then be something like (without dealing with case of an empty array):
let min_array a =
let min = ref a.(0) in
for i = 1 to Array.length a -1 do
if a.(i) < !min then min := a.(i)
done;
!min
The last line is important here, because it gets the value to be returned by the function.
Can then be used like that:
# min_array [| 10 ; 5 ; 7 ; 8 ; 12 |];;
- : int = 5
If you really do want to use lists instead of arrays, just use List.nth a i instead of a.(i) and List.length instead of Array.length.
Edit after question update
As Shawn and Jeffrey Scofield said in their respective comment, you should try to understand a bit better OCaml's syntax. And please don't use ;; in your programs, just keep it for the REPL.
As described in the documentation,
ref returns a fresh reference containing the given value.
Which means that when you write ref (List.nth a i) < min,
you create a fresh reference containing the i-th value the list, then compare it to min (which is also a reference). Luckily, mutable structures are compared by contents, which means that OCaml will access to your fresh reference's value, then access to min's value, and compare them. Thus, it will produce the same result as the direct comparison List.nth a i < !min, with a bit of useless memory allocation/access.
You can do this quite concisely by taking advantage of some features of the OCaml stdlib:
(* 'a list -> 'a option *)
let min_list l =
if List.length l > 0 then
Some (List.fold_left min (List.hd l) l)
else
None
Thanks to the min built-in, this works for lists of any type.
e.g. in a utop shell we can see:
min_list [99; 33; -1];;
- : int option = Some (-1)
min_list [99.1; 33.2; -1.3];;
- : float option = Some (-1.3)
min_list ["z"; "b"; "k"];;
- : string option = Some "b"
Explanation
First we recognise that the list may be empty, in which case we cannot return a meaningful value. This implies the function should return an option type, so either Some <value> or None.
Next we can use List.fold_left to iterate through the list.
Unfortunately the docs for List.fold_left are almost completely unhelpful:
val fold_left : ('a -> 'b -> 'a) -> 'a -> 'b list -> 'a
fold_left f init [b1; ...; bn] is f (... (f (f init b1) b2) ...) bn.
It's as if they assume that if you're using OCaml you're already an elite master of functional programming, who naturally knows what a "fold left" does.
I'm not an elite master of functional programming, but I've been around long enough to know that fold_left is basically the same as the reduce function in Python.
It's a function that iterates through a list, applies a function to each value as it goes, and returns a single value.
So we can start to make sense of the signature of fold_left...
It takes three arguments:
The first arg, f, is a function which itself takes two args - the first or 'left' arg is the 'accumulated' value, and the second arg is the current value from the list as we iterate through. Whatever value you return from this function will be passed back into it as the left 'accumulated' value on the next iteration. When the list is exhausted the accumulated value will be returned from fold_left.
The second arg, init is an initial value. It is passed to f as the left 'accumulated' arg in the first step, when nothing has been otherwise accumulated yet.
Third arg is our list of values
So when we return:
Some (List.fold_left min (List.hd l) l)
...we are passing the min function as f and (List.hd l) as init.
List.hd l just returns the first element of the list l. We could use any element from the list as an initial value, but List.hd exists and gives us the first.
So fold_left is going to iterate through the list and f will return min <accumulated> <current>. So at each iteration step the accumulated value passed forward is the lowest value seen so far.
Non-recursive?
I did wonder if perhaps the fold_left method does not count as non-recursive somehow, since no one else had suggested it. Even though we have not used let rec anywhere, maybe somewhere internally it is secretly recursive?
For fun I decided to try writing the reduce/fold function from scratch:
let reduce f init l =
let acc = ref init in
List.iter (fun el -> acc := f !acc el) l;
!acc
(* we can directly substitute `reduce` for `List.fold_left` *)
let min_list l =
if List.length l > 0 then
Some (reduce min (List.hd l) l)
else
None
...again, no let rec needed so I guess it counts as non-recursive.
I am trying to write a program in sml that takes in the length of a list, the max number that will appear on the list and the list of course. It then calculates the length of the smallest "sub-list" that contains all numbers.
I have tried to use the sliding window approach , with two indexes , front and tail. The front scans first and when it finds a number it writes into a map how many times it has already seen this number. If the program finds all numbers then it calls the tail. The tail scans the list and if it finds that a number has been seen more times than 1 it takes it off.
The code I have tried so far is the following:
structure Key=
struct
type ord_key=int
val compare=Int.compare
end
fun min x y = if x>y then y else x;
structure mymap = BinaryMapFn ( Key );
fun smallest_sub(n,t,listall,map)=
let
val k=0
val front=0
val tail=0
val minimum= n;
val list1=listall;
val list2=listall;
fun increase(list1,front,k,ourmap)=
let
val number= hd list1
val elem=mymap.find(ourmap,number)
val per=getOpt(elem,0)+1
fun decrease(list2,tail,k,ourmap,minimum)=
let
val number=hd list2
val elem=mymap.find(ourmap,number)
val per=getOpt(elem,0)-1
val per1=getOpt(elem,0)
in
if k>t then
if (per1=1) then decrease(tl list2,tail+1,k-1,mymap.insert(ourmap,number,per),min minimum (front-tail))
else decrease(tl list2,tail+1,k,mymap.insert(ourmap,number,per),min minimum (front-tail))
else increase (list1, front,k,ourmap)
end
in
if t>k then
if (elem<>NONE) then increase (tl list1,front+1,k,mymap.insert(ourmap,number,per))
else increase(tl list1,front+1,k+1,mymap.insert(ourmap,number,per))
else (if (n>front) then decrease(list2,tail,k,ourmap,minimum) else minimum)
end
in
increase(list1,front,k,map)
end
fun solve (n,t,acc)= smallest_sub(n,t,acc,mymap.empty)
But when I call it with this smallest_sub(10,3,[1,3,1,3,1,3,3,2,2,1]); it does not work. What have I done wrong??
Example: if input is 1,3,1,3,1,3,3,2,2,1 the program should recognize that the parto of the list that contains all numbers and is the smallest is 1,3,3,2 and 3,2,2,1 so the output should be 4
This problem of "smallest sub-list that contains all values" seems to recur in
new questions without a successful answer. This is because it's not a minimal,
complete, and verifiable example.
Because you use a "sliding window" approach, indexing the front and the back
of your input, a list taking O(n) time to index elements is not ideal. You
really do want to use arrays here. If your input function must have a list, you
can convert it to an array for the purpose of the algorithm.
I'd like to perform a cleanup of the code before answering, because running
your current code by hand is a bit hard because it's so condensed. Here's an
example of how you could abstract out the book-keeping of whether a given
sub-list contains at least one copy of each element in the original list:
Edit: I changed the code below after originally posting it.
structure CountMap = struct
structure IntMap = BinaryMapFn(struct
type ord_key = int
val compare = Int.compare
end)
fun count (m, x) =
Option.getOpt (IntMap.find (m, x), 0)
fun increment (m, x) =
IntMap.insert (m, x, count (m, x) + 1)
fun decrement (m, x) =
let val c' = count (m, x)
in if c' <= 1
then NONE
else SOME (IntMap.insert (m, x, c' - 1))
end
fun flip f (x, y) = f (y, x)
val fromList = List.foldl (flip increment) IntMap.empty
end
That is, a CountMap is an int IntMap.map where the Int represents the
fixed key type of the map, being int, and the int parameter in front of it
represents the value type of the map, being a count of how many times this
value occurred.
When building the initialCountMap below, you use CountMap.increment, and
when you use the "sliding window" approach, you use CountMap.decrement to
produce a new countMap that you can test on recursively.
If you decrement the occurrence below 1, you're looking at a sub-list that
doesn't contain every element at least once; we rule out any solution by
letting CountMap.decrement return NONE.
With all of this machinery abstracted out, the algorithm itself becomes much
easier to express. First, I'd like to convert the list to an array so that
indexing becomes O(1), because we'll be doing a lot of indexing.
fun smallest_sublist_length [] = 0
| smallest_sublist_length (xs : int list) =
let val arr = Array.fromList xs
val initialCountMap = CountMap.fromList xs
fun go countMap i j =
let val xi = Array.sub (arr, i)
val xj = Array.sub (arr, j)
val decrementLeft = CountMap.decrement (countMap, xi)
val decrementRight = CountMap.decrement (countMap, xj)
in
case (decrementLeft, decrementRight) of
(SOME leftCountMap, SOME rightCountMap) =>
Int.min (
go leftCountMap (i+1) j,
go rightCountMap i (j-1)
)
| (SOME leftCountMap, NONE) => go leftCountMap (i+1) j
| (NONE, SOME rightCountMap) => go rightCountMap i (j-1)
| (NONE, NONE) => j - i + 1
end
in
go initialCountMap 0 (Array.length arr - 1)
end
This appears to work, but...
Doing Int.min (go left..., go right...) incurs a cost of O(n^2) stack
memory (in the case where you cannot rule out either being optimal). This is a
good use-case for dynamic programming because your recursive sub-problems have a
common sub-structure, i.e.
go initialCountMap 0 10
|- go leftCountMap 1 10
| |- ...
| `- go rightCountMap 1 9 <-.
`- go rightCountMap 0 9 | possibly same sub-problem!
|- go leftCountMap 1 9 <-'
`- ...
So maybe there's a way to store the recursive sub-problem in a memory array and not
perform a recursive lookup if you know the result to this sub-problem. How to
do memoization in SML is a good question in and of itself. How to do purely
functional memoization in a non-lazy language is an even better one.
Another optimization you could make is that if you ever find a sub-list the
size of the number of unique elements, you need to look no further. This number
is incidentally the number of elements in initialCountMap, and IntMap
probably has a function for finding it.
The Problem Statement: Write a function pair that takes two lists of integers and generates a list of pairs, where each pair is a combination of each element from each list.
For example, pair ([1,2], [3,4,5]) should return
[(1,3), (1,4), (1,5), (2,3), (2,4), (2,5)].
My work so far:
-fun pair(a:int list, b:int list) = if null a then nil else if null b then nil else (hd a, hd b)::pair(a, tl b)#pair(tl a, b);
val pair = fn : int list * int list -> (int * int) list
-pair([1,2],[3,4,5]);
val it = [(1,3),(1,4),(1,5),(2,5),(2,4),(2,5),(2,3),(2,4),(2,5)]
I've tried to trace the function to find out why the (2,5), (2,4), (2,5) are showing up but I'm not seeing it clearly still.
It seems straightforward enough but I can't seem to get the last bits ironed out. Some assistance pinpointing why those elements are being added in the middle would be of help.
Thanks.
Peter
The main problem is that you're recursing over both lists.
If you look at your example,
pair ([1,2], [3,4,5]) -> [(1,3), (1,4), (1,5), (2,3), (2,4), (2,5)]
you will see that it has two sublists,
[(1,3), (1,4), (1,5)]
[(2,3), (2,4), (2,5)]
where the first consists of pairs formed from the first element of [1,2] and every element of [3,4,5], and the second is the second element of [1,2] also paired with every element of [3,4,5].
Note that each sublist contains all of [3,4,5] but only one element of [1,2] - the first is the same as pair ([1], [3,4,5]) and the second is pair ([2], [3,4,5]) - so you should only need to recurse over the first list.
You can create such a list like this:
If any input list is empty, the result is empty.
Otherwise:
Take the first element of a and pair it with every element of b in a list (hint: think about map.)
Recursively make pairs from the tail of a and all of b.
Combine the results of 1 and 2.
With pattern matching:
fun pair ([], _) = []
| pair (_, []) = []
| pair (x::xs, ys) = <something involving x and ys, suitably combined with 'pairs (xs, ys)'>
It might help if you write step 1 as a separate function.
Since this is an exercise, I'm not going to show you the answer to the problem statement.
What you're trying to generate is called the cartesian product of the two lists.
Your current approach (formatted a bit nicer),
fun pair (a, b) =
if null a then nil else
if null b then nil else
(hd a, hd b) :: pair (a, tl b) # pair (tl a, b);
produces duplicate results down the line because you leave out hd b in pair (a, tl b) and you leave out hd a in pair (b, tl a), but on the second iteration of e.g. pair (a, tl b), the first element of a is processed once again for each remaining element of tl b.
You can avoid this duplication of work by addressing each element once. I would recommend that you look at the functions map and concat. The general approach is this: For every element x of a, generate (x,y) for every element y of b. "For every element" is map. And
map (fn x => ...something with (x,y)...) a
produces a list of results, just like you want. But if you repeat the same approach of map (fn y => ...) b as the ...something with (x,y)... part, you'll run into an inconvenient surprise that concat can help you with.
You can solve this exercise without using map and concat and instead using manual recursion, but you might have to split the work into two functions, since you'll want one function that folds over the x of a and, for each x, fold once over b. The function map takes the recursion part that both of these functions would have in common and lets you only write the things they don't have in common.
I want to write a code in Haskell, to return an element just before the occurrence of another element in a list. For ex:
eBefore 3 [1,2,3,4,5] should return 2
I am quiet new to haskell. The code that i've written up till now is :
eBefore :: Eq a => a -> [a] -> Maybe a
eBefore n [] = Nothing
eBefore n (x:xs) = if x == n then Just x else eBefore n xs
I would be highly obliged if some one could help me understand the approach or help me out with the problem. Thank you!
You can match more elaborated patterns:
eBefore n [] = Nothing
eBefore n [_] = Nothing
eBefore n (x1:xs#(x2:_))
| x2 == n = Just x1
| otherwise = eBefore n xs
Here we return Nothing for lists containing zero or one elements because they contain no member with another one preceding them. (x1:xs#(x2:_)) is a pattern that matches a x1:xs, where xs in turn matches x2:_, that is, a list with at least two elements, the first element is bound to x1, the second to x2, the residue is unimportant (matched by _).
We also might write thus:
eBefore n [] = Nothing
eBefore n [_] = Nothing
eBefore n (x1:x2:xs)
| x2 == n = Just x1
| otherwise = eBefore n (x2:xs)
However, this variant might be worse in terms of performance. (x1:x2:xs) is equivalent to (x1:(x2:xs)), and we see that (x2:xs) repeated again as an argument to recursive call. But the compiler may fail to recognize the identity of the two expressions and create a new node. That's a waste. By using the #-notation in the former variant, we give that (x2:_) from the pattern a name, xs, and pass it to the recursive call as a ready whole.
The difficult moment here is what we should return in case n is equal to the head of the list, e. g. eBefore 3 [3,4,5,6,3]. The definition above will skip the first occurrence of 3 and return 6.
So i just got in ML programming and I found this excercise in a book. The excercise says to build a recursive function that takes an integer and a list. If L=[a1,a2,a3] then the desired result is [ai+1,ai+2,...,an,a1,a2,...,ai]. So I wrote a function and after a lot of hours I narrowed the errors down to one which I can't understand. Here is my function:
fun cycle L i =
if i = 0 then L
else (cycle tl(L) (i-1)) # [hd(L)];
I will upload an image with the error that i get so someone can explain to me what the interpreter is trying to say to me.
The numbers next to the "a" just show the order of these elements in the list.So for L=[1,2,3,4,5] and for i = 2, the desire result is the List L=[3,4,5,1,2]. I don't think that the type of list is essential in this problem. Hope this further explanation helped
It's a syntactic problem with the recursive call cycle tl(L) (i-1).
In SML, the syntax for function application is juxtaposition, not parentheses. In your case tl(L) indeed calls the function tl with argument L, but that's equivalent to just tl L. The parentheses are redundant and, as such, ignored.
Now, if you replace the minimal version within your original call, you'll get this: cycle tl L (i-1). It's calling cycle with three arguments, instead of just two.
The correct way of writing it would be: cycle (tl L) (i-1).
Ionuț already gave a sufficient answer to the syntax problem; here are some further suggestions:
Use pattern matching rather than hd and tl.
Consider the base cases; what are the simplest sub-problems you can think of? E.g. cycling the empty list will always give the empty list regardless of n, and cycling L 0 times will always give L back. Having both base cases as patterns helps.
Consider the recursive case; the top element (assuming it exists) is cycled and i is reduced by one, until eventually i is 0 or L is empty. Because the second base case catches the empty list, we can freely assume that L is non-empty here, in which case it will match the pattern x::xs.
fun cycle 0 xs = xs
| cycle i [] = []
| cycle i (x::xs) = cycle (i-1) (xs # [x])
Depending on whether 0 <= i and i <= length xs are preconditions for the function or not, you may want to handle these once before activating the main recursion, e.g. by wrapping the function above:
fun cycle i ys =
let fun fun cycle' 0 xs = xs
| cycle' i [] = []
| cycle' i (x::xs) = cycle' (i-1) (xs # [x])
in
if 0 <= i andalso i <= length xs
then cycle' i ys
else raise Domain
end
The main operation, namely xs # [x] is terribly inefficient, since its running time is proportional to the length of xs and is activated n times. So the running time of cycle becomes O(n • |L|) when something like O(min(n,|L|)) should be achievable.
You could probably make a much faster version if you store the cycled elements in a separate list, without using #, and combine the remaining elements with this list after the elements have been cycled. Depending on what you felt about 0 <= i and i <= length xs, you may run into problems with the following test case:
val cycle_test_1 = (cycle 5 [1,2,3,4] = [2,3,4,1])