Question

I am working on a SML programming assingment for class and stuck on a question. The question is:
“Write an ML function that uses map, foldr, or foldl to compute the intersection of a nonempty list of sets.Here you may assume that sets are denoted as lists. For this problem, you may use auxilary named functions (e.g., isMember). Hint: A nonempty list of sets contains at least one set.”

This is what I have so far can anyone point me in the right direction I am fairly new to SML?

fun member(x,[]) = false
|   member(x,L) =
        if x=hd(L) then true
        else member(x,tl(L));

fun intersect(L1,L2) = if tl(L1) = [] then L1 
else if member(hd(L1),L2) = true then L1
else intersect(tl(L1),L2);

fun combine(L1) = if tl(L1) = [] then hd(L1)
else
    foldr intersect [] L1;

What I want the code to do is start off by executing the combine function with a lists of lists. It checks if there is only one list (i.e. tl(L1) = []) and if it is true then just print the first list. If its false I want to call the foldr function that then calls the intersect function. In theory during the foldr function I want it to check the first list and second list and only keep what values are the same, then check the next list to those kept values and keeping doing this until it has checked every list. After that is done I want it to print each value that was in each list (i.e. intersection of sets).

I know my member function works and the combine function does what its supposed to do, my QUESTION is, what is wrong with the intersection function and can someone explain what the intersection should be doing?

I obviously don’t want the straight up answer, that’s not what I am here for. I need help to get on track for the right answer.

Answer 1

Here is my working answer!

fun member(x,[]) = false
|   member(x,L) =
        if x=hd(L) then true
        else member(x,tl(L));       

fun intersect([],[]) = []
| intersect(L1,[]) = []
| intersect(L1,L2) = if member(hd(L2), L1) then hd(L2)::intersect(L1, tl(L2))
    else intersect(L1, tl(L2));

fun combine(L1) = if L1 = [] then []
else if tl(L1) = [] then hd(L1)
else foldr intersect (hd(L1)) (tl(L1));

Here are some test cases ran.

combine([[1,2],[1,3],[1,4]]);
val it = [1] : int list
combine([[1,2,3],[1,8,9,3],[1,4,3,8,9]]);
val it = [1,3] : int list
combine([[1,2,3,8,9],[1,8,9,3],[1,4,3,8,9]]);
val it = [1,3,8,9] : int list