Prolog List. Check if first and last element in list is similar

Question

Example:

firstlast([1,2,3,4,1]).
true;

firstlast([1,2,3,4]).
false;

firstlast([5,10,4,3]).
false;

exc...

The problem is im only allowed to use recursion with the predicate "firstlast". ? I have really tried to break this, but i cant seem to check / compare the last element with the first.

Any hints ?

Answer 1

UPDATE : Since you are not allowed to use other predicates, try this:

firstlast([H,H]).
firstlast([F,_|T]) :- firstlast([F|T]).

The first predicate deals with the base case, the second one removes the second element in a list of three or more items, and recurses down.

Answer 2

You probably mean that the first and last element are the same. Here is a solution using dcg -notation:

firstlast(Xs) :-
    phrase(([X],...,[X]), Xs).

... --> [] | [_], ... .

I am not sure whether firstlast([1]) should succeed or not ...

Answer 3

Well since you can only use a recursion with firstlast/1 the solution will look like:

firstlast(...) :- ... .
firstlast(...) :- ... .
firstlast(...) :- ... .
....
firstlast(...) :- ... .

some of those will be the rules regarding the base case and some of them the rules that "eat away" the problem. this problem requires one check: compare the first and the last element. so, in your base case you should have only these 2 elements; you dont need anything else. so the solution will ignore all the other elements

last hint: you can access the 2 first elements of a list with the following unification pattern:

foo([H1,H2|T])

Answer 4

So fair i got this:

firstlast([H,_|T]) :-
(T1 = H, T1 = T) -> firstlast([H|T]).

My code compares the last and first element but the recursion is just wrong :/

As answered above it should not be allowed to succed with one element in the list. Though im only allowed to use the predicate "firstlast".