How to get the first, middle and last element of a list scheme and prolog?

Question

I am trying to write a function in Scheme and Prolog that returns first, middle and last item of a list. E.g., find([4,5,8,7,9],L), L = [4,8,9].

I came up with this piece of code for Scheme language, but I am new to Prolog and don't know much, so how I can get the same result in Prolog?

(define (frst L)
   (car L))

(define (last L)
   (if (null? (cdr L))
       (car L)
       (last (cdr L))))

(define (nth L x)
   (if (= x 1)
       (car L)
       (nth (cdr L) (- x 1))))

(define (firstMidLast L)
   (list (frst L)
         (nth L (ceiling (/ (length L) 2)))
         (last L)))

Answer 1

Here's another way to do it!

• The "trick" is to walk down the same list at two different speeds.
• First argument indexing keeps goals list_first_mid_last(+,?,?,?) deterministic.

We define list_first_mid_last/4 like this:

list_first_mid_last([E|Es],E,M,L) :-
    ahead_of_mid_last([E|Es],[E|Es],M,L).

ahead_of_mid_last([],[M|_],M,M).
ahead_of_mid_last([F|Fs],Es,M,L) :-
   more_ahead_of_mid_last(Fs,F,Es,M,L).

more_ahead_of_mid_last([],L,[E|_],E,L).
more_ahead_of_mid_last([F|Fs],_,Es,E,L) :-
   evenmore_ahead_of_mid_last(Fs,F,Es,E,L).

evenmore_ahead_of_mid_last([],L,[E|_],E,L).
evenmore_ahead_of_mid_last([F|Fs],_,[_|Es],M,L) :-
    more_ahead_of_mid_last(Fs,F,Es,M,L).

Let's run a few queries and put Prolog¹ and Scheme² results side-by-side!

%  Prolog                                     % ; Scheme
?- list_first_mid_last([1],F,M,L).            % > (firstMidLast `(1))
F = M, M = L, L = 1.                          % (1 1 1)
                                              %
?- list_first_mid_last([1,2],F,M,L).          % > (firstMidLast `(1 2))
F = M, M = 1, L = 2.                          % (1 1 2)
                                              %
?- list_first_mid_last([1,2,3],F,M,L).        % > (firstMidLast `(1 2 3))
F = 1, M = 2, L = 3.                          % (1 2 3)
                                              %
?- list_first_mid_last([1,2,3,4],F,M,L).      % > (firstMidLast `(1 2 3 4))
F = 1, M = 2, L = 4.                          % (1 2 4)
                                              %
?- list_first_mid_last([1,2,3,4,5],F,M,L).    % > (firstMidLast `(1 2 3 4 5))
F = 1, M = 3, L = 5.                          % (1 3 5)
                                              %
?- list_first_mid_last([1,2,3,4,5,6],F,M,L).  % > (firstMidLast `(1 2 3 4 5 6))
F = 1, M = 3, L = 6.                          % (1 3 6)
                                              %
?- list_first_mid_last([1,2,3,4,5,6,7],F,M,L).% > (firstMidLast `(1 2 3 4 5 6 7))
F = 1, M = 4, L = 7.                          % (1 4 7)

---

^{Footnote 1: Using swi-prolog version 7.3.11 (64-bit).
Footnote 2: Using the scheme interpreter SCM version 5e5 (64-bit).}

Answer 2

Assuming that in the case that the list has even number of items you can choose one of them (in this case the first one), this procedure should work:

Edited per comment from user 'repeat' (i am not versed on Scheme)

find([First|List], [First, Middle, Last]):-
  append(_, [Last], [First|List]),
  length(List, Length),
  NLength is Length >> 1,
  nth0(NLength, [First|List], Middle).

The head of the clause instantiates the First item of the list, then append/3 takes the Last item, length/2 computes the size-1 of the list, >>/2 will divide that size by 2, and nth0/3 will get the Middle item.