I am trying to get a feel for Prolog programming by going through Ulle Endriss' lecture notes. When my solution to an exercise does not behave as expected, I find it difficult to give a good explanation. I think this has to do with my shaky understanding of the way Prolog evaluates expressions.
Exercise 2.6 on page 20 calls for a recursive implementation of a predicate last1 which behaves like the built-in predicate last. My attempt is as follows:
last1([_ | Rest], Last) :- last1(Rest, Last). last1([Last], Last).
It gives the correct answer, but for lists with more than one element, I have to key in the semicolon to terminate the query. This makes last1 different from the built-in last.
?- last1([1], Last).
Last = 1.
?- last1([1, 2], Last).
Last = 2 ;
false.
If I switch the order in which I declared the rule and fact, then I need to key in the semicolon in both cases.
I think I know why Prolog thinks that last1 may have one more solution (thus the semicolon). I imagine it follows the evaluation sequence
last1([1, 2], Last).
==> last1([2], Last).
==> last1([], Last). OR Last = 2.
==> false OR Last = 2.
That seems to suggest that I should look for a way to avoid matching Rest with []. Regardless, I have no explanation why switching the order of declaration ought to have any effect at all.
Question 1: What is the correct explanation for the behavior of last1?
Question 2: How can I implement a predicate last1 which is indistinguishable from the built-in last?