What does the following recursive Prolog call output? - prolog

Question

I'm trying to learn prologue, but man am I having trouble.
I have an example below as well as what it outputs, and I'm clearly stuck on some concepts but not sure what.
output([]).
output([c|R]):- output(R), !, nl.
output([X|R]) :- output(R), write(X).
?- output([a,b,c,d,e]).
Answer:
ed
ba
true.
Correct me if I'm wrong, but here is what I understand so far...
When we call output([a,b,c,d,e]).
prologue looks for a solution using unification,
it tries output([]) and fails, so it proceeds to the second output([c|R]) which then passes the tail of the list recursively into output([c|R]) until it hits the base case of output([]).
Now I get confused...It then hits the cut which locks R to [] and c with a value of e? how does the output afterwards happens? I'm really confused.

Answer 1

I think you're having a fundamental misunderstanding of what Prolog is doing and what unification is about. In Prolog when you make a query such as output([a,b,c,d,e]). Prolog will start from the beginning of your asserted facts and predicates and attempt to unify this term (your query) with a fact or the head of a predicate.
Unification
We need to stop here for a moment and understand what unification is. In Prolog, the operator =/2 is the unification operator and can be used to query the unification of two terms, term1 = term2. This query will succeed if term and term2 can be successfully unified. How can they be successfully unified? This can happen if there is a binding of variables in term1 and term2 such that the terms become, essentially, identical (by "essentially" I mean they might differ only in syntactic representation but are truly identical when in canonical form - see details below on what that is).
Here are examples of unification attempts that fail. You can enter these at a Prolog prompt and it will show immediate failure.
a = e. % This fails because the atom `a` is different than the atom `e1`
% There are no variables here that can change this fact
foo(X) = bar(Y)
% This fails because the functor `foo` is different than
% the functor `bar`. There's no way to get these terms to match
% regardless of how the variables `X` or `Y` might be instantiated
foo(a, Y) = foo(b, Y)
% This fails because no matter how the variable `Y` is instantiated
% the 1st argument of `foo` just cannot match. That is, the atom
% `a` doesn't match the atom `b`.
foo(a, b, X) = foo(a, b)
% This fails because the `foo/3` and `foo/2` have a different
% number of arguments. No instantiation of the variable `X` can
% change that fact
[1,2] = [1,2,3] % Fails because a list of 2 elements cannot match a list of 3 elements
[] = [_|_] % Fails because the empty list cannot match a list of at
% least one element.
[a,b,c] = [x|T] % Fails, regardless of how `T` might be bound, because `[a,b,c]`
% is a list whose first element is `a`
% and `[x|T]` is a list whose first element is `x`. The
% atoms `a` and `x` do not and cannot match.
Here are examples of successful unifications. You can test these as well at a Prolog prompt and you should get success or, if variables are involved, get at least one solution showing binding of variables that causes it to succeed:
a = a. % Trivial case: an atom successfully unifies with itself
X = a. % Succeeds with `X` bound to `a`
foo(X) = foo(a). % Succeeds with `X` bound to `a`
[a,b,c] = [a|T] % Succeeds with `T` bound to `[b,c]` because the first element
% `a` is the same in both cases.
[1,2,3] = [H|T] % Succeeds with `H` bound to 1, and `T` bound to `[2,3]`
% since `[1,2,3]` is equivalent to `[1|[2,3]]` (they are two
% different syntaxes representing the same term)
Just an aside: Prolog list syntax
We're writing lists using a form that's familiar from other languages. So [] is an empty list, and [1,2,3] is a list of the 3 elements 1, 2, and 3. You can also have lists inside of lists, or any terms in a list for that matter. This, for example, is a valid list of 3 elements: [a, [1,foo(a)], bar(x,Y,[])]. The first element is a, the second is a list of two elements, [1, foo(a)], and the third element is bar(x,Y,[]). In Prolog, you can also write a list in a form that describes the first of one or more elements and a tail. For example [H|T] is a list whose first element is H and the rest of the list is T (itself a list). A list of at least two elements could be written as [H|T] and you'd know that T has at least one element. Or you could write it as [H1,H2|T] and explicitly indicate the first two elements and understand that T would be a list of zero or more arguments. The first elements are individual elements of the list, and the tail is a list representing the rest of the list. The following forms all represent the list [a,b,c,d,e]:
[a,b,c,d,e]
[a|[b,c,d,e]]
[a,b|[c,d,e]]
[a,b,c|[d,e]]
[a,b,c,d|[e]]
[a,b,c,d,e|[]]
If you had a list, L, and wanted prolog to ensure that L had at least two arguments, you could unify L with an anonymous list of 2 elements: L = [_,_|_]. This will only succeed if L is a list of at least two elements.
Another aside: canonical form
Prolog, though, has what it calls a canonical form for terms which is its fundamental representation of a given term. You can see the canonical form of a term by calling write_canonical(Term):
| ?- write_canonical([a,b,c]).
'.'(a,'.'(b,'.'(c,[])))
yes
So that's interesting, what on earth is that? It doesn't look like a list at all! It's actually the canonical form in Prolog of what a list really looks like to Prolog (if you want to think of it that way). The fundamental term form in Prolog is a functor and zero or more arguments. The atom a is a term which could be viewed as a functor a with no arguments. The term foo(1,X) has functor foo and arguments 1 and X. The list [a,b,c] written that way is just a convenient syntax for programmers that make it easy to read. A list is actually formed by the functor '.' and two arguments: the head and the tail. So the list [H|T] in general is '.'(H,T) and the empty list [] is just itself, an atom representing the empty list. When Prolog unifies (or attempts to unify) two lists, it's really looking at a list as '.'(H, T) so it matches the '.' functor, then attempts to match arguments. In the case of multiple elements, it's a recursive match since T is itself a list.
Expressions in Prolog such as X + 3 are also a syntactic convenience for the canonical form, '+'(X, 3).
Back to our story
As we were saying, when you query output([a,b,c,d,e])., Prolog tries to unify this with heads of predicate clauses or facts that you have already asserted. Here's what you have asserted:
output([]).
output([c|R]):- output(R), !, nl.
output([X|R]) :- output(R), write(X).
Starting from the top, Prolog attempts this unification:
output([a,b,c,d,e]) = output([])
This fails since there are no variables to change the terms to make them match. It fails because the list [a,b,c,d,e] and the empty list [] cannot match.
On to the next clause:
output([a,b,c,d,e]) = output([c|R])
This can only succeed if the unification [a,b,c,d,e] = [c|R] can succeed with some binding of R. You can look at this as [a|[b,c,d,e,]] = [c|R]. Clearly, for this unification to succeed, the first element of each list must match. But a and c don't match, so this fails.
On to the next one:
output([a,b,c,d,e]) = output([X|R])
Prolog attempts then to unify [a,b,c,d,e] with [X|R], or [a|[b,c,d,e]] with [X|R]... and this succeeds since X and R are variables and they can be bound as X = a and R = [b,c,d,e]. Now the body of the clause can be executed:
output([b,c,d,e]), write(a).
Before we can get to the write(a), the call output([b,c,d,e]) must execute first and succeed. Following the same logic above, the the first and second clauses of the output/1 predicate do not match. But the 3rd clause matches again with [b,c,d,e] = [X|R] resulting in X = b and R = [c,d,e]. Now the body of this clause is executed again (and you must remember we're now one level deep in a recursive call... the above call to output([b,c,d,e]) is pending awaiting the result):
output([c,d,e]), write(b).
Now it gets more interesting. The first clause of output/1 still doesn't match since [c,d,e] = [] fails. But the second clause now does match since [c,d,e] = [c|R] succeeds with the binding R = [d,e]. So that body is executed:
output([d,e]), !, nl.
Now we need to chase down the call to output([d,e]) (we're now another level deep in recursion remember!). This one fails to match the first two clauses but matches the 3rd clause, by [d,e] = [X|R] with bindings X = d and R = [e].
I could keep going but I'm getting tired of typing and I do have a real job I work at and am running out of time. You should get the idea hear and start working through this logic yourself. The big hint moving forward is that when you finally get to output([]) in a recursive call an you match the first clause, you will start "unwinding" the recursive calls (which you need to keep track of if you're doing this by hand) and the write(X) calls will start to be executed as well as the !, nl portion of the second clause in the case where c was matched as the first element.
Have fun...

Answer 2

The main problem with your reasoning is that c is not a variable but an atom. It cannot be unified with any other value.
So with your example input, for the first 2 calls it will not execute output([c|R]) (since a nor b can be unified with c), but it goes on to output([X|R]) instead. Only for the third call, when the head is c, the former clause is called. After this it will call the latter clause another 2 times for d and e, and then it hits the base case.
From that point on we can easily see the output: if first writes 'e', then 'd', then a new line (for the time we matched c), ad then b and a. Finally you get true as output, indicating that the predicate call succeeded.
Also note that due to the cut we only get a single output. If the cut wasn't there, we would also get edcba, since the c case would also be able to match the last clause.