Question

This was given as an example in a professor’s lecture:

append([ ], A, A).
append([A|B], C, [A|D]) :- append(B,C,D).

Build a list:

?- append([a],[b],Y).
Y = [ a,b ]

Break a list into constituent parts:

?- append(X,[b],[a,b]).
X = [ a ]
?- append([a],Y,[a,b]).
Y = [ b ]

I’ve spent 3 hours trying to grasp it and can’t. Hadn’t had any trouble with any prolog concepts preceding this slide. There is no further explanation provided, no other information. This is all. If someone could walk me through how this procedure works, I would love them til death do us part.

Answer 1

The first thing to understand is the :- operator.

this_will_be_true :- if_this_is_true

Basically, whatever is on the right of :- is a precondition. A good example is:

sibling(X, Y) :- parent_child(Z, X), parent_child(Z, Y).

This basically means that X and Y are siblings if there exists a parent Z such that Z is the parent of both X and Y.

append([ ], A, A).

This line basically means that appending something to an empty list returns the something. It’s the base case in the recursion.

append([A|B], C, [A|D]) :- append(B,C,D).

This line means that appending C to an existing list with A and B returns a list with A and D given that appending C to B returns D.

Build a list:

?- append([a],[b],Y).
Y = [ a,b ]

So, what’s going on here is that Prolog returns the only possible value of Y that satisfies the two rules given the two initial values. Let’s think about how this happens. This would need to first be evaluated by the second rule. So [A|B] is [a] and C is [b].

So with [A|B] we have to go back to the first rule because B is empty list (it is [ ]). The first rule basically states that we can write [a] as [a|[ ]] and they are the same thing. So now we can go back to the second rule. A is a, B is [ ], and C is [b].

So now let’s check the precondition of append(B, C, D). This is append([ ], [b], D). Once again, using the first rule, we can see that D is also [b].

So Y, by the second rule definition, is [A|D]. Now that we know D is [b], we know that Y is [a, b].

I’ll only do one of the breaking apart since they’re basically the same thing.

?- append(X,[b],[a,b]).
X = [ a ]

So here, Prolog is going to return the only possible value of X so that the statement returns true. Let’s take a look at the second rule. So we know that [a, b] is [A|D]. This means that A is a and D is [b]. We also know that C is [b]. So now, we need to look at the precondition to figure out what B is. append(B, C, D) translates to append(B, [b], [b]). Now, using the first rule, we know that B has to be [ ]. So now we know that [A|B] is [a|[ ]] which is the same as [a]. Therefore, X must be [a].

I hope this was a detailed enough explanation.