I am trying to understand how to solve the following problem in Prolog. Given the following Prolog code, list all solutions in order for the query c(X,Y,Z).
a(1).
a(2).
b(a).
c(A,B,C) :- a(A),d(B,C).
c(A,B,C) :- b(A),d(B,C).
d(B,C) :- a(B),!,a(C).
d(B,_) :- b(B).
I loaded up SWI Prolog to try it out myself, but I am unsure how it reaches the conclusion:
X = Y, Y = Z, Z = 1 ;
X = Y, Y = 1,
Z = 2 ;
X = 2,
Y = Z, Z = 1 ;
X = Z, Z = 2,
Y = 1 ;
X = a,
Y = Z, Z = 1 ;
X = a,
Y = 1,
Z = 2.
Starting small, I tried to query just d(B,C). with the same code, leading to the following conclusion:
Y = Z, Z = 1 ;
Y = 1,
Z = 2.
I understand how to get the Y=1 and Z=1,2, but I am unsure how the code leads to Y=Z. If anyone could help me with the logic to reach these conclusions, that would be well appreciated!
Let's start by looking at the predicates a/1 and b/1: We expect them to produce two and one answer(s) respectively and indeed that is the case:
?- a(X).
X = 1 ? ;
X = 2
?- b(X).
X = a
Next let's look at the predicate d/2 but without the cut, that is:
d(B,C) :- a(B),a(C).
d(B,_) :- b(B).
If we query this predicate:
?- d(B,C).
Prolog will start with the first rule of the predicate d/2, that is d(B,C) :- a(B),a(C). and tries to prove the first goal a(B). That succeeds with the substitution B = 1. So Prolog goes on to prove the goal a(C). Again that succeeds with the substitution C = 1. Prolog then reports the first solution to the query:
?- d(B,C).
B = C = 1 ?
So we ask if there are any more solutions by pressing ;:
?- d(B,C).
B = C = 1 ?;
Now Prolog tries to find another solution for a(C) and finds the substitution C = 2:
?- d(B,C).
B = C = 1 ?;
B = 1,
C = 2 ?
Again we ask if there's more:
?- d(B,C).
B = C = 1 ?;
B = 1,
C = 2 ? ;
Now Prolog fails to find another solution for a(C), so it backtracks to the first goal and tries to find an alternative solution for a(B) and succeeds with the substitution B = 2. So it goes on to prove the second goal a(C) to which again the substitution C = 1 is found. So Prolog reports this new solution.
?- d(B,C).
B = C = 1 ?;
B = 1,
C = 2 ? ;
B = 2,
C = 1 ?
We ask for yet more solutions:
?- d(B,C).
B = C = 1 ?;
B = 1,
C = 2 ? ;
B = 2,
C = 1 ? ;
So Prolog looks for another solution for a(C) and finds C = 2. So it answers:
?- d(B,C).
B = C = 1 ?;
B = 1,
C = 2 ? ;
B = 2,
C = 1 ? ;
B = C = 2 ?
We press ; again
?- d(B,C).
B = C = 1 ?;
B = 1,
C = 2 ? ;
B = 2,
C = 1 ? ;
B = C = 2 ? ;
And Prolog looks for other solutions for a(C) but can't find any. So it backtracks to the goal a(B) and again fails to produce new answers here. So those are all solutions for the first rule of d/2. Prolog now goes on and tries to find a solution for the second rule d(B,_) :- b(B).. This rule only contains one goal: b(B) and Prolog finds a substitution: B = a, so it answers:
?- d(B,C).
B = C = 1 ?;
B = 1,
C = 2 ? ;
B = 2,
C = 1 ? ;
B = C = 2 ? ;
B = a
?-
And there are no more solutions. Let's observe that while Prolog traverses the proof tree as seen above, every time alternative substitutions can't be ruled out a choice point is created and in the course of the search for answers Prolog can backtrack to these choicepoints to look for alternative solutions. Every time an answer in the above query ended with ? Prolog still had some choice points left to explore and by pressing ; we asked it to explore them.
Now let's move our attention back to your posted version of d/2 with the cut in the first rule:
d(B,C) :- a(B),!,a(C).
d(B,_) :- b(B).
What a cut does is to prune away choice points back to where it was when the predicate that contains it was called. So it is essentially throwing away untried alternatives, thereby committing Prolog to the first substitution(s) it finds. So if we issue the query again with this version:
?- d(B,C).
Prolog again starts with the first rule, looking for a solution for a(B), successfully substituting B = 1 encountering the cut, therefore committing to this solution and then searching for a proof for a(C), again successfully substituting C = 1. Prolog then answers:
?- d(B,C).
B = C = 1 ?
indicating that there are open choice points. Again we want more:
?- d(B,C).
B = C = 1 ? ;
Prolog finds another substitution C=2 for the goal a(C) and reports:
?- d(B,C).
B = C = 1 ? ;
B = 1,
C = 2
?-
But this time the query terminates. That is because the choicepoint for an alternative solution for the goal a(B) was pruned away thereby leaving us without the solutions B = 2, C = 1 and B = C = 2. Furthermore the choicepoint that would have led to the second rule being explored was also pruned away thereby robbing us of the solution B = a. Cuts that prune away correct solutions as seen above are often referred to as red cuts in the literature.
Note that by cutting away actual solutions this program is not logically correct anymore. With the above query you asked for all solutions of the predicate d/2 but got only two of them. However, if you ask for the other solutions specifically, the respective queries still succeed:
?- d(2,1).
yes
?- d(2,2).
yes
?- d(a,_).
yes
?-
Returning to your original question, let's compare the query ?- c(X,Y,Z). with both versions of d/2. To make comparisons easier, the variables X, Y and Z are replaced with A, B and C just like they occur in the predicate definitions:
% cut-free version % version with the cut:
?- c(A,B,C). ?- c(A,B,C).
A = B = C = 1 ? ; A = B = C = 1 ? ;
A = B = 1, A = B = 1,
C = 2 ? ; C = 2 ? ;
A = C = 1,
B = 2 ? ;
A = 1,
B = C = 2 ? ;
A = 1,
B = a ? ;
A = 2, A = 2,
B = C = 1 ? ; B = C = 1 ? ;
A = C = 2, A = C = 2,
B = 1 ? ; B = 1 ? ;
A = B = 2,
C = 1 ? ;
A = B = C = 2 ? ;
A = 2,
B = a ? ;
A = a, A = a,
B = C = 1 ? ; B = C = 1 ? ;
A = a, A = a,
B = 1, B = 1,
C = 2 ? ; C = 2
A = a,
B = 2,
C = 1 ? ;
A = a,
B = C = 2 ? ;
A = B = a
For the sake of brevity I'm not going to iterate through the steps of the proof, but let's observe a few key things: Firstly, in both rules of c/3 the predicate d/2 is called by the same goal d(B,C) and we have already looked at that goal in detail for both versions. Secondly, we can observe that both rules of c/3 are taken into regard in the search for answers with both versions of the predicate by seeing all three substitutions, A = 1, A = 2 and A = a appearing in both answer-sets. So the cut doesn't prune away the choice point here. Thirdly, the three solutions of d/2 that are pruned away by the cut: B = 2, C = 1, B = 2, C = 2 and B = a are missing in the answers for all three substitutions of A in the version with the cut. And finally, the version of c/3 with the cut is also not logically correct, as the most general query does not give you all the answers but you can query for the thrown away solutions directly and Prolog will find them, e.g.:
?- c(1,2,1).
yes
?- % etc.