4 Backtracking
It is helpful to go into the following evaluation (Predicates with Rules) in a little detail:
| (%which () |
| (%computer-literate 'Penelope)) |
The starting goal is:
G0 = (%computer-literate Penelope)
(I’ve taken out the quote because Penelope is the result of evaluating 'Penelope.)
Racklog tries to match this with the head of the first clause of %computer-literate. It succeeds, generating a binding [person . Penelope].
But this means it now has two new goals – subgoals – to solve. These are the goals in the body of the matching clause, with the logic variables substituted by their instantiations:
| G1 = (%knows Penelope TeX) |
| G2 = (%knows Penelope Racket) |
For G1, Racklog attempts matches with the clauses of %knows, and succeeds at the fifth try. (There are no subgoals in this case, because the bodies of these “fact” clauses are empty, in contrast to the “rule” clauses of %computer-literate.) Racklog then tries to solve G2 against the clauses of %knows, and since there is no clause stating that Penelope knows Racket, it fails.
All is not lost though. Racklog now backtracks to the goal that was solved just before, viz., G1. It retries G1, ie, tries to solve it in a different way. This entails searching down the previously unconsidered %knows clauses for G1, ie, the sixth onwards. Obviously, Racklog fails again, because the fact that Penelope knows TeX occurs only once.
Racklog now backtracks to the goal before G1, ie, G0. We abandon the current successful match with the first clause-head of %computer-literate, and try the next clause-head. Racklog succeeds, again producing a binding [person . Penelope], and two new subgoals:
| G3 = (%knows Penelope TeX) |
| G4 = (%knows Penelope Prolog) |
It is now easy to trace that Racklog finds both G3 and G4 to be true. Since both of G0’s subgoals are true, G0 is itself considered true. And this is what Racklog reports. The interested reader can now trace why the following query has a different denouement:
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#f |