Did you know ... | Search Documentation: |
Incremental tabling |
Incremental tabling maintains the consistency of a set of tabled
predicates that depend on a set of dynamic predicates. Both the tabled
and dynamic predicates must have the property incremental
set. See dynamic/1
and table/1.
Incremental tabling causes the engine to connect the answer tries and incremental dynamic predicates in an Incremental Dependency Graph (IDG). Modifications (asserta/1, retract/1, retractall/1 and friends) of an incremental dynamic predicate mark all depending tables as invalid. Subsequent usage of these tables forces re-evaluation.
Re-evaluation of invalidated tables happens on demand, i.e., on access to an invalid table. First the dependency graph of invalid tables that lead to dynamic predicates is established. Next, tables are re-evaluated in bottom-up order. For each re-evaluated table the system determines whether the new table has changed. If the table has not changed, this event is propagated to the affected nodes of the IDG and no further re-evaluation may be needed. Consider the following program:
:- table (p/1, q/1) as incremental. :- dynamic([d/1], [incremental(true)]). p(X) :- q(X). q(X) :- d(X), X < 10. d(1).
Executing this program creates tables for X=1 for p/1 and
q/1 . After calling assert(d(100))
the tables for p/1 and
q/1 have an invalid count of
1. Re-running p(X)
first re-evaluates q/1
(bottom-up) which results to the same table as X=100 does not
lead to a new answer. Re-evaluation clears the invalid count for q/1
and, because the q/1 tables is not changed, decrements the invalid count
of affected tables. This sets the
invalid count for p/1 to zero, completing the re-evaluation.
Note that invalidating and re-evaluation is done at the level of tables. Notably asserting a clause invalidates all affected tables and may lead to re-evaluating of all these tables. Incremental tabling automates manual abolishing of invalid tables in a changing world and avoids unnecessary re-evaluation if indirectly affected tables prove unaffected because the answer set of dependent tables is unaffected by the change. This is the same policy as implemented in XSB Swift, 2014. Future versions may implement a more fine grained approach.