datalog: dl-magic-rewrite worklist now drains across rule chains (239/239)
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Real bug: the worklist used (set! queue (rest queue)) to pop the
head, which left queue bound to a fresh empty list as soon as the
last item was popped. Subsequent (append! queue ...) was a no-op
on the empty list — so when the head's rewrite generated new
(rel, adn) pairs to enqueue, they vanished. Multi-relation
programs (e.g. shortest -> path -> edge, or chained derived
relations) only had their head's rules rewritten; downstream
rules silently dropped.

Fix: use an index-based loop (idx 0 → len queue), with append!
adding to the same list. Items added after the current pointer
are picked up in subsequent iterations.

2 new regression tests:
- 4-level chain (a → r1 → r2 → r3 → r4) under magic returns 2
- shortest-path demo via magic equals dl-query (1 result)
This commit is contained in:
2026-05-08 14:41:05 +00:00
parent e981368dcf
commit 17d6f58cc5
4 changed files with 53 additions and 26 deletions

View File

@@ -242,6 +242,32 @@
(= (len semi) (len magic))))
true)
;; Multi-relation rewrite chain: query r4 → propagate to r3,
;; r2, r1, a. The worklist must process all of them; an
;; earlier bug stopped after only the head pair.
(dl-mt-test! "magic chain through 4 rule levels"
(let
((db (dl-program
"a(1). a(2). r1(X) :- a(X). r2(X) :- r1(X).
r3(X) :- r2(X). r4(X) :- r3(X).")))
(= 2 (len (dl-magic-query db (list (quote r4) (quote X))))))
true)
;; Shortest-path demo via magic — exercises the rewriter
;; against rules that mix recursive positive lits with an
;; aggregate body literal.
(dl-mt-test! "magic on shortest-path demo"
(let
((db (dl-program-data
(quote ((edge a b 5) (edge b c 3) (edge a c 10)))
dl-demo-shortest-path-rules)))
(let
((semi (dl-query db (quote (shortest a c W))))
(magic (dl-magic-query db (quote (shortest a c W)))))
(and (= (len semi) (len magic))
(= (len semi) 1))))
true)
;; Magic over a rule whose body contains an aggregate.
;; The rewriter passes aggregate body lits through unchanged
;; (no propagation generated for them), so semi-naive's count