It is pretty common to unify a variable with a vector or list of a fixed length like {{(== q (list a b))}} and then define more goals on {{a}} and {{b}} with the intention of being able to define the * query's result format. However, if one does the same using a set one gets a StackOverflowError.
Example:
`
(run* [q]
(fresh [a b]
(== q #{a b}) ;; StackOverflowError
;;(== q [a b]) ;; works
;;(== q (list a b)) ;; works, too
(conde [(== a 1)] [(== a 2)] [(== a 3)])
(conde [(== b 1)] [(== b 2)] [(== b 3)])))
`
Basically, what I wanted to express here is that the order of {{a}} and {{b}} is not significant so that * } gives me the unique results.
The backtrace is:
`
logic.clj: 231 clojure.core.logic/walk*/fn
logic.clj: 984 clojure.core.logic/eval8917/fn
protocols.clj: 55 clojure.core.logic.protocols/eval7398/fn/G
logic.clj: 229 clojure.core.logic/walk*
logic.clj: 233 clojure.core.logic/walk*/fn
logic.clj: 984 clojure.core.logic/eval8917/fn
protocols.clj: 55 clojure.core.logic.protocols/eval7398/fn/G
logic.clj: 229 clojure.core.logic/walk*
logic.clj: 233 clojure.core.logic/walk*/fn
logic.clj: 984 clojure.core.logic/eval8917/fn
protocols.clj: 55 clojure.core.logic.protocols/eval7398/fn/G
[...]
`
