考起碩士生的小一數學題(ppp)解法 - 香港高登 http://m2.hkgolden.com/view.aspx?message=6307521&type=SW&page=1

repl.clj

(use 'clojure.core.logic)
(require '[clojure.core.logic.fd :as fd])

(def result
(run* [q]
  (fresh [a b c d e f g h p ab cd ef gh ppp]
    (fd/in a b c d e f g h p (fd/interval 1 9)) ; why not zero?
    (fd/in ab cd ef gh (fd/interval 10 99))
    (fd/in ppp (fd/interval 100 999))
    (fd/eq
      (= ab (+ (* a 10) b))
      (= cd (+ (* c 10) d))
      (= ef (+ (* e 10) f))
      (= gh (+ (* g 10) h))
      (= ppp (+ (* p 100) (* p 10) p))
      (= ppp (+ gh ef))
      (= ef (- ab cd)))
    (fd/distinct [a b c d e f g h p])
    (== q [a b c d e f g h p]) 
    ))
)

(map 
  (fn [[a b c d e f g h p]] 
    (println " " a b "\n"
             "-" c d "\n"
                "  ---\n"
             " " e f "\n"
             "+" g h "\n"
                " ----\n" 
                   p p p)) 
  result)

; result
;   8 5 
; - 4 6 
;  ----
;   3 9 
; + 7 2 
;  ----
; 1 1 1
;   9 5 
; - 2 7 
;  ----
;   6 8 
; + 4 3 
;  ----
; 1 1 1
;   8 6 
; - 5 4 
;  ----
;   3 2 
; + 7 9 
;  ----
; 1 1 1