math-brute-force.clj

;; Solves math excercise: 
;;   9 = 5 [ ] 3 [ ] 4 [ ] 5,
;; , where [ ] should be one of operations: +, -
;; Uses just random brute force :)

user> (def ops #{+ -})
user> (loop [cnt 0
             op1 (first ops)
             op2 (first ops)
             op3 (first ops)]
        (let [eq [9 (op3 (op2 (op1 5 3) 4) 5)]
              opname #(-> % .getClass .getName clojure.main/demunge (.split "/") second)]
          (cond
            (= cnt 20) (println "result not found in" 20 "iterations")
            (apply = eq) (println (str "result found after " cnt " iterations: "
                                        (->> [op1 op2 op3]
                                             (map opname)
                                             (clojure.string/join " "))))
            :else (recur (inc cnt)
                         (rand-nth (seq ops))
                         (rand-nth (seq ops))
                         (rand-nth (seq ops))))))
result found after 5 iterations: + - +









user> ;; Version that uses infix library (https://github.com/rm-hull/infix) to allow using infix notation
user> (def ops #{'+ '-})
#'user/ops
user> (loop [cnt 0
             op1 (first ops)
             op2 (first ops)
             op3 (first ops)]
        (let [eq [9 (eval (ic/rewrite (list 5 op1 3 op2 4 op3 5)))]]
          (cond
            (= cnt 20) (println "result not found in" 20 "iterations")
            (apply = eq) (println (str "result found after " cnt " iterations: "
                                       (clojure.string/join " " [op1 op2 op3])))
            :else (recur (inc cnt)
                         (rand-nth (seq ops))
                         (rand-nth (seq ops))
                         (rand-nth (seq ops))))))
result found after 6 iterations: + - +