Question

我尝试在core.logic

中做这样的事情

(defn count-different-elements-in-list [coll]
  (count (set coll)))

这适用于整数很好

(should= 1 (count-different-elements-in-list '(1 1 1)))
(should= 2 (count-different-elements-in-list '(1 1 2)))
(should= 3 (count-different-elements-in-list '(1 3 2)))

但是现在我正在尝试使用core.logic来解决一些问题而且它会变得很混乱

(run* [a b c]
  ;;the variables get values between 1 and 3
  (fd/in a b c (fd/interval 1 3))
  ;; in the list there should only be 2 different values
  (== 2 (count-different-elements-in-list '(a b c))))

但问题出现了，b c不会作为值传递给函数。它们作为变量传递。有三个变量count-different-elements-in-list返回总是3，core.logic找不到解决方案（空列表）。

但我正在寻找这个结果。

([1 1 2] [1 2 1] [2 1 1] 
 [1 1 3] [1 3 1] [3 1 1]
 [2 2 1] [2 1 2] [1 2 2]
 [2 2 3] [2 3 2] [3 2 2]
 [3 3 1] [3 1 3] [1 3 3]
 [3 3 2] [3 2 3] [2 3 3])

Answer 1

您需要将core.logic/project逻辑变量转换为非关系目标，例如普通函数count-different-elements-in-list。遗憾的是，您不能project有限域逻辑变量，如a，b和c，它们不受单个值的限制。（见：this question）

在您拥有的示例中，您可以为生成的范围和fd/in换出fd/interval和membero。这将删除无约束的有限域变量，保持整数的范围约束，并允许投影。

(def interval (vec (range 1 4)))
(run* [a b c]
  (membero a interval)
  (membero b interval)
  (membero c interval)
  (project [a b c]
    (== 2 (count-different-elements-in-list (list a b c)))))

clojure core.logic计数集合中的元素

1 个答案: