clojure core.logic 计算集合中的元素

Question

我尝试在 core.logic 中做类似的事情

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

这适用于整数

(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)))

但现在我尝试使用 core.logic 来解决一些问题，但事情变得很混乱

(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))))

但问题来了， abc 不会作为值传递给函数。它们作为变量传递。对于三个变量，count- different-elements-in-list 返回始终为 3 并且 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变量逻辑化为非关系目标，例如普通函数count-different-elements-in-list。不幸的是，您不能project有限域逻辑变量，例如a、b和c，它们不限于单个值。（参见：这个问题）

在您的示例中，您可以将fd/in和替换fd/interval为生成的范围 和membero。这将删除不受约束的有限域变量，保留整数的范围约束，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)))))