3

我尝试在 core.logic 中做这样的事情

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

这适用于整数就好了

(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)))

但现在我正在尝试使用 core.logic 来解决一些问题，但它变得一团糟

(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))))

但是问题来了， abc 没有作为值传递给函数。它们作为变量传递。使用三个变量 count-different-elements-in-list 总是返回 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])
4

1 回答 1

4

您需要将core.logic/projectvar 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限定域逻辑变量，例如a、b和c，它们不限于单个值。（见：这个问题）

在您的示例中，您可以将fd/inand换成fd/interval生成的范围 and 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)))))
于 2013-09-26T04:07:30.047 回答