clojure - 如何从 Clojure core.logic 中的 [[:a :A] [:b :B] [:c :C]] 中删除 [:b (lvar)]？

Question

我想[:b :B]从[[:a :A] [:b :B] [:c :C]]以下示例代码中删除，如果我替换:B为(lvar)，它不再起作用：

;; Helper Function
(defne not-membero [x l]
  ([_ []])
  ([_ [?y . ?r]]
   (!= x ?y)
   (not-membero x ?r)))

这些工作：

(run* [q]
  (membero q [[:a :A] [:b :B] [:c :C]])
  (not-membero q [[:b :B]]))
(run* [q]
  (membero q [[:a :A] [:b :B] [:c :C]])
  (!= q [:b :B]))
;; both return [[:a :A] [:c :C]], as expected

这些不（注意lvar）：

(run* [q]
  (membero q [[:a :A] [:b :B] [:c :C]])
  (not-membero q [[:b (lvar)]]))
(run* [q]
  (membero q [[:a :A] [:b :B] [:c :C]])
  (!= q [:b (lvar)]))
;; both return [[:a :A] [:b :B] [:c :C]], unexpected

Answer 1

我相信这在您的示例中不起作用的原因是创建(lvar)的与程序中的任何其他逻辑变量无关/无关。fresh如果您使用逻辑变量，您的程序可以正常工作（至少我认为这是您想要的） ：

(run* [q]
  (fresh [x]
    (membero q [[:a :A] [:b x] [:c :C]])
    (not-membero q [[:b x]])))
=> ([:a :A] [:c :C])
(run* [q]
  (fresh [x]
    (membero q [[:a :A] [:b x] [:c :C]])
    (!= q [:b x])))
=> ([:a :A] [:c :C])

或者，这些在不知道:b元组中的项目的情况下返回相同的结果：

(run* [q]
  (fresh [x]
    (membero q [[:a :A] x [:c :C]])
    (not-membero q [x])))
(run* [q]
  (fresh [x]
    (membero q [[:a :A] x [:c :C]])
    (!= q x)))