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

【问题标题】：How can I remove [:b (lvar)] from [[:a :A] [:b :B] [:c :C]] in Clojure core.logic?如何从 Clojure core.logic 中的 [[:a :A] [:b :B] [:c :C]] 中删除 [:b (lvar)]？
【发布时间】：2016-04-08 20:44:10
【问题描述】：

我想在以下示例代码中从[[:a :A] [:b :B] [:c :C]] 中删除[:b :B]，如果我将:B 替换为(lvar)，它将不再有效：

;; 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

这些没有（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

【问题讨论】：

标签： clojure logic clojure-core.logic

【解决方案1】：

我相信这在您的示例中不起作用的原因是创建的 (lvar) 与程序中的任何其他逻辑变量未绑定/无关。如果您使用 fresh 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，在不知道元组中的 :b 项的情况下，这些返回相同的结果：

(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)))

【讨论】：