序言否定
我正在尝试解决 Prolog 中使用否定的简单查询，但我无法破解它。查询是“查找从未售出的类别”。
知识库如下：
category(stationery, 30, 200, 10, 2).
category(books, 10, 30, 3, 2).
category(consumables, 50, 300, 15, 3).
item(pen, stationery, 10, 150).
item(colgate_small, consumables, 20, 65).
item(colgate_medium, consumables, 45, 70).
item(colgate_big, consumables, 70, 34).
item(juice_small, consumables, 45, 23).
item(juice_medium, consumables, 60, 23).
item(juice_big, consumables, 80, 12).
item(book, stationery, 5, 65).
item(pencil, stationery, 7, 56).
item(newspaper, books, 50, 400).
sale(tom, 1/1/07, pen, 3).
sale(peter, 1/1/07, book, 85).
sale(peter, 1/1/07, juice_small,1).
sale(alice, 7/1/07, pen, 10).
sale(alice, 7/1/07, book, 5).
sale(patrick, 12/1/07, pen, 7).
I am trying to solve a simple query in Prolog that uses negation but I can't crack it. The query is "Find the categories that have never been sold".
The knowledge base is as follows:
category(stationery, 30, 200, 10, 2).
category(books, 10, 30, 3, 2).
category(consumables, 50, 300, 15, 3).
item(pen, stationery, 10, 150).
item(colgate_small, consumables, 20, 65).
item(colgate_medium, consumables, 45, 70).
item(colgate_big, consumables, 70, 34).
item(juice_small, consumables, 45, 23).
item(juice_medium, consumables, 60, 23).
item(juice_big, consumables, 80, 12).
item(book, stationery, 5, 65).
item(pencil, stationery, 7, 56).
item(newspaper, books, 50, 400).
sale(tom, 1/1/07, pen, 3).
sale(peter, 1/1/07, book, 85).
sale(peter, 1/1/07, juice_small,1).
sale(alice, 7/1/07, pen, 10).
sale(alice, 7/1/07, book, 5).
sale(patrick, 12/1/07, pen, 7).
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（5）
Sam Segers 的答案是正确的，尽管它会给您未售出的类别列表。
如果您不需要聚合，而是需要一个谓词，该谓词将回溯所有没有售出任何商品的类别，您可以编写如下内容：
此谓词在回溯时，将生成所有没有售出商品的类别。
Sam Segers answer is correct, though it will give you the list of categories not sold.
If you don't want an aggregation but rather a predicate which will backtrack over all the categories which do not have any items sold you would write something like this:
This predicate, upon backtracking, will yield all the categories for which no items where sold.
可能不是最有效的方法。
但我认为它应该有效。
Might not be the most efficient way.
But I think it should work.
您是否知道 Prolog 的 negation-as-failure （控制）谓词，\+/1？当且仅当目标无法被证明时，它才是
true。使用谓词，任务可以简化为查找已售出的类别，即
，
如果您想要所有这些类别的列表，只需使用 findall 谓词即可。
Are you aware of Prolog's negation-as-failure (control) predicate, \+/1? It is
trueiff the goal cannot be proven.Using the predicate, the task reduces to finding categories that have been sold, i.e.
and
If you want a list of all those categories, simply use the findall predicate.
正如 Sam Segers 提到的，您可以在表达式周围放置 not() 。
您还可以使用 \+ 运算符在逻辑上否定谓词：
As Sam Segers mentioned you can put a not() around an expression.
You can also use the \+ operator to logically negate a predicate:
用法：
Usage: