Question

我用或运算符定义了一个规则，但它返回多个true或false。

isloanaccept(Name,Guarantor,LoanType,LoanAmount,LoanTenure) 
:-  customer(Name,bank(_),customertype(_),
 citizen(Ci),age(Age),credit(C),
 income(I),property(_),bankemployee(_)), 
 Ci == 'malaysian',
 Age >= 18,
 C > 500, 
    I > (LoanAmount / LoanTenure) / 12,
 isguarantor(Guarantor,Name), 
 ispersonalloan(LoanType,LoanAmount,LoanTenure);
 ishouseloan(LoanType,LoanAmount,LoanTenure);
 isbusinessloan(LoanType,LoanAmount,LoanTenure);
 iscarloan(LoanType,LoanAmount,LoanTenure).

实际上，我需要检查贷款类型是否符合特定贷款要求并与一般规则相结合。

换句话说，我需要像这样定义上面的规则。

Ci == 'malaysian', Age >= 18,C > 500, 
I > (LoanAmount / LoanTenure) / 12,
isguarantor(Guarantor,Name) 
    Or with   (ispersonalloan(LoanType,LoanAmount,LoanTenure);
             ishouseloan(LoanType,LoanAmount,LoanTenure);
             isbusinessloan(LoanType,LoanAmount,LoanTenure);
             iscarloan(LoanType,LoanAmount,LoanTenur)

它应该在命令行中返回1个true / false而不是多个语句。

在命令行中检查规则后，每个或者规则返回1布尔值，这是我想要的。我需要这样（一般规则＆amp;（多重或规则））。

如何组合返回1个布尔值的几个或规则？

请帮忙。

感谢。

Answer 1

用once围绕所有“或”目标。

e.g。

once(
 ispersonalloan(LoanType,LoanAmount,LoanTenure);
 ishouseloan(LoanType,LoanAmount,LoanTenure);
 isbusinessloan(LoanType,LoanAmount,LoanTenure);
 iscarloan(LoanType,LoanAmount,LoanTenure)
).

现在，“或”目标成功或失败。

Answer 2

首先，您应该将(和)放在目标上并与;结合使用。因为它目前将其解释为customer(...),...,isguarantor(Guarantor,Name), ispersonalloan(...)，ishouseloan(...)，...，iscarloan(...)的分离。这是因为运营商,和;的优先级不同。

实际上; - 意味着真实的“或”，而不是“相互排斥或”，而不是“在其他情况下”。因此，如果“ishouseloan”可以“与'ispersonalloan'一起成功”，那么你将拥有几个成功的目标。在此示例中，once/1可能有所帮助（以及not(not(...))），但您可以尝试更深入地了解您的任务并指定非常规目标，例如（我对重叠{{1}做一些个人假设}）：

isXXX

在这种情况下，当isloan(LT, Am, T):- (ishouseloan(LT,Am,T) ;iscarloan(LT,AM,T) ;not((ishouseloan(LT,Am,T);iscarloan(LT,AM,T))), (ispersonalloan(LT,Am,T) ;isbusinessloan(LT,Am,T) ) )，LT和Am尚未绑定到特定值并且T可以免费绑定时，您应该能够生成所有贷款变量

Prolog或（;）规则返回多个结果

2 个答案: