我有一个简单的布尔表达式作为lisp列表显示如下: '(或0(AND A1 A2))
前面的列表是((A1和A2)OR 0)的表示。
无论如何,我正在写一个函数来简化这个表达式.. 例如:
像这样调用函数“reduce”:
(减少'(OR 0(AND A1 A2)))
会产生
(AND A1 A2)
我首先尝试创建基本规则,所以我定义了 以下身份:
(AND 1 S)== S,
(OR 0 S)== S,
(AND 0 S)== 0,
(或1 S)== 1,
(不是O)== 1,
(不是1)== 0。*
我正在考虑定义6个函数,每个规则一个,然后 在包装器中逐个调用它们,我是lisp的新手,所以我没有实现它的线索,我在java中做了一次,但我不知道如何使用lisp的语法处理这样的问题所以请帮助我......
答案 0 :(得分:2)
鉴于您的解决方案的复杂性,我的实施方式更短,更具可读性:
(defun reduc (exp)
(if (atom exp)
exp
(flet ((helper (op args n) ; and and or is nearly the same code so we factor it out
(let ((newargs (remove n args)) (cn (- 1 n)))
(cond
((null newargs) n)
((some (lambda (e) (eql cn e)) newargs) cn)
((null (cdr newargs)) (car newargs))
((cons op newargs))))))
(let ((op (car exp)) (args (mapcar #'reduc (cdr exp))))
(ecase op
((not) (if (= 1 (length args))
(let ((arg1 (car args)))
(if (and (numberp arg1) (<= 0 arg1 1)) (- 1 arg1) exp))
(error "'not' must have exactly one parameter")))
((and) (helper op args 1))
((or) (helper op args 0)))))))
测试:
? (reduc '(OR 0 (AND A1 A2)))
(AND A1 A2)
? (reduc '(OR 0 (AND A1 1 A2)))
(AND A1 A2)
? (reduc '(or ERROR (not 0)))
1
? (reduc '(AND ERROR (not 0)))
ERROR
? (reduc '(OR 0 (AND A1 0)))
0
? (reduc '(OR 0 (AND A1 1)))
A1
答案 1 :(得分:0)
我终于想出了这个解决方案。
(defun simplify (EXPR)
(simplify-expr NIL EXPR))
(defun simplify-expr (EXPR1 EXPR2)
(cond
((or (atom EXPR2) (equal EXPR1 EXPR2)) EXPR2)
(T (simplify-expr EXPR2 (simplify-boolean-expr EXPR2)))))
(defun simplify-boolean-expr (EXPR)
(cond
((and (equal (first EXPR) `and) (>= (length EXPR) 3))
(simplify-and-expr (rest EXPR)))
((and (equal (first EXPR) `or) (>= (length EXPR) 3))
(simplify-or-expr (rest EXPR)))
((and (equal (first EXPR) `not) (= (length EXPR) 2))
(simplify-not-expr (rest EXPR)))
(T
(error "~S is not a valid circuit descriptor expression or has an unknown operator." EXPR))))
(defun simplify-and-expr (EXPR)
(let ((SIMPLIFIED_EXPR (remove `T (remove-duplicates EXPR))))
(cond
((null SIMPLIFIED_EXPR) `T)
((member `NIL SIMPLIFIED_EXPR) `NIL)
((null (second SIMPLIFIED_EXPR)) (first SIMPLIFIED_EXPR))
(T (cons `and (simplify-operand SIMPLIFIED_EXPR))))))
(defun simplify-or-expr (EXPR)
(let ((SIMPLIFIED_EXPR (remove `NIL (remove-duplicates EXPR))))
(cond
((null SIMPLIFIED_EXPR) `NIL)
((member `T SIMPLIFIED_EXPR) `T)
((null (second SIMPLIFIED_EXPR)) (first SIMPLIFIED_EXPR))
(T (cons `or (simplify-operand SIMPLIFIED_EXPR))))))
(defun simplify-not-expr (EXPR)
(cond
((equal (first EXPR) `NIL) `T)
((equal (first EXPR) `T) `NIL)
((and (listp (first EXPR)) (equal (first (first EXPR)) `not))
(first (rest (first EXPR))))
(T (cons `not (simplify-operand EXPR)))))
(defun simplify-operand (OPERAND_LIST)
(cond
((null OPERAND_LIST) NIL)
((atom (first OPERAND_LIST))
(cons (first OPERAND_LIST) (simplify-operand (rest OPERAND_LIST))))
(T
(cons (simplify-expr NIL (first OPERAND_LIST)) (simplify-operand (rest OPERAND_LIST))))))
它为(0,1)获取(nil,T)并减少任何布尔表达式,我尝试了它并且它工作正常。