在Z3中是否有一种方法可以证明/显示给定模型是唯一的并且没有其他解决方案存在?
一个演示
的小示例(declare-const a1 Int)
(declare-const a2 Int)
(declare-const a3 Int)
(declare-const b1 Int)
(declare-const b2 Int)
(declare-const b3 Int)
(declare-const c1 Int)
(declare-const c2 Int)
(declare-const c3 Int)
(declare-const ra Int)
(declare-const rb Int)
(declare-const rc Int)
(declare-const r1 Int)
(declare-const r2 Int)
(declare-const r3 Int)
(assert (>= a1 0))
(assert (>= a2 0))
(assert (>= a3 0))
(assert (>= b1 0))
(assert (>= b2 0))
(assert (>= b3 0))
(assert (>= c1 0))
(assert (>= c2 0))
(assert (>= c3 0))
(assert (<= a1 9))
(assert (<= a2 9))
(assert (<= a3 9))
(assert (<= b1 9))
(assert (<= b2 9))
(assert (<= b3 9))
(assert (<= c1 9))
(assert (<= c2 9))
(assert (<= c3 9))
(assert (= ra 38))
(assert (= rb 1))
(assert (= rc 27))
(assert (= r1 55))
(assert (= r2 72))
(assert (= r3 6))
(assert (= ra (- (* a1 a2) a3)))
(assert (= rb (- (- b1 b2) b3)))
(assert (= rc (* (* c1 c2) c3)))
(assert (= r1 (- (* a1 b1) c1)))
(assert (= r2 (* (+ a2 b2) c2)))
(assert (= r3 (- (+ a3 b3) c3)))
(check-sat)
(get-model)
我知道下面的模型是唯一的,但是我可以使用一些Z3选项或添加断言来确保这一点吗?
(model
(define-fun c3 () Int
3)
(define-fun c2 () Int
9)
(define-fun c1 () Int
1)
(define-fun b3 () Int
5)
(define-fun b2 () Int
2)
(define-fun b1 () Int
8)
(define-fun a3 () Int
4)
(define-fun a2 () Int
6)
(define-fun a1 () Int
7)
(define-fun r3 () Int
6)
(define-fun r2 () Int
72)
(define-fun r1 () Int
55)
(define-fun rc () Int
27)
(define-fun rb () Int
1)
(define-fun ra () Int
38)
)
为了澄清,我使用的是Z3通过de JAVA API
Yes:这个想法是断言由找到的模型分配的值是唯一可能的分配,因此它是唯一的。这可以通过添加一个断言来实现,该断言声明所有常量不等于它们分配的模型值。
对于您的示例,这将是:
(assert (not (and
(= c3 3)
(= c2 9)
(= c1 1)
(= b3 5)
(= b2 2)
(= b1 8)
(= a3 4)
(= a2 6)
(= a1 7)
(= r3 6)
(= r2 72)
(= r1 55)
(= rc 27)
(= rb 1)
(= ra 38))))
(check-sat) ; unsat => no other model exists
如果您尝试更改任何值(例如,将c3 = 3更改为c3 = 4),这将再次变得令人满意。这里有一个rise@fun链接到完整的示例:http://rise4fun.com/Z3/nD5n
有关如何使用api以编程方式完成此操作的更多详细信息,请参阅以下问题和回答:
Z3:寻找所有满足的模型
(Z3Py)检验方程
的所有解请注意,根据第二个链接答案中的注释,您不能仅使用SMT-lib前端以编程方式完成此操作,如果您想使此过程自动化,则必须使用API来遍历找到的模型。