Math 利用Z3和SMT-LIB寻找最高回文乘积
基于,我尝试在SMT-LIB中编码,并使用Z3解决它 问题是找到两个三位数的最大回文整数乘积。解决方案是Math 利用Z3和SMT-LIB寻找最高回文乘积,math,z3,smt,Math,Z3,Smt,基于,我尝试在SMT-LIB中编码,并使用Z3解决它 问题是找到两个三位数的最大回文整数乘积。解决方案是993*913=906609 在下面的代码中,我只能将两个三位数的数字编码为回文。这将生成正确但不是最大的604406值 如何更改代码以找到906609的最大值? 我尝试过使用(maximize(*pq)),但它报告了一个错误,不支持说目标函数“(*pq)”。我可以调整a的范围,但我正在寻找一种让Z3为我做这件事的方法 到目前为止,我得到的是: (declare-const a Int) (d
993*913=906609
在下面的代码中,我只能将两个三位数的数字编码为回文。这将生成正确但不是最大的604406
值
如何更改代码以找到906609
的最大值?
我尝试过使用(maximize(*pq))
,但它报告了一个错误,不支持说目标函数“(*pq)”。我可以调整a
的范围,但我正在寻找一种让Z3为我做这件事的方法
到目前为止,我得到的是:
(declare-const a Int)
(declare-const b Int)
(declare-const c Int)
(declare-const p Int)
(declare-const q Int)
(declare-const pq Int)
(define-fun satisfy ((pq Int)) Bool
(and
(<= 1 a 9)
(<= 0 b 9)
(<= 0 c 9)
(<= 100 p 999)
(<= p q 999)
(= pq
(* p q)
(+ (* 100001 a)
(* 10010 b)
(* 1100 c)))))
(assert (satisfy pq))
; Does not work:
;(maximize (* p q))
(check-sat)
(get-model)
一个可能的解决办法是
(declare-const a Int)
(declare-const b Int)
(declare-const c Int)
(declare-const p Int)
(declare-const q Int)
(declare-const pq Int)
(define-fun satisfy ((pq Int)) Bool
(and
(<= 1 a 9)
(<= 0 b 9)
(<= 0 c 9)
(<= 100 p 999)
(<= p q 999)
(= pq
(* p q)
(+ (* 100001 a)
(* 10010 b)
(* 1100 c)))))
(assert (satisfy pq))
(assert (> pq 888888))
(check-sat)
(get-model)
请联机运行此代码。其他可能的解决方案:我们搜索一个编号efggfe
,它是编号9ab
和9cd
的乘积。使用代码
(declare-const a Int)
(declare-const b Int)
(declare-const c Int)
(declare-const d Int)
(declare-const e Int)
(declare-const f Int)
(declare-const g Int)
(assert (and (>= a 0) (<= a 9)))
(assert (and (>= b 0) (<= b 9)))
(assert (and (>= c 0) (<= c 9)))
(assert (and (>= d 0) (<= d 9)))
(assert (and (>= e 0) (<= e 9)))
(assert (and (>= f 0) (<= f 9)))
(assert (and (>= g 0) (<= g 9)))
(assert (= (* (+ 900 (* 10 a) b) (+ 900 (* 10 c) d))
(+ (* 100000 e) (* 10000 f) (* 1000 g) (* 100 g) (* 10 f) e)))
(check-sat)
(get-model)
它对应于编号906609
请联机运行此代码
为了验证906609
是否为最大值,我们运行以下代码
(declare-const a Int)
(declare-const b Int)
(declare-const c Int)
(declare-const d Int)
(declare-const e Int)
(declare-const f Int)
(declare-const g Int)
(assert (and (>= a 0) (<= a 9)))
(assert (and (>= b 0) (<= b 9)))
(assert (and (>= c 0) (<= c 9)))
(assert (and (>= d 0) (<= d 9)))
(assert (and (>= e 0) (<= e 9)))
(assert (and (>= f 0) (<= f 9)))
(assert (and (>= g 0) (<= g 9)))
(assert (= (* (+ 900 (* 10 a) b) (+ 900 (* 10 c) d))
(+ (* 100000 e) (* 10000 f) (* 1000 g) (* 100 g) (* 10 f) e) ) )
(assert (> (+ (* 100000 e) (* 10000 f) (* 1000 g) (* 100 g) (* 10 f) e) 906609))
(check-sat)
请运行最后一个代码,因为这不是我想要的。我可以调整任何变量(p、q、pq、a等),以得到我知道的答案。我要寻找的是一种编码约束的方法,即produkt pq应该是最高的,考虑到其他约束。假设我事先不知道答案,希望自动找到答案。使用maximize
需要构建Z3的优化分支。寻找最大值的另一个可能的解决方案是断言没有更小的解决方案。我从未在Euler 4上使用过这种方法(见我博客上的评论),但我在Euler 4上使用了类似的解决方案。是的,我设法找到了opt分支,而且我对理论C了解太少,无法找到最大化不起作用的线索(它似乎在抱怨乘法不受支持;我想这与我们在复杂性树中的地位很高这一事实有关)。你不是在隐藏一个要求,即数字应该由两个大于900的因子组成吗?(*(+900(*10a)b)(+900(*10c)d))
?是的,这是可行的,但我们仍然在给Z3强有力的提示,我们只能事后才知道。我希望我们能让Z3说“这是最大值”无需重播。是否可以使用push
和pop
进行类似的操作?非常感谢您给出的好答案,但这并不正是我想要的。
(declare-const a Int)
(declare-const b Int)
(declare-const c Int)
(declare-const d Int)
(declare-const e Int)
(declare-const f Int)
(declare-const g Int)
(assert (and (>= a 0) (<= a 9)))
(assert (and (>= b 0) (<= b 9)))
(assert (and (>= c 0) (<= c 9)))
(assert (and (>= d 0) (<= d 9)))
(assert (and (>= e 0) (<= e 9)))
(assert (and (>= f 0) (<= f 9)))
(assert (and (>= g 0) (<= g 9)))
(assert (= (* (+ 900 (* 10 a) b) (+ 900 (* 10 c) d))
(+ (* 100000 e) (* 10000 f) (* 1000 g) (* 100 g) (* 10 f) e)))
(check-sat)
(get-model)
sat
(model
(define-fun g () Int 6)
(define-fun f () Int 0)
(define-fun e () Int 9)
(define-fun d () Int 3)
(define-fun c () Int 1)
(define-fun b () Int 3)
(define-fun a () Int 9) )
(declare-const a Int)
(declare-const b Int)
(declare-const c Int)
(declare-const d Int)
(declare-const e Int)
(declare-const f Int)
(declare-const g Int)
(assert (and (>= a 0) (<= a 9)))
(assert (and (>= b 0) (<= b 9)))
(assert (and (>= c 0) (<= c 9)))
(assert (and (>= d 0) (<= d 9)))
(assert (and (>= e 0) (<= e 9)))
(assert (and (>= f 0) (<= f 9)))
(assert (and (>= g 0) (<= g 9)))
(assert (= (* (+ 900 (* 10 a) b) (+ 900 (* 10 c) d))
(+ (* 100000 e) (* 10000 f) (* 1000 g) (* 100 g) (* 10 f) e) ) )
(assert (> (+ (* 100000 e) (* 10000 f) (* 1000 g) (* 100 g) (* 10 f) e) 906609))
(check-sat)
unsat