（Z3Py）检查方程的所有解

问题描述：

在 Z3Py 中，如何检查给定约束的方程是否只有一个解？

如果有多个解决方案，我该如何列举它们？

解决方案 1：

您可以通过添加新的约束来阻止 Z3 返回的模型来实现这一点。例如，假设在 Z3 返回的模型中，我们有 和x = 0。y = 1然后，我们可以通过添加约束 来阻止此模型Or(x != 0, y != 1)。以下脚本可以解决问题。您可以在线尝试： http: //rise4fun.com/Z3Py/4blB

请注意，以下脚本有几个限制。输入公式不能包含未解释的函数、数组或未解释的排序。

from z3 import *

# Return the first "M" models of formula list of formulas F 
def get_models(F, M):
    result = []
    s = Solver()
    s.add(F)
    while len(result) < M and s.check() == sat:
        m = s.model()
        result.append(m)
        # Create a new constraint the blocks the current model
        block = []
        for d in m:
            # d is a declaration
            if d.arity() > 0:
                raise Z3Exception("uninterpreted functions are not supported")
            # create a constant from declaration
            c = d()
            if is_array(c) or c.sort().kind() == Z3_UNINTERPRETED_SORT:
                raise Z3Exception("arrays and uninterpreted sorts are not supported")
            block.append(c != m[d])
        s.add(Or(block))
    return result

# Return True if F has exactly one model.
def exactly_one_model(F):
    return len(get_models(F, 2)) == 1

x, y = Ints('x y')
s = Solver()
F = [x >= 0, x <= 1, y >= 0, y <= 2, y == 2*x]
print get_models(F, 10)
print exactly_one_model(F)
print exactly_one_model([x >= 0, x <= 1, y >= 0, y <= 2, 2*y == x])

# Demonstrate unsupported features
try:
    a = Array('a', IntSort(), IntSort())
    b = Array('b', IntSort(), IntSort())
    print get_models(a==b, 10)
except Z3Exception as ex:
    print "Error: ", ex

try:
    f = Function('f', IntSort(), IntSort())
    print get_models(f(x) == x, 10)
except Z3Exception as ex:
    print "Error: ", ex

解决方案 2：

Himanshu Sheoran 给出的答案引用了论文https://theory.stanford.edu/%7Enikolaj/programmingz3.html#sec-blocking-evaluations

不幸的是，当时论文中给出的实现中有一个错误，该错误在该答案中被引用。该函数现已得到纠正。

为了方便后人参考，以下是该代码的正确版本：

def all_smt(s, initial_terms):
    def block_term(s, m, t):
        s.add(t != m.eval(t, model_completion=True))
    def fix_term(s, m, t):
        s.add(t == m.eval(t, model_completion=True))
    def all_smt_rec(terms):
        if sat == s.check():
           m = s.model()
           yield m
           for i in range(len(terms)):
               s.push()
               block_term(s, m, terms[i])
               for j in range(i):
                   fix_term(s, m, terms[j])
               yield from all_smt_rec(terms[i:])
               s.pop()   
    yield from all_smt_rec(list(initial_terms))

解决方案 3：

下面的python函数是包含常量和函数的公式模型的生成器。

import itertools
from z3 import *

def models(formula, max=10):
    " a generator of up to max models "
    solver = Solver()
    solver.add(formula)

    count = 0
    while count<max or max==0:
        count += 1

        if solver.check() == sat:
            model = solver.model()
            yield model
            
            # exclude this model
            block = []
            for z3_decl in model: # FuncDeclRef
                arg_domains = []
                for i in range(z3_decl.arity()):
                    domain, arg_domain = z3_decl.domain(i), []
                    for j in range(domain.num_constructors()):
                        arg_domain.append( domain.constructor(j) () )
                    arg_domains.append(arg_domain)
                for args in itertools.product(*arg_domains):
                    block.append(z3_decl(*args) != model.eval(z3_decl(*args)))
            solver.add(Or(block))

x, y = Ints('x y')
F = [x >= 0, x <= 1, y >= 0, y <= 2, y == 2*x]
for m in models(F):
    print(m)

解决方案 4：

参考http://theory.stanford.edu/~nikolaj/programmingz3.html#sec-blocking-evaluations

def all_smt(s, initial_terms):
    def block_term(s, m, t):
        s.add(t != m.eval(t))
    def fix_term(s, m, t):
        s.add(t == m.eval(t))
    def all_smt_rec(terms):
        if sat == s.check():
           m = s.model()
           yield m
           for i in range(len(terms)):
               s.push()
               block_term(s, m, terms[i])
               for j in range(i):
                   fix_term(s, m, terms[j])
               yield from all_smt_rec(terms[i:])
               s.pop()   
    yield from all_smt_rec(list(initial_terms))  

这确实比莱昂纳多自己的回答表现得更好（考虑到他的回答已经很老了）

start_time = time.time()
v = [BitVec(f'v{i}',3) for i in range(6)]
models = get_models([Sum(v)==0],8**5)
print(time.time()-start_time)
#211.6482105255127s

start_time = time.time()
s = Solver()
v = [BitVec(f'v{i}',3) for i in range(6)]
s.add(Sum(v)==0)
models = list(all_smt(s,v))
print(time.time()-start_time)
#13.375828742980957s

据我观察，将搜索空间划分为不相交的模型会产生巨大的差异