Question

在Z3Py中，我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，我该如何枚举它们？

Answer 1

您可以通过添加阻止Z3返回的模型的新约束来实现。例如，假设在Z3返回的模型中，我们有x = 0和y = 1。然后，我们可以通过添加约束Or(x != 0, y != 1)来阻止此模型。以下脚本可以解决问题。您可以在以下网址进行尝试：http://rise4fun.com/Z3Py/4blB

请注意，以下脚本有一些限制。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

Answer 2

下面的python函数是同时包含常量和函数的公式模型的生成器。

search

Answer 3

引用 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])
               for m in all_smt_rec(terms[i:]):
                   yield m
               s.pop()   
    for m in all_smt_rec(list(initial_terms)):
        yield m

这确实从 Leonardo 自己的答案中表现得更好（考虑到他的答案很旧）

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

就我所观察到的，将搜索空间拆分为不相交的模型会产生巨大的差异

（Z3Py）检查方程的所有解

3 个答案: