(Z3Py) checking all solutions for equation

Solution 1:

You can do that by adding a new constraint that blocks the model returned by Z3. For example, suppose that in the model returned by Z3 we have that x = 0 and y = 1. Then, we can block this model by adding the constraint Or(x != 0, y != 1). The following script does the trick. You can try it online at: http://rise4fun.com/Z3Py/4blB

Note that the following script has a couple of limitations. The input formula cannot include uninterpreted functions, arrays or uninterpreted sorts.

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

Solution 2:

The python function below is a generator of models for formulas that contain both constants and functions.

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)

Solution 3:

Referencing 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))  

This indeed performs quite better from Leonardo's own answer (considering his answer is quite old)

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

Splitting the search space into disjoint models creates a huge difference as far as I have observed