问题
I want to verify a formula of the form:
Exists p . ForAll x != 0 . f(x, p) > 0
An implementation (that isn't working) is the following:
def f0(x0, x1, x, y):
return x1 ** 2 * y + x0 ** 2 * x
s = Solver()
x0, x1 = Reals('x0 x1')
p0, p1 = Reals('p0 p1')
s.add(Exists([p0, p1],
ForAll([x0, x1],
f0(x0, x1, p0, p1) > 0
)
))
#s.add(Or(x0 != 0, x1 != 0))
while s.check() == sat:
m = s.model()
m.evaluate(x0, model_completion=True)
m.evaluate(x1, model_completion=True)
m.evaluate(p0, model_completion=True)
m.evaluate(p1, model_completion=True)
print m
s.add(Or(x0 != m[x0], x1 != m[x1]))
The formula isn't satisfied.
With f0() >= 0
, the only output is (0, 0)
.
I want to have f0() > 0
and constrain (x0, x1) != (0, 0)
.
Something I'd expect is: p0, p1 = 1, 1
or 2, 2
for instance, but I don't know how to remove 0, 0
from the possible values for x0, x1
.
回答1:
Following up on Levent's reply. During the first check, Z3 uses a custom decision procedure that works with the quantifiers. In incremental mode it falls back to something that isn't a decision procedure. To force the one-shot solver try the following:
from z3 import *
def f0(x0, x1, x, y):
return x1 * x1 * y + x0 * x0 * x
p0, p1 = Reals('p0 p1')
x0, x1 = Reals('x0 x1')
fmls = [ForAll([x0, x1], Implies(Or(x0 != 0, x1 != 0), f0(x0, x1, p0, p1) > 0))]
while True:
s = Solver()
s.add(fmls)
res = s.check()
print res
if res == sat:
m = s.model()
print m
fmls += [Or(p0 != m[p0], p1 != m[p1])]
else:
print "giving up"
break
回答2:
You'd simply write that as an implication inside the quantification. I think you're also mixing up some of the variables in there. The following seems to capture your intent:
from z3 import *
def f0(x0, x1, x, y):
return x1 * x1 * y + x0 * x0 * x
s = Solver()
p0, p1 = Reals('p0 p1')
x0, x1 = Reals('x0 x1')
s.add(ForAll([x0, x1], Implies(Or(x0 != 0, x1 != 0), f0(x0, x1, p0, p1) > 0)))
while True:
res = s.check()
print res
if res == sat:
m = s.model()
print m
s.add(Or(p0 != m[p0], p1 != m[p1]))
else:
print "giving up"
break
Of course, z3 isn't guaranteed to find you any solutions; though it seems to manage one:
$ python a.py
sat
[p1 = 1, p0 = 1]
unknown
giving up
Once you use quantifiers all bets are off, as the logic becomes semi-decidable. Z3 is doing a good job here and returning one solution, and then it's giving up. I don't think you can expect anything better, unless you use some custom decision procedures.
来源:https://stackoverflow.com/questions/50279196/nonzero-vector-in-quantifier