I have this code in Z3 python:
x = Bool('x')
y = Bool('y')
z = Bool('z')
z == (x xor y)
s = Solver()
s.add(z == True)
print s.check()
But this code reports below error when running:
c.py(4): error: invalid syntax
If I replace xor with and, there is no problem. So this means XOR is not supported?
You should use Xor(a, b). Moreover, to create the Z3 expression that represents the formula a and b, we must use And(a, b). In Python, we can't overload the operators and and or.
Here is an example with the Xor (available online at rise4fun).
x = Bool('x')
y = Bool('y')
z = Xor(x, y)
s = Solver()
s.add(z)
print s.check()
print s.model()
Related
I have the following formula and Python code trying to find the largest n satisfying some property P:
x, u, n, n2 = Ints('x u n n2')
def P(u):
return Implies(And(2 <= x, x <= u), And(x >= 1, x <= 10))
nIsLargest = ForAll(n2, Implies(P(n2), n2 <= n))
exp = ForAll(x, And(P(n), nIsLargest))
s = SolverFor("LIA")
s.reset()
s.add(exp)
print(s.check())
if s.check() == sat:
print(s.model())
My expectation was that it would return n=10, yet Z3 returns unsat. What am I missing?
You're using the optimization API incorrectly; and your question is a bit confusing since your predicate P has a free variable x: Obviously, the value that maximizes it will depend on both x and u.
Here's a simpler example that can get you started, showing how to use the API correctly:
from z3 import *
def P(x):
return And(x >= 1, x <= 10)
n = Int('n')
opt = Optimize()
opt.add(P(n))
maxN = opt.maximize(n)
r = opt.check()
print(r)
if r == sat:
print("maxN =", maxN.value())
This prints:
sat
maxN = 10
Hopefully you can take this example and extend it your use case.
I want to check the value of a, b, c, and if value 'a' equals to 1, 'x' is added one. We continue the process for values 'b' and 'c'.
So if a=1, b=1, c=1, the result of x should be 3.
if a=1, b=1, c=0, so the result of x should be 2.
Any methods to be implemented in z3?
The source code looks like this:
from z3 import *
a, b, c = Ints('a b c')
x, y = Ints('x y')
s = Solver()
s.add(If(a==1, x=x + 1, y = y-1))
s.add(If(b==1, x=x + 1, y = y-1))
s.add(If(c==1, x=x + 1, y = y-1))
s.check()
print s.model()
Any suggestions about what I can do?
This sort of "iterative" processing is usually modeled by unrolling the assignments and creating what's known as SSA form. (Static single assignment.) In this format, every variable is assigned precisely once, but can be used many times. This is usually done by some underlying tool as it is rather tedious, but you can do it by hand as well. Applied to your problem, it'd look something like:
from z3 import *
s = Solver()
a, b, c = Ints('a b c')
x0, x1, x2, x3 = Ints('x0 x1 x2 x3')
s.add(x0 == 0)
s.add(x1 == If(a == 1, x0+1, x0))
s.add(x2 == If(b == 1, x1+1, x1))
s.add(x3 == If(c == 1, x2+1, x2))
# Following asserts are not part of your problem, but
# they make the output interesting
s.add(b == 1)
s.add(c == 0)
# Find the model
if s.check() == sat:
m = s.model()
print("a=%d, b=%d, c=%d, x=%d" % (m[a].as_long(), m[b].as_long(), m[c].as_long(), m[x3].as_long()))
else:
print "no solution"
SSA transformation is applied to the variable x, creating as many instances as necessary to model the assignments. When run, this program produces:
a=0, b=1, c=0, x=1
Hope that helps!
Note that z3 has many functions. One you could use here is Sum() for the sum of a list. Inside the list you can put simple variables, but also expression. Here an example for both a simple and a more complex sum:
from z3 import *
a, b, c = Ints('a b c')
x, y = Ints('x y')
s = Solver()
s.add(a==1, b==0, c==1)
s.add(x==Sum([a,b,c]))
s.add(y==Sum([If(a==1,-1,0),If(b==1,-1,0),If(c==1,-1,0)]))
if s.check() == sat:
print ("solution:", s.model())
else:
print ("no solution possible")
Result:
solution: [y = 2, x = 2, c = 1, b = 0, a = 1]
If your problem is more complex, using BitVecs instead of Ints can make it run a little faster.
edit: Instead of Sum() you could also simply use addition as in
s.add(x==a+b+c)
s.add(y==If(a==1,-1,0)+If(b==1,-1,0)+If(c==1,-1,0))
Sum() makes sense towards readability when you have a longer list of variables, or when the variables already are in a list.
I'm trying to understand traversing quantified formula in z3 (i'm using z3py). Have no idea how to pickup the quantified variables. For example in code shown below i'm trying to print the same formula and getting error.
from z3 import *
def traverse(e):
if is_quantifier(e):
var_list = []
if e.is_forall():
for i in range(e.num_vars()):
var_list.append(e.var_name(i))
return ForAll (var_list, traverse(e.body()))
x, y = Bools('x y')
fml = ForAll(x, ForAll (y, And(x,y)))
same_formula = traverse( fml )
print same_formula
With little search i got to know that z3 uses De Bruijn index and i have to get something like Var(1, BoolSort()). I can think of using var_sort() but how to get the formula to return the variable correctly. Stuck here for some time.
var_list is a list of strings, but ForAll expects a list of constants. Also, traverse should return e when it's not a quantifier. Here's a modified example:
from z3 import *
def traverse(e):
if is_quantifier(e):
var_list = []
if e.is_forall():
for i in range(e.num_vars()):
c = Const(e.var_name(i) + "-traversed", e.var_sort(i))
var_list.append(c)
return ForAll (var_list, traverse(e.body()))
else:
return e
x, y = Bools('x y')
fml = ForAll(x, ForAll (y, And(x,y)))
same_formula = traverse( fml )
print(same_formula)
I am still struggling with the problem of findiong a value so that a * b == b with all value of b. The expected result is a == 1. I have two solutions below.
(A) I implemented this with ForAll quantifier in below code (correct me if there is a solution without using any quantifier). The idea is to prove f and g are equivalent.
from z3 import *
a, b, a1, tmp1 = BitVecs('a b a1 tmp1', 32)
f = True
f = And(f, tmp1 == b)
f = And(f, a1 == a * tmp1)
g= True
g = And(g, a1 == b)
s = Solver()
s.add(ForAll([b, tmp1, a1], f == g))
if s.check() == sat:
print 'a =', s.model()[a]
else:
print 'Unsat'
However, this simple code runs forever without giving back result. I think that is because of the ForAll. Any idea on how to fix the problem?
(B) I tried again with another version. This time I dont prove two formulas to be equivalent, but put them all into one formula f. Logically, I think this is true, but please correct me if I am wrong here:
from z3 import *
a, b, a1, tmp = BitVecs('a b a1 tmp', 32)
f = True
f = And(f, tmp == b)
f = And(f, a1 == a * tmp)
f = And(f, a1 == b)
s = Solver()
s.add(ForAll([b, a1], f))
if s.check() == sat:
print 'a =', s.model()[a]
else:
print 'Unsat'
This time the code does not hang, but immediately returns 'Unsat'. Any idea on how to fix this?
Thanks a lot.
The direct formulation of your problem provides the answer you were expecting: http://rise4fun.com/Z3Py/N07sW
The versions you propose use auxiliary variables, a1, tmp, tmp1 and you
use universal quantification over these variables. This does not correspond
to the formula you intended.
Solver.model() sometimes returns an assignment with a seemingly-needless Var(), whereas I was (perhaps naively) expecting Solver.model() to always return a concrete value for each variable. For example:
#!/usr/bin/python
import z3
x, y = z3.Ints('x y')
a = z3.Array('a', z3.IntSort(), z3.IntSort())
e = z3.Not(z3.Exists([x], z3.And(x != y, a[x] == a[y])))
solver = z3.Solver()
solver.add(e)
print solver.check()
print solver.model()
produces
sat
[k!1 = 0,
a = [else -> k!5!7(k!6(Var(0)))],
y = 1,
k!5 = [else -> k!5!7(k!6(Var(0)))],
k!5!7 = [1 -> 3, else -> 2],
k!6 = [1 -> 1, else -> 0]]
What's going on here? Is Var(0) in a's "else" referring to the 0th argument to the a array, meaning a[i] = k!5!7[k!6[i]]? Is it possible to get a concrete satisfying assignment for a out of Z3, such as a = [1 -> 1, else -> 0]?
This is the intended output. The interpretation for functions and arrays should be viewed as function definitions. Keep in mind that the assertion
z3.Not(z3.Exists([x], z3.And(x != y, a[x] == a[y])))
is essentially a universal quantifier. For quantifier free problems, Z3 does generate the "concrete assignments" suggested in your post. However, this kind of representation is not expressive enough. In the end of the message, I attached an example that cannot be encoded using "concrete assignments".
The following post has additional information about how models are encoded in Z3.
understanding the z3 model
You can find more details regarding the encoding used by Z3 at http://rise4fun.com/Z3/tutorial/guide
Here is an example that produces a model that can't be encoded using "concrete" assignments (available online at http://rise4fun.com/Z3Py/eggh):
a = Array('a', IntSort(), IntSort())
i, j = Ints('i j')
solver = Solver()
x, y = Ints('x y')
solver.add(ForAll([x, y], Implies(x <= y, a[x] <= a[y])))
solver.add(a[i] != a[j])
print solver.check()
print solver.model()