The SMT-LIB 2.0 array initialization and manipulation is a bit cumbersome. As the following code explains, http://rise4fun.com/Z3/kxmrd.
Is there any elegant way to initialise or manipulate arrays, using Python/C/C++/.Net API than SMT-LIB 2.0 ?
You could use regular Python constructs such as for i in range(n) to achieve what you want:
s = Solver()
a = Array('a', IntSort(), IntSort())
xs = [20, 23, 27, 12, 19, 31, 41, 7]
for i in range(len(xs)):
s.add(Select(a, i) == xs[i])
a1 = Array('a1', IntSort(), IntSort())
s.add(a1 == Store(a, 3,9 ))
print s.check()
m = s.model()
for d in m.decls():
print "%s = %s" % (d.name(), m[d])
Run it online here.
It would be nice if you could write something like s.add(a == xs) or s.add(a.startsWith(xs)), but I don't know if that is possible.
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 want to use a Int vector as an array index.
python.
array = [12,45,66,34]
s= Solver()
x = Int('x')
s.add(array[x] == 66)
so the x should be 2..
how can i do it?
Here's one way to do it:
from z3 import *
s = Solver ()
array = [12, 45, 66, 34]
A = Array ('A', IntSort(), IntSort())
i = 0
for elem in array:
A = Store(A, i, elem)
i = i + 1
x = Int ('x')
s.add(x >= 0)
s.add(x < len(array))
s.add(Select(A, x) == 66)
if s.check() == sat:
print s.model()
else:
print "Not found!"
This prints:
$ python a.py
[x = 2]
It is not possible to access integer array element with Z3 ArithRef variables. I suppose, the specified problem is looking for an element in an integer array (index) through Z3 solver.
Another possible way of doing it might be:
array = [12, 45, 66, 34]
x = Int('x')
s= Solver()
for i in range(len(array)):
s.add(Implies(array[i] == 66, x == i))
if s.check() == sat:
print(s.model())
i am looking for some code examples using theory of array in Z3 python, but cannot find any.
please could somebody provide some code examples?
thanks!
Here is an example showing array declarations and accessing items by indices http://rise4fun.com/Z3Py/7jAj:
x = Int('x')
a = Array('a', IntSort(), BoolSort())
b = Array('b', IntSort(), BoolSort())
c = Array('c', BoolSort(), BoolSort())
e = ForAll(x, Or(Not(a[x]), c[b[x]]))
print e
solver = Solver()
solver.add(e)
c = solver.check()
print c
Here is another example using Select and Store on array theory http://rise4fun.com/Z3Py/2CAn:
x = Int('x')
y = Int('y')
a = Array('a', IntSort(), IntSort())
s = Solver()
s.add(Select(a, x) == x, Store(a, x, y) == a)
print s.check()
print s.model()
That said, there are a few array examples floating around StackOverflow. You can try to search on the site using "z3py array" keyword for more information.
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()