Z3(Py) does not “intersect” Skolem functions as I expected.
I will try to explain my doubt using an imaginary problem which is the one that I have been testing on.
Consider a system that has two processes A and B, and an environment chooses between A or B. When process A is “activated”, then the system must solve P1: Exists y. Forall x. (x>=2) --> (y>1) /\ (y<=x). When the process B is “activated” then the system must solve P2: Exists y. Forall x. (x<2) --> (y>1) /\ (y>x). We can see, in Z3-PY, models of both the P1 and the P2 formulae:
#P1
x,y = Ints('x y')
ct_0 = (x >= 2)
ct_1 = (y > 1)
ct_2 = (y <= x)
phi = ForAll([x], Implies(ct_0, And(ct_1,ct_2)))
s = Solver()
s.add(phi)
print(s.check())
print(s.model())
#P2
x,y = Ints('x y')
ct_0 = (x < 2)
ct_1 = (y > 1)
ct_2 = (y > x)
phi = ForAll([x], Implies(ct_0, And(ct_1,ct_2)))
s = Solver()
s.add(phi)
print(s.check())
print(s.model())
Note that the model of P1 is unique: y=2; whereas a model of P2 is y=2, but also y=3, y=4, y=5, ... ad infinitum.
Then, I consider that the "intersection" of both P1 and P2 is the model y=2. In other words, whenever the environment chooses A or B, i.e., whenever the environment forces the system to solve P1 or P2, the system can just respond y=2 and will satisfy the formula, no matter what the environment chose.
I obtained this (unique) “intersection” y=2 by performing the (unique) model of the conjunction of both formulae:
#P1 and P2 intersection
#Note that, this time, I distinguish between ct_k and ctk because I executed them in the same script, no other reason involved.
x,y = Ints('x y')
ct_0 = (x >= 2)
ct_1 = (y > 1)
ct_2 = (y <= x)
phi0 = ForAll([x], Implies(ct_0, And(ct_1,ct_2)))
ct0 = (x < 2)
ct1 = (y > 1)
ct2 = (y > x)
phi1 = ForAll([x], Implies(ct0, And(ct1,ct2)))
phiT = And(phi0,phi1)
s = Solver()
s.add(phiT)
print(s.check())
print(s.model())
for i in range(0, 5):
if s.check() == sat:
m = s.model()[y]
print(m)
s.add(And(y != m))
This outputs [y = 2] as expected.
Now, consider I have a problem: P1 and P2 are no longer Exists-Forall formulae, but Forall-Exists formulae: this means that we will not have models of y, but Skolem functions (as solved in this question What does a model mean in a universally quantified formula? Is it a function?). I implement that as follows:
#P1 Skolem
x = Int('x')
skolem = Function('skolem', IntSort(), IntSort())
ct_0 = (x >= 2)
ct_1 = (skolem(x) > 1)
ct_2 = (skolem(x) <= x)
phi = ForAll([x], Implies(ct_0, And(ct_1,ct_2)))
s = Solver()
s.add(phi)
print(s.check())
print(s.model())
#P2 Skolem
x = Real('x')
skolem = Function('skolem', RealSort(), RealSort())
ct_0 = (x < 2)
ct_1 = (skolem(x) > 1)
ct_2 = (skolem(x) > x)
phi = ForAll([x], Implies(ct_0, And(ct_1,ct_2)))
s = Solver()
s.add(phi)
As we can see, both of them return [skolem = [else -> 2]] as the first option, and then they offer other (different) options, such as [x = 0, skolem = [else -> If(1 <= Var(0), 5, 2)]] for P2.
My question is: how can I perform the “intersection” of these two formulae? In general, how can I perform intersection of n skolem functions (in Z3)? I tried as follows (using the same idea as before when seeking y=2:
#P1 and P2 Skolem intersection
#Once again, I distinguish ct_k and ctk for execution reasons.
x,y = Ints('x y')
ct_0 = (x >= 2)
ct_1 = (skolem(x) > 1)
ct_2 = (skolem(x) <= x)
phi0 = ForAll([x], Implies(ct_0, And(ct_1,ct_2)))
ct0 = (x < 2)
ct1 = (skolem(x) > 1)
ct2 = (skolem(x) > x)
phi1 = ForAll([x], Implies(ct0, And(ct1,ct2)))
phiT = And(phi0,phi1)
s = Solver()
s.add(phiT)
print(s.check())
print(s.model())
for i in range(0, 5):
if s.check() == sat:
m = s.model()
print(m)
s.add(skolem(x) != i)
And results were as follows:
[skolem = [else -> 3/2]]
[skolem = [else -> 3/2]]
[x = 0, skolem = [else -> 2]]
[x = 0, skolem = [else -> 2]]
[x = 0, skolem = [else -> 3/2]]
[x = 0, skolem = [else -> 3/2]]
What does this mean? Why does it not output just [skolem = [else -> 2]]? In mean, in the same way that y=2 was the model with the other quantifier alternation.
Also, how does it offer 3/2 if we are in the domain of integers?
PS: In the Forall-Exists version of P1, if we print several Skolem functions, then the result is as follows:
[skolem = [else -> 2]]
[skolem = [else -> 2]]
[x = 0, skolem = [else -> If(2 <= Var(0), 2, 1)]]
[x = 0, skolem = [else -> If(2 <= Var(0), 2, -1)]]
[x = 0, skolem = [else -> If(2 <= Var(0), 2, -1)]]
[x = 0, skolem = [else -> If(2 <= Var(0), 2, -1)]]
How is this so?
In one of your programs, you have the declaration:
skolem = Function('skolem', RealSort(), RealSort())
I think you confused yourself by somehow leaving this in? It's hard to tell when you don't post code segments that are fully loadable by themselves. In any case, the following:
from z3 import *
skolem = Function('skolem', IntSort(), IntSort())
#P1 and P2 Skolem intersection
#Once again, I distinguish ct_k and ctk for execution reasons.
x,y = Ints('x y')
ct_0 = (x >= 2)
ct_1 = (skolem(x) > 1)
ct_2 = (skolem(x) <= x)
phi0 = ForAll([x], Implies(ct_0, And(ct_1,ct_2)))
ct0 = (x < 2)
ct1 = (skolem(x) > 1)
ct2 = (skolem(x) > x)
phi1 = ForAll([x], Implies(ct0, And(ct1,ct2)))
phiT = And(phi0,phi1)
s = Solver()
s.add(phiT)
for i in range(0, 5):
if s.check() == sat:
m = s.model()
print(m)
s.add(skolem(x) != i)
Prints:
[skolem = [else -> 2]]
[x = 0,
skolem = [else ->
If(And(1 <= Var(0), 2 <= Var(0)),
2,
If(1 <= Var(0), 4, 2))]]
[x = 0,
skolem = [else ->
If(And(1 <= Var(0), 2 <= Var(0)),
2,
If(1 <= Var(0), 4, 2))]]
[x = 0,
skolem = [else ->
If(And(1 <= Var(0), 2 <= Var(0)),
2,
If(1 <= Var(0), 5, 3))]]
[x = 0,
skolem = [else ->
If(And(1 <= Var(0), 2 <= Var(0)),
2,
If(1 <= Var(0), 6, 4))]]
which looks fine to me. Does this help resolve your issue?
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 find a maximal interval in which an expression e is true for all x. A way to write such a formula should be: Exists d : ForAll x in (-d,d) . e and ForAll x not in (-d,d) . !e.
To get such a d, the formula f in Z3 (looking at the one above) could be the following:
from z3 import *
x = Real('x')
delta = Real('d')
s = Solver()
e = And(1/10000*x**2 > 0, 1/5000*x**3 + -1/5000*x**2 < 0)
f = ForAll(x,
And(Implies(And(delta > 0,
-delta < x, x < delta,
x != 0),
e),
Implies(And(delta > 0,
Or(x > delta, x < -delta),
x != 0),
Not(e))
)
)
s.add(Not(f))
s.check()
print s.model()
It prints: [d = 2]. This is surely not true (take x = 1). What's wrong?
Also: by specifying delta = RealVal('1'), a counterexample is x = 0, even when x = 0 should be avoided.
Your constants are getting coerced to integers. Instead of writing:
1/5000
You should write:
1.0/5000.0
You can see the generated expression by:
print s.sexpr()
which would have alerted you to the issue.
NB. Being explicit about types is always important when writing constants. See this answer for a variation on this theme that can lead to further problems: https://stackoverflow.com/a/46860633/936310
Please consider a C program that, given x, will return y and z such that y + z * 2 = x, for the smallest possible y. Roughly, I could create a nested loop:
for(y = 0; y < x; ++ y){
for(z = 0; z < x; ++z){
if(y + 2 * z == x){
printf("%d + 2 * %d = %d", y, z, x);
}
}
}
How could I translate this kind of nested loop in the functional way? Is it feasible? Is it reasonable or am I just misjudging the approach? My best attempt so far:
let foo x =
let rec aux (y, z, q) =
match (y + z * 2) with
r when r = q -> (y, z)
|_ -> aux(y + 1, z + 1, q) //How to check different values of z
aux(0, 0, x) //for each value of y?
It will not work, since it will just increment both y and z. How can I check different values of z, for every value of y?
You have to add those checks in the match.
See here what your code is missing:
let foo x =
let rec aux (y, z, q) =
match (y + z * 2) with
| r when r = q -> (y, z)
| _ when y = q -> failwith "not found !"
| _ when z = q -> aux (y + 1, 0, q)
| _ -> aux (y, z + 1, q)
aux (0, 0, x)
And here's a different approach, equally functional but without recursion:
let foo2 x =
let s =
{0 .. x} |> Seq.collect (fun y ->
{0 .. x} |> Seq.collect (fun z ->
seq [y, z]))
Seq.find (fun (y, z) -> y + z * 2 = x) s
which in F# can be written using seq expressions:
let foo3 x =
let s = seq {
for y in {0 .. x} do
for z in {0 .. x} do
yield (y, z)}
Seq.find (fun (y, z) -> y + z * 2 = x) s
and it resembles your original C program.
I write the following Z3 python code
x, y = Ints('x y')
F = (x == y & 16) # x has the value of (y & 16)
print F
But I get bellow error:
TypeError: unsupported operand type(s) for &: 'instance' and 'int'
How to do bitwise arithmetic (& in this case) in Z3 equation?
Thanks.
x and y should be bit-vectors:
x, y = BitVecs('x y', 32)
F = (x == y & 16) # x has the value of (y & 16)
print F
See Bitvectors section under http://rise4fun.com/Z3/tutorial/guide