Here is my code:
from z3 import *
a = BitVecVal(5,3)
b = BitVecVal(8,3)
print a
print b
s = Solver()
s.add(a<b)
print s.check()
print s.model()
The output is
5
0
sat
Could someone help me understand why 'a' is smaller than 'b'?
< is signed comparison.
5 = -3
8 = 0
so, yes, -3 < 0.
Related
I have a problem where I want to limit the range of a real variable between the maximum and minimum value of another set of real variables.
s = Solver()
y = Real('y')
Z = RealVector('z', 10)
s.add(And(y >= min(Z), y <= max(Z)))
Is there a way to do this in z3py?
You can use Axel's solution; though that one requires you to create an extra variable and also asserts more constraints than needed. Moreover, it doesn't let you use min and max as simple functions. It might be easier to just program this in a functional way, like this:
# Return minimum of a vector; error if empty
def min(vs):
m = vs[0]
for v in vs[1:]:
m = If(v < m, v, m)
return m
# Return maximum of a vector; error if empty
def max(vs):
m = vs[0]
for v in vs[1:]:
m = If(v > m, v, m)
return m
Another difference is that in the functional style we throw an error if the vector is empty. In the other style, the result will essentially be unconstrained. (i.e., min/max can take any value.) You should consider which semantics is right for your application, in case the vector you're passing might be empty. (At the least, you should change it so it prints out a nicer error message. Currently it'll throw an IndexError: list index out of range error if given an empty vector.)
Now you can say:
s = Solver()
y = Real('y')
Z = RealVector('z', 10)
s.add(And(y >= min(Z), y <= max(Z)))
print (s.check())
print (s.model())
This prints:
sat
[z__7 = -1,
z__0 = -7/2,
z__4 = -5/2,
z__5 = -2,
z__3 = -9/2,
z__2 = -4,
z__8 = -1/2,
y = 0,
z__9 = 0,
z__6 = -3/2,
z__1 = -3]
You could benefit from Hakan Kjellerstrand's collection of useful z3py definitions:
from z3 import *
# Functions written by Hakan Kjellerstrand
# http://hakank.org/z3/
# The following can be used by importing http://www.hakank.org/z3/z3_utils_hakank.py
# v is the maximum value of x
def maximum(sol, v, x):
sol.add(Or([v == x[i] for i in range(len(x))])) # v is an element in x)
for i in range(len(x)):
sol.add(v >= x[i]) # and it's the greatest
# v is the minimum value of x
def minimum(sol, v, x):
sol.add(Or([v == x[i] for i in range(len(x))])) # v is an element in x)
for i in range(len(x)):
sol.add(v <= x[i]) # and it's the smallest
s = Solver()
y = Real('y')
zMin = Real('zMin')
zMax = Real('zMax')
Z = RealVector('z', 10)
maximum(s, zMin, Z)
minimum(s, zMax, Z)
s.add(And(y >= zMin, y <= zMax))
print(s.check())
print(s.model())
Consider a set of constraints F = [a + b > 10, a*a + b + 10 < 50].
When I run it using:
s = Solver()
s.add(F)
s.check()
I get sat solution.
If I run it with:
s = Solver()
s.check(F)
I get an unknown solution. Can someone explain why this is happening?
Let's see:
from z3 import *
a = Int('a')
b = Int('b')
F = [a + b > 10, a*a + b + 10 < 50]
s = Solver()
s.add(F)
print (s.check())
print (s.model())
This prints:
sat
[b = 15, a = -4]
That looks good to me.
Let's try your second variant:
from z3 import *
a = Int('a')
b = Int('b')
F = [a + b > 10, a*a + b + 10 < 50]
s = Solver()
print (s.check(F))
print (s.model())
This prints:
sat
[b = 7, a = 4]
That looks good to me too.
So, I don't know how you're getting the unknown answer. Maybe you have an old version of z3; or you've some other things in your program you're not telling us about.
The important thing to note, however, is that s.add(F); s.check() AND s.check(F) are different operations:
s.add(F); s.check() means: Assert the constraints in F; check that they are satisfiable.
s.check(F) means: Check that all the other constraints are satisfiable, assuming F is. In particular, it does not assert F. (This is important if you do further asserts/checks later on.)
So, in general these two different ways of using check are used for different purposes; and can yield different answers. But in the presence of no other assertions around, you'll get a solution for both, though of course the models might be different.
Aside One reason you can get unknown is in the presence of non-linear constraints. And your a*a+b+10 < 50 is non-linear, since it does have a multiplication of a variable by itself. You can deal with that either by using a bit-vector instead of an Int (if applicable), or using the nonlinear-solver; which can still give you unknown, but might perform better. But just looking at your question as you asked it, z3 is just fine handling it.
To find out what is going on within s.check(F), you can do the following:
from z3 import *
import inspect
a = Int('a')
b = Int('b')
F = [a + b > 10, a*a + b + 10 < 50]
s = Solver()
print (s.check(F))
print (s.model())
source_check = inspect.getsource(s.check)
print(source_check)
The resulting output:
sat
[b = 10, a = 1]
def check(self, *assumptions):
"""Check whether the assertions in the given solver plus the optional assumptions are consistent or not.
>>> x = Int('x')
>>> s = Solver()
>>> s.check()
sat
>>> s.add(x > 0, x < 2)
>>> s.check()
sat
>>> s.model().eval(x)
1
>>> s.add(x < 1)
>>> s.check()
unsat
>>> s.reset()
>>> s.add(2**x == 4)
>>> s.check()
unknown
"""
s = BoolSort(self.ctx)
assumptions = _get_args(assumptions)
num = len(assumptions)
_assumptions = (Ast * num)()
for i in range(num):
_assumptions[i] = s.cast(assumptions[i]).as_ast()
r = Z3_solver_check_assumptions(self.ctx.ref(), self.solver, num, _assumptions)
return CheckSatResult(r)
The semantics of assumptions vs. assertions are discussed here and here. But if have to admit that they are not really clear to me yet.
F(x1) > a;
F(x2) < b;
∀t, F'(x) >= 0 (derivative) ;
F(x) = ∑ ci*x^i; (i∈[0,n] ; c is a constant)
Your question is quite ambiguous, and stack-overflow works the best if you show what you tried and what problems you ran into.
Nevertheless, here's how one can code your problem for a specific function F = 2x^3 + 3x + 4, using the Python interface to z3:
from z3 import *
# Represent F as a function. Here we have 2x^3 + 3x + 4
def F(x):
return 2*x*x*x + 3*x + 4
# Similarly, derivative of F: 6x^2 + 3
def dF(x):
return 6*x*x + 3
x1, x2, a, b = Ints('x1 x2 a b')
s = Solver()
s.add(F(x1) > a)
s.add(F(x2) < b)
t = Int('t')
s.add(ForAll([t], dF(t) >= 0))
r = s.check()
if r == sat:
print s.model()
else:
print ("Solver said: %s" % r)
Note that I translated your ∀t, F'(x) >= 0 condition as ∀t. F'(t) >= 0. I assume you had a typo there in the bound variable.
When I run this, I get:
[x1 = 0, x2 = 0, b = 5, a = 3]
This method can be generalized to arbitrary polynomials with constant coefficients in the obvious way, but that's mostly about programming and not z3. (Note that doing so in SMTLib is much harder. This is where the facilities of host languages like Python and others come into play.)
Note that this problem is essentially non-linear. (Variables are being multiplied with variables.) So, SMT solvers may not be the best choice here, as they don't deal all that well with non-linear operations. But you can deal with those problems as they arise later on. Hope this gets you started!
I am using z3 to write a static checker. I have the following problem:
>>> from z3 import *
>>> s = Solver()
>>> s.add(FreshInt() + FreshInt() > 0)
>>> s.check()
sat
>>> s.model()
[]
As you can see, fresh variables are not shown in the model. I also cannot get their value:
>>> a = FreshInt()
>>> s.add(a > 3)
>>> s.check()
sat
>>> s.model()
[]
>>> s.model()[a]
I've looked in the docs but I cannot find a way to change this behaviour. I could generate unique variables myself, but it would be nice if z3 can take care of that for me. Can someone point me in the right direction? Or is it not possible to change this in z3py?
FreshInt/FreshReal etc. are intended for creating internal variables that are not user-visible. You should instead use Int('name') and Real('name') to create user level variables that will be shown in models.
If you really want to see the value, you can add an observer function and use it like this:
from z3 import *
def observeInt(s, a):
obs = Int('observer')
s.add(obs == a)
# might want to check the following really returns sat!
s.check()
print s.model()[obs]
s = Solver()
a = FreshInt()
s.add(a + FreshInt() > 0)
s.add(a > 12)
print s.check()
observeInt(s, a)
This prints:
sat
13
This is not cheap obviously (as it involves a call to check), but it is safe and so long as it's used in debugging situations to strong-arm z3 as you put it, it should do the trick.
You could circumvent this restriction as follows:
freshIntIdx = 0
def myFreshInt():
global freshIntIdx
freshIntIdx += 1;
return Int('fi' + str(freshIntIdx))
a = myFreshInt()
b = myFreshInt()
s = Solver()
s.add(a + b > 5, a > 0, b > 0, a + b < 10)
print(s.check())
m = s.model()
print("a = %s" % m[a])
print("b = %s" % m[b])
I'm trying to implement a function that computes the Relu derivative for each element in a matrix, and then return the result in a matrix. I'm using Python and Numpy.
Based on other Cross Validation posts, the Relu derivative for x is
1 when x > 0, 0 when x < 0, undefined or 0 when x == 0
Currently, I have the following code so far:
def reluDerivative(self, x):
return np.array([self.reluDerivativeSingleElement(xi) for xi in x])
def reluDerivativeSingleElement(self, xi):
if xi > 0:
return 1
elif xi <= 0:
return 0
Unfortunately, xi is an array because x is an matrix. reluDerivativeSingleElement function doesn't work on array. So I'm wondering is there a way to map values in a matrix to another matrix using numpy, like the exp function in numpy?
Thanks a lot in advance.
That's an exercise in vectorization.
This code
if x > 0:
y = 1
elif xi <= 0:
y = 0
Can be reformulated into
y = (x > 0) * 1
This is something that will work for numpy arrays, since boolean expressions involving them are turned into arrays of values of these expressions for elements in said array.
I guess this is what you are looking for:
>>> def reluDerivative(x):
... x[x<=0] = 0
... x[x>0] = 1
... return x
>>> z = np.random.uniform(-1, 1, (3,3))
>>> z
array([[ 0.41287266, -0.73082379, 0.78215209],
[ 0.76983443, 0.46052273, 0.4283139 ],
[-0.18905708, 0.57197116, 0.53226954]])
>>> reluDerivative(z)
array([[ 1., 0., 1.],
[ 1., 1., 1.],
[ 0., 1., 1.]])
Basic function to return derivative of relu could be summarized as follows:
f'(x) = x > 0
So, with numpy that would be:
def relu_derivative(z):
return np.greater(z, 0).astype(int)
def dRelu(z):
return np.where(z <= 0, 0, 1)
Here z is a ndarray in my case.
def reluDerivative(self, x):
return 1 * (x > 0)
You are on a good track: thinking on vectorized operation. Where we define a function, and we apply this function to a matrix, instead of writing a for loop.
This threads answers your question, where it replace all the elements satisfy the condition. You can modify it into ReLU derivative.
https://stackoverflow.com/questions/19766757/replacing-numpy-elements-if-condition-is-met
In addition, python supports functional programming very well, try to use lambda function.
https://www.python-course.eu/lambda.php
This works:
def dReLU(x):
return 1. * (x > 0)
As mentioned by Neil in the comments, you can use heaviside function of numpy.
def reluDerivative(self, x):
return np.heaviside(x, 0)
If you want to use pure Python:
def relu_derivative(x):
return max(sign(x), 0)
If you want it with the derivative you can use:
def relu(neta):
relu = neta * (neta > 0)
d_relu = (neta > 0)
return relu, d_relu
When x is larger than 0, the slope is 1.
When x is smaller than or equal to 0, the slope is 0.
if (x > 0):
return 1
if (x <= 0):
return 0
This can be written more compact:
return 1 * (x > 0)