z3py

Over the years I keep track of solving technology - and I maintain a blog post about applying them to a specific puzzle - the "crossing ladders". To get to the point, I accidentally found out about z3, and tried putting it to use in the specific problem. I used the Python bindings, and wrote this: $ cat laddersZ3.py #!/usr/bin/env python from z3 import * a = Int('a') b = Int('b') c = Int('c') d = Int('d') e = Int('e') f = Int('f') solve( a>0, a<200, b>0, b<200, c>0, c<200, d>0, d<200, e>0, e<200, f>0, f<200, (e+f)**2 + d**2 == 119**2, (e+f)**2 + c**2 == 70**2, e**2 + 30**2 == a**2, f**2 + 30*

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

Let's say, I want to restrict each character of the string to the charset: [a-zA-Z0-9_] using Z3 constraints, can I use a boolean operator to specify that? As an example: input = [BitVec("input%s" % i, 8) for i in range(10)] for i in range(10): s.add(input[i] >= 0x30 and input[i] <= 0x39) s.add(input[i] >= 0x41 and input[i] <= 0x5A) s.add(input[i] >= 0x61 and input[i] <= 0x7A) Is this correct? Any other efficient way to define constraints? Usually in Python, I could do something like: import string charset = string.uppercase + string.lowercase + string.digits + "_" for i in charset: ... Can

I'm experimenting with z3 in python. I have the following model: (set-option :produce-models true) (set-logic QF_AUFBV ) (declare-fun a () (Array (_ BitVec 32) (_ BitVec 8) ) ) (declare-fun another () (Array (_ BitVec 32) (_ BitVec 8) ) ) (assert (and (= false (= (_ bv77 32) (concat (select a (_ bv3 32) ) (concat (select a (_ bv2 32) ) (concat (select a (_ bv1 32) ) (select a (_ bv0 32) ) ) ) ) ) ) (= false (= (_ bv12 32) (concat (select another (_ bv3 32) ) (concat (select another (_ bv2 32) ) (concat (select another (_ bv1 32) ) (select another (_ bv0 32) ) ) ) ) ) ) ) ) I can load it and

Given a bit vector in Z3 , I am wondering how can I sum up each individual bit of this vector? E.g., a = BitVecVal(3, 2) sum_all_bit(a) = 2 Is there any pre-implemented APIs/functions that support this? Thank you! Nikolaj Bjorner It isn't part of the bit-vector operations. You can create an expression as follows: def sub(b): n = b.size() bits = [ Extract(i, i, b) for i in range(n) ] bvs = [ Concat(BitVecVal(0, n - 1), b) for b in bits ] nb = reduce(lambda a, b: a + b, bvs) return nb print sub(BitVecVal(4,7)) Of course, log(n) bits for the result will suffice if you prefer. The page: https:/

Suppose I have a list of 10 variables v = [Real('v_%s' % (i+1)) for i in range(10)] and I want to add a simple constraint like this s = Solver() for i in range(10): s.add(v[i] == i) if s.check() == sat: print(s.model()) So a satisfying model is v_1 = 0, v_2 = 1 .... v_10 = 9 . However the output of print(s.model()) is totoally unordered which makes me confused when I have lots of variables in a bigger model. For this example, the output of my computer is v_5, v_7, v_4, v_2, v_1, v_3, v_6, v_8, v_9, v_10 , but I want to ouput the variables of this model in order like v_1, v_2, ..., v_10 . Can

I am using the Python bindings for the Z3 theorem prover (Z3Py). I have N boolean variables, x1,..,xN. I want to express the constraint that exactly K out of N of them should be true. How can I do that, in Z3Py? Is there any built-in support for that? I checked the online documentation, but the Z3Py docs don't have any mention of any API for that. For one-out-of-N constraints, I know I can separately express that at least one is true (assert Or(x1,..,xN)) and that at most one is true (assert Not(And(xi,xj)) for all i,j). I also know of other ways to manually express the 1-out-of-N and K-out-of