I've the following SMT-Lib2 script:
(set-option :produce-models true)
(declare-fun s0 () Int)
(declare-fun table0 (Int) (_ BitVec 8))
(assert (= (table0 0) #x00))
(assert
(let ((s3 (ite (or (< s0 0) (<= 1 s0)) #x01 (table0 s0))))
(let ((s5 (ite (bvuge s3 #x02) #b1 #b0)))
(= s5 #b1))))
(check-sat)
(get-model)
With Z3 v3.2 running on the Mac, I get:
sat
(model
;; universe for (_ BitVec 8):
;; bv!val!2 bv!val!3 bv!val!0 bv!val!1
;; -----------
;; definitions for universe elements:
(declare-fun bv!val!2 () (_ BitVec 8))
(declare-fun bv!val!3 () (_ BitVec 8))
(declare-fun bv!val!0 () (_ BitVec 8))
(declare-fun bv!val!1 () (_ BitVec 8))
;; cardinality constraint:
(forall ((x (_ BitVec 8)))
(and (= x bv!val!2) (= x bv!val!3) (= x bv!val!0) (= x bv!val!1)))
;; -----------
(define-fun s0 () Int
(- 1))
(define-fun table0 ((x!1 Int)) (_ BitVec 8)
(ite (= x!1 0) bv!val!0
(ite (= x!1 (- 1)) bv!val!3
bv!val!0)))
)
Which states that s0 = -1 is a model. However, with s0 = -1, we have s3 = 1 and s5 = #b0, which makes the assertion false. In fact, I'm quite sure the benchmark as stated is unsatisfiable.
One thing I noticed in the Z3 output is the quantified formula it gives for the cardinality constraint. It says:
;; cardinality constraint:
(forall ((x (_ BitVec 8)))
(and (= x bv!val!2) (= x bv!val!3) (= x bv!val!0) (= x bv!val!1)))
The assertion is a conjunction, which sounds rather weird; shouldn't that be a disjunction? I'm not sure if this is the root-cause of the problem, but it sure sounds fishy.
There are two problems in Z3.
First, you are correct, there is a typo in the model printer. It should be a "or" instead of an "and". The second problem is that Z3 did not install the bit-vector theory and treated (_ BitVec 8) as a uninterpreted sort. This was a bug in the preprocessor that is used to decide in which logic the problem is in. You can workaround this bug by adding the following command in the beginning of the file:
(set-option :auto-config false)
These bugs have been fixed, and the fix will be available in the next release.
Related
If possible I'd like a second opinion on my code.
The constraints of the problem are:
a,b,c,d,e,f are non-zero integers
s1 = [a,b,c] and s2 = [d,e,f] are sets
The sum s1_i + s2_j for i,j = 0..2 has to be a perfect square
I don't understand why but my code returns model not available. Moreover, when commenting out the following lines:
(assert (and (> sqrtx4 1) (= x4 (* sqrtx4 sqrtx4))))
(assert (and (> sqrtx5 1) (= x5 (* sqrtx5 sqrtx5))))
(assert (and (> sqrtx6 1) (= x6 (* sqrtx6 sqrtx6))))
(assert (and (> sqrtx7 1) (= x7 (* sqrtx7 sqrtx7))))
(assert (and (> sqrtx8 1) (= x8 (* sqrtx8 sqrtx8))))
(assert (and (> sqrtx9 1) (= x9 (* sqrtx9 sqrtx9))))
The values for d, e, f are negative. There is no constraint that requires them to do so. I'm wondering if perhaps there are some hidden constraints that sneaked in and mess up the model.
A valid expected solution would be:
a = 3
b = 168
c = 483
d = 1
e = 193
f = 673
Edit: inserting (assert (= a 3)) and (assert (= b 168)) results in the solver finding the correct values. This only puzzles me further.
Full code:
(declare-fun sqrtx1 () Int)
(declare-fun sqrtx2 () Int)
(declare-fun sqrtx3 () Int)
(declare-fun sqrtx4 () Int)
(declare-fun sqrtx5 () Int)
(declare-fun sqrtx6 () Int)
(declare-fun sqrtx7 () Int)
(declare-fun sqrtx8 () Int)
(declare-fun sqrtx9 () Int)
(declare-fun a () Int)
(declare-fun b () Int)
(declare-fun c () Int)
(declare-fun d () Int)
(declare-fun e () Int)
(declare-fun f () Int)
(declare-fun x1 () Int)
(declare-fun x2 () Int)
(declare-fun x3 () Int)
(declare-fun x4 () Int)
(declare-fun x5 () Int)
(declare-fun x6 () Int)
(declare-fun x7 () Int)
(declare-fun x8 () Int)
(declare-fun x9 () Int)
;all numbers are non-zero integers
(assert (not (= a 0)))
(assert (not (= b 0)))
(assert (not (= c 0)))
(assert (not (= d 0)))
(assert (not (= e 0)))
(assert (not (= f 0)))
;both arrays need to be sets
(assert (not (= a b)))
(assert (not (= a c)))
(assert (not (= b c)))
(assert (not (= d e)))
(assert (not (= d f)))
(assert (not (= e f)))
(assert (and (> sqrtx1 1) (= x1 (* sqrtx1 sqrtx1))))
(assert (and (> sqrtx2 1) (= x2 (* sqrtx2 sqrtx2))))
(assert (and (> sqrtx3 1) (= x3 (* sqrtx3 sqrtx3))))
(assert (and (> sqrtx4 1) (= x4 (* sqrtx4 sqrtx4))))
(assert (and (> sqrtx5 1) (= x5 (* sqrtx5 sqrtx5))))
(assert (and (> sqrtx6 1) (= x6 (* sqrtx6 sqrtx6))))
(assert (and (> sqrtx7 1) (= x7 (* sqrtx7 sqrtx7))))
(assert (and (> sqrtx8 1) (= x8 (* sqrtx8 sqrtx8))))
(assert (and (> sqrtx9 1) (= x9 (* sqrtx9 sqrtx9))))
;all combinations of sums need to be squared
(assert (= (+ a d) x1))
(assert (= (+ a e) x2))
(assert (= (+ a f) x3))
(assert (= (+ b d) x4))
(assert (= (+ b e) x5))
(assert (= (+ b f) x6))
(assert (= (+ c d) x7))
(assert (= (+ c e) x8))
(assert (= (+ c f) x9))
(check-sat-using (then simplify solve-eqs smt))
(get-model)
(get-value (a))
(get-value (b))
(get-value (c))
(get-value (d))
(get-value (e))
(get-value (f))
Nonlinear integer arithmetic is undecidable. This means that there is no decision procedure that can decide arbitrary non-linear integer constraints to be satisfiable. This is what z3 is telling you when it says "unknown" as the answer your query.
This, of course, does not mean that individual cases cannot be answered. Z3 has certain tactics it applies to solve such formulas, but it is inherently limited in what it can handle. Your problem falls into that category: One that Z3 is just not capable of solving.
Z3 has a dedicated NRA (non-linear real arithmetic) tactic that you can utilize. It essentially treats all variables as reals, solves the problem (nonlinear real arithmetic is decidable and z3 can find all algebraic real solutions), and then checks if the results are actually integer. If not, it tries another solution over the reals. Sometimes this tactic can handle non-linear integer problems, if you happen to hit the right solution. You can trigger it using:
(check-sat-using qfnra)
Unfortunately it doesn't solve your particular problem in the time I allowed it to run. (More than 10 minutes.) It's unlikely it'll ever hit the right solution.
You really don't have many options here. SMT solvers are just not a good fit for nonlinear integer problems. In fact, as I alluded to above, there is no tool that can handle arbitrary nonlinear integer problems due to undecidability; but some tools fare better than others depending on the algorithms they use.
When you tell z3 what a and b are, you are essentially taking away much of the non-linearity, and the rest becomes easy to handle. It is possible that you can find a sequence of tactics to apply that solves your original, but such tricks are very brittle in practice and not easily discovered; as you are essentially introducing heuristics into the search and you don't have much control over how that behaves.
Side note: Your script can be improved slightly. To express that a bunch of numbers are all different, use the distinct predicate:
(assert (distinct (a b c)))
(assert (distinct (d e f)))
Is there an easier way to get the immediate answers when doing bit-vector ops eg. a = 1 million, b = 0, what is a & b (answer: 0)
This method works but have to introduce dummy variable to store answer:
(declare-const a (_ BitVec 64))
(declare-const b (_ BitVec 64))
(declare-const ans (_ BitVec 64))
(assert (= a (_ bv1000000 64)))
(assert (= b (_ bv0000000 64)))
(assert (= ans (bvand a b)))
(check-sat)
(get-model)
This method is what I'd like but my code gives a demorgan identity:
(declare-const a (_ BitVec 64))
(declare-const b (_ BitVec 64))
(simplify (bvand a b))
You can use the model to evaluate arbitrary expressions, for instance like this:
(declare-const a (_ BitVec 64))
(declare-const b (_ BitVec 64))
(assert (= a (_ bv1000000 64)))
(assert (= b (_ bv0000000 64)))
(check-sat)
(eval (bvand a b))
says
sat
#x0000000000000000
I didn't test, but something like (apply (then propagate-values simplify)) should do the trick
I have the following expression
(declare-fun x00 () Real)
(declare-fun x01 () Real)
(declare-fun x10 () Real)
(declare-fun x11 () Real)
(declare-fun t0init () Real)
(declare-fun z0init0 () Real)
(declare-fun z0init1 () Real)
(assert (>= t0init 0))
(assert (= (+ x00 z0init0) x10))
(assert (= (+ x01 z0init1) x11))
(assert (< (+ (* 1 x00)(* 0 x01)) 0.0))
(assert (= (+ (* 0 x00)(* 1 x01)) 0.0))
(assert (< (+ (* 1 x10)(* 0 x11)) 0.0))
(assert (= (+ (* 0 x10)(* 1 x11)) 0.0))
...
(assert (< (+ (* 1 x40)(* 0 x41)) 0.0))
(assert (= (+ (* 0 x40)(* 1 x41)) 0.0))
(assert (= (+ (* 1 z4end0)(* 0 z4end1)) (* t4end 1)))
(assert (= (+ (* 0 z4end0)(* 1 z4end1)) (* t4end -2)))
and I would like to express as a simple formula in order to express the following:
(assert exists (x00 x01) ("the above expression"))
and then perform a quantifier elimination.
Is there anyone who knows how to proceed?
I know how to do it with z3py but I need some faster solution.
Thank you very much for any hint.
One possible solution is as follows
(declare-fun x00 () Real)
(declare-fun x01 () Real)
(declare-fun x10 () Real)
(declare-fun x11 () Real)
(declare-fun t0init () Real)
(declare-fun z0init0 () Real)
(declare-fun z0init1 () Real)
(define-fun conjecture () Bool
(and (>= t0init 0) (= (+ x00 z0init0) x10) (= (+ x01 z0init1) x11)))
(assert (exists ((x00 Real) (x01 Real)) conjecture))
(check-sat)
and the corresponding output is
sat
I am not sure if the quantifier elimination that you need will work with Z3. Maybe for your problem "Redlog" of "Reduce" is the better option. All the best.
Another question from a Z3 newbie. Can options change the behavior of Z3? I might expect them to affect termination, or change sat or unsat into unknown but not sat into unsat or vice versa.
This example:
(set-option :smt.macro-finder true)
(declare-datatypes () ((Option (none) (some (Data Int)))))
(define-sort Set () (Array Option Option))
(declare-fun filter1 (Option) Option)
(declare-fun filter2 (Option) Option)
(declare-var s1 Set)
(declare-var s2 Set)
(declare-var x1 Option)
(declare-var x2 Option)
(declare-var x3 Option)
(declare-var x4 Option)
(assert (not (= x1 none)))
(assert (not (= x2 none)))
(assert (not (= x3 none)))
(assert (not (= x4 none)))
(assert (= (select s1 x1) x2))
(assert (= (select s2 x3) x4))
(assert (forall ((x Option)) (= (filter1 x) (ite (or (= none x) (= (Data x) 1)) x none))))
(assert (forall ((x Option)) (= (filter2 x) (ite (or (= none x) (= (Data x) 2)) x none))))
(assert (= ((_ map filter1) s1) s2))
(assert (= ((_ map filter2) s1) s2))
(check-sat)
(get-model)
returns sat with the first line and unsat without it.
Is this a bug or am I missing something fundamental?
This is a bug. The two quantifiers are essentially providing "definitions" for filter1 and filter2.
The option smt.macro-finder is used to eliminate functions symbols by expanding these definitions. It is essentially performing "macro expansion". However, there is a bug in the macro expander. It does not expand the occurrences of filter1 and filter2 in the map constructs: (_ map filter1) and (_ map filter2).
This bug will be fixed.
In the meantime, we should not use the map construct and smt.macro-finder option simultaneously.
I have the following piece of code:
(declare-const L4 (_ BitVec 6))
(declare-const L1 (_ BitVec 6))
(declare-const L0 (_ BitVec 6))
(declare-const l2 (_ BitVec 6))
(assert (= l2 (_ bv8 6)))
;; All is encoding the set that contains {0, 1, 2, 3, 4, 5}
(define-const All (_ BitVec 6) #b111111)
;; Empty is encoding the empty set
(define-const Empty (_ BitVec 6) #b000000)
(define-fun LT_l ((S (_ BitVec 6)) (l (_ BitVec 6))) Bool
;; True if for all x in S x < l
(= (bvand (bvshl All l) S) Empty))
(define-fun is_in ((e (_ BitVec 6)) (S (_ BitVec 6))) Bool
;; True if e is an element of the "set" S.
(not (= (bvand (bvshl (_ bv1 6) e) S) Empty)))
(define-fun is_minimal ((e (_ BitVec 6)) (S (_ BitVec 6))) Bool
;; True if e is the minimal element of S
(and (is_in e S) ;; S contains e
;; (1 << e) - 1 represents the set of elements that are smaller than e
(= (bvand (bvsub (bvshl (_ bv1 6) e) (_ bv1 6)) S) Empty)))
;; To encode that forall x in L0 and forall y in L1. x < y
(define-fun LT ((L0 (_ BitVec 6)) (L1 (_ BitVec 6))) Bool
; True if forall x in L0 and forall y in L1, x < y
(or (= L0 Empty)
(= L1 Empty)
(exists ((min (_ BitVec 6))) (and (is_minimal min L1) (LT_l L0 min)))))
(assert (not (= L0 Empty)))
(assert (not (= L1 Empty)))
(assert (not (= L4 Empty)))
(assert (LT_l L4 l2))
(assert (LT L0 L1))
(check-sat)
(get-model)
(assert (LT L1 L0))
(check-sat)
When I run this code I get the model is:
sat
(model
(define-fun min!0 () (_ BitVec 6)
#b000011)
(define-fun L1 () (_ BitVec 6)
#b001000)
(define-fun L0 () (_ BitVec 6)
#b000100)
(define-fun L4 () (_ BitVec 6)
#b000100)
(define-fun l2 () (_ BitVec 6)
#b001000)
)
unsat
Why is the result of min is:
(define-fun min!0 () (_ BitVec 6)
#b000011)
and not b001000 since the smallest value of L1 is this and not b000011.
Someone can explain me?
Finally, I define the function Lt_l that checks if for all x in S x < l, but now I wanted to do GT_l that checks if for all x in S l < x. I have the following code:
(define-fun GT_l ((S (_ BitVec 6)) (l (_ BitVec 6))) Bool
(= (bvand (bvneg (bvshl (_ bv0 6) l)) S) Empty))
But this does not work why?
Thanks
In your example, you are representing sets using bit-vectors. For example, the bit-vector #b101000 represents the set {5, 3}. The output (define-fun L1 () (_ BitVec 6) #b001000) is essentially saying that L1 is the "set" {3}. One possible confusion is that bit-vectors are being used to represent sets and elements. The bit-vector min!0 represents an element. The output (define-fun min!0 () (_ BitVec 6) #b000011) is saying that min!0 is the value 3, and it is indeed the "minimal value" in L1.