Bad performance when using `str.indexof` with `int2bv` - z3

I am trying to constrain a bitvector to be equal to the index of a null byte in a string. However, I am getting performance issues with int2bv.
As a simple example of the issue, this problem (without int2bv) solves instantly:
; Solve time: 0m0.065s
(declare-const s String)
(declare-const len Int)
(assert (= (str.indexof s "\00") len))
(assert (>= len 256)) ; 0x100
(assert (<= len 4095)) ; 0xFFF
(check-sat)
(get-model)
On the other hand, the "same" problem (it allows strictly more solutions) using int2bv takes almost 2 minutes:
; Solve time: 1m54.861s
(declare-const s String)
(declare-const len Int)
(assert (= (str.indexof s "\00") len))
(assert (>= len 0))
(assert (not (= (bvand ((_ int2bv 12) len) #xF00) #x000)))
(check-sat)
(get-model)
Is there a better way to encode this kind of bitvector string length operation in z3? These slides on z3strBV claim to have better support, but their extensions don't seem to be implemented in the standard z3 distribution.
I am using z3 version 4.8.0 - 64 bit and running with smt.string_solver=z3str3.

Through trial and error I found that the option smt.str.use_binary_search=true drops the solve time for the second problem to under 2 seconds! Not sure why the default value for this option is false, though.

Related

Z3 should be able to verify this program?

I am a beginner user of Z3. Recently, I have been using z3 to verify some problems. Here is one problem I tried: (x < 0 && y < 0) implies x/y >= 0, below are the programs I wrote:
(declare-const x Int)
(declare-const y Int)
(define-fun assumption() Bool
(and (< x 0) (< y 0))
)
(define-fun predicate() Bool
(<= 0 (div x y))
)
(assert (not (=> assumption predicate)))
(check-sat)
When I use 'z3 -smt2 filename' to verify this program, it works and return unsat. However, when I later modify the program to:
(declare-const x Int)
(declare-const y Int)
(define-fun assumption() Bool
(and (< x 0) (< y 0))
)
(define-fun predicate() Bool
(<= 0 (div (* -1 x) (* -1 y)))
)
(assert (not (=> assumption predicate)))
(check-sat)
This cause z3 to timeout when I changed 'x/y' to (-1 * x)/(-1 * y) and somehow I need to add -1 there. I am confused why this happened and why multiplied by a constant makes this problem more complex.
Can somebody help figure out why this happened?
Thank you!
There is very limited support for non-linear arithmetic, such as division where the divisor is a variable.
So Z3 will do a best effort, but in no way be sure to provide decisions on every formula with non-linear arithmetical operations. So for example, you can enter Diophantine equations into Z3, but not expect it to provide sat/unsat answers.
Rather than trying some infinite search, Z3 may prefer to give up and return unknown.
(Note: This is just a comment instead of an answer. I have to do so because I don't have sufficient reputation to add a comment.)
Here is the quote from the Z3 guide about integer division, modulo and reminder operators:
Z3 also has support for division, integer division, modulo and remainder operators. Internally, they are all mapped to multiplication.
(declare-const a Int)
(declare-const r1 Int)
(declare-const r2 Int)
(declare-const r3 Int)
(declare-const r4 Int)
(declare-const r5 Int)
(declare-const r6 Int)
(assert (= a 10))
(assert (= r1 (div a 4))) ; integer division
(assert (= r2 (mod a 4))) ; mod
(assert (= r3 (rem a 4))) ; remainder
(assert (= r4 (div a (- 4)))) ; integer division
(assert (= r5 (mod a (- 4)))) ; mod
(assert (= r6 (rem a (- 4)))) ; remainder
(declare-const b Real)
(declare-const c Real)
(assert (>= b (/ c 3.0)))
(assert (>= c 20.0))
(check-sat)
(get-model)
In Z3, division by zero is allowed, but the result is not specified. Division is not a partial function. Actually, in Z3 all functions are total, although the result may be underspecified in some cases like division by zero.
It is unclear what is the behavior when the arguments are negative. In other words, the mapping to multiplication for these operations is not clear.

Different check-sat answers when asserting same property in between

Given the following input
(set-option :auto_config false)
(set-option :smt.mbqi false)
(declare-fun len (Int) Int)
(declare-fun idx (Int Int) Int)
(declare-const x Int)
(define-fun FOO () Bool
(forall ((i Int)) (!
(implies
(and (<= 0 i) (< i (len x)))
(exists ((j Int)) (!
(implies
(and (<= 0 j) (< j (len x)))
(> (idx x j) 0))))))))
(assert FOO)
; (push)
(assert (not FOO))
(check-sat)
; (pop)
; (push)
(assert (not FOO))
(check-sat)
; (pop)
Z3 4.3.2 x64 reports unsat for the first check-sat (as expected), but unknown for the second. If the commented push/pops are uncommented, both check-sats yield unknown.
My guess is that this is either a bug, or a consequence of Z3 switching to incremental mode when it reaches the second check-sat. The latter could also explain why both check-sats yield unknown if push/pop is used because Z3 will (as far as I understand) switch to incremental mode on first push.
Question: Is it a bug or an expected consequence?
Good example.
It is a limitation of how Z3 processes the formulas:
1. When using push/pop it does not detect the contradiction among the asserted formulas, instead it converts formulas to negation normal form and skolemizes quantified formulas.
2. When calling check-sat the second time, it does not keep track that the state was not retracted from a previous unsatisfiable state.
It isn't an unsoundness bug, but sure the behavior is not what a user would expect.
In addition to Nikolaj's answer: Yes, this is because Z3 switches to a different solver, which will give up earlier. We can get the same effect by setting (set-option :combined_solver.ignore_solver1 true).

Should Z3 prove monotonicity of <= (with respect to multiplication)?

This is the reduction of a more interesting problem, in which the missing property was (for positive k,M and N), that ((k % M) * N) < M*N. Below is an encoding of the simpler problem that a <= b ==> (a*c) <= (b*c). Such a query succeeds (we get unsat), but if the expression b is replaced by b+1 (as in the second query below) then we get unknown, which seems surprising. Is this the expected behaviour? Are there options to improve the handling of such inequalities? I tried with and without configuration options, and various versions of Z3, including the current unstable branch. Any tips would be much appreciated!
(declare-const a Int)
(declare-const b Int)
(declare-const c Int)
(assert (> a 0))
(assert (> b 0))
(assert (> c 0))
(assert (<= a b))
(assert (not (<= (* a c) (* b c))))
(check-sat)
(assert (<= a (+ b 1)))
(assert (not (<= (* a c) (* (+ b 1) c))))
(check-sat)
This falls into nonlinear integer arithmetic (which has an undecidable decision problem, see, e.g., How does Z3 handle non-linear integer arithmetic? ), so it's actually not too surprising Z3 returns unknown for some examples, although I guess a bit surprising that it toggled between unsat and unknown for quite similar examples.
If it works for your application, you can try a type coercion: encode the constants as Real instead of Int. This will allow you to use Z3's complete solver for nonlinear real arithmetic and returns unsat with check-sat.
Alternatively, you can force Z3 to use the nonlinear solver even for the integer encoding with (check-sat-using qfnra-nlsat) as in the following based on your example (rise4fun link: http://rise4fun.com/Z3/87GW ):
(declare-const a Int)
(declare-const b Int)
(declare-const c Int)
(assert (> a 0))
(assert (> b 0))
(assert (> c 0))
(assert (<= a b))
(assert (not (<= (* a c) (* b c))))
;(check-sat)
(check-sat-using qfnra-nlsat) ; unsat
(assert (<= a (+ b 1)))
(assert (not (<= (* a c) (* (+ b 1) c))))
; (check-sat)
(check-sat-using qfnra-nlsat) ; unsat
Some more questions and answers on similar subjects:
Combining nonlinear Real with linear Int
z3 fails with this system of equations
Using Z3Py online to prove that n^5 <= 5 ^n for n >= 5
Can z3 always give result when handling nonlinear real arithmetic
Z3 Theorem Prover: Pythagorean Theorem (Non-Linear Artithmetic)

Why operators '/' and 'div' in Z3 give different results?

I was trying to represent a real number with two integer numbers as using them as the numerator and the denominator of the real number. I wrote the following program:
(declare-const a Int)
(declare-const b Int)
(declare-const f Real)
(assert (= f (/ a b)))
(assert (= f 0.5))
(assert (> b 2))
(assert (> b a))
(check-sat)
(get-model)
The program returned SAT result as follows:
sat
(model
(define-fun f () Real
(/ 1.0 2.0))
(define-fun b () Int
4)
(define-fun a () Int
2)
)
However, if I write '(assert (= f (div a b)))' instead of '(assert (= f (/ a b)))', then the result is UNSAT. Why does not div return the same result?
Moreover, and the main concern for me, I did not find a way to use operator '/' in z3 .Net API. I can see only function MkDiv, which actually for operator 'div'. Is there a way so that I can apply operator '/' in the case of z3 .Net API? Thank you in advance.
Strictly speaking neither of these formulas is SMT-LIB2 compliant, because / is a function that takes two Real inputs and produces a Real output, whereas div is a function that takes two Int inputs and produces an Int (see SMT-LIB Theories). Z3 is more relaxed and automatically converts those objects. If we enable the option smtlib2_compliant=true then it will indeed report an error in both cases.
The reason for the div version being unsatisfiable is that there is indeed no solution where f is an integer according to (= f (/ a b)), but there is indeed no integer that satisfies (= f 0.5)

Z3 Integer division not showing correct answer

I'm writing some codes by calling Z3 to calculate division but I found the model result is not correct. Basically what I want to do is to the get value of a and b satisfying a/b == 1. So I manually wrote an input file like following to check whether it's my code's problem or Z3's.
(declare-const a Int)
(declare-const b Int)
(assert (= (div a b) 1))
(assert (not (= b 0)))
(check-sat)
(get-model)
Result from this in my machine is a =77 b = 39 instead of some equalized value of a and b. Is this a bug or did I do something wrong?
Using / instead of div will yield the desired behavior (rise4fun link: http://rise4fun.com/Z3/itdK ):
(declare-const a Int)
(declare-const b Int)
(assert (not (= b 0)))
(push)
(assert (= (div a b) 1)) ; gives a=2473,b=1237
(check-sat)
(get-model)
(pop)
(push)
(assert (= (/ a b) 1))
(check-sat)
(get-model) ; gives a=-1,b=-1
(pop)
However, there may be some confusion here, I didn't see / defined in the integer theory ( http://smtlib.cs.uiowa.edu/theories/Ints.smt2 ) only div (but it would appear to be assumed in the QF_NIA logic http://smtlib.cs.uiowa.edu/logics/QF_NIA.smt2 since / is mentioned to be excluded from QF_LIA http://smtlib.cs.uiowa.edu/logics/QF_LIA.smt2 ), so I was a little confused, or maybe it's related to the recent real/int typing issues brought up here: Why does 0 = 0.5?

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