Z3中smt2文件和OCaml的不同结果

我在SMT2和OCaml中也有同样的问题。使用SMT2文件，我可以在约3分钟内获得unsat结果。但是，OCaml中的相同问题仍然存在。请告知

SMT2问题的原因：

(declare-fun x0 () (_ BitVec 32))
(declare-fun x1 () (_ BitVec 32))
(declare-fun x2 () (_ BitVec 32))
(declare-fun y0 () (_ BitVec 32))
(declare-fun y1 () (_ BitVec 32))
(declare-fun y2 () (_ BitVec 32))

(assert (not (=> 
(and (= (bvadd x2 x1 x0) (bvadd y2 y1 y0))
     (= (bvadd x2 (bvmul #x00000002 x1) (bvmul #x00000003 x0)) 
        (bvadd y2 (bvmul #x00000002 y1) (bvmul #x00000003 y0)))
     (= (bvadd x2 (bvmul #x00000003 x1) (bvmul #x00000006 x0))
        (bvadd y2 (bvmul #x00000003 y1) (bvmul #x00000006 y0))))
     (= (bvadd x2 (bvmul #x00000004 x1) (bvmul #x0000000a x0))
        (bvadd y2 (bvmul #x00000004 y1) (bvmul #x0000000a y0))))))
(check-sat)

OCaml中的相同问题：

let cfg = [("model", "true"); ("proof", "false")] in
let ctx = (mk_context cfg) in
let bv_sort = BitVector.mk_sort ctx 32 in
let c2 = Expr.mk_numeral_int ctx 2 bv_sort in
let c3 = Expr.mk_numeral_int ctx 3 bv_sort in
let c4 = Expr.mk_numeral_int ctx 4 bv_sort in
let c10 = Expr.mk_numeral_int ctx 10 bv_sort in
let c6 = Expr.mk_numeral_int ctx 6 bv_sort in
let x0 = Expr.mk_const ctx (Symbol.mk_string ctx "x0") bv_sort in
let x1 = Expr.mk_const ctx (Symbol.mk_string ctx "x1") bv_sort in
let x2 = Expr.mk_const ctx (Symbol.mk_string ctx "x2") bv_sort in
let y0 = Expr.mk_const ctx (Symbol.mk_string ctx "y0") bv_sort in
let y1 = Expr.mk_const ctx (Symbol.mk_string ctx "y1") bv_sort in
let y2 = Expr.mk_const ctx (Symbol.mk_string ctx "y2") bv_sort in
let ex1 = mk_add ctx (mk_add ctx x0 x1) x2 in
let ey1 = mk_add ctx (mk_add ctx y0 y1) y2 in
let ex2 = mk_add ctx (mk_add ctx (mk_mul ctx c3 x0) (mk_mul ctx x1 c2)) x2 in
let ey2 = mk_add ctx (mk_add ctx (mk_mul ctx c3 y0) (mk_mul ctx y1 c2)) y2 in
let ex3 = mk_add ctx (mk_add ctx (mk_mul ctx c6 x0) (mk_mul ctx x1 c3)) x2 in
let ey3 = mk_add ctx (mk_add ctx (mk_mul ctx c6 y0) (mk_mul ctx y1 c3)) y2 in
let ex4 = mk_add ctx (mk_add ctx (mk_mul ctx c10 x0) (mk_mul ctx x1 c4)) x2 in
let ey4 = mk_add ctx (mk_add ctx (mk_mul ctx c10 y0) (mk_mul ctx y1 c4)) y2 in
let left = Boolean.mk_and ctx [(mk_eq ctx ex1 ey1);(mk_eq ctx ex2 ey2);(mk_eq ctx ex3 ey3)] in
let right = mk_eq ctx ex4 ey4 in
let valid = Boolean.mk_implies ctx left right in
let sat = Boolean.mk_not ctx valid in

print_endline (Z3.Expr.to_string sat);
let solver = (mk_solver ctx None) in
Solver.add solver [sat];
let q = (check solver []) in
match q with
| SATISFIABLE -> print_endline "sat"
| UNSATISFIABLE -> print_endline "unsat"
| UNKNOWN -> print_endline "unknow";

---

这两个输入不完全相同，因为其中一个输入的所有加法器的参数顺序相反，这对性能有影响，因为SAT解算器中的启发式进入不同的轨迹。通过改变参数的顺序，然后使用qfbv策略，我们可以说服OCaml API表现出与SMT2版本相同的性能，例如：

let _ =
  let cfg = [("model", "true"); ("proof", "false")] in
  let ctx = (mk_context cfg) in
  let bv_sort = BitVector.mk_sort ctx 32 in
  let c2 = Expr.mk_numeral_int ctx 2 bv_sort in
  let c3 = Expr.mk_numeral_int ctx 3 bv_sort in
  let c4 = Expr.mk_numeral_int ctx 4 bv_sort in
  let c10 = Expr.mk_numeral_int ctx 10 bv_sort in
  let c6 = Expr.mk_numeral_int ctx 6 bv_sort in
  let x0 = Expr.mk_const ctx (Symbol.mk_string ctx "x0") bv_sort in
  let x1 = Expr.mk_const ctx (Symbol.mk_string ctx "x1") bv_sort in
  let x2 = Expr.mk_const ctx (Symbol.mk_string ctx "x2") bv_sort in
  let y0 = Expr.mk_const ctx (Symbol.mk_string ctx "y0") bv_sort in
  let y1 = Expr.mk_const ctx (Symbol.mk_string ctx "y1") bv_sort in
  let y2 = Expr.mk_const ctx (Symbol.mk_string ctx "y2") bv_sort in
  let ex1 = mk_add ctx x2 (mk_add ctx x1 x0) in
  let ey1 = mk_add ctx y2 (mk_add ctx y1 y0) in
  let ex2 = mk_add ctx x2 (mk_add ctx (mk_mul ctx x1 c2) (mk_mul ctx c3 x0)) in
  let ey2 = mk_add ctx y2 (mk_add ctx (mk_mul ctx y1 c2) (mk_mul ctx c3 y0) ) in
  let ex3 = mk_add ctx x2 (mk_add ctx (mk_mul ctx x1 c3) (mk_mul ctx c6 x0)) in
  let ey3 = mk_add ctx y2 (mk_add ctx (mk_mul ctx y1 c3) (mk_mul ctx c6 y0)) in
  let ex4 = mk_add ctx x2 (mk_add ctx (mk_mul ctx x1 c4) (mk_mul ctx c10 x0)) in
  let ey4 = mk_add ctx y2 (mk_add ctx (mk_mul ctx y1 c4) (mk_mul ctx c10 y0)) in
  let left = Boolean.mk_and ctx [(mk_eq ctx ex1 ey1);
                                 (mk_eq ctx ex2 ey2);
                                 (mk_eq ctx ex3 ey3)] in
  let right = mk_eq ctx ex4 ey4 in
  let valid = Boolean.mk_implies ctx left right in
  let sat = Boolean.mk_not ctx valid in  
  print_endline (Z3.Expr.to_string sat);
  let solver = (mk_solver_s ctx "QF_BV") in 
  (add solver [sat]) ;
  let q = (check solver []) in
  match q with
    | SATISFIABLE -> print_endline "sat"
    | UNSATISFIABLE -> print_endline "unsat"
    | UNKNOWN -> print_endline "unknown";

另请注意，为了解决名为“QF_BV”的逻辑的SMT2问题，Z3将应用名为“qfbv”的策略。在这个特殊的例子中，我们正在为逻辑“QF_BV”构造一个求解器，但是如果我们想要显式地构造一个策略，那么

(mk_tactic ctx "QF_BV")

将失败，并出现“无效参数”异常

以下职位可能与绩效差异有关：

---

能否尝试创建位向量解算器而不是默认解算器？SMT-LIB2模式下的Z3将检测到您的问题仅使用QF_BV逻辑（纯位向量逻辑）。根据您使用的OCAMLAPI版本的不同，它被称为：“mk_solver_for_logic ctx”QF_BV”。新ocaml绑定中的等价物是mk_solver_s ctx“QF_BV”。我试过了，但没有帮助。