I can't create function for finding fibonacci numbers in CLIPS
i tryed
(deffunction fibo (?a )
( if (> ?a 0) then (+ fibo(- ?a 1) fibo(- ?a 2)) else (0)))
but it's doesn't work
Here is a function of calculating Fibonacci numbers recursively in CLIPS:
(deffunction fibo (?a)
(if (or (= ?a 0) (= ?a 1)) then
?a
else
(+ (fibo(- ?a 1)) (fibo(- ?a 2)))))
I don't understand, what your code does but there is error in it (you've missed brackets):
(deffunction fibo (?a )
( if (> ?a 0) then (+ ( fibo(- ?a 1) ) ( fibo(- ?a 2) )) else (0)))
Related
We're trying to build a solver for a weighted mean function. The code is written in SMT2Lib, and ran by CVC5 or Z3.
However, after adding an assert that indicates the non-negativity of the denominator, the code does not terminate. ("(assert (> tau0_max 0))")
Are there any alternatives to the parameters or the way we can declare the functions differently in order to approach this problem?
(Ints are not suitable to our problem. We'd like to use Rationals, but instead use Reals as they're offered by SMT2Lib).
(set-option :produce-assignments true)
(set-option :produce-models true)
(set-option :produce-proofs true)
(set-logic ALL)
;define data points x_0 x_1 x_2
(declare-fun x (Int) Real)
(declare-fun n () Int)
(define-fun sqr ((x Real)) Real (* x x))
(define-fun max ((x Real) (y Real)) Real (ite (> x y) x y))
;(define-fun c0 ((i Int)) Real 1.0)
(define-fun miu () Real (/ (+ (x 0) (+ (x 1) (+ (x 2) (+ (x 3) (+ (x 4) (+ (x 5) (+ (x 6) (+ (x 7) (+ (x 8) (+ (x 9) (x 10))))))))))) 11.0))
(define-fun miu_c0 () Real (/ (+ (x 0) (+ (x 1) (+ (x 2) (+ (x 3) (+ (x 4) (+ (x 5) (+ (x 6) (+ (x 7) (+ (x 8) (+ (x 9) (+ (x 10) (x 11)))))))))))) 12.0))
(define-fun tau0 ((i Int)) Real (sqr (- (x i) miu_c0)))
(define-fun tau0_max () Real (max (tau0 0) (max (tau0 1) (max (tau0 2) (max (tau0 3) (max (tau0 4) (max (tau0 5) (max (tau0 6) (max (tau0 7) (max (tau0 8) (max (tau0 9) (max (tau0 10) (tau0 11)))))))))))))
**(assert (> tau0_max 0))**
(define-fun c1 ((i Int)) Real (- 1.0 (/ (tau0 i) tau0_max)))
(define-fun miu_c1 () Real (/ (+ (* (c1 0) (x 0)) (+ (* (c1 1) (x 1)) (+ (* (c1 2) (x 2)) (+ (* (c1 3) (x 3)) (+ (* (c1 4) (x 4)) (+ (* (c1 5) (x 5)) (+ (* (c1 6) (x 6)) (+ (* (c1 7) (x 7)) (+ (* (c1 8) (x 8)) (+ (* (c1 9) (x 9)) (+ (* (c1 10) (x 10)) (* (c1 11) (x 11))))))))))))) 12.0))
(define-fun tau1 ((i Int)) Real (sqr (- (x i) miu_c1)))
(define-fun tau1_max () Real (max (tau1 0) (max (tau1 1) (max (tau1 2) (max (tau1 3) (max (tau1 4) (max (tau1 5) (max (tau1 6) (max (tau1 7) (max (tau1 8) (max (tau1 9) (max (tau1 10) (tau1 11)))))))))))))
(define-fun c2 ((i Int)) Real (* (c1 i) (- 1.0 (/ (tau1 i) tau1_max))))
(assert (= (c2 0) 1.0))
(check-sat)
(get-model)
I am encountering a problem in Z3 for which I can't seem to find where it originates from and how to fix it. My goal that for a given certain iteration (for-loop) that is composed of an if-then-else statement at each step; is it possible to achieve a given value k after the loop has finished. This is done without knowing the structure of if. In other words, i need to check every possible mapping of the function (true or false) for each step. More precisely in smt2 format:
(declare-fun a(Int) Int)
(declare-fun b(Int) Int)
(assert (= 1 (a 0)))
(assert (= 1 (b 0)))
(assert (xor (and ( = (a 1) (+ (a 0) (* 2 (b 0)))) (= (b 1) (+ 1 (b 0)))) (and (= (b 1) (+ (a 0) (b 0))) (= (a 1) (+ (b 0) 1)))))
(assert (xor (and ( = (a 2) (+ (a 1) (* 2 (b 1)))) (= (b 2) (+ 2 (b 1)))) (and (= (b 2) (+ (a 1) (b 1))) (= (a 2) (+ (b 1) 2)))))
(assert (xor (and ( = (a 3) (+ (a 2) (* 2 (b 2)))) (= (b 3) (+ 3 (b 2)))) (and (= (b 3) (+ (a 2) (b 2))) (= (a 3) (+ (b 2) 3)))))
(assert (xor (and ( = (a 4) (+ (a 3) (* 2 (b 3)))) (= (b 4) (+ 4 (b 3)))) (and (= (b 4) (+ (a 3) (b 3))) (= (a 4) (+ (b 3) 4)))))
(assert (xor (and ( = (a 5) (+ (a 4) (* 2 (b 4)))) (= (b 5) (+ 5 (b 4)))) (and (= (b 5) (+ (a 4) (b 4))) (= (a 5) (+ (b 4) 5)))))
(assert (xor (and ( = (a 6) (+ (a 5) (* 2 (b 5)))) (= (b 6) (+ 6 (b 5)))) (and (= (b 6) (+ (a 5) (b 5))) (= (a 6) (+ (b 5) 6)))))
(assert (xor (and ( = (a 7) (+ (a 6) (* 2 (b 6)))) (= (b 7) (+ 7 (b 6)))) (and (= (b 7) (+ (a 6) (b 6))) (= (a 7) (+ (b 6) 7)))))
(assert (xor (and ( = (a 8) (+ (a 7) (* 2 (b 7)))) (= (b 8) (+ 8 (b 7)))) (and (= (b 8) (+ (a 7) (b 7))) (= (a 8) (+ (b 7) 8)))))
(assert (xor (and ( = (a 9) (+ (a 8) (* 2 (b 8)))) (= (b 9) (+ 9 (b 8)))) (and (= (b 9) (+ (a 8) (b 8))) (= (a 9) (+ (b 8) 9)))))
(assert (xor (and ( = (a 10) (+ (a 9) (* 2 (b 9)))) (= (b 10) (+ 10 (b 9)))) (and (= (b 10) (+ (a 9) (b 9))) (= (a 10) (+ (b 9) 10)))))
(assert (= (b 10) 461))
(check-sat)
(get-model)
The xor operator is used to check if the statement for then holds or the statement in else holds, but not both. So the the variable a or b is bounded to follow only one valid path. Somehow the values sometimes don't seem to obey this rule or they do not change, and I am unable to tell why is this happening. As for example take this output for a, for the step 2 and 3 the value doesn't change, which should not be possible:
(define-fun a ((x!0 Int)) Int
(ite (= x!0 0) 1
(ite (= x!0 1) 3
(ite (= x!0 2) 7 <--- should not be possible but keeps happening
(ite (= x!0 3) 7 <---
(ite (= x!0 4) 29
[...]
I don't know if i either am encountering a bug or my logic involved in the solution for this problem is flawed. I tried to use Bounded Model Checking. I would appreciate any help!
Issue: the problem is either in your understanding of how the loop should behave, or in the encoding of the formula implementing the logic of the loop. Since you didn't provide the original pseudo-code, I cannot guess any further.
Let's take this:
(assert (xor
(and
(= (a 1) (+ (a 0) (* 2 (b 0))))
(= (b 1) (+ 1 (b 0)))
)
(and
(= (b 1) (+ (a 0) (b 0)))
(= (a 1) (+ (b 0) 1))
)
)
)
The expression that is being unrolled is:
( -- #then-branch
a' := a + 2 * b
/\
b' := K + b
)
xor
( -- #else-branch
a' = K + b
/\
b' = a + b
)
where K depends on the current iteration, starting with 1.
Q: is the solution provided by the SMT solver feasible? YES!
(the portion you have shared with us..)
a_0 := 1
b_0 := 1
-- execute #then-branch (K = 1)
a_1 := a_0 + 2 * b_0 = 1 + 2 * 1 = 3
b_1 := K + b_0 = 1 + 1 = 2
-- execute #then-branch (K = 2)
a_2 := a_1 + 2 * b_1 = 3 + 2 * 2 = 7
b_2 := K + b_1 = 2 + 2 = 4
-- execute #else-branch (K = 3)
a_3 := K + b_2 = 3 + 4 = 7
b_3 := a_2 + b_2 = 7 + 4 = 11
-- execute #then-branch (K = 4)
a_4 := a_3 + 2 * b_3 = 7 + 2 * 11 = 29
b_4 := K + b_3 = 4 + 11 = 15
...
here is the Z3 context:
(forall ((X (Array Int (Array Int Real))) (i Int) (j Int))
(let ((a!1 (* (- 1) (to_int (select (select X i) j)))))
(= (+ (testArr X i j) a!1) 0))))
this means:
forall X,i,j. testArr(X,i,j) == X[i][j]
now, i try to prove two similar assert:
1 (=> (= (select (select Z i1) j1) (select (select X i) j))
(= (select (select Z i1) j1) (testArr X i j)))
and
2 (=> (= v (select (select X i) j))
(= v (to_real (testArr X i j))))
the second assert Z3 return valid, but the 1st assert Z3 return invalid,and in other example Z3 sometimes return unknown.
why??
I have such an expression
z3::expr expr = (exists ((flag Bool))
(exists ((w Int))
(exists ((i Int))
(exists ((counter Int))
(and (= i (+ x y z))
flag
(not (= i 0))
(= i counter)
(not (= w counter))
(>= i 1)
(= w 0)))))))
I use C++ API to do quantifier elimination and simplification.
z3::goal g(ctx);
g.add(expr);
z3::apply_result result = (qe & simplify & propagate_ineqs)(g);
I have defined the tactics.
I want to get such result:
(or (>= (+ x y z) 1 )
(<= (+ x y z) -1 ))
but I get this output which is appropriate for my application:
(let ((a!1 (or (<= (+ (* (- 1) x)
(* (- 1) y)
(* (- 1) z))
(- 1))
(<= (+ x y z) (- 1)))))
(and (or (<= (+ x y z) (- 1))
(>= (+ x y z) 1))
a!1
(>= (+ x y z) 1)))
what tactics should I use to make it work as I want?
I solved the issue by setting these options in the context, before defining the goal:
ctx.set(":pp-min-alias-size", 1000000);
ctx.set(":pp-max-depth", 1000000);
it didn't simplify as I wanted, but removed the "let"s and that's better than nothing!
I'm trying to implement the String.Containts function.
I have written some simple tests case, but the following one (which should return UNSAT) return SAT.
The test try to match the substring 'bd' in the string 'abcd' by comparig all possible substrings with the string wanted (text taken from the Z3 output):
{(exists ((i Int))
(let ((a!1 (forall ((k Int))
(and (>= k i)
(<= k (+ i 1))
(= (select stringToScan k) (select substring (- k i)))
(= (select stringToScan 0) #x0061)
(= (select stringToScan 1) #x0062)
(= (select stringToScan 2) #x0063)
(= (select stringToScan 3) #x0064)
(= (select stringToSearch 0) #x0062)
(= (select stringToSearch 1) #x0064)))))
(and (>= i 0)
(< i 2)
a!1
(= (select substring 0) (select stringToSearch 0))
(= (select substring 1) (select stringToSearch 1)))))}
I have tried various implementation, but without any success.
If you assert the formula, it returns UNSAT.
http://rise4fun.com/Z3/szN
The part:
(forall ((k Int))
(and (>= k i)
(<= k (+ i 1)) ...)))
is false becuase you can set k to i + 2 or i - 1.
You probably mean to write an implication instead of a conjunction.
Sometimes using arrays for strings is not the best way to perform the encoding.
The automata toolkit (see: http://rise4fun.com/Rex ) uses sequences.