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I want to check the value of a, b, c, and if value 'a' equals to 1, 'x' is added one. We continue the process for values 'b' and 'c'.
So if a=1, b=1, c=1, the result of x should be 3.
if a=1, b=1, c=0, so the result of x should be 2.
Any methods to be implemented in z3?
The source code looks like this:
from z3 import *
a, b, c = Ints('a b c')
x, y = Ints('x y')
s = Solver()
s.add(If(a==1, x=x + 1, y = y-1))
s.add(If(b==1, x=x + 1, y = y-1))
s.add(If(c==1, x=x + 1, y = y-1))
s.check()
print s.model()
Any suggestions about what I can do?
This sort of "iterative" processing is usually modeled by unrolling the assignments and creating what's known as SSA form. (Static single assignment.) In this format, every variable is assigned precisely once, but can be used many times. This is usually done by some underlying tool as it is rather tedious, but you can do it by hand as well. Applied to your problem, it'd look something like:
from z3 import *
s = Solver()
a, b, c = Ints('a b c')
x0, x1, x2, x3 = Ints('x0 x1 x2 x3')
s.add(x0 == 0)
s.add(x1 == If(a == 1, x0+1, x0))
s.add(x2 == If(b == 1, x1+1, x1))
s.add(x3 == If(c == 1, x2+1, x2))
# Following asserts are not part of your problem, but
# they make the output interesting
s.add(b == 1)
s.add(c == 0)
# Find the model
if s.check() == sat:
m = s.model()
print("a=%d, b=%d, c=%d, x=%d" % (m[a].as_long(), m[b].as_long(), m[c].as_long(), m[x3].as_long()))
else:
print "no solution"
SSA transformation is applied to the variable x, creating as many instances as necessary to model the assignments. When run, this program produces:
a=0, b=1, c=0, x=1
Hope that helps!
Note that z3 has many functions. One you could use here is Sum() for the sum of a list. Inside the list you can put simple variables, but also expression. Here an example for both a simple and a more complex sum:
from z3 import *
a, b, c = Ints('a b c')
x, y = Ints('x y')
s = Solver()
s.add(a==1, b==0, c==1)
s.add(x==Sum([a,b,c]))
s.add(y==Sum([If(a==1,-1,0),If(b==1,-1,0),If(c==1,-1,0)]))
if s.check() == sat:
print ("solution:", s.model())
else:
print ("no solution possible")
Result:
solution: [y = 2, x = 2, c = 1, b = 0, a = 1]
If your problem is more complex, using BitVecs instead of Ints can make it run a little faster.
edit: Instead of Sum() you could also simply use addition as in
s.add(x==a+b+c)
s.add(y==If(a==1,-1,0)+If(b==1,-1,0)+If(c==1,-1,0))
Sum() makes sense towards readability when you have a longer list of variables, or when the variables already are in a list.
Assume that I have the following statements:
A p B, A p C, B p C ( p is a symmetric property, i.e. B p A, C p A and C p B)
A v 2, B v 1, C v 1,
I want to use a rule to do something like:
?a p all(?b)
if ?b v 1
than ?a q 'Yes'
that means that you can infer (A q 'Yes'), but B can't since B p A and A v 2(although B p C and C v 1).
[rule: (?a eg:p ?b), (?b eg:v 1) -> (?a eg:q 'Yes')]
I've used the above rule in Jena, but I got A,B,C eg:q 'Yes', which is wrong.
Any help will be greatly appreciated.
Update (originally posted as an answer)
the meaning of (?a p all(?b)) is that I like to get a set which all ?mem in this set fulfill the (?a p ?mem). And all member must fulfill (?mem v 1) to infer (?a q 'Yes').
For example,
A p B and A p C,so I get a set which contains (B, C).since B and C v 1,so A q 'Yes.
B p A and B p C,so I get a set(A, C),but A v 2,so can't infer that B q 'Yes'.
Problem Solved
Thanks to Joshua Taylor.
Firstly, these two rules can't use at the same time.The rule2 should be used after rule1.
And, the rule2 should be [rule2: (?s ?p ?o) noValue(?s, connectedToNonOne) -> (?s q 'Yes')].
but I got A,B,C eg:q 'Yes', which is wrong.
The rule you have actually written in Jena says
For any two individuals X and Y, if (X p Y) and (Y v 1) then (X q 'Yes').
From the rule you've written, this is correct, by:
(A p C), (C v 1) → (A q 'Yes')
(B p C), (C v 1) → (B q 'Yes')
(C p B), (B v 1) → (C q 'Yes')
What you're actually trying to say is:
For any individual X, if for every individual Y, (X p Y) implies (Y v 1), then (X q 'Yes').
In first order logic, your original rule could be written as:
∀ x,y ([p(x,y) ∧ v(y,1)] → q(x,'yes')
What you're actually trying to capture would be:
∀x[(∀y[p(x,y) → v(y,1)]) → q(x,'yes')]
That's harder to capture in Jena rules, because to check whether (∀y[p(x,y) → v(y,1)]) holds or not, all Jena can do is check whether there are currently any counterexamples. If one were added later, you might have incorrect inferences.
Using the builtins available in the rule reasoner, you could do something with noValue and notEqual along the lines of:
#-- If an individual is disqualified by being
#-- connected to a something that is connected
#-- to something that is not equal to 1, then
#-- add a connectedToNonOne triple.
[rule1:
(?x p ?y), (?y v ?z), notEqual(?z,1)
->
(?x connectedToNonOne true)]
#-- Mark everything that is *not* disqualified
#-- with `q 'Yes'`.
[rule2:
noValue(?x, connectedToNonOne)
->
(?x q 'Yes')
Suppose you have a function defined by intervals, such as
f(x):=block(if x<0 then x^2 else x^3);
When we differentiate it with
diff(f(x),x);
we get
d/dx (if x<0 then x^2 else x^3)
whereas I'd like to get
(if x<0 then 2*x else 3*x^2)
Is there a way to obtain such result?
This may help in a simple case:
(%i1) f(x):= charfun(x<0)*x^2 + charfun(x>=0)*x^3$
(%i2) gradef(charfun(y), 0)$
(%i3) diff(f(x),x);
2
(%o3) 2 x charfun(x < 0) + 3 x charfun(x >= 0)
charfun, gradef
You can try also Pw.mac package from Richard Hennessy.
Here's a different approach using a simplification rule for "if" expressions. The unsolved part here is to detect discontinuities and generate delta functions for those locations. If you want to ignore those, you can define FOO to return 0. Note that I didn't attempt to implement the function discontinuities; that part is unsolved here. I can give it a try if there is interest.
(%i1) display2d : false $
(%i2) matchdeclare ([aa, bb, cc], all, xx, symbolp) $
(%i3) 'diff (if aa then bb else cc, xx) $
(%i4) tellsimpafter (''%, apply ("if", [aa, diff (bb, xx), true, diff (cc, xx)]) + FOO (aa, bb, cc, xx)) $
(%i5) FOO (a, b, c, x) := 'lsum ((ev (c, x = d) - ev (b, x = d)) * delta (d, x), d, discontinuities (a, x)) $
(%i6) diff (if x > 0 then x^2 else x^3, x);
(%o6) (if x > 0 then 2*x else 3*x^2)+'lsum((d^3-d^2)*delta(d,x),d,
discontinuities(x > 0,x))
Building on slitinov's answer I wrote this quite naive implementation for functions with more than two "pieces":
gradef(charfun(dummy),0)$
/* piecewise function definition */
itv: [[x<0],[x>=0,x<1], [x>=1]]; /* intervals */
fi: [ 1, x^2+1, 2*x ]; /* local functions */
/* creation of global function f and its derivative df */
f:0;
for i: 1 thru 3 do f:f+charfun(apply("and",itv[i]))*fi[i];
df:diff(f,x);
/* display of local functions and derivatives */
for i: 1 thru 3 do (
apply(assume,itv[i]),
newline(),
print(itv[i]),
print("f = ",ev(f)),
print("df = ",ev(df)),
apply(forget,itv[i])
);
plot2d([f,df],[x,-2,3],[y,-1,5],[style,[lines,4,3],[lines,2,2]]);
I am still struggling with the problem of findiong a value so that a * b == b with all value of b. The expected result is a == 1. I have two solutions below.
(A) I implemented this with ForAll quantifier in below code (correct me if there is a solution without using any quantifier). The idea is to prove f and g are equivalent.
from z3 import *
a, b, a1, tmp1 = BitVecs('a b a1 tmp1', 32)
f = True
f = And(f, tmp1 == b)
f = And(f, a1 == a * tmp1)
g= True
g = And(g, a1 == b)
s = Solver()
s.add(ForAll([b, tmp1, a1], f == g))
if s.check() == sat:
print 'a =', s.model()[a]
else:
print 'Unsat'
However, this simple code runs forever without giving back result. I think that is because of the ForAll. Any idea on how to fix the problem?
(B) I tried again with another version. This time I dont prove two formulas to be equivalent, but put them all into one formula f. Logically, I think this is true, but please correct me if I am wrong here:
from z3 import *
a, b, a1, tmp = BitVecs('a b a1 tmp', 32)
f = True
f = And(f, tmp == b)
f = And(f, a1 == a * tmp)
f = And(f, a1 == b)
s = Solver()
s.add(ForAll([b, a1], f))
if s.check() == sat:
print 'a =', s.model()[a]
else:
print 'Unsat'
This time the code does not hang, but immediately returns 'Unsat'. Any idea on how to fix this?
Thanks a lot.
The direct formulation of your problem provides the answer you were expecting: http://rise4fun.com/Z3Py/N07sW
The versions you propose use auxiliary variables, a1, tmp, tmp1 and you
use universal quantification over these variables. This does not correspond
to the formula you intended.
In Lisp, I can have:
(a b c d e f g)
which means
look up b, c, d, e, f, g
look up a; apply value of a to above
Then, I can also have:
`(a b c d e f g)
which is the equiv to
(list 'a 'b 'c 'd 'e 'f 'g)
Now, in lua, I can have:
[snipplet 1]
foo = {
Foo,
{Cat, cat},
{Dog, dog}
};
Which ends up most likely expanding into:
{ nil, { nil, nil}, {nil, nil}}
Whereas what I really want is something like:
[snipplet 2]
{ "Foo", {"Cat", "cat"}, {"Dog", "dog"}}
Is there some backquote-like method in lua?
I find [snipplet 1] to be more visually pleasing than [snipplet 2], but what I mean is snipplet 2.
Thanks!
There's no syntax for that but you can try this run-time solution:
setmetatable(_G,{__index=function (t,k) return k end})
This makes all undefined variables behave as if they contained strings with their names.
Consider a function that parses a string containing the Lisp syntax and constructs a table from that.