问题
In Complex numbers in Z3 Leonardo de Moura was able to introduce and to compute with complex numbers in Z3.
Using the code proposed by Leonardo I am introducing and computing with quaternion numbers in Z3 according with the code presented here . Using this "quaternion " code I am solving the following problem:
x = Quaternion("x")
s = Tactic('qfnra-nlsat').solver()
s.add(x*x + 30 == 0, x.i3 > 0, x.i2 >0, x.i1 > 0)
print(s.check())
m = s.model()
print m
and the corresponding output is:
sat
[x.r = 0, x.i1 = 1, x.i2 = 1, x.i3 = 5.2915026221?]
This result was verified using Maple.
Other example:
x = Quaternion("x")
y = Quaternion("y")
z = Quaternion("z")
s = Tactic('qfnra-nlsat').solver()
s.add(x*y + 30 + x + y*z == 0, x - y + z == 10)
print(s.check())
m = s.model()
print m
and the output is:
sat
[y.r = 1/8,
z.r = 2601/64,
y.i1 = 1/2,
z.i1 = 45/8,
y.i2 = -1/2,
z.i2 = -45/8,
y.i3 = -1/2,
z.i3 = -45/8,
x.i3 = 41/8,
x.i2 = 41/8,
x.i1 = -41/8,
x.r = -1953/64]
Other example:
Proving that
x * y != y * x
Code:
x = Quaternion("x")
y = Quaternion("y")
a1, b1, c1, d1 = Reals('a1 b1 c1 d1')
a2, b2, c2, d2 = Reals('a2 b2 c2 d2')
x.r = a1
x.i1 = b1
x.i2 = c1
x.i3 = d1
y.r = a2
y.i1 = b2
y.i2 = c2
y.i3 = d2
print simplify((x * y - y * x).r)
print simplify((x * y - y * x).i1)
print simplify((x * y - y * x).i2)
print simplify((x * y - y * x).i3)
Output:
0
2·c2·d1 + -2·c1·d2
-2·b2·d1 + 2·b1·d2
2·b2·c1 + -2·b1·c2
Other example : Proving that the quaternions
A = (1+ I)/sqrt(2),
B =(1 + J)/sqrt(2),
C = (1 + K)/sqrt(2)
generate a representation of the Braid Group, it is to say, we have that
ABA = BAB, ACA = CAC, BCB = CBC.
Code:
A = Quaternion('A')
B = Quaternion('B')
C = Quaternion('C')
A.r = 1/Sqrt(2)
A.i1 = 1/Sqrt(2)
A.i2 = 0
A.i3 = 0
B.r = 1/Sqrt(2)
B.i1 = 0
B.i2 = 1/Sqrt(2)
B.i3 = 0
C.r = 1/Sqrt(2)
C.i1 = 0
C.i2 = 0
C.i3 = 1/Sqrt(2)
print simplify((A*B*A-B*A*B).r)
print simplify((A*B*A-B*A*B).i1)
print simplify((A*B*A-B*A*B).i2)
print simplify((A*B*A-B*A*B).i3)
print "Proved : ABA = BAB:"
print simplify((A*C*A-C*A*C).r)
print simplify((A*C*A-C*A*C).i1)
print simplify((A*C*A-C*A*C).i2)
print simplify((A*C*A-C*A*C).i3)
print "Proved : ACA = CAC:"
print simplify((B*C*B-C*B*C).r)
print simplify((B*C*B-C*B*C).i1)
print simplify((B*C*B-C*B*C).i2)
print simplify((B*C*B-C*B*C).i3)
print "Proved : BCB = CBC:"
Output:
0
0
0
0
Proved : ABA = BAB.
0
0
0
0
Proved : ACA = CAC.
0
0
0
0
Proved : BCB = CBC.
Other example: Proving that
x / x = 1
for all invertible quaternion:
Code:
x = Quaternion("x")
a, a1, a2, a3 = Reals('a a1 a2 a3')
x.r = a
x.i1 = a1
x.i2 = a2
x.i3 = a3
s = Solver()
s.add(Or(a != 0, a1 != 0, a2 != 0, a3 != 0), Not((x/x).r == 1))
print s.check()
s1 = Solver()
s1.add(Or(a != 0, a1 != 0, a2 != 0, a3 != 0), Not((x/x).i1 == 0))
print s1.check()
s2 = Solver()
s2.add(Or(a != 0, a1 != 0, a2 != 0, a3 != 0), Not((x/x).i2 == 0))
print s2.check()
s3 = Solver()
s3.add(Or(a != 0, a1 != 0, a2 != 0, a3 != 0), Not((x/x).i3 == 0))
print s3.check()
Output:
unsat
unsat
unsat
unsat
Please let me know what do you think about the "quaternion" code and how the "quaternion" code can be improved. Many thanks.
来源：https://stackoverflow.com/questions/16595447/how-to-compute-with-quaternion-numbers-in-z3