Reputation: 31800
I'm attempting to solve some simple Boolean satisfiability problems in Sympy. Here, I tried to solve a constraint that contains the Or
logic operator:
from sympy import *
a,b = symbols("a b")
print(solve(Or(Eq(3, b*2), Eq(3, b*3))))
# In other words: (3 equals b*2) or (3 equals b*3)
# [1,3/2] was the answer that I expected
Surprisingly, this leads to an error instead:
TypeError: unsupported operand type(s) for -: 'Or' and 'int'
I can work around this problem using Piecewise
, but this is much more verbose:
from sympy import *
a,b = symbols("a b")
print(solve(Piecewise((Eq(3, b*2),Eq(3, b*2)), (Eq(3, b*3),Eq(3, b*3)))))
#prints [1,3/2], as expected
Unfortunately, this work-around fails when I try to solve for two variables instead of one:
from sympy import *
a,b = symbols("a b")
print(solve([Eq(a,3+b),Piecewise((Eq(b,3),Eq(b,3)), (Eq(b,4),Eq(b,4)))]))
#AttributeError: 'BooleanTrue' object has no attribute 'n'
Is there a more reliable way to solve constraints like this one in Sympy?
Upvotes: 1
Views: 2855
Reputation: 91470
To expand on zaq's answer, SymPy doesn't recognize logical operators in solve
, but you can use the fact that
a*b = 0
is equivalent to
a = 0 OR b = 0
That is, multiply the two equations
solve((3 - 2*b)*(3 - 3*b), b)
As an additional note, if you wanted to use AND instead of OR, you can solve for a system. That is,
solve([eq1, eq2])
is equivalent to solving
eq1 = 0 AND eq2 = 0
Upvotes: 5
Reputation:
Every equation can be expressed as something being equated to 0. For example, 3-2*b = 0 instead of 3 = 2*b. (In Sympy, you don't even have to write the =0
part, it's assumed.) Then you can simply multiply equations to express the OR logic:
>>> from sympy import *
>>> a,b = symbols("a b")
>>> solve((3-b*2)*(3-b*3))
[1, 3/2]
>>> solve([a-3-b, (3-b*2)*(3-b*3)])
[{b: 1, a: 4}, {b: 3/2, a: 9/2}]
Upvotes: 4