Hello everyone,
I wanted to know if the SAGE builtin ANF2CNF converter (I am using the
sparse strategy implementation) can be modified so as to display a limited
number of clauses of the given ANF expression, say 'n' number of clauses (n
must be input to the converter, along with the ANF expre
ation.
On Wednesday, March 13, 2013 3:36:09 PM UTC+5:30, akhil wrote:
>
> Hello,
>
>
> I am using SAGE builtin anf2cnf.py converter, but the DIMACS output
> returned is incorrect, when verified by multiplying the CNF clauses. Sample
> code:
>
> B. = BooleanPo
Hello,
I am using SAGE builtin anf2cnf.py converter, but the DIMACS output
returned is incorrect, when verified by multiplying the CNF clauses. Sample
code:
B. = BooleanPolynomialRing()
from sage.sat.converters.polybori import CNFEncoder
from sage.sat.solvers.dimacs import DIMACS
fn = tmp_file
keys in l}
The result will be the dictionary d1 containing only (key,value) pairs
corresponding to keys x1,x100.
Regards,
AKHIL.
>
>
>
>
>
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Thanks a lot for your prompt help.
Regards,
AKHIL.
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To po
ease help.
Thanks and regards,
AKHIL.
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To post t
x3, x5:x0})
x0 + x1*x2 + x3#Answer returned by SAGE
If you have a single substitution to make, e.g replace only x5 by x0, there
is no need to pass a dictionary.
sage:f.subs(x5 = x0)
x0 + x4
I hope it is clear.
Regards,
AKHIL.
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Output:
x0*x1*x2 + x1*x2*x3
True
The above code was written in online notebook version of SAGE.
Thanks,
AKHIL.
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Thanks a lot Mr. Martin and Mr. Michael for your help.
On Sunday, September 9, 2012 12:01:45 PM UTC+5:30, akhil wrote:
> Hello,
>
>
> Given a set of m linear equations in n unknowns, how do I use SAGE to give
> me the coefficient matrix* A*(size m *n) and column vector *b*(si
Hello,
Given a set of m linear equations in n unknowns, how do I use SAGE to get
the coefficient matrix A (of size m * n) and the vector b, such that Ax =
b, where x is the vector containing unknowns to be solved ?
Thanks a lot.
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Hello,
Given a set of m linear equations in n unknowns, how do I use SAGE to give
me the coefficient matrix* A*(size m *n) and column vector *b*(size m*1)
such that Ax = b; where *x*(size n*1) is the vector of unknowns ?
Thanks a lot.
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