In his answer to a earlier question file:///C:/Users/cyril/Downloads/sagemath%20ask/How%20to%20transform%20a%20derived%20set%20of%20inequation%20in%20the%20good%20polyhedron%20format%20-%20ASKSAGE%20Sage%20Q&A%20Forum.html, SLelievre validated the following code.
def one_dimension_less(representationH): x = list(var('x_%i' % i) for i in (0..2)) sol = [solve_ineq(ieq, [x[0]]) for ieq in Ineq] sol_r = flatten([x[1] for x in sol]) ineq_ge = [z for z in sol_r if z.rhs() == x[0]] ineq_le = [z for z in sol_r if z.lhs() == x[0]] ineq_a = [z for z in sol_r if z.lhs() != x[0] and z.rhs() != x[0]] ineq_aa = [(z.lhs()).full_simplify() >= 0 for z in ineq_a] result = flatten(ineq_aa + [ [(ineq_le[j].rhs()-ineq_ge[i].lhs()).full_simplify() >= 0 for i in range(len(ineq_ge))] for j in range(len(ineq_le))]) # result1 = [result[i].factor() for i in range(len(result))] return result
But I realized that I should be able to choose the eliminated variable. So I code as follow :
def elim_x(representationH,elim_var,k):
x = vector(SR, SR.var('x_', 3))
ieqs = [[el.A()*x + el.b() >= 0] for el in DH]
sol = [solve_ineq(ieq, [elim_var]) for ieq in ieqs]
sol_r = flatten([x[1] for x in sol])
ineq_ge = [z for z in sol_r if z.rhs() == elim_var]
ineq_le = [z for z in sol_r if z.lhs() == elim_var]
ineq_a = [z for z in sol_r if z.lhs() != elim_var and z.rhs() != x[0]]
ineq_aa = [(z.lhs()).full_simplify() >= 0 for z in ineq_a]
result = flatten(ineq_aa + [[(ineq_le[j].rhs() - ineq_ge[i].lhs()).full_simplify() >= 0
for i in range(len(ineq_ge))] for j in range(len(ineq_le))])
reslhs=[el.lhs() for el in result]
if k==0 :
return result
if k==1 :
return reslhs
which generate no apparent error, and call this function by :
D = polytopes.dodecahedron()
DH = D.Hrepresentation()
el0=elim_x(DH,[x[0]],1)
el0[0]
but this call generate an error of the type :
'sage.symbolic.expression.Expression' object is not subscriptable
I am sorry not to understand by myself why since it is just a way to give a variable already inside the first function.
