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 | 1 | initial version |
Not sure I completely understand the question.
The following should at least be a step in the right direction.
Define one_dimension_less as intended in the question.
def one_dimension_less(representationH):
x = vector(SR, SR.var('x_', 3))
ieqs = [[el.A()*x + el.b() >= 0] for el in DH]
sol = [solve_ineq(ieq, [x[0]]) for ieq in ieqs]
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))])
return result
Use it:
sage: D = polytopes.dodecahedron()
sage: DH = D.Hrepresentation()
sage: PD1 = one_dimension_less(DH)
sage: PD1
[-x_2*(sqrt(5) + 1) - 2*x_1 + 4 >= 0,
-x_2*(sqrt(5) + 1) + 2*x_1 + 4 >= 0,
x_2*(sqrt(5) + 1) - 2*x_1 + 4 >= 0,
x_2*(sqrt(5) + 1) + 2*x_1 + 4 >= 0,
-x_1*(sqrt(5) + 1) + 4 >= 0,
4 >= 0,
-(2*x_1*(sqrt(5) + 2) + x_2*(sqrt(5) + 1) - 4*sqrt(5) - 8)/(sqrt(5) + 3) >= 0,
-(2*x_1*(sqrt(5) + 2) - x_2*(sqrt(5) + 1) - 4*sqrt(5) - 8)/(sqrt(5) + 3) >= 0,
4 >= 0,
x_1*(sqrt(5) + 1) + 4 >= 0,
(2*x_1*(sqrt(5) + 2) - x_2*(sqrt(5) + 1) + 4*sqrt(5) + 8)/(sqrt(5) + 3) >= 0,
(2*x_1*(sqrt(5) + 2) + x_2*(sqrt(5) + 1) + 4*sqrt(5) + 8)/(sqrt(5) + 3) >= 0,
-(2*x_1*(sqrt(5) + 2) + x_2*(sqrt(5) + 1) - 4*sqrt(5) - 8)/(sqrt(5) + 3) >= 0,
(2*x_1*(sqrt(5) + 2) - x_2*(sqrt(5) + 1) + 4*sqrt(5) + 8)/(sqrt(5) + 3) >= 0,
-2*(x_2*(sqrt(5) + 1) - 2*sqrt(5) - 2)/(sqrt(5) + 3) >= 0,
8/(sqrt(5) + 1) >= 0,
-(2*x_1*(sqrt(5) + 2) - x_2*(sqrt(5) + 1) - 4*sqrt(5) - 8)/(sqrt(5) + 3) >= 0,
(2*x_1*(sqrt(5) + 2) + x_2*(sqrt(5) + 1) + 4*sqrt(5) + 8)/(sqrt(5) + 3) >= 0,
8/(sqrt(5) + 1) >= 0,
2*(x_2*(sqrt(5) + 1) + 2*sqrt(5) + 2)/(sqrt(5) + 3) >= 0]
Examine one of the inequalities:
sage: ieq = PD1[0]
sage: ieq
-x_2*(sqrt(5) + 1) - 2*x_1 + 4 >= 0
Grab its left hand side:
sage: lhs = ieq.lhs()
sage: lhs
-x_2*(sqrt(5) + 1) - 2*x_1 + 4
Get the constant coefficient:
sage: lhs(x_0=0, x_1=0, x_2=0)
4
Same with a substitution dictionary:
sage: lhs({x_0: 0, x_1: 0, x_2: 0})
4
Get the coefficients of the variables:
sage: lhs.coefficient(x_0)
0
sage: lhs.coefficient(x_1)
-2
sage: lhs.coefficient(x_2)
-sqrt(5) - 1
All in one go:
sage: x = vector(SR.var('x_', 3))
sage: A = vector(lhs.coefficient(xi) for xi in x)
sage: A
(0, -2, -sqrt(5) - 1)
sage: b = lhs({xi: 0 for xi in x})
sage: b
4
Reform the inequation:
sage: ieq_tuple = (b,) + tuple(A)
sage: ieq_tuple
(4, 0, -2, -sqrt(5) - 1)
Make that into a function:
def tuple_from_ieq(ieq, x):
r"""
Return the tuple for an inequation.
INPUT:
- ``ieq`` -- an inequation, assumed to be of the form
``a_0 * x_0 + ... + a_n * x_n + b >= 0``
- ``x`` -- a multidimensional variable
as a list, tuple or vector ``(x_0, ..., x_n)``
OUTPUT: the tuple ``(b, a_0, ..., a_n)``.
"""
b = lhs({xi: 0 for xi in x})
A = vector(lhs.coefficient(xi) for xi in x)
return (b,) + tuple(A)
Use it:
sage: x = vector(SR.var('x_', 3))
sage: tuple_from_ieq(PD1[0])
(4, 0, -2, -sqrt(5) - 1)
| | 2 | No.2 Revision |
Not sure I completely understand the question.
The following should at least be a step in the right direction.
Define one_dimension_less as intended in the question.
def one_dimension_less(representationH):
x = vector(SR, SR.var('x_', 3))
ieqs = [[el.A()*x + el.b() >= 0] for el in DH]
sol = [solve_ineq(ieq, [x[0]]) for ieq in ieqs]
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))])
return result
Use it:
sage: D = polytopes.dodecahedron()
sage: DH = D.Hrepresentation()
sage: PD1 = one_dimension_less(DH)
sage: PD1
[-x_2*(sqrt(5) + 1) - 2*x_1 + 4 >= 0,
-x_2*(sqrt(5) + 1) + 2*x_1 + 4 >= 0,
x_2*(sqrt(5) + 1) - 2*x_1 + 4 >= 0,
x_2*(sqrt(5) + 1) + 2*x_1 + 4 >= 0,
-x_1*(sqrt(5) + 1) + 4 >= 0,
4 >= 0,
-(2*x_1*(sqrt(5) + 2) + x_2*(sqrt(5) + 1) - 4*sqrt(5) - 8)/(sqrt(5) + 3) >= 0,
-(2*x_1*(sqrt(5) + 2) - x_2*(sqrt(5) + 1) - 4*sqrt(5) - 8)/(sqrt(5) + 3) >= 0,
4 >= 0,
x_1*(sqrt(5) + 1) + 4 >= 0,
(2*x_1*(sqrt(5) + 2) - x_2*(sqrt(5) + 1) + 4*sqrt(5) + 8)/(sqrt(5) + 3) >= 0,
(2*x_1*(sqrt(5) + 2) + x_2*(sqrt(5) + 1) + 4*sqrt(5) + 8)/(sqrt(5) + 3) >= 0,
-(2*x_1*(sqrt(5) + 2) + x_2*(sqrt(5) + 1) - 4*sqrt(5) - 8)/(sqrt(5) + 3) >= 0,
(2*x_1*(sqrt(5) + 2) - x_2*(sqrt(5) + 1) + 4*sqrt(5) + 8)/(sqrt(5) + 3) >= 0,
-2*(x_2*(sqrt(5) + 1) - 2*sqrt(5) - 2)/(sqrt(5) + 3) >= 0,
8/(sqrt(5) + 1) >= 0,
-(2*x_1*(sqrt(5) + 2) - x_2*(sqrt(5) + 1) - 4*sqrt(5) - 8)/(sqrt(5) + 3) >= 0,
(2*x_1*(sqrt(5) + 2) + x_2*(sqrt(5) + 1) + 4*sqrt(5) + 8)/(sqrt(5) + 3) >= 0,
8/(sqrt(5) + 1) >= 0,
2*(x_2*(sqrt(5) + 1) + 2*sqrt(5) + 2)/(sqrt(5) + 3) >= 0]
Examine one of the inequalities:
sage: ieq = PD1[0]
sage: ieq
-x_2*(sqrt(5) + 1) - 2*x_1 + 4 >= 0
Grab its left hand side:
sage: lhs = ieq.lhs()
sage: lhs
-x_2*(sqrt(5) + 1) - 2*x_1 + 4
Get the constant coefficient:
sage: lhs(x_0=0, x_1=0, x_2=0)
4
Same with a substitution dictionary:
sage: lhs({x_0: 0, x_1: 0, x_2: 0})
4
Get the coefficients of the variables:
sage: lhs.coefficient(x_0)
0
sage: lhs.coefficient(x_1)
-2
sage: lhs.coefficient(x_2)
-sqrt(5) - 1
All in one go:
sage: x = vector(SR.var('x_', 3))
sage: A = vector(lhs.coefficient(xi) for xi in x)
sage: A
(0, -2, -sqrt(5) - 1)
sage: b = lhs({xi: 0 for xi in x})
sage: b
4
Reform the inequation:
sage: ieq_tuple = (b,) + tuple(A)
sage: ieq_tuple
(4, 0, -2, -sqrt(5) - 1)
Make that into a function:
def tuple_from_ieq(ieq, x):
r"""
Return the tuple for an inequation.
INPUT:
- ``ieq`` -- an inequation, assumed to be of the form
``a_0 * x_0 + ... + a_n * x_n + b >= 0``
- ``x`` -- a multidimensional variable
as a list, tuple or vector ``(x_0, ..., x_n)``
OUTPUT: the tuple ``(b, a_0, ..., a_n)``.
"""
b = lhs({xi: 0 for xi in x})
A = vector(lhs.coefficient(xi) for xi in x)
return (b,) + tuple(A)
Use it:
sage: x = vector(SR.var('x_', 3))
sage: tuple_from_ieq(PD1[0])
t = tuple_from_ieq(PD1[0], x)
sage: t
(4, 0, -2, -sqrt(5) - 1)
Convert to algebraic numbers:
sage: tt = vector(AA, t)
sage: tt
(4, 0, -2, -3.236067977499790?)
sage: Polyhedron(ieqs=[tt])
A 3-dimensional polyhedron in AA^3 defined as
the convex hull of 1 vertex, 1 ray, 2 lines
