def test_shape_4():
# xi has incompatible shape
A = sp.array([[-1.0, 0], [0, -1], [1, 1]])
b = sp.array([[0.0, 0, 1]])
xi = sp.array([0.25, 0.1, 0.1])
with pytest.raises(ValueError):
artools.out_region(xi, A, b)
def test_on_2():
# on an extreme point
A = sp.array([[-1.0, 0], [0, -1], [1, 1]])
b = sp.array([0.0, 0, 1])
xi = sp.array([1.0, 0])
assert artools.out_region(xi, A, b) is False
def test_on_1():
# on a 2-D hyperplane
A = sp.array([[-1.0, 0], [0, -1], [1, 1]])
b = sp.array([0.0, 0, 1])
xi = sp.array([0.5, 0.5])
assert artools.out_region(xi, A, b) is False
def test_out_3():
# simple 3-D case
A = sp.array([[-1.0, 0, 0], [0, -1, 0], [0, 0, -1], [1, 1, 1]])
b = sp.array([0.0, 0, 0, 1])
xi = sp.array([0.7, 0.7, 0.7])
assert artools.out_region(xi, A, b)
def test_out_2():
# positive space
A = sp.array([[-1.0, 0], [0, -1], [1, 1]])
b = sp.array([0.0, 0, 1])
xi = sp.array([2.0, 2])
assert artools.out_region(xi, A, b)
def test_out_1():
# negative space
A = sp.array([[-1.0, 0], [0, -1], [1, 1]])
b = sp.array([0.0, 0, 1])
xi = sp.array([-1.0, -1])
assert artools.out_region(xi, A, b)
def test_in_3():
# simple 3-D case
A = sp.array([[-1.0, 0, 0], [0, -1, 0], [0, 0, -1], [1, 1, 1]])
b = sp.array([0.0, 0, 0, 1])
xi = sp.array([0.25, 0.1, 0.25])
assert artools.out_region(xi, A, b) is False
def test_in_2():
# simple 2-D triangle case
A = sp.array([[-1.0, 0], [0, -1], [1, 1]])
b = sp.array([0.0, 0, 1])
xi = sp.array([0.25, 0.1])
assert artools.out_region(xi, A, b) is False
def test_shape_2():
# b is a 2-D row vector
A = sp.array([[-1.0, 0], [0, -1], [1, 1]])
b = sp.array([[0.0, 0, 1]])
xi = sp.array([-0.25, -0.1])
assert artools.out_region(xi, A, b)
def test_tol_in_1():
# slightly in the region at the origin
A = sp.array([[-1.0, 0], [0, -1], [1, 1]])
b = sp.array([0.0, 0, 1])
tol = 1e-8
xi = sp.array([0.0 + tol, 0.0 + tol])
assert artools.out_region(xi, A, b, tol=tol) is False
def test_tol_out_1():
# slightly out of the region at the origin, and not within the accepted
# tolerance
A = sp.array([[-1.0, 0], [0, -1], [1, 1]])
b = sp.array([0.0, 0, 1])
tol = 1e-10
xi = sp.array([0.0 - 2 * tol, 0.0 + tol])
assert artools.out_region(xi, A, b, tol=tol)
def test_tol_in_2():
# slightly out of the region at the origin, but still within the accepted
# tolerance
A = sp.array([[-1.0, 0], [0, -1], [1, 1]])
b = sp.array([0.0, 0, 1])
tol = 1e-8
xi = sp.array([0.0 - tol, 0.0 - tol])
assert artools.out_region(xi, A, b, tol=tol) is False
def con2vert(A, b):
'''
Compute the V-representation of a convex polytope from a set of hyperplane
constraints. Solve the vertex enumeration problem given inequalities of the
form A*x <= b
Parameters:
A
b
Returns:
Vs (L x d) array. Each row in Vs represents an extreme point
of the convex polytope described by A*x <= b.
Method adapted from Michael Kelder's con2vert() MATLAB function
http://www.mathworks.com/matlabcentral/fileexchange/7894-con2vert-constraints-to-vertices
'''
# check if b is a column vector with ndim=2, or (L,) array with ndim=1 only
if b.ndim == 2:
b = b.flatten()
# attempt to find an interior point in the feasible region
c = scipy.linalg.lstsq(A, b)[0]
# if c is out of the region or on the polytope boundary, try to find a new
# c
num_tries = 0
while artools.out_region(c, A, b) or sp.any(sp.dot(A, c) - b == 0.0):
num_tries += 1
if num_tries > 20:
raise Exception("con2vert() failed to find an interior point"
"after 20 tries. Perhaps your constraints are"
"badly formed or the region is unbounded.")
def tmp_fn(xi):
# find the Chebyshev centre, xc, of the polytope (the
# largest inscribed ball within the polytope with centre at xc.)
d = sp.dot(A, xi) - b
# ensure point actually lies within region and not just on the
# boundary
tmp_ks = sp.nonzero(d >= -1e-6)
# print sum(d[tmp_ks]) #sum of errors
# return max(d)
return sum(d[tmp_ks])
# print "c is not in the interior, need to solve for interior point!
# %f" % (tmp_fn(c))
# ignore output message
c_guess = sp.rand(A.shape[1])
solver_result = scipy.optimize.fmin(tmp_fn, c_guess, disp=False)
c = solver_result
plt.plot(c[0], c[1], "ks")
# calculate D matrix?
b_tmp = b - sp.dot(A, c) # b_tmp is like a difference vector?
D = A / b_tmp[:, None]
# find indices of convex hull belonging to D?
k = scipy.spatial.ConvexHull(D).simplices
# Generate some kind of offset list of vertices offset from c vector
G = sp.zeros((len(k), D.shape[1]))
for idx in range(0, len(k)):
# F is a matrix with rows beloning to the indices of k referencing
# rows in matrix D??
F = D[k[idx, :], :]
# f is simply an nx1 column vector of ones?
f = sp.ones((F.shape[0], 1))
# solve the least squares problem F\f in MATLAB notation for a vector
# that becomes a row in matrix G?
G[idx, :] = scipy.linalg.lstsq(F, f)[0].T
# find vertices from vi = c + Gi
Vs = G + sp.tile(c.T, (G.shape[0], 1))
Vs = artools.unique_rows(Vs)[0]
return Vs
