import numpy, scipy, gridlock, fdfd_tools
from fdfd_tools import bloch
from numpy.linalg import norm
import logging

logging.basicConfig(level=logging.DEBUG)
logger = logging.getLogger(__name__)


dx = 40
x_period = 400
y_period = z_period = 2000
g = gridlock.Grid([numpy.arange(-x_period/2, x_period/2, dx),
                   numpy.arange(-1000, 1000, dx),
                   numpy.arange(-1000, 1000, dx)],
                  shifts=numpy.array([[0,0,0]]),
                  initial=1.445**2,
                  periodic=True)

g.draw_cuboid([0,0,0], [200e8, 220, 220], eps=3.47**2)

#x_period = y_period = z_period = 13000
#g = gridlock.Grid([numpy.arange(3), ]*3,
#                  shifts=numpy.array([[0, 0, 0]]),
#                  initial=2.0**2,
#                  periodic=True)

g2 = g.copy()
g2.shifts = numpy.zeros((6,3))
g2.grids = [numpy.zeros(g.shape) for _ in range(6)]

epsilon = [g.grids[0],] * 3
reciprocal_lattice = numpy.diag(1000/numpy.array([x_period, y_period, z_period])) #cols are vectors

#print('Finding k at 1550nm')
#k, f = bloch.find_k(frequency=1/1550,
#                    tolerance=(1/1550 - 1/1551),
#                    direction=[1, 0, 0],
#                    G_matrix=reciprocal_lattice,
#                    epsilon=epsilon,
#                    band=0)
#
#print("k={}, f={}, 1/f={}, k/f={}".format(k, f, 1/f, norm(reciprocal_lattice @ k) / f ))

print('Finding f at [0.25, 0, 0]')
for k0x in [.25]:
    k0 = numpy.array([k0x, 0, 0])

    kmag = norm(reciprocal_lattice @ k0)
    tolerance = 1e-4 * (1/1550)   # df = f * dn_eff / n
    logger.info('tolerance {}'.format(tolerance))

    n, v = bloch.eigsolve(4, k0, G_matrix=reciprocal_lattice, epsilon=epsilon, tolerance=tolerance)
    v2e = bloch.hmn_2_exyz(k0, G_matrix=reciprocal_lattice, epsilon=epsilon)
    v2h = bloch.hmn_2_hxyz(k0, G_matrix=reciprocal_lattice, epsilon=epsilon)
    ki = bloch.generate_kmn(k0, reciprocal_lattice, g.shape)

    z = 0
    e = v2e(v[0])
    for i in range(3):
        g2.grids[i] += numpy.real(e[i])
        g2.grids[i+3] += numpy.imag(e[i])

    f = numpy.sqrt(numpy.real(numpy.abs(n))) # TODO
    print('k0x = {:3g}\n eigval = {}\n f = {}\n'.format(k0x, n, f))
    n_eff = norm(reciprocal_lattice @ k0) / f
    print('kmag/f = n_eff =  {} \n wl = {}\n'.format(n_eff, 1/f ))