import numpy as np

from .. import util
from ..constants import tol_path as tol


def line_line(origins, directions, plane_normal=None):
    """
    Find the intersection between two lines.
    Uses terminology from:
    http://geomalgorithms.com/a05-_intersect-1.html

    line 1:    P(s) = p_0 + sU
    line 2:    Q(t) = q_0 + tV

    Parameters
    ---------
    origins : (2, d) float
      Points on lines (d in [2,3])
    directions : (2, d) float
      Direction vectors
    plane_normal : (3, ) float
      If not passed computed from cross

    Returns
    ---------
    intersects : bool
      Whether the lines intersect.
        In 2D, false if the lines are parallel
        In 3D, false if lines are not coplanar
    intersection : (d,) float or None
      Point of intersection
    """
    # check so we can accept 2D or 3D points
    origins, is_2D = util.stack_3D(origins, return_2D=True)
    directions, is_2D = util.stack_3D(directions, return_2D=True)

    # unitize direction vectors
    directions /= util.row_norm(directions).reshape((-1, 1))

    # exit if values are parallel
    if np.sum(np.abs(np.diff(directions, axis=0))) < tol.zero:
        return False, None

    # using notation from docstring
    q_0, p_0 = origins
    v, u = directions
    w = p_0 - q_0

    # recompute plane normal if not passed
    if plane_normal is None:
        # the normal of the plane given by the two direction vectors
        plane_normal = np.cross(u, v)
        plane_normal /= np.linalg.norm(plane_normal)

    # vectors perpendicular to the two lines
    v_perp = np.cross(v, plane_normal)
    v_perp /= np.linalg.norm(v_perp)

    # if the vector from origin to origin is on the plane given by
    # the direction vector, the dot product with the plane normal
    # should be within floating point error of zero
    w_norm = np.linalg.norm(w)
    if w_norm > tol.zero and abs(np.dot(plane_normal, w / w_norm)) > tol.zero:
        # not coplanar
        return False, None

    # value of parameter s where intersection occurs
    s_I = np.dot(-v_perp, w) / np.dot(v_perp, u)
    # plug back into the equation of the line to find the point
    intersection = p_0 + s_I * u

    return True, intersection[: (3 - is_2D)]