import numpy as np
from scipy import linalg
from utils.simple_functions import choose

CLOSED_THRESHOLD = 0.01

def bezier(points):
    n = len(points) - 1
    return lambda t : sum([
        ((1-t)**(n-k))*(t**k)*choose(n, k)*point
        for k, point in enumerate(points)
    ])

def partial_bezier_points(points, a, b):
    """
    Given an array of points which define 
    a bezier curve, and two numbers 0<=a<b<=1,
    return an array of the same size, which 
    describes the portion of the original bezier
    curve on the interval [a, b].

    This algorithm is pretty nifty, and pretty dense.
    """
    a_to_1 = np.array([
        bezier(points[i:])(a)
        for i in range(len(points))
    ])
    return np.array([
        bezier(a_to_1[:i+1])((b-a)/(1.-a))
        for i in range(len(points))
    ])


# Linear interpolation variants

def interpolate(start, end, alpha):
    return (1-alpha)*start + alpha*end

def mid(start, end):
    return (start + end)/2.0

def inverse_interpolate(start, end, value):
    return np.true_divide(value - start, end - start)

def match_interpolate(new_start, new_end, old_start, old_end, old_value):
    return interpolate(
        new_start, new_end, 
        inverse_interpolate(old_start, old_end, old_value)
    )

def clamp(lower, upper, val):
    if val < lower:
        return lower
    elif val > upper:
        return upper
    return val

# Figuring out which bezier curves most smoothly connect a sequence of points

def get_smooth_handle_points(points):
    points = np.array(points)
    num_handles = len(points) - 1
    dim = points.shape[1]    
    if num_handles < 1:
        return np.zeros((0, dim)), np.zeros((0, dim))
    #Must solve 2*num_handles equations to get the handles.
    #l and u are the number of lower an upper diagonal rows
    #in the matrix to solve.
    l, u = 2, 1    
    #diag is a representation of the matrix in diagonal form
    #See https://www.particleincell.com/2012/bezier-splines/
    #for how to arive at these equations
    diag = np.zeros((l+u+1, 2*num_handles))
    diag[0,1::2] = -1
    diag[0,2::2] = 1
    diag[1,0::2] = 2
    diag[1,1::2] = 1
    diag[2,1:-2:2] = -2
    diag[3,0:-3:2] = 1
    #last
    diag[2,-2] = -1
    diag[1,-1] = 2
    #This is the b as in Ax = b, where we are solving for x,
    #and A is represented using diag.  However, think of entries
    #to x and b as being points in space, not numbers
    b = np.zeros((2*num_handles, dim))
    b[1::2] = 2*points[1:]
    b[0] = points[0]
    b[-1] = points[-1]
    solve_func = lambda b : linalg.solve_banded(
        (l, u), diag, b
    )
    if is_closed(points):
        #Get equations to relate first and last points
        matrix = diag_to_matrix((l, u), diag)
        #last row handles second derivative
        matrix[-1, [0, 1, -2, -1]] = [2, -1, 1, -2]
        #first row handles first derivative
        matrix[0,:] = np.zeros(matrix.shape[1])
        matrix[0,[0, -1]] = [1, 1]
        b[0] = 2*points[0]
        b[-1] = np.zeros(dim)
        solve_func = lambda b : linalg.solve(matrix, b)
    handle_pairs = np.zeros((2*num_handles, dim))
    for i in range(dim):
        handle_pairs[:,i] = solve_func(b[:,i])
    return handle_pairs[0::2], handle_pairs[1::2]

def diag_to_matrix(l_and_u, diag):
    """
    Converts array whose rows represent diagonal 
    entries of a matrix into the matrix itself.
    See scipy.linalg.solve_banded
    """
    l, u = l_and_u
    dim = diag.shape[1]
    matrix = np.zeros((dim, dim))
    for i in range(l+u+1):
        np.fill_diagonal(
            matrix[max(0,i-u):,max(0,u-i):],
            diag[i,max(0,u-i):]
        )
    return matrix

def is_closed(points):
    return np.linalg.norm(points[0] - points[-1]) < CLOSED_THRESHOLDp