Как преобразование перспективы просто работает в NumPy без использования какой-либо функции OpenCV

Question

import numpy as np
import matplotlib.image as mtimg
import matplotlib.pyplot as mtplot

def get_image_size(src):
    ### give the corner points of the object in the image 
    ### return the points of the scanned image

    (tl, tr, br, bl) = src

    # compute the width of the new image, which will be the
    # maximum distance between bottom-right and bottom-left
    # x-coordiates or the top-right and top-left x-coordinates
    widthA = np.sqrt(((br[0] - bl[0]) ** 2) + ((br[1] - bl[1]) ** 2))
    widthB = np.sqrt(((tr[0] - tl[0]) ** 2) + ((tr[1] - tl[1]) ** 2))
    maxWidth = max(int(widthA), int(widthB))

    # compute the height of the new image, which will be the
    # maximum distance between the top-right and bottom-right
    # y-coordinates or the top-left and bottom-left y-coordinates
    heightA = np.sqrt(((tr[0] - br[0]) ** 2) + ((tr[1] - br[1]) ** 2))
    heightB = np.sqrt(((tl[0] - bl[0]) ** 2) + ((tl[1] - bl[1]) ** 2))
    maxHeight = max(int(heightA), int(heightB))

    # now that we have the dimensions of the new image, construct
    # the set of destination points to obtain a "birds eye view",
    # (i.e. top-down view) of the image, again specifying points
    # in the top-left, top-right, bottom-right, and bottom-left
    # order

    dst = np.array([
        [0, 0],
        [maxWidth - 1, 0],
        [maxWidth - 1, maxHeight - 1],
        [0, maxHeight - 1]])

    return dst

Функция get_image_size возвращает точку dst

def four_point_transform(src, dst):  
    # compute the perspective transform matrix and then apply it
    # the transform is found by solving ax = b

    # for perspective projection, the 3x3 matrix has 8 unknows
    #      [t11 t12 t13] [s1]   [v1]
    #  T = [t21 t22 t23] [s2] = [v2]
    #      [t31 t32  1 ] [ 1]   [v3]
    #
    #  where d1 = v1/v3 and d2 = v2/v3.
    #  
    #  t11*s1 + t12*s2 + t13 = v1 = d1*v3 = d1*(t31*s1+t32*s2+1)
    #  t21*s1 + t22*s2 + t23 = v2 = d2*v3 = d2*(t31*s1+t32*s2+1)
    #  t31*s1 + t32*s2 + 1   = v3

    A = np.zeros((8, 8))
    b = np.zeros((8, 1))

    # assigning values
    # ******* your code here ********






    # now put the solution x into a 3x3 matrix T
    # the t33 element = 1
    x = np.linalg.solve(A, b)
    x = np.concatenate((x, [[1]]), axis=0)
    T = x.reshape(3,3)

    # We need the inverse transformation, so R = np.inv(T)
    R = np.linalg.inv(T)

    # the r33 should be 1, so normalize it.
    R = R/R[2][2]

    return R

Функция four_point_transform заключается в том, что я хочу найти матрицу перспективы

def perspective_projection(size, image, R):
    ### Convert the original 3D image to 2D
    # img is the dst picture
    img = np.zeros((size[0], size[1], 3))
    [w, h, c] = np.array(image).shape
    v = np.zeros((3,1))
    v[2] = 1

    # transformation is done by finding the 'conor' of points in dst picture
    # *********** your code here *************

    warped = img.astype(np.uint8)

    # return the warped image
    return warped

Функция перспективного проецирования, которую я хочуполучить деформированное изображение

Как выполнить функцию "four_point_transform" и функцию перспективного проецирования, чтобы получить деформированное изображение?

Это оригинальное изображение Это моя цель