使用OpenCV實現迷宮解密的全過程

import cv2
import numpy as np
 
img = cv2.imread('../photos/1.png')
# cv2.imshow('maze',img)
# cv2.waitKey(0)
# cv2.destroyAllWindows()
 
#對影象進行二值化
# Binary conversion
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
 
#反轉tholdolding將給我們一個二進位制的影象與白色的牆壁和黑色的背景。
ret, thresh = cv2.threshold(gray, 127, 255, cv2.THRESH_BINARY_INV)
cv2.imshow('THRESH_BINARY_INV',thresh)
cv2.waitKey(0)
cv2.destroyAllWindows()

#對二值化處理的影象進行輪廓檢測並標註
# Contours
contours, hierarchy = cv2.findContours(thresh, cv2.RETR_EXTERNAL,
                                       cv2.CHAIN_APPROX_NONE)
print('len(contours):',len(contours))
 
# dc=cv2.drawContours(thresh, contours, 0, (255, 255, 255), -1)
#用不同顏色來標註
dc=cv2.drawContours(thresh, contours, 0, (255, 255, 255), 5)
 
# TODO 大迷宮的len(contours): 26
dc=cv2.drawContours(dc, contours, 1, (0, 0, 0), 5)
cv2.imshow('drawContours',dc)
cv2.waitKey(0)
cv2.destroyAllWindows()

#對影象閾值進行處理
ret, thresh = cv2.threshold(dc, 240, 255, cv2.THRESH_BINARY)
 
# ret, thresh = cv2.threshold(thresh, 240, 255, cv2.THRESH_BINARY)
cv2.imshow('thresh2',thresh)
cv2.waitKey(0)
cv2.destroyAllWindows()

# Dilate
ke = 10
# kernel = np.ones((19, 19), np.uint8)
kernel = np.ones((ke, ke), np.uint8)
dilation = cv2.dilate(thresh, kernel, iterations=1)
cv2.imshow('dilation',dilation)
cv2.waitKey(0)
cv2.destroyAllWindows()

# Erosion
erosion = cv2.erode(dilation, kernel, iterations=1)
cv2.imshow('erosion',erosion)
cv2.waitKey(0)
cv2.destroyAllWindows()

#找到兩個影象的差異
diff = cv2.absdiff(dilation, erosion)
 
# 分迷宮的通道
b, g, r = cv2.split(img)
mask_inv = cv2.bitwise_not(diff)
 
# masking out the green and red colour from the solved path
r = cv2.bitwise_and(r, r, mask=mask_inv)
g = cv2.bitwise_and(g, g, mask=mask_inv)
 
res = cv2.merge((b, g, r))
cv2.imshow('Solved Maze', res)
 
cv2.waitKey(0)
cv2.destroyAllWindows()

import cv2
import numpy as np
 
img = cv2.imread('../photos/1.png')
cv2.imshow('maze',img)
cv2.waitKey(0)
cv2.destroyAllWindows()
 
 
 
#對影象進行二值化
# Binary conversion
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
 
#反轉tholdolding將給我們一個二進位制的影象與白色的牆壁和黑色的背景。
ret, thresh = cv2.threshold(gray, 127, 255, cv2.THRESH_BINARY_INV)
cv2.imshow('THRESH_BINARY_INV',thresh)
cv2.waitKey(0)
cv2.destroyAllWindows()
 
#對二值化處理的影象進行輪廓檢測並標註
# Contours
contours, hierarchy = cv2.findContours(thresh, cv2.RETR_EXTERNAL,
                                       cv2.CHAIN_APPROX_NONE)
print('len(contours):',len(contours))
 
# dc=cv2.drawContours(thresh, contours, 0, (255, 255, 255), -1)
#用不同顏色來標註
dc=cv2.drawContours(thresh, contours, 0, (255, 255, 255), 5)
 
# TODO 大迷宮的len(contours): 26
dc=cv2.drawContours(dc, contours, 1, (0, 0, 0), 5)
cv2.imshow('drawContours',dc)
cv2.waitKey(0)
cv2.destroyAllWindows()
 
#對影象閾值進行處理
ret, thresh = cv2.threshold(dc, 240, 255, cv2.THRESH_BINARY)
 
# ret, thresh = cv2.threshold(thresh, 240, 255, cv2.THRESH_BINARY)
cv2.imshow('thresh2',thresh)
cv2.waitKey(0)
cv2.destroyAllWindows()
 
 
'''
擴張
擴張是數學形態領域的兩個基本操作者之一，另一個是侵蝕。它通常應用於二進位制影象，但有一些版本可用於灰度影象。
操作者對二進位制影象的基本效果是逐漸擴大前景畫素區域的邊界（通常為白色畫素）。因此，前景畫素的面積大小增加，而這些區域內的孔變小。
'''
# Dilate
ke = 10
# kernel = np.ones((19, 19), np.uint8)
kernel = np.ones((ke, ke), np.uint8)
dilation = cv2.dilate(thresh, kernel, iterations=1)
cv2.imshow('dilation',dilation)
cv2.waitKey(0)
cv2.destroyAllWindows()
 
# Erosion
#侵蝕是第二個形態運運算元。它也適用於二進位制影象。操作者對二進位制影象的基本效果是消除前景畫素區域的邊界（通常為白色畫素）。
# 因此，前景畫素的面積縮小，並且這些區域內的孔變大。
erosion = cv2.erode(dilation, kernel, iterations=1)
cv2.imshow('erosion',erosion)
cv2.waitKey(0)
cv2.destroyAllWindows()
 
#找到兩個影象的差異
diff = cv2.absdiff(dilation, erosion)
cv2.imshow('diff',diff)
cv2.waitKey(0)
cv2.destroyAllWindows()
 
# 分迷宮的通道
b, g, r = cv2.split(img)
mask_inv = cv2.bitwise_not(diff)
#為了在原始迷宮影象上顯示解決方案，首先將原來的迷宮分割成r，g，b元件。現在通過反轉diff影象建立一個掩碼。
# 使用在最後一步中建立的掩碼的原始迷宮的按位元和r和g分量。這一步將從迷宮解決方案的影象部分去除紅色和綠色成分。
# 最後一個是合併所有元件，我們將使用藍色標記的解決方案。
# masking out the green and red colour from the solved path
r = cv2.bitwise_and(r, r, mask=mask_inv)
g = cv2.bitwise_and(g, g, mask=mask_inv)
 
res = cv2.merge((b, g, r))
cv2.imshow('Solved Maze', res)
 
cv2.waitKey(0)
cv2.destroyAllWindows()