迷宫解密

Author: makelove

my04-Maze-Solver迷宫解密/SampleImages/1.png

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+# -*- coding: utf-8 -*-
+# @Time    : 2017/7/30 18:16
+# @Author  : play4fun
+# @File    : aStar1.py.py
+# @Software: PyCharm
+
+"""
+aStar1.py: 不行！？？
+"""
+
+import sys
+
+# from Queue import Queue
+from multiprocessing import Queue
+from PIL import Image
+
+start = (400, 984)
+end = (398, 25)
+
+
+def iswhite(value):
+    if value == (255, 255, 255):
+        return True
+
+
+def getadjacent(n):
+    x, y = n
+    return [(x - 1, y), (x, y - 1), (x + 1, y), (x, y + 1)]
+
+
+def BFS(start, end, pixels):
+    queue = Queue()
+    queue.put([start])  # Wrapping the start tuple in a list
+
+    while not queue.empty():
+
+        path = queue.get()
+        pixel = path[-1]
+
+        if pixel == end:
+            return path
+
+        for adjacent in getadjacent(pixel):
+            x, y = adjacent
+            if iswhite(pixels[x, y]):
+                pixels[x, y] = (127, 127, 127)  # see note
+                new_path = list(path)
+                new_path.append(adjacent)
+                queue.put(new_path)
+
+    print("Queue has been exhausted. No answer was found.")
+
+
+if __name__ == '__main__':
+
+    # invoke: python mazesolver.py <mazefile> <outputfile>[.jpg|.png|etc.]
+    base_img = Image.open(sys.argv[1])
+    base_pixels = base_img.load()
+    print(base_pixels)
+
+    path = BFS(start, end, base_pixels)
+    if path is None:
+        print('path is None')
+        exit(-1)
+    print('path:',path)
+
+    path_img = Image.open(sys.argv[1])
+    path_pixels = path_img.load()
+
+    for position in path:
+        x, y = position
+        path_pixels[x, y] = (255, 0, 0)  # red
+
+    path_img.save(sys.argv[2])

my04-Maze-Solver迷宫解密/SampleImages/2.png

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+#include<iostream>
+#include<cstdlib>
+#include<opencv2/core/core.hpp>
+#include<opencv2/highgui/highgui.hpp>
+#include<opencv2/imgproc/imgproc.hpp>
+
+using namespace std;
+using namespace cv;
+Mat kernel = Mat::ones(15, 15, CV_8UC1);
+class Morph{
+private:
+	int dilationElem,erodeElem;
+	int dilationSize,erodeSize;
+	
+public:
+	Morph(){
+		dilationElem=0;
+		erodeElem=0;
+		dilationSize=2;
+		erodeSize=2;
+	}
+	Mat dilateImage(Mat input){
+		Mat temp,element;
+		int dilationType;
+		if(dilationElem==0)
+			dilationType=MORPH_RECT;
+		else if(dilationElem==1)
+			dilationType=MORPH_CROSS;
+		else if(dilationElem==2)
+			dilationType=MORPH_ELLIPSE;
+		element= getStructuringElement(dilationType,Size(2*dilationSize+1,2*dilationSize+1),Point(dilationSize,dilationSize));
+		dilate(input,temp,kernel);
+		return temp;
+	}
+	Mat erodeImage(Mat input){
+		Mat temp,element;
+		int erodeType;
+		if(erodeElem==0)
+			erodeType=MORPH_RECT;
+		else if(erodeElem==1)
+			erodeType=MORPH_CROSS;
+		else if(erodeElem==2)
+			erodeType=MORPH_ELLIPSE;
+		element= getStructuringElement(erodeType,Size(2*erodeSize+1,2*erodeSize+1),Point(erodeSize,erodeSize));
+		dilate(input,temp,kernel);
+		return temp;
+	}
+};
+int main(int argc, char **argv){
+	if(argc!=2){
+		cout<<"Wait for an image"<<endl;
+		return -1;
+	}
+	vector<vector<Point> > contours;
+	Mat inputMaze,gray,binary,dilation,erosion,imgDiff,BGRcomp[3],imgDiff_inv,output,red,green;
+	Morph mp;
+	inputMaze=imread(argv[1],CV_LOAD_IMAGE_COLOR);
+	cvtColor(inputMaze,gray,CV_BGR2GRAY);
+	threshold(gray,binary,127,255,CV_THRESH_BINARY_INV);
+
+	findContours(binary,contours,CV_RETR_EXTERNAL, CV_CHAIN_APPROX_NONE);
+	drawContours(binary, contours, 0, CV_RGB(255,255,255), CV_FILLED);
+	threshold(binary,binary,240,255,CV_THRESH_BINARY);
+
+	dilation=mp.dilateImage(binary);
+	erosion=mp.erodeImage(dilation);
+	absdiff(dilation,erosion,imgDiff);
+	bitwise_not(imgDiff,imgDiff_inv);
+	split(inputMaze,BGRcomp);
+	namedWindow("diff",WINDOW_AUTOSIZE);
+	imshow("diff",imgDiff);
+	bitwise_and(BGRcomp[2],BGRcomp[2],red,imgDiff_inv);
+	bitwise_and(BGRcomp[1],BGRcomp[1],green,imgDiff_inv);
+	BGRcomp[2]=red.clone();
+	BGRcomp[1]=green.clone();
+	merge(BGRcomp,3,output);
+	namedWindow("SolvedMaze",WINDOW_AUTOSIZE);
+	imshow("SolvedMaze",output);
+	//imwrite("OutputMaze.jpg",output);
+	waitKey(0);
+	return 0;
+}

my04-Maze-Solver迷宫解密/SampleImages/3.jpg

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+
+
+'''
+源文件是使用opencv2.4，
+改成opencv3.2有点问题。
+https://ishankgulati.github.io/posts/Maze-Solver/
+'''
+import cv2
+import numpy as np
+
+
+
+img = cv2.imread('SampleImages/1.png')
+# img = cv2.imread('SampleImages/2.png')
+# img = cv2.imread('SampleImages/3.jpg')#不行，得修改
+# img = cv2.imread('SampleImages/huge_maze.jpg')#不行，得修改
+cv2.imshow('maze',img)
+cv2.waitKey(0)
+
+# Binary conversion
+gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
+ret, thresh = cv2.threshold(gray, 127, 255, cv2.THRESH_BINARY_INV)#反转tholdolding将给我们一个二进制的图像与白色的墙壁和黑色的背景。
+cv2.imshow('THRESH_BINARY_INV',thresh)
+cv2.waitKey(0)
+
+# Contours
+image,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)
+dc=cv2.drawContours(thresh, contours, 0, (255, 255, 255), 5)#用不同颜色来标注
+dc=cv2.drawContours(dc, contours, 1, (0, 0, 0), 5)# TODO 大迷宫的len(contours): 26
+cv2.imshow('drawContours',dc)
+cv2.waitKey(0)
+
+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)
+
+# Dilate
+'''
+扩张
+
+扩张是数学形态领域的两个基本操作者之一，另一个是侵蚀。它通常应用于二进制图像，但有一些版本可用于灰度图像。操作者对二进制图像的基本效果是逐渐扩大前景像素区域的边界（通常为白色像素）。因此，前景像素的面积大小增加，而这些区域内的孔变小。
+'''
+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)
+
+# Erosion
+#侵蚀是第二个形态运算符。它也适用于二进制图像。操作者对二进制图像的基本效果是消除前景像素区域的边界（通常为白色像素）。因此，前景像素的面积缩小，并且这些区域内的孔变大。
+erosion = cv2.erode(dilation, kernel, iterations=1)
+cv2.imshow('erosion',erosion)
+cv2.waitKey(0)
+
+#找到两个图像的差异
+diff = cv2.absdiff(dilation, erosion)
+cv2.imshow('diff',diff)
+cv2.waitKey(0)
+
+# splitting the channels of maze
+b, g, r = cv2.split(img)
+mask_inv = cv2.bitwise_not(diff)
+#为了在原始迷宫图像上显示解决方案，首先将原来的迷宫分割成r，g，b组件。现在通过反转diff图像创建一个掩码。使用在最后一步中创建的掩码的原始迷宫的按位和r和g分量。这一步将从迷宫解决方案的图像部分去除红色和绿色成分。最后一个是合并所有组件，我们将使用蓝色标记的解决方案。
+# 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.imwrite('SampleImages/Solved-Maze-1.png',res)
+
+cv2.waitKey(0)
+cv2.destroyAllWindows()