如何識別矩形的其他角落？

Question

我在MATLAB上工作的附加圖像。粗糙的矩形可能在沿其周邊的某些點處有斷裂（噪音）。給出矩形的兩個角點（顯示爲藍色和紅色）的位置。我怎樣才能準確識別矩形的其他兩個角？

Answer 1

我找到了一個解決方案，但我錯過了「準確」的部分。

關鍵因素（在我的解決方案中）是使用morphological operations關閉形狀，然後使用Suever建議的函數corner。
我用'square'面膜而不是'disk'，以保持角落鋒利。

這是我的代碼：

%Read input image from imgur hosting site. 
I = imread('https://i.stack.imgur.com/g2iTN.jpg'); 

%Convert image to binary 
I = im2bw(I); 

%Add margins of 10 pixels from each size 
J = padarray(I, [10, 10]); 

%Dilate input image with 9x9 square "mask" 
se0 = strel('square', 9); 
J = imdilate(J, se0); 

%Erode J image with 8x8 square "mask" (keep lines a bit more fat then original lines). 
se1 = strel('disk', 4); 
J = imerode(J, se1); 

%Use corner function to detect 4 corners (I had to plyed with Quality and Sensitivity parameters). 
C = corner(J, 4, 'QualityLevel', 0.5, 'SensitivityFactor', 0.1); 


%Plot corners on image J 
figure;imshow(J);hold on 
plot(C(:,1), C(:,2), 'r*'); 

%Plot corners on image I 
C = C - 10; %Subtract 10 from C, because J is padded with 10 pixels. 
figure;imshow(I);hold on 
plot(C(:,1), C(:,2), 'r*'); 

---

輸出數字：

Ĵ

我

我的解決方案足夠準確嗎？

---

Hough變換方法：
該解決方案几乎完成 - 所有的左邊是找到交叉點。

%Read input image from imgur hosting site. 
I = imread('https://i.stack.imgur.com/g2iTN.jpg'); 

%Convert image to binary 
I = im2bw(I); 

%Compute the Hough transform of the binary image 
[H,theta,rho] = hough(I); 

%Find the peaks in the Hough transform matrix, H, using the houghpeaks function. 
P = houghpeaks(H,2,'threshold',ceil(0.3*max(H(:)))); 

%Find lines in the image using the houghlines function. 
lines = houghlines(I,theta,rho,P,'FillGap',50,'MinLength',20); 

%Create a plot that displays the original image with the lines superimposed on it. 
figure, imshow(I), hold on 
for k = 1:length(lines) 
    xy = [lines(k).point1; lines(k).point2]; 
    plot(xy(:,1),xy(:,2),'LineWidth',2,'Color','green'); 
end 

%Angle of top and bottom edges. 
theta0 = mean([lines(1).theta, lines(2).theta]); 

%Leave lines with theta that is close to perpendicular with the two lines found. 
perpendicular_idx = abs((mod(theta+360 - theta0, 360)-90)) < 10; 
perpendicular_idx = perpendicular_idx | abs((mod(theta+360+180 - theta0, 360)-90)) < 10; 
H1 = H; 
H1(:, ~perpendicular_idx) = 0; 

%Find the peaks in the Hough transform matrix, H, using the houghpeaks function. 
P1 = houghpeaks(H1,2,'threshold',ceil(0.3*max(H1(:)))); 

%Find lines in the image using the houghlines function. 
lines1 = houghlines(I,theta,rho,P1,'FillGap',20,'MinLength',20); 

for k = 1:length(lines1) 
    xy = [lines1(k).point1; lines1(k).point2]; 
    plot(xy(:,1),xy(:,2),'LineWidth',2,'Color','red'); 
end 

%Angle of left and right edges. 
theta1 = mean([lines1(1).theta, lines1(2).theta]); 

---

---

查找線交點：

假設的平方的形狀是梯形（不是矩形）。

我用直線的Parametric Equation。

%In image axis system, the X axis goes from top to bottom, and Y axis goes from left to right. 

%y 
%^ 
%| 
%|  a     b 
%|  -------------------- 
%|  |     | 
%|  |     | 
%|  |     | 
%|  -------------------- 
%|  c     d 
%| 
% -------------------------------->x 

%Coordinatates of two given corners 
h = size(I, 1); 

%Use h-y, to convert the coordinates system from image system (y axis direction is down) to mathematical (y direction is up). 
b = [420; h-15]; %(X, Y) coordinate of top right corner (center of blue circle). 
c = [5; h-101]; %(X, Y) coordinate bottom left corner (center of red circle). 

%Remark: I modified the coordinates a little (the center of your drawn circles do not look in place). 

%Finding a coordinate 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
%t - distance parameter (scalar) 
%Lines equations: 
% top_xy = b + u*t; 
% left_xy = c + v*t; 

%Use 90 degrees minus theta because image coordinate system is rotated in 90 degrees. 
%Direction vector of top lines 
u = [cos(deg2rad(90-theta0)); sin(deg2rad(90-theta0))]; 

%Direction vector of left line 
v = [cos(deg2rad(90-lines1(2).theta)); sin(deg2rad(90-lines1(2).theta))]; 

%Finding top-left corner (intersection of top line and left line): 
% b + u*t0 = c + v*t1 
% 
% u*t0 - v*t1 = c - b 
% 
% [u, -v]*t = c - b 
% 
% A = [u, -v] 
% 
% A*t = (c - b) 
% 
% t = inv(A)*(c - b) 


%Assignment: 
A = [u, -v]; 

t = inv(A)*(c - b); 

a = b + u*t(1); 

plot(round(a(1)), round(h - a(2)), 'x', 'LineWidth', 2, 'Color', 'yellow'); 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 


%Finding d coordinate 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
%t - distance parameter (scalar) 
%Lines equations: 
% bottom_xy = c + u*t; 
% right_xy = b + v*t; 

%Direction vector of top lines 
u = [cos(deg2rad(90-theta0)); sin(deg2rad(90-theta0))]; 

%Direction vector of left line 
v = [cos(deg2rad(90-lines1(1).theta)); sin(deg2rad(90-lines1(1).theta))]; 

%Finding top-left corner (intersection of top line and left line): 
% c + u*t0 = b + v*t1 
% 
% u*t0 - v*t1 = b - c 
% 
% [u, -v]*t = b - c 
% 
% A = [u, -v] 
% 
% A*t = (b - c) 
% 
% t = inv(A)*(b - c) 


%Assignment: 
A = [u, -v]; 

t = inv(A)*(b - c); 

d = c + u*t(1); 

plot(round(d(1)), round(h - d(2)), 'x', 'LineWidth', 2, 'Color', 'yellow'); 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 


%Plot b and c coordinates 
plot(b(1), h - b(2), 'x', 'LineWidth', 2, 'Color', 'blue'); 
plot(c(1), h - c(2), 'x', 'LineWidth', 2, 'Color', 'red'); 

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解決方案：
左上像素座標：[66, 6]
右下像素座標：[50, 419]