I'm trying to get a Hough transform to work in MATLAB, but I'm having problems. I have a really bad way of detecting peaks that needs to be fixed, but before that I need to be able to reverse the hough transform to create the lines again properly. This is the type of stuff I'm getting right now:

enter image description here

looks like its rotated by 90 degrees, but I'm not sure why. I'm not sure if it's my Hough space that's wrong, or if it's the way I de-Hough and draw the lines. Also could someone help improve my peak detection? The images used in the code are here

%% load a sample image; convert to grayscale; convert to binary

%create 'x' image (works well)
a = eye(255);
b = flipud(eye(255));
x = a + b;
x(128,128) = 1;

%image = rgb2gray(imread('up.png')) < 255;
%image = rgb2gray(imread('hexagon.png')) < 255;
%image = rgb2gray(imread('traingle.png')) < 255;
%%% these work
%image = x;
%image = a;
image = b;    

%% set up variables for hough transform
theta_sample_frequency = 0.01;                                             
[x, y] = size(image);
rho_limit = norm([x y]);                                                
rho = (-rho_limit:1:rho_limit);
theta = (0:theta_sample_frequency:pi);
num_thetas = numel(theta);
num_rhos = numel(rho);
hough_space = zeros(num_rhos, num_thetas);

%% perform hough transform
for xi = 1:x
    for yj = 1:y
        if image(xi, yj) == 1 
            for theta_index = 1:num_thetas
                th = theta(theta_index);
                r  = xi * cos(th) + yj * sin(th);
                rho_index = round(r + num_rhos/2);                      
                hough_space(rho_index, theta_index) = ...
                     hough_space(rho_index, theta_index) + 1;

%% show hough transform
imagesc(theta, rho, hough_space);
title('Hough Transform');
xlabel('Theta (radians)');
ylabel('Rho (pixels)');

%% detect peaks in hough transform
r = [];
c = [];
[max_in_col, row_number] = max(hough_space);
[rows, cols] = size(image);
difference = 25;
thresh = max(max(hough_space)) - difference;
for i = 1:size(max_in_col, 2)
   if max_in_col(i) > thresh
       c(end + 1) = i;
       r(end + 1) = row_number(i);

%% plot all the detected peaks on hough transform image
hold on;
plot(theta(c), rho(r),'rx');
hold off;

%% plot the detected line superimposed on the original image
hold on;

for i = 1:size(c,2)
    th = theta(c(i));
    rh = rho(r(i));
    m = -(cos(th)/sin(th));
    b = rh/sin(th);
    x = 1:cols;
    plot(x, m*x+b);
    hold on;

Linked: How to do De-Houghing of a Hough transform'ed Image?

  • $\begingroup$ did you manage to solve your problem? i am facing a similar issue. thanks $\endgroup$ Commented Nov 4, 2015 at 11:35
  • $\begingroup$ transformation de hough sous matlab pour détecter plusieurs ellipses $\endgroup$
    – user21253
    Commented May 30, 2016 at 9:54

3 Answers 3


Firstly, Matlab has a Built in Hough Transform: no need to reinvent the wheel.

[H,T,R] = hough(BW,'RhoResolution',0.5,'Theta',-90:0.5:89.5);

Although your image doesn't necessarily require edge detection you could improve the processing time and effectiveness of the algorithm my using it. Your triangle has fat regions of white and black. Ideally, the triangle would be 1px thick marking the edges of the triangle. Use Canny Edge Detection

BW = edge(Image,'canny');

the result is \theta is in the range $-90 < \theta < 90$ degrees while your plot is $0 < \theta < 180$ (or $0 < \theta < \pi$) so subtract $90$ degrees ($\pi/2$) .

There is the potential for you to choose the wrong peak because there are neighboring peaks that may be larger in the accumulator matrix. While there are many algorithms here is one that I've seen used in Hough Transforms in the past:

1) Define a region shape (typically its square) 
2) Define an accumulator threshold  
3) Select one pixel in the accumulator matrix
4) If it is a peak (i.e., larger than neighboring values above a threshold)
       a) look at the points within the region shape.
       b) If there is a larger value
              this is not a peak
              this is a peak
 5) Move to next pixel in accumulator matrix.

Look into HoughLines for displaying the hough transform lines, results:


Effects of Using Canny Edge Detector

Edge Detection can potentially turn each side of the triangle into two lines.

The goal of canny edge detection is to produce maximally thin/narrow edges by using nonmaximal supression

Canny Edge Detection in a Nutshell (Source: Digital Image Processing, Gonazalez)

1) Smooth input Image using a Gaussian Filter
2) Compute the Gradient magnitude and angle (Sobel, Perwitt or robert cross filters)
3) Apply Nonmaxima suppression (this is where the thinning happens) 
   a) Figure out which direction the edge is
   b) If the edge's magnitude is smaller than one of its two neighbors in the direction of the edge
          set the edge point to zero
          leave it alone
4) Use double thresholding and connectivity analysis to detect and link edges
  • $\begingroup$ thanks for the reply. I'm doing it from scratch to understand it better. canny edge detection still gives 2 triangles. one for the inner edge, and out for the outer edge. I learned the theory from wikipedia, which states that theta is 0:pi. I know that the built in function uses -pi/2:pi/2, but there shouldn't be a real difference? $\endgroup$
    – waspinator
    Commented Apr 10, 2012 at 18:13
  • $\begingroup$ Right off the bat, the range shouldnt make a difference. (can you tell the difference between a line that has been rotated 180 degrees?) HOWEVER, this does make a difference if you are using the hough transform for image correction algorithms. (It would mean the difference between a rightside up and upsidedown image) $\endgroup$
    – CyberMen
    Commented Apr 10, 2012 at 18:38
  • $\begingroup$ Wouldn't edge detection produce 2 lines where you want to find only 1? Something that finds the center of a thick line would be better. $\endgroup$
    – endolith
    Commented Apr 10, 2012 at 19:31
  • $\begingroup$ @endolith Included a bit of discussion of edge detection in the original post $\endgroup$
    – CyberMen
    Commented Apr 11, 2012 at 12:50
  • $\begingroup$ 'no need to reinvent the wheel'? Tell that to my instructor ;-) $\endgroup$ Commented Oct 20, 2012 at 6:56
    if image(xi, yj) == 1 

needs to be changed to

    if image(yj, xi) == 1 

for the lines to work out in the dehough


The answer using the 3 loops is less then optimal and could be improved, here is more of an intuitive approach/point of view:

Every pair of valid points sets a unique a & b of y = ax + b. A line will have lots of pairs with the same a & b value, so a long line will have be present as a peak. This is also true for polar r & teta coordinates.

Instead of treating each point separately, use pairs of dots. If you can store all (usually sparse) points in a separate list its better, but it is not a must.

  1. Take each pair & compute its a & b. (rounded to discrete values)
  2. Go to the specific place in the array and add 1.

Long line --> lots of pairs with the same a,b.
Sporadic points --> small counting in specific cells --> more like clutter.

Another way to look at it is from a Radon/projective point of view.

  • A line will project strongly to a perpendicular "collecting line" so having high score.
  • If the angle between the line and the "collecting line" is not 90 deg or the collecting line does not collect all the line points projected into this "collecting line", there will be lower score.

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