Draw hough transform lines over an image

Question

I am trying to implement Hough line transform by myself, but I just couldn't do the last step which is drawing high voted thetas/rhos values. I tried to do its math on my own but It remained to get wrong outputs. When I checked some implementations of others, they were always using this approach to transform from Polar to cartesian coordinates in order to find two points.

for r,theta in lines[0]: 

# Stores the value of cos(theta) in a 
a = np.cos(theta) 

# Stores the value of sin(theta) in b 
b = np.sin(theta) 

# x0 stores the value rcos(theta) 
x0 = a*r 

# y0 stores the value rsin(theta) 
y0 = b*r 

# x1 stores the rounded off value of (rcos(theta)-1000sin(theta)) 
x1 = int(x0 + 1000*(-b)) 

# y1 stores the rounded off value of (rsin(theta)+1000cos(theta)) 
y1 = int(y0 + 1000*(a)) 

# x2 stores the rounded off value of (rcos(theta)+1000sin(theta)) 
x2 = int(x0 - 1000*(-b)) 

# y2 stores the rounded off value of (rsin(theta)-1000cos(theta)) 
y2 = int(y0 - 1000*(a)) 

# cv2.line draws a line in img from the point(x1,y1) to (x2,y2). 
# (0,0,255) denotes the colour of the line to be  
#drawn. In this case, it is red.  
cv2.line(img,(x1,y1), (x2,y2), (0,0,255),2)

The previous code from GeeksForGeeks. What I didn't get are those equations x1 = int(x0 + 1000*(-b)) & y2 = int(y0 - 1000*(a)). By translating these equations as a mathematic form : x1 = r cos(theta) + r (-sin(theta))|| y1 = r sin(theta) + r (cos(theta))

These two equations are just strange to me. 'r cos(theta)' is normal as we transfer from polar to cartesian. However, the next part isn't clear. Could anyone explain the original math behind it?

I'm not sure about your math, but why are you multiplying everything by 1000? — Eric
No .. no, I already implemented it by another equation. I just wanted to understand the implementation above as it has been used by almost hough line implementations [just like the link I've mentioned above]. Also, I wrote this code as it's, and then it worked as well. That's why I am so confused about its math. — Mohamed Nagy Mostafa

Ahmed M.Naguib Ahmed M.Naguib · Accepted Answer · 2019-09-04T09:52:11

(r cos, r sin) is the point closest to the origin on the line. It is not a line equation. To draw a line, u need 2 points. Here, he just moved along line 1000 and -1000 to get two points that are guaranteed to be outside a medium sized image

There are many ways to get line equation. Easiest is: r = x cos + y sin

If x1 = x - t sin
y1 = (1 / sin) (r - x cos + t sin cos)
= y + t cos

He used t=+/-1000 for image size

Draw hough transform lines over an image

1 Answers