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Accumulator Array
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at A(1,2)
– for y = ax + b we apply 1,2, to x, y, resulting in b = 2 – 1a
at B(2,4)
– for y = ax + b we apply 2,4, to x, y, resulting in b = 4 – 2a
at C(3,6)
– for y = ax + b we apply 3,6, to x, y, resulting in b = 6 -3a
When a = 1 the corresponding value of b = 1
When a = 2 the corresponding value of b = 0
When a = 3 the corresponding value of b = -1
When a = 4 the corresponding value of b = -2
When a = 5 the corresponding value of b = -3
b = 2 – 1a
Accumulator Array
b = 4 – 2a
When a = 1 the corresponding value of b = 2
When a = 2 the corresponding value of b = 0
When a = 3 the corresponding value of b = -2
When a = 4 the corresponding value of b = -4
When a = 5 the corresponding value of b = -6
Since 2 is the highest value of all the numbers in the Accumulator Array, this
indicates that A(1,2), and B(2,4) are valid points.
We can now draw a line connecting A and B. The resulting line is y = 2x + 0.
However, this method has its drawbacks. If the line is horizontal, then “a” is 0, and
if the line is vertical, then “a” is infinite.
Hough transform
• Paul Hough [1962], patented by IBM
• How to determine a line ?
(a, b)
(ak, bk)
Tow point A(xi, yi) and B(xj, yj)
determine a line
 yi = axi + b
yj = axj + b
yi = Axi + B
 Line with parameters (a, b)
A line is determined by
slope-intercept (a, b)
Problem with y=ax+b
• Solution for line function with slope a = 
• Polar coordinates representation of a line
translating Cartesian coordinates (x, y) to Polar coordinates (ρ,θ)
(x, y)
(  cos  ,  sin  )
Inner product = 0
(  cos  ,  sin  )
 ( x   cos  , y   sin  )  0
 x cos   y sin   
Hough transform
3 points, A, B and C in polar coordinates, these 3 points will have
three curves that intersect at (ρ0,θ0).
Implementing Hough transform
1. Input as Cartesian coordinate entries,
“ImgArr[i][j]”
2. Calculate the given point in polar space, we get
the curve calculate the values of  =
xcosθ+ysinθfor all discrete θ
3. Process the results enter into an accumulator
array whose size are the number of angles θ
and values 
4. Updating the accumulator array
5. Detect a peak position to the accumulative array
to find potential locations of straight lines
Implementing Hough transform
1. decide on a discrete set of values of θ
and  to use
   0 and -90θ180
 angles=[-90:180]*pi/180; % 弳度
2. For each edge point, calculate the values
of  = xcosθ+ysinθfor all discrete θ
 [x,y]=find(im); % im is binary, find nonzero
 r=floor(x*cos(angles)+y*sin(angles));
% floor: take integers
% what’s the dimension of r ?
Implementing Hough transform
(cont.)
3. Create an accumulator array whose size
are the number of angles θ and values 
rmax=max(r( find(r>0) ));
acc=zeros(rmax+1, length(angles));
for extra 0
4. Updating the accumulator array as we go
– for i=1:size(r,1)
– for j=1:size(r,2)
–
if r(i,j)>=0
–
acc( r(i,j)+1, j) =acc ( r(i,j)+1, j) +1;
– end; end; end;
Implementing Hough transform
(cont.)
• Exercise#1:
– c=imread(‘cameraman.tif’);
– edge=edge(c, ‘canny’);
– According to the previous slides, write a
Hough transform MATLAB function
– M=max(acc(:));
– [r, theta]=find(acc==M)
Plot the detected line
• r=152, theta=169=> index in [-90:180]
=> θ=169-1-90=78
y
Θ=78o
Exercise#2:
Calculate the coordinate of
the green points
x
MATLAB line function
• line([x1, x2], [y1, y2])
– Which plots in usual Cartesian coordinate
(x1, y1)
x
Exercise#3:
Plot your calculated
line on the image
(x2, y2)
y