Accumulator Array 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

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. Accumulator Array 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 ? Tow point A(x i, y i ) and B(x j, y j ) determine a line  y i = ax i + b y j = ax j + b  Line with parameters (a, b) (a k, b k ) (a, b) A line is determined by slope-intercept (a, b) y i = Ax i + B

Problem with y=ax+b Solution for line function with slope a =  Polar coordinates representation of a line Inner product = 0 (x, y) translating Cartesian coordinates (x, y) to Polar coordinates (ρ,θ)

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. I nput 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)); 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; for extra 0

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] x y => θ= =78 Θ=78 o Exercise#2: Calculate the coordinate of the green points

MATLAB line function line([x1, x2], [y1, y2]) –Which plots in usual Cartesian coordinate x y (x1, y1) (x2, y2) Exercise#3: Plot your calculated line on the image