Boundary Finding
Examples
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Department of Electrical Engineering and Computer Science
Example 1 – Iterative polygonal fit
• Linear
approximation:
approximate a
segment by a
line between its
endpoints; split
at point with
largest error,
repeat until
approximation is
good
2
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clear all
close all
I = imread('cameraman.tif');
E = edge(I, 'canny');
imshow(E, 'InitialMagnification', 300);
hold on
% Turn hold on so we can "plot" on top of image
% Click on the first point along the contour.
fprintf('Pick first point on a contour\n');
while true
p1 = round(ginput(1)); % Get x,y coords
if ~E(p1(2),p1(1))
fprintf('Try again\n');
else
break;
end
end
plot(p1(1),p1(2), 'g*');
Matlab
program for
polygonal fit
(1 of 4)
% Trace the boundary, starting from that point. Use 'W' as initial step
% direction. Matlab's "bwtraceboundary" seems to have problems occasionally
% when the neighbor to the west is occupied. So if it is, let's move p1 to
% that point.
while E(p1(2),p1(1)-1)
% Check if neighbor to the left is occupied
p1(1) = p1(1)-1;
end
pts = bwtraceboundary(E, [p1(2),p1(1)], 'W', 8, Inf, 'clockwise');
N = size(pts, 1);
% Number of points
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% Click on the 2nd point along the contour.
fprintf('Pick 2nd point on the contour\n');
while true
p2 = round(ginput(1)); % Get x,y coords
if ~E(p2(2),p2(1))
fprintf('Try again\n');
else
break;
end
end
plot(p2(1),p2(2), 'g*');
Matlab
program for
polygonal fit
(2 of 4)
% Get the indices of the 1st point in our list.
indices1 = find( (pts(:,1)==p1(2)) & (pts(:,2)==p1(1)) );
% Get the indices of the 2nd point in our list.
indices2 = find( (pts(:,1)==p2(2)) & (pts(:,2)==p2(1)) );
assert(~isempty(indices2), 'Can''t find point2 along the contour');
% Find the pair of indices that are closest to each other.
for i=1:length(indices1)
for j=1:length(indices2)
d(i,j) = abs(indices2(j) - indices1(i));
end
end
[i,j] = find(d == min(d(:)));
if indices1(i)<indices2(j)
i1 = indices1(i);
i2 = indices2(j);
else
i1 = indices2(j);
i2 = indices1(i);
end
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4
% Get points from p1 to p2.
pts = pts(i1:i2, :);
plot(pts(:,2), pts(:,1), 'g*');
% Fit line segments to the points in the contour.
% "lineSegs" consists of
%
[x0, y0, x1, y1]
lineSegs = fitLineSegments(pts, [], 2.0);
Each row of
Matlab
program for
polygonal fit
(3 of 4)
% Draw final line segments
figure, imshow(I, 'InitialMagnification', 300);
for iSeg=1:size(lineSegs,1)
p0 = lineSegs(iSeg,1:2);
p1 = lineSegs(iSeg,3:4);
line([p0(1) p1(1)], [p0(2) p1(2)], 'Color', 'r', 'LineWidth', 3.0);
end
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Department of Electrical Engineering and Computer Science
function lineSegs = fitLineSegments(pts, lineSegs, THRESH)
% Input array pts is size (N,2), where each row is the coordinates of a
% point on the contour, (r,c).
N = size(pts,1);
% Number of points along contour
p1 = pts(1,:); % First point on contour
p3 = pts(end,:); % Last point on contour
% Draw line from p1 to p3
line([p1(2) p3(2)], [p1(1) p3(1)], 'Color', 'r', 'LineWidth', 3.0);
pause
Matlab
program for
polygonal fit
(4 of 4)
% Find the point p2 that is the furthest from the line from p1 to p3.
v = [p3(2)-p1(2); -(p3(1)-p1(1))]; % A vector perpendicular to that line
v = v/norm(v);
% Get vectors from the first endpoint to all other points.
r = pts - repmat(p1,N,1);
% The distance from each point to the line is just abs(dot(v,r))
d = abs( v(1)*r(:,1) + v(2)*r(:,2) );
[dmax,i2] = max(d);
% Point p2 is the furthest from the line
% If dmax is less than the threshold, then we are done.
if dmax < THRESH
lineSegs = [lineSegs; p1(2) p1(1) p3(2) p3(1)];
else
% Recursively fit line segments to each piece.
lineSegs = [lineSegs;
fitLineSegments(pts(1:i2,:), [], THRESH);
fitLineSegments(pts(i2:end,:), [], THRESH)];
end
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return
Colorado School of Mines
Department of Electrical Engineering and Computer Science
Example 2 – find a square target
• Find the square
in the image
• Squares are
frequently used as
“fiducial” targets
“ARToolkit”, www.hitl.washington.edu/artoolkit
Example video http://www.youtube.com/watch?v=5M-oAmBDcZk
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Department of Electrical Engineering and Computer Science
7
A simple method
• Get the boundary of a
connected component
• Find the point that is
furthest from the
centroid – assume that
is a corner
• Then find the point
furthest from that
detected corner –
assume that is the
opposite corner
• Then find the points
that are furthest from
the line connecting the
two opposite corners
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• Assume that the square is
not too large and not too
small
Department of Electrical Engineering and Computer Science
8
• Threshold image
• Find connected components
clear all
close all
I = imread('square.jpg'); % open file
I = rgb2gray(I);
imshow(I,[]);
BW = im2bw(I, graythresh(I));
BW = ~BW;
% invert the image so we look for a white square
figure, imshow(BW);
% Find connected components
[L,nBlobs] = bwlabel(BW);
blobs = regionprops(L);
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% Look at the blobs and find the first one that looks like a square.
fFoundSquare = false;
% This will be set to true if a square is found
for i=1:nBlobs
if blobs(i).Area > 40000 || blobs(i).Area < 2000
continue;
• For each blob,
end
% Find a point on the boundary of this blob
[rows,cols] = find(L==i);
[r0,i0] = min(rows);
c0 = cols(i0);
•
get its boundary
Find point
furthest from
centroid
% Get coordinates (row,col) along boundary
pts = bwtraceboundary(BW, [r0 c0], 'N');
N = size(pts,1);
c = mean(pts);
% Number of points along the boundary
% Get centroid
% Find the point furthest from centroid
dp = pts - repmat(c,N,1);
d = dp(:,1).^2 + dp(:,2).^2;
[~,i1] = max(d);
% Assume that this is a corner
p1 = pts(i1,:);
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Department of Electrical Engineering and Computer Science
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•
•
Find point furthest from first corner
Find the points that are furthest from the
line connecting the two opposite corners
% Get vectors from the first corner to all other points
r = pts - repmat(p1,N,1);
% Find the point furthest from first corner
d = r(:,1).^2 + r(:,2).^2;
[~,i3] = max(d);
% Assume that this is the opposite corner
p3 = pts(i3,:);
% Find the points that are the furthest from the line from i1 to i3.
v = [p3(2)-p1(2); -(p3(1)-p1(1))]; % A vector perpendicular to that line
% The signed distance from each point p to the line is just dot(v,r)
d = v(1)*r(:,1) + v(2)*r(:,2);
[~,i2] = max(d);
% Point 2 is to the right of 1->3
p2 = pts(i2,:);
[~,i4] = min(d);
% Point 3 is to the left of 1->3
p4 = pts(i4,:);
% So the order of point is 1,2,3,4 in counterclockwise order
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Department of Electrical Engineering and Computer Science
11
•
•
As partial verification, check side lengths
Break out of loop if a square is found
% Verify that this is a square - there are many ways to do this.
% We will just check to see if the sides are all about the same length.
s12 = norm(p1-p2);
s23 = norm(p2-p3);
s34 = norm(p3-p4);
s41 = norm(p4-p1);
smax = max([s12,s23,s34,s41]);
sthresh = 0.7;
% minimum ratio of lengths
if (s12/smax > sthresh) && (s23/smax > sthresh) && ...
(s34/smax > sthresh) && (s41/smax > sthresh)
fFoundSquare = true;
break;
end
end
hold on
plot(pts(:,2), pts(:,1), 'g*');
hold off
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Department of Electrical Engineering and Computer Science
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• Choose the first point to be the one closest to the top left
• Draw a colored marker on each corner
c = [p1; p2; p3; p4];
% Arbitrarily choose the first point to be the point closest to the top left
d = c(:,1).^2 + c(:,2).^2;
[~,i0] = min(d);
for i=1:4
i1 = i0+(i-1);
if i1>4 i1=i1-4; end
corners(i,:) = c(i1, :);
end
figure(1)
mycolors = ['r', 'g', 'b', 'y'];
hold on
for i=1:4
plot(corners(i,2), corners(i,1), 's', 'Color', mycolors(i));
end
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Example 3 – Hough peaks
• The left image below is a noisy 200x200 image of two
squares, rotated at 45 degrees. An edge detection
operation is performed to yield the binary edge image on
the right.
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Department of Electrical Engineering and Computer Science
Example 3 – Hough peaks (continued)
• Recall the normal representation of a line:
x cos θ + y sin θ = ρ
θ
x
ρ
y
• Where are the peaks in the Hough transform; i.e., what
are the values of (ρ,θ) at the peaks?
• Describe the relative height of the peaks to each other.
15
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Department of Electrical Engineering and Computer Science
Example 3 - Solution
• The geometry is as shown below.
– The lines at +45° have ρ equal to 50√2, 100√2, or 150√2.
– The lines at -45° have ρ equal to 50√2, 0, or -50√2. (Or
equivalently, you could have one line at θ = -45°, ρ = 50√2,
another at θ = -45°, ρ = 0, and the third at θ = 135°, ρ = 50√2.)
• The peaks at (+45°, 100√2), (-45°, 0) are twice as big as
the others.
ρ
+45°
-45°
x
x
ρ
y
y
16
Colorado School of Mines
Department of Electrical Engineering and Computer Science
Example 4 – Generalized Hough
• We’ll first create an “R-table” from the template object
• Then try to find instances of the object in the test image
Template object
Test image
Image: “part.jpg”
Image: “parts.jpg”
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Department of Electrical Engineering and Computer Science
% Create an R-table from the template image T.
% Assume that the object of interest is centered in the image.
clear all, close all;
T = imread('part.jpg');
DY = conv2(T,fspecial('prewitt'));
DX = conv2(T,(-1.*(fspecial('prewitt')')));
GMAG = sqrt(DX.^2 + DY.^2);
[iHeight,iWidth] = size(GMAG);
ixCen = round(iWidth/2);
iyCen = round(iHeight/2);
iThetaMax = 64;
% Center of object
% Angles are quantized into this many bins
Rcount = zeros(1,iThetaMax);
gthresh = 37;
% This is the count of vectors for each angle
% A threshold for gradient magnitude
for ix=1:iWidth
for iy=1:iHeight
if GMAG(iy,ix) > gthresh
theta = atan2(DY(iy,ix),DX(iy,ix));
iTheta = round((180/pi)*theta*iThetaMax/360)+1;
if (iTheta <= 0)
iTheta = iTheta + iThetaMax;
end
% Calculate vector to object center from this edge point
rx = ixCen - ix;
ry = iyCen - iy;
% Store in R-table
n = Rcount(iTheta) + 1;
Rx(iTheta,n) = rx;
Ry(iTheta,n) = ry;
Rcount(iTheta) = n;
% # vectors stored
end
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end
end
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Department of Electrical Engineering and Computer Science
% Perform the Generalized Hough Transform to find instances of an object in image I.
% Assume that the R-table has already been created.
I = imread('parts.jpg');
DY = conv2(I,fspecial('prewitt'));
DX = conv2(I,(-1.*(fspecial('prewitt')')));
GMAG = sqrt(DX.^2 + DY.^2);
[iHeight,iWidth] = size(GMAG);
H = zeros(iHeight,iWidth);
% zero the accumulator array
iThetaMax = length(Rcount);
% Number of discrete angles in the R-table
gthresh = 37;
% A threshold for gradient magnitude
for ix=1:iWidth
for iy=1:iHeight
if GMAG(iy,ix) > gthresh
theta = atan2(DY(iy,ix),DX(iy,ix));
iTheta = round((180/pi)*theta*iThetaMax/360)+1;
if (iTheta <= 0)
iTheta = iTheta + iThetaMax;
end
% Access R-table to get all the vectors stored at this angle
for i=1:Rcount(iTheta)
rx = Rx(iTheta,i);
ry = Ry(iTheta,i);
x0 = rx + ix;
y0 = ry + iy;
% Estimated location of object origin
if (x0 > 0 && x0 <= iWidth)
if (y0 > 0 && y0 <= iHeight)
H(y0,x0) = H(y0,x0) + 1;
end
end
% increment accumulator array
end
end
end
end
imshow(H,[])
Display the accumulator
array
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Department
of Electrical Engineering and Computer Science
19