qiang wang

8/10/2019 Qiang Wang

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CIS581-Presentation

Contour finding

Presented by: Wang , Qiang

Supervised by: Dr. Longin Jan Latecki


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Contour findingIntroduction

Output: Sequence of boundary pixels of thelargest object

Input: Binary image containing segmented objectsas black pixels

Apply to MPEG-7 Shape 1 data set. (1400 images)


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IntroductionSample images:


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Pre-processing: erase the noiseEdge detectionContour detectionDelete the redundant points

Simplify the contours

Five steps:Processing


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Pre-processingPre-processing: Erasing the noise

for num_bmp=1:1400

tmp=imread(strcat(int2str(num_bmp),'.bmp'));tmp=medfilt2(tmp,[10 10]);

x = zeros(size(tmp,1)+6,size(tmp,2)+6);x(4:size(tmp,1)+3,4:size(tmp,2)+3)=tmp;

x=x~=0;x=im2bw(x);

End% some other choices of filters:% tmp=filter2(1/25*ones(5,5),tmp);% tmp=ordfilt2(tmp,36,ones(6,6));


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Another way to erase the noise: Object Labeling

Find out all those pixels connected to each other, and label them withthe same number.

1 1 1 0 0 0 0 01 1 1 0 1 1 0 0

1 1 1 0 1 1 0 01 1 1 0 0 0 1 01 1 1 0 0 0 1 01 1 1 0 0 0 1 01 1 1 0 0 1 1 01 1 1 0 0 0 0 0

[labeled,numObjects] = bwlabel(bw,4)

1 1 1 0 0 0 0 01 1 1 0 2 2 0 0

1 1 1 0 2 2 0 01 1 1 0 0 0 01 1 1 0 0 0 3 01 1 1 0 0 0 3 01 1 1 0 0 3 3 01 1 1 0 0 0 0 0


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Edge detectionEdge detection

Fortunately there is a very useful function in matlab:

img_edge=edge(x);


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Edge detection


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Contour detectionProblem: after edge detection, we just get animage with all those points on the contour as 1while all the others as 0,we need to get thecoordinates of those 1 pixels (vertices).

Basic idea: just travel around all the contour,find out all the pixels of 1.


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Contour detectionHow to travel??

Need a starting point;Need to decide the direction to travel;Need to avoid any circle in traveling.

(The last one is the most tricky thing)


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Contour detectionSolution:

A matrixand

a vector


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Contour detectionThe matrix : Anti-clockwise

[1 1 0 -1 -1 -1 0 1 for x0 1 1 1 0 -1 -1 -1]; for y


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Contour detectionCode: [a b]=find(img_edge);counter=size(find(img_edge),1);point_x(1)=a(1);point_y(1)=b(1); % find out the starting point

neighbor_offset=[1 1 0 -1 -1 -1 0 1;0 1 1 1 0 -1 -1 -1];new_neighbor=[8 8 2 2 4 4 6 6];neighbor=1;

for i=1:counter %try to find out all those pixels on the contourwhile img_edge(point_x(i)+neighbor_offset(1,neighbor),point_y(i)+neighbor_offset(2,neighbor))==0

neighbor=mod(neighbor,8)+1;end

point_x(i+1)=point_x(i)+neighbor_offset(1,neighbor);point_y(i+1)=point_y(i)+neighbor_offset(2,neighbor);neighbor=new_neighbor(neighbor);

if and(point_x(1)==point_x(i+1),point_y(1)==point_y(i+1))break

endend


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Contour detection


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redundant pointsDelete the redundant pointsWhen there are many points on exact the same straight line, we just

need to keep the starting and ending points.

for i=1:size(point_x,2)-2;if or(and(point_x(i)==point_x(i+1),point_x(i+1)==point_x(i+2)),

and(point_y(i)==point_y(i+1),point_y(i+1)==point_y(i+2)))point_x(i+1)=[];point_y(i+1)=[];

endend


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Simplify the contourfunction K=evo(Z)n=length(Z);LM=norm(Z(2)-Z(n));LR=norm(Z(1)-Z(2));

LL=norm(Z(n-1)-Z(1));His(1)=LL+LR-LM;for j=2:n-1

LM=norm(Z(j-1)-Z(j+1));LR=norm(Z(j)-Z(j+1));LL=norm(Z(j-1)-Z(j));His(j)=LL+LR-LM;

end;LM=norm(Z(n-1)-Z(1));LR=norm(Z(n)-Z(1));LL=norm(Z(n-1)-Z(n));His(n)=LL+LR-LM;K=His;


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Simplify the contour


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Conclusioninput image: 512 * 512 pixels Output file: 50 points

input size: 250 KB Output size: < 400 bytes

Based on the output: we can do Image compression

Image processingFeature extraction and analysis

...


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