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CIS 663 Biometrics - Syracuse University

Question 3. Perform a singularity detection in the following data. Use the definitions used in week 3 live session slides. Your answer should include all missing values in the table and the type of singularity detected.

k

θ

δ

Δ

0

80

 

 

1

90

 

 

2

260

 

 

3

50

 

 

4

110

 

 

5

270

 

 

6

130

 

 

7

180

 

 

Use the following two equations to fill in the column titled δ and the column titled Δ.

δ (k)= θ ((k +1)mod N )- θ (k)

            δ (k)     if |δ (k)|< Π/2
Δ(k) =   δ (k)+Π  if δ (k) ≤ -Π/2
           δ (k)-Π   if δ (k) ≥ Π/2

Once you have filled in the table give the type of singularity you believe is being represented based on:

                         360, then whorl

                         180, then loop

k∈(0...7} Δ(k) = - 180, then delta

                          0, then singularity

Answer:

k

θ

Δ

0

80

560

0

1

90

1170

1170

2

260

4940

9880

3

50

1250

3750

4

110

3410

13640

5

270

9990

49950

6

130

5590

33540

7

180

8820

61740

At the point when the wavelet is the nth subordinate of a gaussian, the maxima bends are associated and experience the majority of the better scales. The rot rate of the maxima along the bends demonstrate the request of the separated singularities.

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Question 4. Determine the 3x3 binary pixel grid for:

a) A bifurcation point;

b) A non-minutiae point
(In live session we displayed and discussed the grid for the termination case).

For each specify:
- The values of b0 , ... , b7 for each case.
- What are their crossing numbers?

542_figure.jpg

Crossing Number = ∑i∈(1...7)|bi - b(i+1)mod8|

Answer:

The most regularly utilized strategy for particulars extraction in this classification is the Crossing Number (CN) idea. An enormous number of methods for particulars extraction accessible in the writing have a place with this class. This strategy is supported over different techniques for its computational proficiency and natural straightforwardness. This technique includes the utilization of the skeleton picture where the edge stream example is eight-associated. The particulars are extricated by checking the nearby neighbouhood of each edge pixel in the picture utilizing a 3X3 window.

B2

B3

B4

B1

B

B5

B0

B7

B6

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Question 5. The following image shows the values in grayscale. Perform the necessary steps to detect minutiae points. You don't need to detect any minutiae centered at the edge. Show your steps. Your result will include the coordinate of detected minutiae points and their types.

Answer:

%Program Description

%This program extracts the ridges and bifurcation from a fingerprint image

%Read Input Image

binary_image=im2bw(imread('input_1.tif'));

%Small region is taken to show output clear

binary_image = binary_image(120:400,20:250);

figure;imshow(binary_image);title('Input image');

%Thinning

thin_image=~bwmorph(binary_image,'thin',Inf);

figure;imshow(thin_image);title('Thinned Image');

%Minutiae extraction

s=size(thin_image);

N=3;%window size

n=(N-1)/2;

r=s(1)+2*n;

c=s(2)+2*n;

double temp(r,c);  

temp=zeros(r,c);bifurcation=zeros(r,c);ridge=zeros(r,c);

temp((n+1):(end-n),(n+1):(end-n))=thin_image(:,:);

outImg=zeros(r,c,2);%For Display

outImg(:,:,1) = temp .* 204;

outImg(:,:,2) = temp .* 204;

outImg(:,:,3) = temp .* 204;

for x=(n+1+10):(s(1)+n-10)

    for y=(n+1+10):(s(2)+n-10)

        e=1;

        for k=x-n:x+n

            f=1;

            for l=y-n:y+n

                mat(e,f)=temp(k,l);

                f=f+1;

            end

            e=e+1;

        end;

         if(mat(2,2)==0)

            ridge(x,y)=sum(sum(~mat));

            bifurcation(x,y)=sum(sum(~mat));

         end

    end;

end;

% RIDGE END FINDING

[ridge_x ridge_y]=find(ridge==2);

len=length(ridge_x);

%For Display

for i=1:len

    outImg((ridge_x(i)-3):(ridge_x(i)+3),(ridge_y(i)-3),2:3)=0;

    outImg((ridge_x(i)-3):(ridge_x(i)+3),(ridge_y(i)+3),2:3)=0;

    outImg((ridge_x(i)-3),(ridge_y(i)-3):(ridge_y(i)+3),2:3)=0;

    outImg((ridge_x(i)+3),(ridge_y(i)-3):(ridge_y(i)+3),2:3)=0;

 

    outImg((ridge_x(i)-3):(ridge_x(i)+3),(ridge_y(i)-3),1)=204;

    outImg((ridge_x(i)-3):(ridge_x(i)+3),(ridge_y(i)+3),1)=204;

    outImg((ridge_x(i)-3),(ridge_y(i)-3):(ridge_y(i)+3),1)=204;

    outImg((ridge_x(i)+3),(ridge_y(i)-3):(ridge_y(i)+3),1)=204;

end

%BIFURCATION FINDING

[bifurcation_x bifurcation_y]=find(bifurcation==4);

len=length(bifurcation_x);

%For Display

for i=1:len

    outImg((bifurcation_x(i)-3):(bifurcation_x(i)+3),(bifurcation_y(i)-3),1:2)=0;

    outImg((bifurcation_x(i)-3):(bifurcation_x(i)+3),(bifurcation_y(i)+3),1:2)=0;

    outImg((bifurcation_x(i)-3),(bifurcation_y(i)-3):(bifurcation_y(i)+3),1:2)=0;

    outImg((bifurcation_x(i)+3),(bifurcation_y(i)-3):(bifurcation_y(i)+3),1:2)=0;

 

    outImg((bifurcation_x(i)-3):(bifurcation_x(i)+3),(bifurcation_y(i)-3),3)=204;

    outImg((bifurcation_x(i)-3):(bifurcation_x(i)+3),(bifurcation_y(i)+3),3)=204;

    outImg((bifurcation_x(i)-3),(bifurcation_y(i)-3):(bifurcation_y(i)+3),3)=204;

    outImg((bifurcation_x(i)+3),(bifurcation_y(i)-3):(bifurcation_y(i)+3),3)=204;

end

figure;imshow(outImg);title('Minutiae');

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