Biometrics & Security

Tutorial 9

• 1 (a) What is palmprint and palmprint authentication? (P10: 9-10)

• 1 (b) What kinds of features can be extracted from a palm? (P10:11)

• 1 (c) How to get Datum Point in P10: 14?

50 50 50

10 10 10

100 100 100

10 10 10

50 50 50

150

150

30

30

300

• 1 (d) Given two lines in a palmprint image, please explain how to get their Euclidean distance (P10: 21).

• If we mark each line as orientation features, how to understand their matching score: S(C1, C2) in P10:33?

i

j

(x 1 (i),y 1 (i))

(x 2 (i),y 2 (i))

(x 1 (j),y 1 (j))

(x 2 (i),y 2 (i))

1 2

• C1=[1 0 1 0]

• C2=[0 1 0 1]

• C3=[1 0 1 0]

• N=4

• H(C1,C2)=4

• S(C1,C2)=0

• H(C1,C3)=0

• S(C1,C3)=1

• 1 (e) What are the advantages and disadvantages of palmprint recognition? (P10: 58)

• 2. In line feature based offline palmprint recognition, four different directional templates are defined. Please try to explain why they can detect the corresponding directional line segments (See P10:19). Can you design another set of four different directional templates to determine line segments?

A line segment

10 10 90 90 10 10

10 10 90 90 10 10

10 10 90 90 10 10

10 10 90 90 10 10

Its cross-section 10 10 90 90 10 10

Our line detector

1112111

1112111

1112111

1112111

1112111

Its cross-section -1 -1 1 2 1 -1 -1

The inner product of two vectors increases with their similarity !

• Line detectorZero sumSimilar cross-section to a line segmentLarger than a line segmentThe better similarity, the better detection

• How many directions we need?The more directions, the more accuracy, the

less efficiencyA tradeoff is made according to experience

• 3. In P10:28-29, a segmentation approach, Tangent Points of the Finger Hole, is introduced. The line passing though the two points, (x1, y1) and (x2, y2), satisfies the inequality , for all i and j. Please understand this method and check the determinant rule.

cxmFyF jiji

cmxy

1 1F y mF x c

1 1F y mF x c

1 1F y mF x c

2 2F y mF x c

2 2F y mF x c

2 2F y mF x c

Y

X

y mx c 1 1,x y 2 2,x y

1 1F y mF x c

1 1F y mF x c

2 2F y mF x c

2 2F y mF x c

Y

X

y mx c 1 1,x y 2 2,x y

• 4. Assume that there is a Palmprint Code with 256 bytes by using texture feature. Please compute the number of features represented by the Palmprint Code (See P10:36). (256 * 8 / 2 == 1,024)

R e a l p a rt o f G a b o r F ilte r Im a gina ry p a rt o f G a b o r F ilte r

O rgina l im a ge

R e a l p a rt o f F e a tu re Im a gina ry p a rt o f F e a tu re

• Tut 7, 2. There are four steps in the Daugman’s approach (P8: 32-36). The third step generates IrisCode with 512 bytes. If 2 bits represent a feature, please compute the total number of features. (512*(8/2)=2,048)

N o rm a liz e d im a ge

Iris C o d e

10 1 0 1 0 10 1 0 1 0

15 1 5 1 5 15 1 5 1 5

20 2 0 2 0 20 2 0 2 0

G a b o r w a v e le t

1 0 1 0 10 1 0 10 10

2 0 2 0 20 2 0 20 20

2 0 2 0 20 2 0 20 2 0

0 0 00 0 0 00 0 0 0 0

0 1 01 0 1 01 0 1 0 1

0 1 01 0 1 01 0 1 0 1

T hre s ho ld

• 5. In texture features matching stage, hamming distance is used to measure the similarity of two palmprints, as shown in P10:37.

• What conclusion can you get from the normalized distance D0?

• Hint: Consider the normalized distance with the experimental results in P10:38.

• (The smaller in the normalized distance, the closer in two palmprints)

• C1=[1 0 1 0]

• C2=[0 1 0 1]

• C3=[1 0 1 0]

• N=4

• H(C1,C2)=4

• S(C1,C2)=0

• H(C1,C3)=0

• S(C1,C3)=1

1 0 1

0 1 0

1 0 1

1 0 1

0 1 0

1 0 1

0 0 0

1 1 1

1 1 1

RP RQ

IP IQ

0 0 0

1 1 0

1 0 1

N=3

D0 = 0.11

, , 0

, , 2

R R

I I

P i j Q i j

P i j Q i j