biometrics & security tutorial 9. 1 (a) what is palmprint and palmprint authentication? (p10:...
TRANSCRIPT
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