Abstract
2DPalmHash Code (2DPHC) was proposed as a cancelable code for secure palmprint verification. In order to relieve the vertical and horizontal dislocation problems, palmprint codes, including 2DPHC, need to be shifted both in horizontal and vertical directions and matched repeatedly, which leads to high computational complexity. However, according to our analysis, horizontal-shift matching can be ignored. Therefore, the multiple-shift matching of 2DPHC can be greatly simplified. Simplified 2DPHC (S2DPHC) has three-fold advantages: (1) reduces matching complexity; (2) enhances changeability performance; (3) improves verification performance. Furthermore, the superiorities of S2DPHC over 2DPHC in terms of changeability and verification performances are validated via rigorously analysis and extensive experimentation.
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Acknowledgments
The authors would like to thank Biometric Research Center at Hong Kong Polytechnic University for providing us with the palmprint database. This work was supported by Basic Science Research Program through the National Research Foundation (NRF) of Korea funded by Ministry of Science, ICT and Future Planning (2013006574), National Natural Science Foundation of China (61305010, 61262019), Institute of BioMed-IT, Energy-IT and Smart-IT Technology (BEST), a Brain Korea 21 Plus Program, Yonsei University (2014-11-0007), Science and Technology Project of Education Department of Jiangxi Province(GJJ150715), Voyage Project of Jiangxi Province (201450), and Open Foundation of Key Laboratory of Jiangxi Province for Image Processing and Pattern Recognition (TX201604002).
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Appendix
Appendix
1.1 Proof of property 1
Since Z = XY,
Then \( {f}_Z(z)={\displaystyle {\int}_{-\infty}^{\infty}\frac{1}{\left|x\right|}}{f}_{X,Y}\left(x,\frac{z}{x}\right)dx={\displaystyle {\int}_{-\infty}^{\infty}\frac{1}{\left|x\right|}}{f}_X(x){f}_Y\left(\frac{z}{x}\right)dx \),\( {f}_Z\left(-z\right)={\displaystyle {\int}_{-\infty}^{\infty}\frac{1}{\left|x\right|}}{f}_X(x){f}_Y\left(\frac{-z}{x}\right)dx={\displaystyle {\int}_{-\infty}^{\infty}\frac{1}{\left|x\right|}}{f}_X(x){f}_Y\left(\frac{z}{x}\right)dx, \)Therefore, f Z (z) = f Z (−z). ■
1.2 Proof of property 2
Since Z = X + Y,then f Z (z) = ∫ ∞− ∞ f X (z − y)f X (y)dy, f Z (−z) = ∫ ∞− ∞ f X (−z − y)f X (y)dy. \( \begin{array}{l}{f}_Z(z)-{f}_Z\left(-z\right)\\ {}={\displaystyle {\int}_{-\infty}^{\infty }{f}_X\left(z-y\right)}{f}_Y(y) dy-{\displaystyle {\int}_{-\infty}^{\infty }{f}_X\left(-z-y\right)}{f}_Y(y) dy\\ {}={\displaystyle {\int}_{-\infty}^{\infty}\left[{f}_X\left(z-y\right)-{f}_X\left(-z-y\right)\right]}{f}_Y(y) dy\end{array} \), then g Y (y) = f X (z − y) − f X (−z − y) = f X (z − y) − f X (z + y), g Y (−y) = f X (z + y) − f X (−z + y) = f X (z + y) − f X (z − y).
Since g Y (y) = −g Y (−y), then g Y (y) is an odd function with respect of y.
Since f Y (y) is an even function with respect of y, so g Y (y)f Y (y) = [f X (z–y)–f X (−z–y)]f Y (y) is an odd function with respect of y, then f Z (z) − f Z (−z) = ∫ ∞− ∞ [f X (z − y) − f X (−z − y)]f Y (y)dy = 0.Therefore, f Z (z) = f Z (−z). ■
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Leng, L., Teoh, A.B.J. & Li, M. Simplified 2DPalmHash code for secure palmprint verification. Multimed Tools Appl 76, 8373–8398 (2017). https://doi.org/10.1007/s11042-016-3458-3
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DOI: https://doi.org/10.1007/s11042-016-3458-3