An Introduction to the IrisCode Theory

Abstract

IrisCode is the most successful iris recognition method. Developed for over 18 years, IrisCode still dominates the market even though numerous iris recognition algorithms have been proposed in the academics. Currently, more than 60 million people have been mathematically enrolled by this algorithm. Its computational advantages, including high matching speed, predictable false acceptance rates, and robustness against local brightness and contrast variations, play a significant role in its commercial success. To further these computational advantages, researchers have modified this algorithm to enhance iris recognition performance and recognize other biometric traits (e.g., palm print). Many scientific papers on iris recognition have been published, but its theory is almost completely ignored. In this chapter, we will report our most recent theoretical work on the IrisCode.

Appendices

Appendices

16.1.1 Appendix A

This appendix proves that $\left\| {{{M}_R}} \right\| = \left\| {{{M}_I}} \right\|$ when $k\to \infty$,where $k = \omega \beta$.

Considering ${{\left\| {{{M}_R}} \right\|}^2} - {{\left\| {{{M}_I}} \right\|}^2}$

$$ = \int {{{{\int {\left( {{{e}^{{ - \frac{{{{\rho}^2}}}{{{{\alpha}^2}}}}}}{{e}^{{ - \frac{{{{\varphi}^2}}}{{{{\beta}^2}}}}}}\cos \left( {\omega \varphi } \right)} \right)} }}^2}\textit{d}\,\,\rho \textit{d}\varphi - } \int {{{{\int {\left( {{{e}^{{ - \frac{{{{\rho}^2}}}{{{{\alpha}^2}}}}}}{{e}^{{ - \frac{{{{\varphi}^2}}}{{{{\beta}^2}}}}}}\sin \left( {\omega \varphi } \right)} \right)} }}^2}\textit{d}\,\,\rho \textit{d}\varphi } $$

$$ = \int {\int {{{e}^{{ - \frac{{2{{\rho}^2}}}{{{{\alpha}^2}}}}}}{{e}^{{ - \frac{{2{{\varphi}^2}}}{{{{\beta}^2}}}}}}\left( {{{{\cos }}^2}\left( {\omega \varphi } \right) - {{{\sin }}^2}\left( {\omega \varphi } \right)} \right)\textit{d}\,\,\rho \textit{d}\varphi } } $$

$$ = \int {{{e}^{{ - \frac{{2{{\rho}^2}}}{{{{\alpha}^2}}}}}}} \textit{d}\,\,\rho \int {{{e}^{{ - \frac{{2{{\varphi}^2}}}{{{{\beta}^2}}}}}}\left( {{{{\cos }}^2}\left( {\omega \varphi } \right) - {{{\sin }}^2}\left( {\omega \varphi } \right)} \right)\textit{d}\varphi } $$

$$ = \frac{{\alpha \sqrt {{2\pi }} }}{2}\int {{{e}^{{ - \frac{{2{{\varphi}^2}}}{{{{\beta}^2}}}}}}\left( {{{{\cos }}^2}\left( {\omega \varphi } \right) - {{{\sin }}^2}\left( {\omega \varphi } \right)} \right)\textit{d}\varphi } $$

Let $\gamma = \frac{k}{\beta}\varphi$. Thus,

$$ = \frac{{\alpha \beta \sqrt {{2\pi }} }}{{2k}}\int {{{e}^{{ - \frac{{2{{\gamma}^2}}}{{{{k}^2}}}}}}\left( {{{{\cos }}^2}\left( \gamma \right) - {{{\sin }}^2}\left( \gamma \right)} \right)\textit{d}\gamma } $$

$$ = \frac{{\alpha \beta \sqrt {{2\pi }} }}{{2k}}\int {{{e}^{{ - \frac{{2{{\gamma}^2}}}{{{{k}^2}}}}}}\cos \left( {2\gamma } \right)\textit{d}\gamma } $$

Let $2\gamma = \tau$

$$ = \frac{{\alpha \beta \sqrt {{2\pi }} }}{{4k}}\int {{{e}^{{ - \frac{{{{\tau}^2}}}{{2{{k}^2}}}}}}\cos \left( \tau \right)\textit{d}\tau } $$

$$ = \frac{{\alpha \beta \sqrt {{2\pi }} }}{{4k}}\sqrt {{2\pi }} k{{e}^{{ - \frac{{{{k}^2}}}{2}}}} $$

$$ = \frac{1}{2}\alpha \beta \pi {{e}^{{ - \frac{{{{k}^2}}}{2}}}}. $$

${{\left\| {{{M}_R}} \right\|}^2} - {{\left\| {{{M}_I}} \right\|}^2}$ is always greater than zero because $\alpha$, $\beta$, and $k$ are greater than zero. Note that $\mathop{{\lim }}\limits_{{k\to \infty }} \left( {{{{\left\| {{{M}_R}} \right\|}}^2} - {{{\left\| {{{M}_I}} \right\|}}^2}} \right) = \mathop{{\lim }}\limits_{{k\to \infty }} \frac{1}{2}\alpha \beta \pi {{e}^{{ - \frac{{{{k}^2}}}{2}}}} = 0$.

16.1.2 Appendix B

This appendix shows that the phase distance can be calculated through bitwise hamming distance.

Let two winning indexes be j − 1 and j − 1 + k, where $1 \leq j \leq j + k < 2n$. Their phase distance is $\min (k,\;2n - k)$. By using the coding scheme given in Fig. 16.4, they are represented by the jth and j + kth column vectors of matrix B. We would like to prove that

$$ \sum\limits_{{i = 1}}^n {{{b}_{{i,j}}}\otimes {{b}_{{i,j + k}}}} = \min \left( {k,\;2n - k} \right). $$

Because all b _i,j are either one or zero, $\sum\nolimits_{{i = 1}}^n {{{b}_{{i,j}}}\otimes {{b}_{{i,j + h}}}} = \sum\nolimits_{{i = 1}}^n {{{b}_{{i,j}}} - {{b}_{{i,j + k}}}}$.

Case 1::

If $j \leq n$ and $j + k \leq n$:

From the definition of A, we know $\sum\nolimits_{{i = 1}}^n {\left| {{{b}_{{i,j}}} - {{b}_{{i,j + k}}}} \right| = k}$.

Case 2::

If $j > n$ and $j + k > n$:

As in Case 1, we know $\sum\nolimits_{{i = 1}}^n {\left| {{{b}_{{i,j}}} - {{b}_{{i,j + k}}}} \right| = k}$.

Case 3::

If $j \leq n$ and $j + k > n$ and $k \leq n$:

$$ \mathrm{Consider}\quad {{b}_{{i,j}}} = 1\quad \mathrm{and}\quad {{b}_{{i,j + k}}} = 1. $$

(16.20)

From the definition of A, we have $1 \leq i < j$ and $j + k - n \leq i \leq n$.

Then, $j + k - n \leq i < j$.

The number of i that satisfies condition (16.20) is

$$ \max \left( {0,\;j - \left( {j + k - n} \right)} \right). $$

(16.21)

Since $k \leq n$, $\max \left( {0,\;j - \left( {j + k - n} \right)} \right) = n - k$.

$$ \mathrm{Consider}\quad {{b}_{{i,j}}} = 0\quad \mathrm{and}\quad {{b}_{{i,j + k}}} = 0. $$

(16.22)

From the definition of A, we have $i \geq j$ and $i < j + k - n$.

Then $j \leq i < j + k - n$.

The number of i that satisfies condition (16.22) is

$$ \max \left( {0,\;j + k - n - j} \right). $$

(16.23)

Since $k \leq n$, $\max \left( {0,\;k - n} \right)$=0.

Thus, $\sum\nolimits_{{i = 1}}^n {\left| {{{b}_{{i,j}}} - {{b}_{{i,j + k}}}} \right|} = n - \left( {n - k} \right) = k$.

Case 4::

If $j \leq n$ and $j + k > n$ and $k > n$:

$$ \mathrm{Consider}\quad {{b}_{{i,j}}} = 1\quad \mathrm{and}\quad {{b}_{{i,j + k}}} = 1. $$

(16.24)

From (16.21), the number of i that satisfies condition (16.24) is $\max \left( 0,\;j\ - \right.$ $\left.\left( {j + k - n} \right)\right)$.

Since $k > n$, $\max \left( {0,\;n - k} \right) = 0$.

$$ \mathrm{Consider}\quad {{b}_{{i,j}}} = 0\quad \mathrm{and}\quad {{b}_{{i,j + k}}} = 0. $$

(16.25)

From (16.23), the number of i that satisfies condition (16.25) is $\max \left( 0,\;j + \right.$ $\left. k - n - j \right)$.

Since k > n, $\max \left( {0,\;k - n} \right) = k - n$.

Thus, $\sum\nolimits_{{i = 1}}^n {\left| {{{b}_{{i,j}}} - {{b}_{{i,j + k}}}} \right|} = n - (k - n) = 2n - k$.

Thus, $\sum\nolimits_{{i = 1}}^n {{{b}_{{i,j}}}\otimes {{b}_{{i,j + k}}}} = k$ for Cases 1–3, and $\sum\nolimits_{{i = 1}}^n {{{b}_{{i,j}}}\otimes {{b}_{{i,j + k}}}} = 2n - k$ for Case 4. Since $2n - k \geq k$ for Cases 1–3 and $2n - k < k$ for Case 4, $\sum\nolimits_{{i = 1}}^n {{{b}_{{i,j}}}\otimes {{b}_{{i,j + k}}}} = \min \left( {k,\;2n - k} \right).$