CN102521590B - Method for identifying left and right palm prints based on directions - Google Patents

Abstract

The invention discloses a method for identifying left and right palm prints based on directions. The method comprises the following steps of: (1) dividing a palm print image into a foreground area block and a background area block of the palm print; (2) performing two-dimensional Fourier transformation on the foreground area block of the image, thereby obtaining a direction theta (x, y) of the foreground area image; (3) calculating a direction histogram g(i) of the foreground area according to the direction theta (x, y) of the foreground area image, and performing one-dimensional mean value smoothing on the direction histogram g(i), thereby obtaining a smoothed histogram; (4) calculating the peak values of all peaks in the smoothed histogram and selecting the direction of the highest peak as the direction of a palm print outside area; and (5) judging left and right palms according to the direction of the palm print outside area, judging as the left palm when the direction is more than 90 degrees and judging as the right palm when the direction is less than 90 degrees. According to the method, the left and right palms are judged by utilizing the palm print direction, the accuracy is very high and can reach 98.92%, the number of the palm print identification is efficiently reduced, the efficiency of the whole identifying process is greatly increased, and the identifying accuracy is also increased.

Description

A kind of left and right palm grain identification method based on direction

Technical field

The present invention relates to the palmmprint recognition technology, particularly the left and right palm grain identification method based on direction.Belong to Digital Image Processing and biometrics identification technology field.

Background technology

As a kind of important biometrics identification technology, palmmprint is identified in the recent period just the interest that causes people gradually.On the depth & wideth of research, with the fingerprint recognition with long research history, to compare, the research of palmmprint identification can only be said also in initial period.But due to the outstanding characteristics that palmmprint itself has, make the palmmprint recognition technology have huge potentiality, the research of people to the palmmprint recognition technology, also become more and more deep.

For high-resolution palm print, itself and fingerprint image have a lot of similarities, therefore the general technology that is similar to fingerprint image processing and coupling of using is identified high-resolution palm print, but because palmmprint has area and the more unique point that is far longer than fingerprint, so recognition speed can reduce greatly.But, for palmmprint, it has the feature that fingerprint does not have, palmmprint has the difference of the left and right palm, if can carry out the differentiation of efficiently and accurately to the left and right palm, just can make the palmmprint that need to be identified reduce by half.Therefore, before palmmprint is identified, the left and right palm is differentiated, can effectively be reduced the number of being identified palmmprint, improved greatly the efficiency of whole identifying, simultaneously the also accuracy rate effect of haveing a certain upgrade to identifying.Therefore, palmmprint is carried out to the left and right palm and differentiate, just become and there is very much Practical significance.

Summary of the invention

The object of the invention is to propose a kind of for the palm recognition methods of left and right fast and effectively of Palm Print Recognition System automatically.

The present invention utilizes the palmmprint direction to be differentiated the left and right palm, and method is simple, efficient, and has very high accuracy rate.Concrete thought is: because the exterior lateral area of palmmprint generally has more consistent direction (as shown in Figure 1), other region directions are relatively mixed and disorderly, therefore after the direction of whole palmmprint being carried out to statistics with histogram, higher direction corresponding to peak in histogram is exactly generally the direction of exterior lateral area.Therefore, can obtain by the histogram of direction the principal direction of exterior lateral area.And because the palmprint image gathered does not have apparent in view rotation, therefore the direction of left and right palm exterior lateral area has obvious difference, the level of take is to the right the X-axis positive dirction, be the Y-axis positive dirction straight down, left palm exterior lateral area direction is 135 degree left and right, and right palm exterior lateral area direction is 45 degree left and right.Therefore, can utilize the principal direction of the exterior lateral area of trying to achieve to be differentiated the left and right palm.

Technical scheme of the present invention is: a kind of left and right palm grain identification method based on direction comprises the steps:

1) palmprint image is divided into to foreground area and the background area that has palmmprint; Background area is other outer zone of palmmprint, the color of foreground area will be deeper than background area, the purpose that palmprint image is cut apart is to eliminate the impact that background area produces the direction statistics, to exist foreground area and the background area of palmprint image to distinguish, follow-up direction calculating be only carried out in foreground area.

2) 2 dimension Fourier transforms are carried out in the foreground area piecemeal of image, obtain the direction θ (x, y) of foreground region image;

3) calculate the direction histogram g (i) of foreground area according to the direction θ (x, y) of foreground region image, and direction histogram g (i) is carried out to the mean value smoothing of one dimension, obtain the histogram after level and smooth

4) calculate the position at all peaks in smoothed histogram, select the direction of the direction at top as the palmmprint exterior lateral area;

5) according to the discriminating direction left and right palm of palmmprint exterior lateral area, direction is greater than the left palm of being judged to be of 90 degree, is less than the right palm of being judged to be of 90 degree.

Further, using foreground area and the background area of gray variance as the Image Segmentation Methods Based on Features palmprint image.

The method that described palmprint image is cut apart is:

A, palmprint image is divided into to the image block of w * w;

B, calculate the feature of the variance of gray scale in each image block as this piece, just obtain a gray variance image.Use gray variance as feature, can the realization of efficiently and accurately before background segment.

C, use automatic Selection of Image Threshold calculate segmentation threshold, and the image block that variance is more than or equal to this threshold value is defined as foreground area, and the image block that variance is less than to this threshold value is defined as background area.

The computing method of the gray variance image of described palmprint image are:

At first the gray average m (x, y) of computed image piece:

$m(x,y)=\frac{1}{w^2}\underset{i=(x-1)w+1}{\overset{\mathrm{xw}}{\mathrm{\Σ}}}\underset{j=(y-1)w+1}{\overset{\mathrm{yw}}{\mathrm{\Σ}}}I(i,j)$

The coordinate that wherein (x, y) is variance image, I (i, j) means the gray scale of point (i, j).

Then calculate variance image v (x, y):

$v(x,y)=\underset{i=(x-1)w+1}{\overset{\mathrm{xw}}{\mathrm{\Σ}}}\underset{j=(y-1)w+1}{\overset{\mathrm{yw}}{\mathrm{\Σ}}}{(I(i,j)-m(x,y))}^2$

Variance image is normalized to t[0,255] and be converted to integer:

$\tilde{v}(x,y)=[\frac{v(x,y)-v_{\min }}{v_{\max }-v_{\min }}\×255]$

V wherein _maxMean the variance maximal value, v _minMean the variance minimum value, [] rounds under meaning.

Described automatic threshold is chosen: utilize the automatic selected threshold of OTSU algorithm.

The method of concrete computed segmentation threshold value is:

According to the gray variance image

The counting statistics histogram:

$h\left(i\right)=\mathrm{card}\left(\right\{(x,y)|\tilde{v}(x,y)=i\}),i=\mathrm{0,1,2},...,255$

Wherein card () means set mid point number.

Second step, calculate inter-class variance:

g(t)＝w ₀(p ₀-p) ²+w ₁(p ₁-p) ²

P=w wherein ₀p ₀+ w ₁p ₁,

T ∈ [0,255] is the threshold value independent variable.

Final threshold value is:

$T=\underset{t\∈\left[\mathrm{0,255}\right]}{\arg \max }\left\{g\left(t\right)\right\}$

The described method according to Threshold segmentation foreground area and background area is:

The threshold value T that utilization obtains is right

Carry out binaryzation, obtain cutting apart figure:

$l(x,y)=\left\{\begin{array}{cc}1& \tilde{v}(x,y)\&GreaterEqual;T\\ 0& \tilde{v}(x,y)<T\end{array}\right.,$

Wherein 1 means foreground area, and 0 means background area.

The direction of described acquisition image block is:

A, palmprint image is divided into to the image block of w ' * w ', the image block that comprises foreground area is carried out 2 dimension Fourier transforms.

B, near response initial point in 2 dimension Fourier transform domains is set to 0; Impact with removal of images piece medium and low frequency part.

C, find the position (u of the maximum point of response ₀, v ₀), wherein, u and v mean the coordinate system on Fourier transform domain, level is to the right u axle positive dirction, is v axle positive dirction straight up.

The direction of D, computed image piece: the direction of image block is quantized to 0 degree to 179 degree.

$\mathrm{\&theta;}(x,y)=\arctan \left(\frac{u_0}{v_0}\right),$

Wherein to take level be x axle positive dirction to the right to the coordinate axis of image block direction, is y axle positive dirction straight down.

The described method according to image block direction calculating statistic histogram is:

g(i)＝card({(x，y)|θ(x，y)＝i})，i＝0，1，2，...，179

Wherein card () means set mid point number, the direction that θ (x, y) is image block.

Described histogram is carried out to the one dimension mean value smoothing, level and smooth method is:

$\tilde{g}\left(i\right)=\frac{1}{2n+1}\underset{j=-n}{\overset{n}{\mathrm{\&Sigma;}}}g(i+j),i=\mathrm{0,1,2},...,179$

Wherein: $g\left(k\right)=\left\{\begin{array}{cc}g(k-180)& k\&GreaterEqual;180\\ g(k+180)& k<0\end{array}\right.,$ N is smoothing parameter, k=i+j, and k, i, j are the independents variable of representative function.

Being calculated as of described histogram peak:

Have

I=0,1,2 ..., 179, think

It is the peak value at a peak.

Wherein $\tilde{g}\left(k\right)=\left\{\begin{array}{cc}\tilde{g}(k-180)& k\&GreaterEqual;180\\ \tilde{g}(k+180)& k<0\end{array}\right.,$ N ' is for judging the width parameter at peak.

Beneficial effect:

Method of the present invention can fast and effeciently be identified the left and right palmmprint, the accuracy of identification is 98.92%, before palmmprint is identified, the left and right palm is differentiated, can effectively reduce the number of being identified palmmprint, improve greatly the efficiency of whole identifying, simultaneously the also accuracy rate effect of haveing a certain upgrade to identifying.Therefore, palmmprint is carried out to the left and right palm and differentiate, just become and there is very much Practical significance.

The accompanying drawing explanation

Fig. 1 is the palmprint image schematic diagram.

Fig. 2 is divided into palmprint image according to the inventive method the process flow diagram of foreground area and background area;

Fig. 3 is that the method according to this invention is carried out image block the process flow diagram of Fourier transform;

Fig. 4 is the process flow diagram that the method according to this invention calculates directional image;

Fig. 5 is the process flow diagram that the method according to this invention calculates direction histogram;

Fig. 6 is that histogram peak is chosen schematic diagram.

Specific implementation method

Below by specific embodiment, the invention will be further described by reference to the accompanying drawings.

Fig. 1 is a palmprint image, and the arrow direction is palmmprint exterior lateral area direction, can see that the palmmprint direction is basically identical.

The embodiment of the present invention comprises the following steps:

1, at first carry out palmprint image (Fig. 2 cutting apart a):

1) variance is calculated: what palmprint image was divided into to w * w does not overlap mutually fritter, and w can be according to the streakline width value in palmprint image.

Calculate the gray variance of each piece, gray variance represents a piece, and all these gray variances have just formed variance image:

$v(x,y)=\underset{i=(x-1)w+1}{\overset{\mathrm{xw}}{\mathrm{\&Sigma;}}}\underset{j=(y-1)w+1}{\overset{\mathrm{yw}}{\mathrm{\&Sigma;}}}{(I(i,j)-m(x,y))}^2$

$m(x,y)=\frac{1}{w^2}\underset{i=(x-1)w+1}{\overset{\mathrm{xw}}{\mathrm{\&Sigma;}}}\underset{j=(y-1)w+1}{\overset{\mathrm{yw}}{\mathrm{\&Sigma;}}}I(i,j)$

Wherein I (i, j) means the gray scale of point (i, j), the coordinate that (x, y) is variance image.

Because gradation of image is generally 0-255, in order conveniently to calculate and to be convenient to, mean, the variogram obtained is normalized to [0,255] and is converted to integer:

$\tilde{v}(x,y)=[\frac{v(x,y)-v_{\min }}{v_{\max }-v_{\min }}\&times;255]$

V wherein _maxMean the variance maximal value, v _minMean the variance minimum value, [] rounds under meaning.

Fig. 2 (b) has provided a variance image that Fig. 2 a obtains after above-mentioned calculating.

2) automatic threshold is chosen: utilize the automatic selected threshold of OTSU algorithm, be specially:

At first calculate

Statistic histogram:

$h\left(i\right)=\mathrm{card}\left(\right\{(x,y)|\tilde{v}(x,y)=i\}),i=\mathrm{0,1,2},...,255$

Wherein card () means set mid point number.

Then calculate inter-class variance:

g(t)＝w ₀(p ₀-p) ²+w ₁(p ₁-p) ²

P=w wherein ₀p ₀+ w ₁p ₁,

T ∈ [0,255] is the threshold value independent variable.

Final segmentation threshold is:

$T=\underset{t\&Element;\left[\mathrm{0,255}\right]}{\arg \max }\left\{g\left(t\right)\right\}$

3) front background segment: utilize the segmentation threshold T obtained, right

Carry out binaryzation, obtain cutting apart figure:

$l(x,y)=\left\{\begin{array}{cc}1& \tilde{v}(x,y)\&GreaterEqual;T\\ 0& \tilde{v}(x,y)<T\end{array}\right.,$

1 expression prospect wherein, 0 means background.

Fig. 2 (c) has provided a segmentation result figure, and wherein white is foreground area, and black is background area.2, direction calculating:

The fritter that whole palmmprint is divided into to w ' * w ', because Fourier transform needs some powers that tile size is 2, so be taken as 32 * 32 in specific implementation, the image block that then each is dropped on to foreground area carries out 2 dimension Fourier transforms.Near response initial point in 2 dimension Fourier transform domains is set to 0, with the impact of removal of images piece medium and low frequency part, and then find out the position that responds maximum point, be designated as: (u ₀, v ₀), (level of take is to the right u axle positive dirction, is v axle positive dirction straight up) the direction of image block is:

$\mathrm{\&theta;}(x,y)=\arctan \left(\frac{u_0}{v_0}\right),$

Wherein to take level be x axle positive dirction to the right to the direction coordinate axis, is y axle positive dirction straight down.

Fig. 3 has provided the schematic flow sheet of image block Fourier transform, and Fig. 3 a is an image block, and Fig. 3 b is the image block after Fourier transform, and wherein the position of Fig. 3 b circle is the peak response position.Fig. 4 has provided the direction of palmprint image, and wherein Fig. 4 a is original palmprint image, and Fig. 4 b has shown the direction of Fig. 4 a.

3, direction histogram:

To direction θ obtained above (x, y) counting statistics histogram:

g(i)＝card({(x，y)|θ(x，y)＝i})，i＝0，1，2，...，179

Wherein card () means set mid point number.

Again histogram is carried out smoothly:

$\tilde{g}\left(i\right)=\frac{1}{2n+1}\underset{j=-n}{\overset{n}{\mathrm{\&Sigma;}}}g(i+j),i=\mathrm{0,1,2},...,179$

Wherein: $g\left(k\right)=\left\{\begin{array}{cc}g(k-180)& k\&GreaterEqual;180\\ g(k+180)& k<0\end{array}\right.,$ N is smoothing parameter, is taken as 5 in specific implementation.

Fig. 5 has provided the direction histogram after a palmprint image directional image and its correspondence level and smooth.Fig. 5 a is the palmprint image direction, and Fig. 5 b is the direction histogram after level and smooth.

4, outside principal direction is calculated:

Calculate the position at all peaks in smoothed histogram.

Provide the definition of histogram peak:

$\&ForAll;j\&Element;[-n^{\&prime;},n^{\&prime;}],$ Have $\tilde{g}\left(i\right)\&GreaterEqual;\tilde{g}(i+j),i=\mathrm{0,1,2},...,179$

Wherein $\tilde{g}\left(k\right)=\left\{\begin{array}{cc}\tilde{g}(k-180)& k\&GreaterEqual;180\\ \tilde{g}(k+180)& k<0\end{array}\right.,$ N ' is for judging the width parameter at peak, and this parameter is excessive or too smallly all can cause in histogram the peak value judgement inaccurate, and through overtesting, this parameter is taken as 5.

Find out peaks all in histogram according to above-mentioned definition, select the highest peak, the direction that it is corresponding, be the direction of exterior lateral area.Fig. 6 has provided the schematic diagram of a detection peak, and its orbicular spot and square mean histogrammic peak, and square means top, corresponding 33 degree in top in this figure.

5, the left and right palm is differentiated

After obtaining the principal direction of exterior lateral area, just can be differentiated the left and right palm.Exterior lateral area principal direction is greater than the left palm of being judged to be of 90 degree, is less than the right palm of being judged to be of 90 degree.In embodiment, exterior lateral area principal direction is 33 degree, is judged as the right palm.

Claims (7)

1. the left and right palm grain identification method based on direction, comprise the steps:

(1) palmprint image is divided into to foreground area piecemeal and the background area piecemeal of palmmprint;

(2) 2 dimension Fourier transforms are carried out in the foreground area piecemeal of image, obtain the direction θ (x, y) of foreground region image;

(3) calculate the direction histogram g (i) of foreground area according to the direction θ (x, y) of foreground region image, and the mean value smoothing that direction histogram g (i) is carried out to one dimension, smoothed histogram obtained

The histogrammic method of described calculated direction is:

g(i)=card({(x,y)|θ(x,y)=i})，i=0,1,2,…,179

Wherein, card () means set mid point number, the direction that θ (x, y) is image block; The coordinate that (x, y) is variance image;

Describedly the direction histogram carried out to level and smooth method be:

$\tilde{g}\left(i\right)=\frac{1}{2n+1}\underset{j=-n}{\overset{n}{\mathrm{\&Sigma;}}}g(i+1),i=\mathrm{0,1,2},...,179$

Wherein: $g\left(k\right)=\left\{\begin{array}{cc}g(k-180)& k\&GreaterEqual;180\\ g(k+180)& k<0\end{array}\right.,$ N is smoothing parameter, k=i+j, and k, i, j are the independents variable of representative function;

(4) calculate the peak value at all peaks in smoothed histogram, select the direction of the direction at top as the palmmprint exterior lateral area;

The computing method of described smoothed histogram peak value are:

Have

I=0,1,2 ..., 179,

Be the peak value at a peak,

Wherein $\tilde{g}\left(k\right)=\left\{\begin{array}{cc}\tilde{g}(k-180)& k\&GreaterEqual;180\\ \tilde{g}(k+180)& k<0\end{array}\right.,$ N ' is the width parameter at judgement peak, k=i+j, and k, i, j are the independents variable of representative function;

(5) according to the discriminating direction left and right palm of palmmprint exterior lateral area, direction is greater than the left palm of being judged to be of 90 degree, is less than the right palm of being judged to be of 90 degree.

2. the left and right palm grain identification method based on direction according to claim 1, is characterized in that, usings foreground area and the background area of gray variance as the Image Segmentation Methods Based on Features palmprint image.

3. the left and right palm grain identification method based on direction according to claim 1 and 2, is characterized in that, the described method of cutting apart palmprint image is:

A, palmprint image is divided into to the image block of w * w;

B, calculate the variance of gray scale in each image block, form the gray variance image

C, use automatic Selection of Image Threshold computed segmentation threshold value T, the image block that variance is more than or equal to this threshold value is defined as foreground area, and the image block that variance is less than to this threshold value is defined as background area.

4. the left and right palm grain identification method based on direction according to claim 3, is characterized in that, the method for described calculating gray variance image is:

The first step, the gray average m (x, y) of computed image piece:

$m(x,y)=\frac{1}{w^2}\underset{i=(x-1)w+1}{\overset{\mathrm{xw}}{\mathrm{\&Sigma;}}}\underset{j=(y-1)w+1}{\overset{\mathrm{yw}}{\mathrm{\&Sigma;}}}I(i,j)$

Wherein, the coordinate that (x, y) is variance image, I (i, j) means the gray scale of point (i, j)

Second step, calculate variance image v (x, y):

$v(x,y)=\underset{i=(x-1)w+1}{\overset{\mathrm{xw}}{\mathrm{\&Sigma;}}}\underset{j=(y-1)w+1}{\overset{\mathrm{yw}}{\mathrm{\&Sigma;}}}{(I(i,j)-m(x,y))}^2$

The 3rd step normalizes to variance image v (x, y) [0,255] and is converted to integer:

$\tilde{v}(x,y)=[\frac{v(x,y)-v_{\min }}{v_{\max }-v_{\min }}\&times;255]$

Wherein, v _maxMean the variance maximal value, v _minMean the variance minimum value, [] rounds under meaning.

5. the left and right palm grain identification method based on direction according to claim 3, is characterized in that, the method for described computed segmentation threshold value is:

The first step, according to the gray variance image

The counting statistics histogram:

$h\left(i\right)=\mathrm{card}\left(\right\{(x,y)|\tilde{v}(x,y)=i\}),i=\mathrm{0,1,2},...,255$

Wherein card () means set mid point number;

Second step, calculate inter-class variance:

g(t)=w ₀(p ₀-p) ²+w ₁(p ₁-p) ²

Wherein, p=w ₀p ₀+ w ₁p ₁, $w_0=\frac{N_0}{N},$ $w_1=\frac{N_1}{N},$ $p_0=\frac{\underset{i=0}{\overset{t}{\mathrm{\&Sigma;}}}\mathrm{ih}\left(i\right)}{N_0},$ $p_1=\frac{\underset{i=t+1}{\overset{255}{\mathrm{\&Sigma;}}}\mathrm{ih}\left(i\right)}{N_1},$ $N_0=\underset{i=0}{\overset{t}{\mathrm{\&Sigma;}}}h\left(i\right),$ $N_1=\underset{i=t+1}{\overset{255}{\mathrm{\&Sigma;}}}h\left(i\right),$ $N=\underset{i=0}{\overset{255}{\mathrm{\&Sigma;}}}h\left(i\right),$ T ∈ [0,255] is the threshold value independent variable;

The 3rd step, calculated threshold T:

$T=\underset{t\&Element;\left[\mathrm{0,255}\right]}{\arg \max }\left\{g\left(t\right)\right\}.$

6. the left and right palm grain identification method based on direction according to claim 3, is characterized in that, the method for cutting apart foreground area and background area according to segmentation threshold T is:

Utilize segmentation threshold T, to the gray variance image

Carry out binaryzation, obtain cutting apart figure:

$l(x,y)=\left\{\begin{array}{cc}1& \tilde{v}(x,y)\&GreaterEqual;T\\ 0& \tilde{v}(x,y)<T\end{array}\right.,$

Wherein 1 means foreground area, and 0 means background area.

7. the left and right palm grain identification method based on direction according to claim 1: it is characterized in that, the computing method of described image block direction are:

A, palmprint image is divided into to the image block of w ' * w ', the image block that comprises foreground area is carried out 2 dimension Fourier transforms;

B, near response initial point in 2 dimension Fourier transform domains is set to 0;

C, definition u and v are the coordinate system on Fourier transform domain, find the position (u of the maximum point of response ₀, v ₀), find the position (u that responds maximum point ₀, v ₀);

The direction of D, computed image piece:

$\mathrm{\&theta;}(x,y)=\arctan \left(\frac{u_0}{v_0}\right),$

Wherein to take level be x axle positive dirction to the right to the coordinate axis of image block direction, is y axle positive dirction straight down.

Images