CN104504654B - High-resolution image reconstruction method based on directivity gradient - Google Patents

Abstract

The invention discloses a high-resolution image reconstruction method based on a directivity gradient. The high-resolution image reconstruction method comprises the following steps: 1) obtaining a main texture direction [Theta] for inputting a low-resolution image I belongs to R<m*n>; 2) utilizing a bicubic interpolation method to interpolate the low-resolution image I into a high-dimensionality interpolation image IL; 3) carrying out variable initialization and parameter selection; 4) obtaining a fuzzy operator matrix Fh and a conjugate fuzzy operator matrix FHh; 5) updating the value of xij; 6) updating the value of IH; 7 updating the value of Xij; and 8) judging whether convergence is carried out. A regular item of the directivity gradient is introduced to better show and dig direction texture information implied in the image so as to introduce prior information with better value into an ill-posed high-resolution reconstruction problem, and the high-resolution image with a better effect is obtained.

Description

High-definition picture reconstructing method based on directional gradient

Technical field

The invention belongs to image processing field, it is related to a kind of utilization orientation gradient and convex Optimization Solution algorithm from low resolution The method reconstructing high-definition picture in rate image.

Background technology

Reconstruct a high-resolution image to be always at computer vision and image from single width low-resolution image One important topic in reason field, is also the problem of a great challenge.

Common technological means can be divided into three major types: interpolation class method, machine learning class method and based on sparse about The optimal reconfiguration method of bundle item.Realizing of interpolation class method is fairly simple, but the high-definition picture generally yielding is very fuzzy.Machine Device study class method is by learning to a large amount of low resolution and high-definition picture feature, thus obtain corresponding to accordingly Relation is instructing the reconstruct of high-definition picture.This kind of method depends on selected training image storehouse, and amount of calculation can be very Greatly.An other class method is the optimal reconfiguration method based on sparse constraint item, due to directly reconstructing from low-resolution image High-resolution image is belonging to the ill-conditioning problem of shortage of data, so needing to add some regularization term to overcome such disease State condition.At present some regularization term (as the total variation of image, wavelet basiss sparse constraint etc.) are both for pervasive image, and Depth is not had to excavate potential prior information, such as the image texture information of directivity in some particular type images.

Content of the invention

The invention aims to solving there is no depth excavation direction present in existing high-definition picture reconfiguration technique Property texture information problem it is proposed that a kind of high-definition picture reconstructing method based on directional gradient, the method can be deeper The useful information that degree excavation image is implied is as priori so that the image resolution ratio reconstructing is higher.

The purpose of the present invention is achieved through the following technical solutions:

A kind of high-definition picture reconstructing method based on directional gradient, comprises the following steps:

Step one: obtain input low-resolution imageMain grain direction θ.

Step 11: calculate horizontal gradient and the vertical gradient of low-resolution image i.Horizontal gradient δ₁I (i, j)=i (i+ 1, j)-i (i, j), i=1 ..., m-1, j=1 ..., n, and δ₁The m behavior 0 of i.Vertical gradient δ₂I (i, j)=i (i, j+ 1)-i (i, j), i=1 ..., m, j=1 ..., n-1, and δ₂The n-th of i is classified as 0.

Step 12: for the angle [alpha] of a determination₀And length β (integer typically between 2 to 6 for the value of β), calculate Low-resolution image i is in α₀Under direction gradient value.

$\mathrm{di}(i,j)=\left(\begin{array}{cc}\mathrm{\&beta;}& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}\cos {\mathrm{\&alpha;}}_0& \sin {\mathrm{\&alpha;}}_0\\ -{\sin \mathrm{\&alpha;}}_0& \cos {\mathrm{\&alpha;}}_0\end{array}\right)\left(\begin{array}{c}{\mathrm{\&delta;}}_{1}i(i,j)\\ {\mathrm{\&delta;}}_{2}i(i,j)\end{array}\right)$

Low-resolution image i is in α₀Under direction gradient value dtv (i) be above formula component sum it may be assumed that

Dtv (i)=∑_{I, j}Di (i, j)

Step 13: determine main grain direction θ of low-resolution image i.Make α₀Change from [- π, π], be spaced apart-π/ 24 that is to say, that ${\mathrm{\&alpha;}}_0=-\mathrm{\&pi;},-\frac{23\mathrm{\&pi;}}{24},...,-\frac{\mathrm{\&pi;}}{24},0,\frac{\mathrm{\&pi;}}{24},...,\frac{23\mathrm{\&pi;}}{24},\mathrm{\&pi;},$ Calculate different α₀Corresponding dtv (i) value, and Find the α corresponding to dtv (i) value of minimum₀Value, by this angle value determine based on texture direction θ.

Step 2: using bicubic interpolation method, low-resolution image i is interpolated to the interpolation image i of a panel height dimension_l, If the image pixel point quantity after interpolation is original k²Times, that is,K value refers to the multiple of increase resolution, one As round numbers, such as: 4,6,8 etc..

Step 3: the selection of initialization of variable and parameter.

Initializing variableT=0, wherein i=1 ..., km, j=1 ..., kn, x_ij WithIt is all the column vector that dimension is 2, (0) represents initial value, and (t) represents the iteration updated value of the t time. Represent the high-definition picture wanting to obtain.Choose the value of some parameters, including λ, μ, (λ is that error in data bound term is normal to beta, gamma Number, μ, γ all represent iteration step length, wherein: μ-iteration step length 1, r- iteration step length 2, and β is the parameter in step one, representative be The length of major axis in ELLIPTIC REVOLUTION coordinate transform), fuzzy operatorP is an odd number, maximum iteration time t, iteration Stop discriminant value tol, λ according to circumstances depending on, typically select the numerical value between 1-100, μ typically takes 10^-3, with step one, γ is general for β Take 10^-3, h typically takes 13 × 13 gaussian random matrix, and maximum iteration time t typically takes 100, tol typically to take 10^-4.

Step 4: obtain fuzzy operator matrix fh and conjugate ambiguity operator matrix fhh.

Step 41: initialization matrixFor 0 matrix, by matrixIt is filled into P row is arrived in the 1st of matrix fh, and the 1st arrives p row, then it is done with fast Fourier transform obtains fuzzy operator matrix fh.

Step 42: first matrix fh is done a conjugate transpose, the non-conjugated transposition that tries again obtains conjugate ambiguity Operator Moment Battle array fhh.

Step 5: update x_ijValue.

Step 51: calculateHorizontal gradient and vertical gradient.Horizontal gradient ${\mathrm{\&delta;}}_1{i}_h^{\left(t\right)}(i,j)=i_h^{\left(t\right)}(i+1,j)-i_h^{\left(t\right)}(i,j),$ I=1 ..., km-1, j=1 ..., kn, andKth m behavior 0.Vertical gradient ${\mathrm{\&delta;}}_2{i}_h^{\left(t\right)}(i,j)=i_h^{\left(t\right)}(i,j+1)-i_h^{\left(t\right)}(i,j),$ I=1 ..., km, j=1 ..., kn-1, andKth n It is classified as 0.

Step 52: calculateDirection gradient value under θ (value that step one determines).

${\mathrm{\&lambda;}}_{\mathrm{ij}}\left(i_h^{\left(t\right)}\right)=\left(\begin{array}{cc}\mathrm{\&beta;}& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}\cos \mathrm{\&theta;}& \sin \mathrm{\&theta;}\\ -\sin \mathrm{\&theta;}& \cos \mathrm{\&theta;}\end{array}\right)\left(\begin{array}{c}{\mathrm{\&delta;}}_1{i}_h^{\left(i\right)}(i,j)\\ {\mathrm{\&delta;}}_2{i}_h^{\left(t\right)}(i,j)\end{array}\right)$

Step 53: calculateValue, $x_{\mathrm{ij}}^{(t+1)}=\mathrm{sgn}\left({\mathrm{\&lambda;}}_{\mathrm{ij}}\left(i_h^{\left(t\right)}\right)\right).*\max \left(\right|{\mathrm{\&lambda;}}_{\mathrm{ij}}\left(i_h^{\left(t\right)}\right)|-\frac{1}{\mathrm{\&mu;}},0),$ Wherein sgn For sign function, that is, as s ＞ 0, sgn (s)=1, as s=0, sgn (s)=0, as s ＜ 0, sgn (s)=- 1.Max is Take max function .* is the point multiplication operation between vector.

Step 6: update i_hValue.

Step 61: step 2 is obtainedIt is filled into one (km+p-1) × (kn+p-1) and tie up 0 matrix 1st arrives km row and 1 arrives kn row, is designated as i '_l.

Step 62: to i '_lDo fast Fourier transform, then put and be multiplied by conjugate ambiguity operator matrix fhh, then this result is done Inversefouriertransform obtains a matrix, and the middle k m × kn submatrix finally taking out this matrix is designated as γ (i_l).

Step 63: two dimensions of initialization are the 0 matrix a of km × kn¹And a², and calculate:

$\left(\begin{array}{c}{a}_{\mathrm{ij}}^1\\ a_{\mathrm{ij}}^2\end{array}\right)=\left(\begin{array}{cc}\cos \mathrm{\&theta;}& -\sin \mathrm{\&theta;}\\ \sin \mathrm{\&theta;}& \cos \mathrm{\&theta;}\end{array}\right)\left(\begin{array}{cc}\mathrm{\&beta;}& 0\\ 0& 1\end{array}\right)(x_{\mathrm{ij}}^{(t+1)}-{\tilde{x}}_{\mathrm{ij}}^{\left(t\right)}).$

Step 64: to matrix a¹Carry out following computing:

${\&dtri;}_1{a}_{\mathrm{ij}}^1=\left\{\begin{array}{cc}-a_{i-1j}^1& i=\mathrm{km}\\ a_{\mathrm{ij}}^1-a_{i-1j}^1& i=2,...,\mathrm{km}-1\\ a_{\mathrm{ij}}^1& i=1\end{array}\right..$

To matrix a²Carry out following computing:

${\&dtri;}_2{a}_{\mathrm{ij}}^2=\left\{\begin{array}{cc}-a_{\mathrm{ij}-1}^2& j=\mathrm{kn}\\ a_{\mathrm{ij}}^2-a_{\mathrm{ij}-1}^2& i=2,...,\mathrm{kn}-1\\ a_{\mathrm{ij}}^2& j=1\end{array}\right..$

Step 65: calculate item on the right of system of linear equations, $\mathrm{eq}\_r=\mathrm{\&gamma;}\left(i_l\right)+\mathrm{\&mu;\&lambda;}({\&dtri;}_1{a}_{\mathrm{ij}}^1+{\&dtri;}_2{a}_{\mathrm{ij}}^2).$

Step 66: solve following linear operator equation, θ with conjugate gradient method^*(θ(i_h))+μλυ^*(υ(i_h))= Eq_r, the solution obtained isθ, θ^*, υ, υ^*Represent four kinds of linear operators, they are to matrix i_hComputing have as follows Description,

θ: first to matrix i_hThe padding of execution step 61, then fast Fourier transform is carried out to it, then dot product mould Paste operator matrix fh, finally carries out inversefouriertransform to it.

θ^*: to θ (i_h) carry out fast Fourier transform, then put and be multiplied by conjugate ambiguity operator matrix fhh, it is carried out instead Km × kn the submatrix taking out centre after Fourier transformation obtains θ^*(θ(i_h)).

υ: first to matrix i_hAsk horizontal gradient and vertical gradient computing in execution step 51 obtain δ₁i_hAnd δ₂i_h, Operation is weighted to two matrixes and finally gives matrix b¹And b², so υ (i_h) result that obtains of computing is two matrixes.

$\left(\begin{array}{c}{b}_{\mathrm{ij}}^1\\ b_{\mathrm{ij}}^2\end{array}\right)=\left(\begin{array}{cc}\mathrm{\&beta;}& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}\cos \mathrm{\&theta;}& \sin \mathrm{\&theta;}\\ -\sin \mathrm{\&theta;}& \cos \mathrm{\&theta;}\end{array}\right)\left(\begin{array}{c}{\mathrm{\&delta;}}_1{i}_h(i,j)\\ {\mathrm{\&delta;}}_2{i}_h(i,j)\end{array}\right)$

υ^*: two dimensions of initialization are the 0 matrix ab of km × kn first¹And ab², and calculate:

$\left(\begin{array}{c}{\mathrm{ab}}_{\mathrm{ij}}^1\\ a{b}_{\mathrm{ij}}^2\end{array}\right)=\left(\begin{array}{cc}\cos \mathrm{\&theta;}& -\sin \mathrm{\&theta;}\\ \sin \mathrm{\&theta;}& \cos \mathrm{\&theta;}\end{array}\right)\left(\begin{array}{cc}\mathrm{\&beta;}& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}{b}_{\mathrm{ij}}^1\\ b_{\mathrm{ij}}^2\end{array}\right)$

Again to matrix ab¹In steps performed 64Computing obtainsTo matrix ab²In steps performed 64 Computing obtainsFinally

Step 7: updateValue.

Step 71: calculateDirection gradient value under θ, method is the same with 52 with step 51, finally obtains

Step 72: calculateValue, ${\tilde{x}}_{\mathrm{ij}}^{(t+1)}={\tilde{x}}_{\mathrm{ij}}^{\left(t\right)}+\mathrm{\&gamma;}({\mathrm{\&lambda;}}_{\mathrm{ij}}\left(i_h^{(t+1)}\right)-x_{\mathrm{ij}}^{(t+1)}).$

Step 8: discriminate whether to restrain.

CalculateBecome 2 norm value after one-dimensional vector, this value is less than or equal to tol or t >=t if judging, Iteration stopping simultaneously obtains high-definition pictureOtherwise iterationses t increases by 1 and return to step five.

The present invention passes through incoming direction gradient regular terms, preferably represents and excavate implicit direction texture in the picture Information, thus introducing more valuable prior information for the high-resolution reconstruction problem of morbid state, obtains better high-resolution Rate image.So the present invention is be applied to reconstruct high-definition picture from the low resolution image have certain orientation texture information Situation.The method is widely used in the field such as computer vision and image procossing tool.

Brief description

Fig. 1 is the high-resolution reconstruction method flow chart based on directional gradient of the present invention；

Fig. 2 is an original low resolution 270*270 image；

Fig. 3 is the curve chart with direction change for the direction gradient value of image shown in Fig. 2；

Fig. 4 is the 810*810 high-definition picture directly being obtained by bicubic interpolation；

Fig. 5 is the 810*810 high score being obtained using directional gradient high-definition picture reconstructing method of the present invention Resolution image.

Specific embodiment

Below in conjunction with the accompanying drawings technical scheme is further described, but is not limited thereto, every to this Inventive technique scheme is modified or equivalent, without deviating from the spirit and scope of technical solution of the present invention, all should cover In protection scope of the present invention.

The invention provides a kind of high-definition picture reconstructing method based on directional gradient, as shown in figure 1, include with Lower step:

Execution step one: obtain the main grain direction of input low resolution 270*270 image (as shown in Figure 2).

Execution step 11: calculate the horizontal gradient δ of this image₁I and vertical gradient δ₂i；

Execution step 12: order ${\mathrm{\&alpha;}}_0=-\mathrm{\&pi;},-\frac{23\mathrm{\&pi;}}{24},...,-\frac{\mathrm{\&pi;}}{24},0,\frac{\mathrm{\&pi;}}{24},...,\frac{23\mathrm{\&pi;}}{24},\mathrm{\&pi;},\mathrm{\&beta;}=3,$ Calculate every respectively One α₀Corresponding direction gradient value:

$\mathrm{di}(i,j)=\left(\begin{array}{cc}\mathrm{\&beta;}& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}\cos {\mathrm{\&alpha;}}_0& \sin {\mathrm{\&alpha;}}_0\\ -{\sin \mathrm{\&alpha;}}_0& \cos {\mathrm{\&alpha;}}_0\end{array}\right)\left(\begin{array}{c}{\mathrm{\&delta;}}_{1}i(i,j)\\ {\mathrm{\&delta;}}_{2}i(i,j)\end{array}\right),$ Dtv (i)=∑_{I, j}Di (i, j)；

Execution step 13: find the corresponding α of minimum dtv (i)₀Value, is designated as θ.θ=- π/3 in the present example, as Fig. 3 institute Show.

Execution step two: using bicubic interpolation method, low resolution 270*270 image interpolation is become the high score of 810*810 Resolution image, is designated asAs shown in Figure 4.

Execution step three: initializing variableT=0, wherein i=1 ..., 810, j=1 ..., 810, x_ijWithIt is all the column vector of 2 dimensions,Represent the high resolution graphics wanting to obtain Picture.Choose the value of some parameters, including λ=10, μ=10^-3, γ=10^-3, fuzzy operatorChoose is one Average is 0, and variance is 1.8 Gaussian matrix.Maximum iteration time t=100, iteration stopping discriminant value tol=10^-4.

Execution step four: obtain fuzzy operator matrix fh and conjugate ambiguity operator matrix fhh.

Execution step 41: initialization matrixFor 0 matrix, by matrixIt is filled into square 1 to 13rd row of battle array fh, the 1 to 13rd row, then it is done with fast Fourier transform obtain fuzzy operator matrix fh；

Execution step 42: first matrix fh is done a conjugate transpose, the non-conjugated transposition that tries again obtains conjugate ambiguity and calculates Submatrix fhh.

Execution step five: update x_ijValue.

Execution step 51: calculateHorizontal gradient and vertical gradient.Horizontal gradient ${\mathrm{\&delta;}}_1{i}_h^{\left(t\right)}(i,j)=i_h^{\left(t\right)}(i+1,j)-i_h^{\left(t\right)}(i,j),$ I=1 ..., 809, j=1 ..., 810, andThe 810th behavior 0.Vertical gradient ${\mathrm{\&delta;}}_2{i}_h^{\left(t\right)}(i,j)=i_h^{\left(t\right)}(i,j+1)-i_h^{\left(t\right)}(i,j),$ I=1 ..., 810, j=1 ..., 809, andThe 810th It is classified as 0；

Execution step 52: calculateDirection gradient value under θ=- π/3,

${\mathrm{\&lambda;}}_{\mathrm{ij}}\left(i_h^{\left(t\right)}\right)=\left(\begin{array}{cc}3& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}\cos \mathrm{\&theta;}& \sin \mathrm{\&theta;}\\ -\sin \mathrm{\&theta;}& \cos \mathrm{\&theta;}\end{array}\right)\left(\begin{array}{c}{\mathrm{\&delta;}}_1{i}_h^{\left(i\right)}(i,j)\\ {\mathrm{\&delta;}}_2{i}_h^{\left(t\right)}(i,j)\end{array}\right);$

Execution step 53: calculateValue, $x_{\mathrm{ij}}^{(t+1)}=\mathrm{sgn}\left({\mathrm{\&lambda;}}_{\mathrm{ij}}\left(i_h^{\left(t\right)}\right)\right).*\max \left(\right|{\mathrm{\&lambda;}}_{\mathrm{ij}}\left(i_h^{\left(t\right)}\right)|-\mathrm{1000,0}),$ Wherein sgn is sign function, that is, as s ＞ 0, sgn (s)=1, and as s=0, sgn (s)=0, as s ＜ 0, sgn (s)=- 1.Max is to take max function, and .* is the point multiplication operation between vector.

Execution step six: update i_hValue.

Execution step 61: step 2 is obtainedBe filled into one 823 × 823 dimension 0 matrix the 1st is arrived 810 row and 1 to 810 row, are designated as i '_l；

Execution step 62: to i '_lDo fast Fourier transform, then put and be multiplied by conjugate ambiguity operator matrix fhh, then to this knot Fruit is cooked inversefouriertransform and obtains a matrix, and centre 810 × 810 submatrix finally taking out this matrix is designated as γ (i_l)；

Execution step 63: two dimensions of initialization are 810 × 810 0 matrix a¹And a², and calculate:

$\left(\begin{array}{c}{a}_{\mathrm{ij}}^1\\ a_{\mathrm{ij}}^2\end{array}\right)=\left(\begin{array}{cc}\cos \mathrm{\&theta;}& -\sin \mathrm{\&theta;}\\ \sin \mathrm{\&theta;}& \cos \mathrm{\&theta;}\end{array}\right)\left(\begin{array}{cc}3& 0\\ 0& 1\end{array}\right)(x_{\mathrm{ij}}^{(t+1)}-{\tilde{x}}_{\mathrm{ij}}^{\left(t\right)});$

Execution step 64: to matrix a¹Carry out following computing:

${\&dtri;}_1{a}_{\mathrm{ij}}^1=\left\{\begin{array}{cc}-a_{i-1j}^1& i=810\\ a_{\mathrm{ij}}^1-a_{i-1j}^1& i=2,...,809\\ a_{\mathrm{ij}}^1& i=1\end{array}\right.;$

To matrix a²Carry out following computing:

${\&dtri;}_2{a}_{\mathrm{ij}}^2=\left\{\begin{array}{cc}-a_{\mathrm{ij}-1}^2& j=810\\ a_{\mathrm{ij}}^2-a_{\mathrm{ij}-1}^2& i=2,...,809\\ a_{\mathrm{ij}}^2& j=1\end{array}\right.;$

Execution step 65: calculate item on the right of system of linear equations, $\mathrm{eq}\_r=\mathrm{\&gamma;}\left(i_l\right)+\mathrm{\&mu;\&lambda;}({\&dtri;}_1{a}_{\mathrm{ij}}^1+{\&dtri;}_2{a}_{\mathrm{ij}}^2);$

Execution step 66: use conjugate gradient method, specifically use the pcg function in matlab software to solve following linear operator Equation, θ^*(θ(i_h))+10^-2υ^*(υ(i_h))=eq_r, solve unknown matrix i_h, obtain i_hThe t+1 time iterative value

Execution step seven: updateValue.

Execution step 71: calculateDirection gradient value under θ=- π/3, method is the same with 52 with step 51, finally Obtain

Execution step 72: calculateValue, ${\tilde{x}}_{\mathrm{ij}}^{(t+1)}={\tilde{x}}_{\mathrm{ij}}^{\left(t\right)}+\mathrm{\&gamma;}({\mathrm{\&lambda;}}_{\mathrm{ij}}\left(i_h^{(t+1)}\right)-x_{\mathrm{ij}}^{(t+1)}).$

Execution step eight: calculateBecome 2 norm value after one-dimensional vector, this value is less than or equal to if judging 10^-4Or t >=100, iteration stopping simultaneously obtains high-definition picture(as shown in Figure 5), otherwise iterationses t increase by 1 and Return execution step five.

Claims (5)

1. a kind of high-definition picture reconstructing method based on directional gradient is it is characterised in that methods described step is as follows:

Step one, acquisition input low-resolution imageMain grain direction θ:

Step 11, the horizontal gradient δ of calculating low-resolution image i₁I and vertical gradient δ₂I, horizontal gradient δ₁I (i, j)=i (i+1, j)-i (i, j), i=1 ..., m-1, j=1 ..., n, and δ₁The m behavior 0 of i；Vertical gradient δ₂I (i, j)=i (i, j+1)-i (i, j), i=1 ..., m, j=1 ..., n-1, and δ₂The n-th of i is classified as 0；

Step 12, the angle [alpha] for a determination₀With length β, calculate low-resolution image i in α according to below equation₀Under side To Grad:

$dtv\left(i\right)={\&sigma;}_{i,j}di(i,j),di(i,j)=\left(\begin{array}{cc}\mathrm{\&beta;}& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}{\mathrm{cos\&alpha;}}_0& {\mathrm{sin\&alpha;}}_0\\ -{\mathrm{sin\&alpha;}}_0& {\mathrm{cos\&alpha;}}_0\end{array}\right)\left(\begin{array}{c}{\mathrm{\&delta;}}_{1}i(i,j)\\ {\mathrm{\&delta;}}_{2}i(i,j)\end{array}\right);$

Step 13, determine main grain direction θ of low-resolution image i: make α₀From [- π, π], change, is spaced apart-pi/2 4, meter Calculate different α₀Corresponding dtv (i) value, and find the α corresponding to dtv (i) value of minimum₀Value, based on the determination of this angle value The direction θ of texture；

Step 2, using bicubic interpolation method, low-resolution image i is interpolated to the interpolation image i of a panel height dimension_lIf inserting Image pixel point quantity after value is original k²Times, that is,

The selection of step 3, initialization of variable and parameter:

Initializing variableT=0 wherein i=1 ..., km, j=1 ..., kn, x_ijWithAll It is the column vector that dimension is 2, (0) represents initial value, (t) represents the iteration updated value of the t time；Representative is wanted The high-definition picture arriving；

Choose the value of following parameter: λ, μ, beta, gamma, fuzzy operatorP is an odd number, maximum iteration time t, iteration Stop discriminant value tol, wherein: λ is error in data bound term constant, select the numerical value between 1-100, μ-iteration step length 1, take 10^-3, R- iteration step length 2, takes 10^-3；

Step 4, acquisition fuzzy operator matrix fh and conjugate ambiguity operator matrix fhh；

Step 5, renewal x_ijValue:

Step 51, calculatingHorizontal gradientAnd vertical gradient

Step 52, calculate according to the following formulaDirection gradient value under θ:

${\mathrm{\&lambda;}}_{ij}\left(i_h^{\left(t\right)}\right)=\left(\begin{array}{cc}\mathrm{\&beta;}& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}\cos \mathrm{\&theta;}& \sin \mathrm{\&theta;}\\ -\sin \mathrm{\&theta;}& \cos \mathrm{\&theta;}\end{array}\right)\left(\begin{array}{c}{\mathrm{\&delta;}}_1{i}_h^{\left(t\right)}(i,j)\\ {\mathrm{\&delta;}}_2{i}_h^{\left(t\right)}(i,j)\end{array}\right);$

Step 53, calculatingValue,Wherein sgn is symbol Number function, that is, as s ＞ 0, sgn (s)=1, as s=0, sgn (s)=0, as s ＜ 0, sgn (s)=- 1；Max is to take Big value function .* is the point multiplication operation between vector；

Step 6, renewal i_hValue:

Step 61, step 2 obtainedBe filled into one (km+p-1) × (kn+p-1) dimension 0 matrix the 1st is arrived Km row and 1 arrives kn row, is designated as i '_ll；

Step 62, to i '_llDo fast Fourier transform, then put and be multiplied by conjugate ambiguity operator matrix fhh, then do inversefouriertransform Obtain a matrix, the middle k m × kn submatrix finally taking out this matrix is designated as γ (i_l)；

Step 63, two dimensions of initialization are the 0 matrix a of km × kn¹And a², and calculate:

$\left(\begin{array}{c}{a}_{ij}^1\\ a_{ij}^2\end{array}\right)=\left(\begin{array}{cc}cos\mathrm{\&theta;}& -sin\mathrm{\&theta;}\\ sin\mathrm{\&theta;}& \cos \mathrm{\&theta;}\end{array}\right)\left(\begin{array}{c}\mathrm{\&beta;}0\\ 01\end{array}\right)(x_{ij}^{(t+1)}-{\tilde{x}}_{ij}^{\left(t\right)});$

Step 64, to matrix a¹Carry out following computing:

${\&dtri;}_1{a}_{ij}^1=\left\{\begin{array}{cc}-a_{i-1j}^1& i=km\\ a_{ij}^1-a_{i-1j}^1& i=2,\mathrm{...},km-1\\ a_{ij}^1& i=1\end{array}\right.;$

To matrix a²Carry out following computing:

${\&dtri;}_2{a}_{ij}^2=\left\{\begin{array}{cc}-a_{ij-1}^2& j=kn\\ a_{ij}^2-a_{ij-1}^2& j=2,\mathrm{...},kn-1\\ a_{ij}^2& j=1\end{array}\right.;$

Item on the right of step 65, calculating system of linear equations,

Step 66, solve following linear operator equation, θ with conjugate gradient method^*(θ(i_h))+μλυ^*(υ(i_h))=eq_r, The solution obtained isθ,θ^*,υ,υ^*Represent four kinds of linear operators；

Step 7, renewalValue；

Step 8, discriminate whether restrain:

CalculateBecome 2 norm value after one-dimensional vector, this value is less than or equal to tol or t >=t, iteration if judging Stop and obtain high-definition pictureOtherwise iterationses t increases by 1 and return to step five.

2. the high-definition picture reconstructing method based on directional gradient according to claim 1 is it is characterised in that described The specifically comprising the following steps that of step 4

Step 41, initialization matrixFor 0 matrix, by matrixIt is filled into matrix fh The 1st arrive p row, the 1st arrives p row, then it is done with fast Fourier transform obtains fuzzy operator matrix fh；

Step 42, first matrix fh is done a conjugate transpose, the non-conjugated transposition that tries again obtains conjugate ambiguity operator matrix fhh.

3. the high-definition picture reconstructing method based on directional gradient according to claim 1 is it is characterised in that described In step 51, horizontal gradientI=1 ..., km-1, j=1 ..., kn, AndKth m behavior 0；Vertical gradientI=1 ..., km, j= 1 ..., kn-1, andKth n be classified as 0.

4. the high-definition picture reconstructing method based on directional gradient according to claim 1 is it is characterised in that described In step 66, θ, θ^*,υ,υ^*To matrix i_hComputing have described below:

θ: first to matrix i_hThe padding of execution step 61, then fast Fourier transform is carried out to it, then dot product obscures and calculates Submatrix fh, finally carries out inversefouriertransform to it；

θ^*: to θ (i_h) carry out fast Fourier transform, then put and be multiplied by conjugate ambiguity operator matrix fhh, it is carried out in anti-Fu Km × kn the submatrix taking out centre after leaf transformation obtains θ^*(θ(i_h))；

υ: first to matrix i_hAsk horizontal gradient and vertical gradient computing in execution step 51 obtain δ₁i_hAnd δ₂i_h, to two Individual matrix is weighted operation and finally gives matrix b¹And b², so υ (i_h) result that obtains of computing is two matrixes:

$\left(\begin{array}{c}{b}_{ij}^1\\ b_{ij}^2\end{array}\right)=\left(\begin{array}{cc}\mathrm{\&beta;}& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}\cos \mathrm{\&theta;}& \sin \mathrm{\&theta;}\\ -\sin \mathrm{\&theta;}& \cos \mathrm{\&theta;}\end{array}\right)\left(\begin{array}{c}{\mathrm{\&delta;}}_1{i}_h(i,j)\\ {\mathrm{\&delta;}}_2{i}_h(i,j)\end{array}\right);$

υ^*: two dimensions of initialization are the 0 matrix ab of km × kn first¹And ab², and calculate:

$\left(\begin{array}{c}{\mathrm{ab}}_{ij}^1\\ {\mathrm{ab}}_{ij}^2\end{array}\right)=\left(\begin{array}{cc}\cos \mathrm{\&theta;}& -\sin \mathrm{\&theta;}\\ \sin \mathrm{\&theta;}& \cos \mathrm{\&theta;}\end{array}\right)\left(\begin{array}{cc}\mathrm{\&beta;}& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}{b}_{ij}^1\\ b_{ij}^2\end{array}\right),$

Again to matrix ab¹In steps performed 64₁Computing obtains₁ab¹, to matrix ab²In steps performed 64₂Computing obtains Arrive₂ab², last υ^*(υ(i_h))=₁ab¹+▽₂ab².

5. the high-definition picture reconstructing method based on directional gradient according to claim 1 is it is characterised in that described The specifically comprising the following steps that of step 7

Step 71, according to step 51 and 52 method, calculateDirection gradient value under θ, obtains

Step 72, calculatingValue,