CN103679721B - Use the image outline method for simplifying of nearest neighbor method Hough transform - Google Patents

Abstract

A kind of image outline method for simplifying using nearest neighbor method Hough transform, first carries out morphologic filtering operation to bianry image；Then application moment characteristics calculates its barycenter and main shaft equation, and uses minimum distance method to calculate the distance of profile point and main shaft, thus obtains the datum mark that profile simplifies；Finally use nearest neighbor method Hough transform to ask for profile and simplify line segment, the concept of neighborhood is introduced in parameter space, the curved line number calculate, being verified in defined neighborhood, so that it is determined that image space can be approximated to be the point of straight line, the Origin And Destination of line segment, and one closed curve being made up of line segment of formation that is connected in turn by the point asked for is asked for according to ultimate range method.The present invention can simplify image outline well, irregular for image curved boundary is simplified to a series of line segment, and according to head and the tail, these line segments are linked in sequence into a closed figure.Use the simplification profile that obtains of the method and former profile sufficient approximation.

Description

Use the image outline method for simplifying of nearest neighbor method Hough transform

Technical field

The present invention relates to Image Processing and Pattern Recognition technical field, be primarily adapted for use in target in image Detection and identification, specifically use the image outline method for simplifying of nearest neighbor method Hough transform,

Background technology

Image outline is to analyze to obtain target shape feature in Image Processing and Pattern Recognition technical field Important evidence.

Target to be identified in the remote sensing images of naval battle field not only comprise Ship Target (aircraft carrier, cruiser, Destroyer and escort vessel), it is also possible to comprise other targets with vision significant properties, such as island, reef Stone etc., these targets have different shape facilities.In the target information of naval battle field, Ship Target Being to pay close attention to target, its detection efficiency, precision directly affect the decision-making of commander.In order to more preferably Complete the identification to Ship Target, need to be analyzed, the feature of Ship Target by Ship Target Identify from target complex to be identified.

Naval vessel, as the naval target identified main in military operation, is different from other targets, they Warship bow, warship body, warship stern have obvious shape facility, can be by naval vessel mesh when carrying out target recognition Target warship bow, warship body and warship stern shape are abstract for triangle, rectangle and the simple geometric figure such as trapezoidal (table 1).But the objective contour to be identified obtained by image segmentation is frequently not smooth, regular, And target area to be identified there may be hole, these factors are unfavorable for that these clarifications of objective are extracted, Eventually affect the accuracy of identification of target.

Summary of the invention

The purpose of the present invention: find out a kind of method simplifying image outline, efficiently solve because image divides The problem that the target feature shape that cut edge edge is unsmooth, irregularly cause extracts difficulty.

The present invention proposes to use the image outline method for simplifying of nearest neighbor method Hough transform, the method straight line Replace the curve unsmooth, irregular in objective contour, and these line segments are connected into and former profile The closed area of sufficient approximation.Profile geometries feature after simplification is obvious, it is easy to carry out target shape Shape characteristic matching, the identification for target provides foundation;It specifically comprises the following steps that

(1) bianry image is carried out morphologic filtering process, smoothed image edge, fill up vacancy；

(2) calculate the moment characteristics of target area, obtain center-of-mass coordinate and the main shaft equation of target, by meter Calculate the distance relation of profile point and target main shaft, use minimum distance method to obtain the datum mark that profile simplifies；

(3) use nearest neighbor method Hough transform to ask for profile and simplify line segment, parameter space introduces neighborhood Concept, the curved line number being verified in defined neighborhood, so that it is determined that can approximate in image space For the coordinate of point on straight line, ask for the Origin And Destination of line segment according to ultimate range method, and will ask The point taken is connected in turn and forms a closed curve being made up of line segment, reaches to simplify image outline Purpose.

Feature of the present invention: the parameter space at Hough transformation introduces neighborhood concepts.Described neighborhood concepts Point (the θ being mapped in parameter space with image space cathetus Lj_j,ρ_jCentered by), define a neighborhood U₁(θ_j,Δθ)*U₂(ρ_j, Δ ρ), by by the image space corresponding to all curves of this neighborhood Objective contour on point all regard straight line L as_jOn point.Straight line L is determined by the distribution of point set_j's Origin And Destination, and connect into the line segment of approximate target profile, finally give through loop computation The simplification profile of image.

Described morphological image is filtered into:

If bianry image is that (u, v), the size of the filter window of definition Morphologic filters is B (2p_m+1)×(2p_m+ 1), order:

In formula 1, (i, j) be bianry image B (u, v) in a point.

Step one: in order to remove the weak thin straight line in image target area, isolated miscellaneous point, the completeest The morphological erosion becoming image processes:

In formula 2, i ∈ [u-p_m,u+p_m],j∈[v-p_m,v+p_m]；(u is v) rotten through morphology to E Image after erosion process.

Step 2: after morphological erosion processes, the pixel in object edge is also corroded. In order to reduce target area deformation, fill up hole present in target area, it is ensured that object edge complete Whole property, will carry out morphological dilations process to target area.

In formula 3, condition I be E (u, v)=1；Condition II be B (u, v)=1, and exist a bit (p,q)(p∈[u-p_m,u+p_m],q∈[v-p_m,v+p_m]) meet E (p, q)=1；(u v) is warp to D The image that morphological dilations processes.

After morphological dilations processes, object edge and shape are recovered, in target area Weak thin straight line, isolated miscellaneous point is removed, hole is filled up.

The described profile simplification datum mark that solves is: (see figure 2)

Step one: calculate target centroid

Assume that on image, target has homogenous properties, and its distributed function f (x, y), then the matter of target The heart (X₀,Y₀) coordinate representation is:

$m_{00}=\underset{i}{\mathrm{\Σ}}\underset{j}{\mathrm{\Σ}}f(i,j)---\left(4\right)$

$m_{10}=\underset{i}{\mathrm{\Σ}}\underset{j}{\mathrm{\Σi}}f(i,j)---\left(5\right)$

$m_{01}=\underset{i}{\mathrm{\Σ}}\underset{j}{\mathrm{\Σj}}f(i,j)---\left(6\right)$

$X_0=\frac{m_{10}}{m_{00}}---\left(7\right)$

$Y_0=\frac{m_{01}}{m_{00}}---\left(8\right)$

Formula 4 in formula 8, m₀₀It it is the zeroth order square of target；m₁₀、m₀₁It it is the first moment of target；(X₀,Y₀) Barycenter for target.

Step 2: calculate target main shaft equation

After obtaining target centroid, calculate target main shaft equation.The central moment of target is tried to achieve in normalization:

$U_{\mathrm{xx}}=\frac{\underset{i=1}{\overset{N}{\mathrm{\Σ}}}{\left(x_i{-X}_0\right)}^2}{U_{00}}---\left(9\right)$

$U_{\mathrm{yy}}=\frac{\underset{i=1}{\overset{N}{\mathrm{\Σ}}}{\left(y_i{-X}_0\right)}^2}{U_{00}}---\left(10\right)$

$U_{\mathrm{xy}}=\frac{\underset{i=1}{\overset{N}{\mathrm{\Σ}}}(x_i-X_0)(y_i-Y_0)}{U_{00}}---\left(11\right)$

In formula 9,10,11, N represents the pixel number of given image-region, x_iAnd y_iIt is respectively i-th The abscissa of individual pixel and vertical coordinate；U₀₀=m₀₀；X₀And Y₀It is respectively the abscissa of target centroid And vertical coordinate；U_xx、U_yy、U_xyFor target's center's square.

The target slice major axes orientation that can be calculated is:

$\mathrm{\φ}=\left\{\begin{array}{c}\arctan (2*U_{\mathrm{xy}},U_{\mathrm{xx}}-U_{\mathrm{yy}}+\sqrt{{(U_{\mathrm{xx}}-U_{\mathrm{yy}})}^2+4*U_{\mathrm{xy}}^2})u_{\mathrm{xy}}\≥U_{\mathrm{yy}}\\ \arctan (U_{\mathrm{yy}}-U_{\mathrm{xx}}+\sqrt{{(U_{\mathrm{yy}}-U_{\mathrm{xx}})}^2+4*U_{\mathrm{xy}}^2,2*U_{\mathrm{xy}}})U_{\mathrm{xx}}\≤U_{\mathrm{yy}}\end{array}\right.---\left(12\right)$

In formula 12, φ is the angle of target main shaft and coordinate axes,When image wraps During containing geography information, φ can also represent the course of Ship Target.

The main shaft equation calculating target according to formula 7,8,12 is:

Y=xtan φ+Y₀-X₀Tan φ (13)

Step 3: ask and simplify the datum mark calculated

Extract single pixel profile of target in image with Robert algorithm, then the point set of objective contour is {(x₁,y₁),(x₂,y₂),...,(x_i,y_i) (i=1,2...., n), wherein n be in profile number a little.

Objective contour point to the distance of target main shaft is:

$\mathrm{\Δ}{h}_i=\frac{|x_i\tan \mathrm{\φ}-y_i+(Y_0-X_0\tan \mathrm{\φ})|}{\sqrt{1+{\left(\tan \mathrm{\φ}\right)}^2}}---\left(14\right)$

WhenTime corresponding contour line on the intersection point that point is profile and target main shaft (x_j,y_j), by point all of on contour line with (x_j,y_j) it is that starting point sorts in clockwise direction, Obtain point set:

P={P₁,P₂,...,P_i(i=1,2...., n), P_i=(x_i,y_i) (15)

The profile of described target simplifies (see figure 3)

Defining a two-dimensional array P and deposit the coordinate figure of profile point, first unit of array is profile base P on schedule₁Coordinate, second unit deposits P₂Coordinate, follow-up location contents the like, until P_n。P₁To P_nRankine-Hugoniot relations with P₁For starting point, by profile point at the clock-wise order of image space；

Define a two-dimensional array ThetaR (θ_n-1,ρ_n-1) deposit C₁With C_jThe θ of intersection point_j、ρ_jValue；C₁、 C_jRepresented curve it is mapped in parameter space for image space midpoint；θ_j、ρ_jFor C₁With C_j? The horizontal stroke of the aerial intersection point of parameter, vertical coordinate；

Define an one-dimension array A (θ_n-1,ρ_n-1) as accumulator, to intersection point phase same in parameter space Hand over number of times to count, initialize A (θ_n-1,ρ_n-1) each unit is zero；

Define an one-dimension arrayAs accumulator, deposit satisfied (θ_j,ρ_j) neighborhood pass The coordinate of the profile point of system, initializesEach unit is zero；

Step one: calculate C₁, C_jIntersection point (θ in parameter space_j,ρ_j), and be stored in ThetaR(θ_j,ρ_j)；(see figure 4)

$\left\{\begin{array}{c}{\mathrm{\ρ}}_j=x_1\cos {\mathrm{\θ}}_j+y_i\sin {\mathrm{\θ}}_j\\ {\mathrm{\ρ}}_j=x_j\cos {\mathrm{\θ}}_j+y_j\sin {\mathrm{\θ}}_j\end{array}\right.---\left(16\right)$

$\left\{\begin{array}{c}{\mathrm{\θ}}_j=\arctan \left(\frac{x_1-x_j}{y_j-y_1}\right)\\ {\mathrm{\ρ}}_j=x_1\cos \left[\arctan \right(\frac{x_1-x_j}{y_j-y_1}\left)\right]+y_1\sin \left[\arctan \right(\frac{x_1-x_j}{y_j-y_1}\left)\right]\end{array}\right.---\left(17\right)$

Step 2: calculate A (θ_n-1,ρ_n-1) each unit numerical value, by formula 15 P={P₁,P₂,...,P_i(i=1,2...., n), P_i=(x_i,y_i) (except P on 49 sequence number traversal contour lines₁ Outer all of point, calculates curve and curve C that they are mapped in parameter space₁Intersection point.

When there is θ_j=θ_q=γ, ρ_j=ρ_q=s and j ＜ q (j, q=1,2 ..., n-1) time, will (θ_j,ρ_j) the accumulator element value at place adds 1, such as formula 18:

$\left\{\begin{array}{c}A({\mathrm{\θ}}_j,{\mathrm{\ρ}}_j)=A({\mathrm{\θ}}_j,{\mathrm{\ρ}}_j)+1\\ A({\mathrm{\θ}}_q,{\mathrm{\ρ}}_q)=0\end{array}\right.---\left(18\right)$

As curve corresponding to k objective contour point and curve C₁(γ, time s), and j is for occurring to meet at same point It is identical that (γ, s) corresponding smallest sequence number profile point, then have A (θ_j,ρ_j)=k, other k-1 some correspondence is cumulative The value of device is zero.

Step 3: whether the curve calculating the corresponding parameter space of the point on contour line passes through (θ_j,ρ_j) place Neighborhood.

Calculate as accumulator A (θ_j,ρ_jDuring)=α >=1, with point (θ in parameter space_j,ρ_jNeighbour centered by) Territory comprises the bar number of curve:

If any point coordinate is P on contour line_m=(x_m,y_m), then the equation of its corresponding parameter space For:

$\stackrel{\‾}{\mathrm{\ρ}}=x_m\cos \stackrel{\‾}{\mathrm{\θ}}+y_m\sin \stackrel{\‾}{\mathrm{\θ}}---\left(19\right)$

In formula 19, $\stackrel{\‾}{\mathrm{\θ}}\∈[{\mathrm{\θ}}_j-\mathrm{\Δ\θ},{\mathrm{\θ}}_j+\mathrm{\Δ\θ}].$

To formula 19 derivation, then have:

$\frac{d\stackrel{\‾}{\mathrm{\ρ}}}{d\stackrel{\‾}{\mathrm{\θ}}}=\sqrt{x_m^2+y_m^2}\sin (\stackrel{\‾}{\mathrm{\θ}}+\arctan (-\frac{y_m}{x_m}))---\left(20\right)$

When $\stackrel{\‾}{\mathrm{\θ}}\∈[{\mathrm{\θ}}_j-\mathrm{\Δ\θ},{\mathrm{\θ}}_j+\mathrm{\Δ\θ}]$ Time, $\frac{d\stackrel{\‾}{\mathrm{\ρ}}}{d\stackrel{\‾}{\mathrm{\θ}}}>0,$ Then have:

${\stackrel{\‾}{\mathrm{\ρ}}}_{\min }=x_m\cos ({\mathrm{\θ}}_j-\mathrm{\Δ\θ})+y_m\sin ({\mathrm{\θ}}_j-\mathrm{\Δ\θ})---\left(21\right)$

${\stackrel{\‾}{\mathrm{\ρ}}}_{\max }=x_m\cos ({\mathrm{\θ}}_j+\mathrm{\Δ\θ})+y_m\sin ({\mathrm{\θ}}_j+\mathrm{\Δ\θ})---\left(22\right)$

When $\stackrel{\&OverBar;}{\mathrm{\&theta;}}\&Element;[{\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;},{\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;}]$ Time, $\frac{d\stackrel{\&OverBar;}{\mathrm{\&rho;}}}{d\stackrel{\&OverBar;}{\mathrm{\&theta;}}}<0,$ Then have:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\min }=x_m\cos ({\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;})+y_m\sin ({\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;})---\left(23\right)$

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\max }=x_m\cos ({\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;})+y_m\sin ({\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;})---\left(24\right)$

When $\stackrel{\&OverBar;}{\mathrm{\&theta;}}\&Element;[{\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;},{\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;}]$ Time, exist $\frac{d\stackrel{\&OverBar;}{\mathrm{\&rho;}}}{d\stackrel{\&OverBar;}{\mathrm{\&theta;}}}=0,$ Now $\stackrel{\&OverBar;}{\mathrm{\&theta;}}=\mathrm{\&beta;},$ The second order of formula 19 Derivative is:

$\frac{d^2\stackrel{\&OverBar;}{\mathrm{\&rho;}}}{d{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}^2}=\sqrt{x_m^2+y_m^2}\sin (\stackrel{\&OverBar;}{\mathrm{\&theta;}}+\arctan \left(\frac{x_m}{y_m}\right))---\left(25\right)$

Judged by formula 19,25 $\stackrel{\&OverBar;}{\mathrm{\&theta;}}\&Element;[{\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;},{\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;}]$ On intervalExtreme value.

(1) when $\stackrel{\&OverBar;}{\mathrm{\&theta;}}\&Element;[{\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;},{\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;}]$ Time, $\frac{d^2\stackrel{\&OverBar;}{\mathrm{\&rho;}}}{d{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}^2}<0,$ Then have:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\min }=x_m\cos \mathrm{\&beta;}+y_m\sin \mathrm{\&beta;}---\left(26\right)$

As [x_mcos(θ_j+Δθ)+y_msin(θ_j+Δθ)]-[x_mcos(θ_j-Δθ)+y_msin(θ_j-Δ θ)] ＞ 0 time:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\max }=x_m\cos ({\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;})+y_m\sin ({\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;})---\left(27\right)$

As [x_mcos(θ_j+Δθ)+y_msin(θ_j+Δθ)]-[x_mcos(θ_j-Δθ)+y_msin(θ_j-Δ θ)] ＜ 0 time:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\max }=x_m\cos ({\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;})+y_m\sin ({\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;})---\left(28\right)$

(2) when $\stackrel{\&OverBar;}{\mathrm{\&theta;}}\&Element;[{\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;},{\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;}]$ Time, $\frac{d^2\stackrel{\&OverBar;}{\mathrm{\&rho;}}}{d{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}^2}>0,$ Then have:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\min }=x_m\cos \mathrm{\&beta;}+y_m\sin \mathrm{\&beta;}---\left(29\right)$

As [x_mcos(θ_j+Δθ)+y_msin(θ_j+Δθ)]-[x_mcos(θ_j-Δθ)+y_msin(θ_j-Δ θ)] ＞ 0 time:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\min }=x_m\cos ({\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;})+y_m\sin ({\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;})---\left(30\right)$

As [x_mcos(θ_j+Δθ)+y_msin(θ_j+Δθ)]-[x_mcos(θ_j-Δθ)+y_msin(θ_j-Δ θ)] ＜ 0 time:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\min }=x_m\cos ({\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;})+y_m\sin ({\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;})---\left(31\right)$

Can obtain to formula 31 according to formula 19Interval be:

$\stackrel{\&OverBar;}{\mathrm{\&rho;}}\&Element;[{\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\min },{\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\max }]---\left(32\right)$

When meeting formula 33, then put p_mIt is required point.

$[{\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\min },{\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\max }]\&cap;[{\mathrm{\&rho;}}_j-\mathrm{\&Delta;\&rho;},{\mathrm{\&rho;}}_j+\mathrm{\&Delta;\&rho;}]\&NotEqual;\mathrm{\&phi;}---\left(33\right)$

Step 4: ask for through (θ_j,ρ_j) set of neighborhood profile point.

When pm value in neighborhood meets formula 33, point (θ_j,ρ_j) value of corresponding accumulator elementFor:

$\stackrel{\&OverBar;}{A}({\mathrm{\&theta;}}_j,{\mathrm{\&rho;}}_j)=\stackrel{\&OverBar;}{A}({\mathrm{\&theta;}}_j,{\mathrm{\&rho;}}_j)+1---\left(34\right)$

Calculate all as A (θ_j,ρ_jDuring)=α >=1, with (θ_j,ρ_jCentered by) defined adjacent under study for action In territory, meet the point of above-mentioned requirements.TakeTime, with (θ_j,ρ_j) it is The set of the profile point tried to achieve at center is to meet and simplifies the point required.

The present invention uses the image outline method for simplifying of nearest neighbor method Hough transform.Hough transformation the most former Reason is the duality utilizing point with line, the curve negotiating curve representation shape given by original image space Formula becomes a point of parameter space；The test problems of given curve in original image is converted into searching Spike problem in parameter space, namely detection overall permanence is converted into detection local characteristics, such as Straight line, ellipse, circle, camber line etc..Being different from tradition Hough transformation application, the present invention utilizes application suddenly The ultimate principle of husband's conversion, introduces the concept of neighborhood at parameter space, and in image space is straight The point that line is mapped as in parameter space, sets up a sufficiently small neighborhood centered by this is put, works as figure When the curve that point in image space is mapped in parameter space all passes this neighborhood, it is believed that image These points in space are all on same straight line, so that it is determined that can be approximated to be one in image space The set of the point of straight line.Then ask for the Origin And Destination of line segment according to ultimate range method, and will ask for Point be connected in turn formation one closed curve being made up of line segment, thus reach simplify image wheel Wide purpose.The method advantage is can to simplify image outline easily, by irregular for image curve Border is simplified to a series of line segment, and according to head and the tail, these line segments are linked in sequence into a closed figure. Use the simplification profile that obtains of the method and original image profile sufficient approximation.

Accompanying drawing explanation

Accompanying drawing 1 objective contour simplifies algorithm block diagram；

Accompanying drawing 2 objective contour datum mark computing block diagram；

Accompanying drawing 3 nearest neighbor method Hough transform algorithm block diagram；

Accompanying drawing 4 parameter space based on neighborhood image；

Accompanying drawing 5 warship shape facility statistical table.

Detailed description of the invention

1, morphological image filtering

If bianry image is that (u, v), the size of the filter window of definition Morphologic filters is B (2p_m+1)×(2p_m+ 1), order:

Wherein, (i, j) be bianry image B (u, v) in a point.

Step one: in order to remove the weak thin straight line in image target area, isolated miscellaneous point, the completeest The morphological erosion becoming image processes:

In formula 36

i∈[u-p_m,u+p_m],j∈[v-p_m,v+p_m]；(u v) is the figure after morphological erosion processes to E Picture.

Step 2: after morphological erosion processes, the pixel in object edge is also corroded. In order to reduce target area deformation, fill up hole present in target area, it is ensured that object edge complete Whole property, will carry out morphological dilations process to target area.

FormulaIn, condition I is E (u, v)=1；Condition II be B (u, v)=1, and exist a bit (p,q)(p∈[u-p_m,u+p_m],q∈[v-p_m,v+p_m]) meet E (p, q)=1；(u v) is warp to D The image that morphological dilations processes.

After morphological dilations processes, object edge and shape are recovered, in target area Weak thin straight line, isolated miscellaneous point is removed, hole is filled up.

2, solve profile and simplify datum mark (see figure 2)

Step one: calculate target centroid

Assume that on image, target has homogenous properties, and its distributed function f (x, y), then the matter of target The heart (X₀,Y₀) coordinate representation is:

$m_{00}=\underset{i}{\mathrm{\&Sigma;}}\underset{j}{\mathrm{\&Sigma;}}f(i,j)---\left(38\right)$

$m_{10}=\underset{i}{\mathrm{\&Sigma;}}\underset{j}{\mathrm{\&Sigma;i}}f(i,j)---\left(39\right)$

$m_{01}=\underset{i}{\mathrm{\&Sigma;}}\underset{j}{\mathrm{\&Sigma;j}}f(i,j)---\left(40\right)$

$X_0=\frac{m_{10}}{m_{00}}---\left(41\right)$

$Y_0=\frac{m_{01}}{m_{00}}---\left(42\right)$

In above formula, m₀₀It it is the zeroth order square of target；m₁₀、m₀₁It it is the first moment of target；(X₀,Y₀) it is The barycenter of target.

Step 2: calculate target main shaft equation

After obtaining target centroid, calculate target main shaft equation.The central moment of target is tried to achieve in normalization:

$U_{\mathrm{xx}}=\frac{\underset{i=1}{\overset{N}{\mathrm{\&Sigma;}}}{\left(x_i{-X}_0\right)}^2}{U_{00}}---\left(43\right)$

$U_{\mathrm{yy}}=\frac{\underset{i=1}{\overset{N}{\mathrm{\&Sigma;}}}{\left(y_i{-X}_0\right)}^2}{U_{00}}---\left(44\right)$

$U_{\mathrm{xy}}=\frac{\underset{i=1}{\overset{N}{\mathrm{\&Sigma;}}}(x_i-X_0)(y_i-Y_0)}{U_{00}}---\left(45\right)$

In above formula, N represents the pixel number of given image-region, x_iAnd y_iIt is respectively ith pixel point Abscissa and vertical coordinate；U₀₀=m₀₀；X₀And Y₀It is respectively abscissa and the vertical coordinate of target centroid； U_xx、U_yy、U_xyFor target's center's square.

The target slice major axes orientation that can be calculated is:

$\mathrm{\&phi;}=\left\{\begin{array}{c}\arctan (2*U_{\mathrm{xy}},U_{\mathrm{xx}}-U_{\mathrm{yy}}+\sqrt{{(U_{\mathrm{xx}}-U_{\mathrm{yy}})}^2+4*U_{\mathrm{xy}}^2})u_{\mathrm{xy}}\&GreaterEqual;U_{\mathrm{yy}}\\ \arctan (U_{\mathrm{yy}}-U_{\mathrm{xx}}+\sqrt{{(U_{\mathrm{yy}}-U_{\mathrm{xx}})}^2+4*U_{\mathrm{xy}}^2,2*U_{\mathrm{xy}}})U_{\mathrm{xx}}\&le;U_{\mathrm{yy}}\end{array}\right.---\left(46\right)$

In formula 46, φ is the angle of target main shaft and coordinate axes,When image wraps During containing geography information, φ can also represent the course of Ship Target.

The main shaft equation calculating target according to formula 41,42,46 is:

Y=xtan φ+Y₀-X₀Tan φ (47)

Step 3: ask and simplify the datum mark calculated

Extract single pixel profile of target in image with Robert algorithm, then the point set of objective contour is {(x₁,y₁),(x₂,y₂),...,(x_i,y_i) (i=1,2...., n), wherein n be in profile number a little.

Objective contour point to the distance of target main shaft is:

$\mathrm{\&Delta;}{h}_i=\frac{|x_i\tan \mathrm{\&phi;}-y_i+(Y_0-X_0\tan \mathrm{\&phi;})|}{\sqrt{1+{\left(\tan \mathrm{\&phi;}\right)}^2}}---\left(48\right)$

WhenTime corresponding contour line on the intersection point that point is profile and target main shaft (x_j,y_j), by point all of on contour line with (x_j,y_j) it is that starting point sorts in clockwise direction, Obtain point set:

P={P₁,P₂,...,P_i(i=1,2...., n), P_i=(x_i,y_i) (49)

3, the profile of target simplifies (see figure 3)

Defining a two-dimensional array P and deposit the coordinate figure of profile point, first unit of array is profile base P on schedule₁Coordinate, second unit deposits P₂Coordinate, follow-up location contents the like, until P_n。P₁To P_nRankine-Hugoniot relations with P₁For starting point, by profile point at the clock-wise order of image space；

Define a two-dimensional array ThetaR (θ_n-1,ρ_n-1) deposit C₁With C_jThe θ of intersection point_j、ρ_jValue；C₁、 C_jRepresented curve it is mapped in parameter space for image space midpoint；θ_j、ρ_jFor C₁With C_j? The horizontal stroke of the aerial intersection point of parameter, vertical coordinate；

Define an one-dimension array A (θ_n-1,ρ_n-1) as accumulator, to intersection point phase same in parameter space Hand over number of times to count, initialize A (θ_n-1,ρ_n-1) each unit is zero；

Define an one-dimension arrayAs accumulator, deposit satisfied (θ_j,ρ_j) neighborhood pass The coordinate of the profile point of system, initializesEach unit is zero；

Step one: calculate C₁, C_jIntersection point (θ in parameter space_j,ρ_j), and be stored in ThetaR(θ_j,ρ_j)；(see figure 4)

$\left\{\begin{array}{c}{\mathrm{\&rho;}}_j=x_1\cos {\mathrm{\&theta;}}_j+y_i\sin {\mathrm{\&theta;}}_j\\ {\mathrm{\&rho;}}_j=x_j\cos {\mathrm{\&theta;}}_j+y_j\sin {\mathrm{\&theta;}}_j\end{array}\right.---\left(50\right)$

$\left\{\begin{array}{c}{\mathrm{\&theta;}}_j=\arctan \left(\frac{x_1-x_j}{y_j-y_1}\right)\\ {\mathrm{\&rho;}}_j=x_1\cos \left[\arctan \right(\frac{x_1-x_j}{y_j-y_1}\left)\right]+y_1\sin \left[\arctan \right(\frac{x_1-x_j}{y_j-y_1}\left)\right]\end{array}\right.---\left(51\right)$

Step 2: calculate A (θ_n-1,ρ_n-1) each unit numerical value, by formula 49 P={P₁,P₂,...,P_i(i=1,2...., n), P_i=(x_i,y_i) (except P on 49 sequence number traversal contour lines₁ Outer all of point, calculates curve and curve C that they are mapped in parameter space₁Intersection point.

When there is θ_j=θ_q=γ, ρ_j=ρ_q=s and j ＜ q (j, q=1,2 ..., n-1) time, will (θ_j,ρ_j) the accumulator element value at place adds 1, such as formula 52:

$\left\{\begin{array}{c}A({\mathrm{\&theta;}}_j,{\mathrm{\&rho;}}_j)=A({\mathrm{\&theta;}}_j,{\mathrm{\&rho;}}_j)+1\\ A({\mathrm{\&theta;}}_q,{\mathrm{\&rho;}}_q)=0\end{array}\right.---\left(52\right)$

As curve corresponding to k objective contour point and curve C₁(γ, time s), and j is for occurring to meet at same point It is identical that (γ, s) corresponding smallest sequence number profile point, then have A (θ_j,ρ_j)=k, other k-1 some correspondence is cumulative The value of device is zero.

Step 3: whether the curve calculating the corresponding parameter space of the point on contour line passes through (θ_j,ρ_j) place Neighborhood.

Calculate as accumulator A (θ_j,ρ_jDuring)=α >=1, with point (θ in parameter space_j,ρ_jNeighbour centered by) Territory comprises the bar number of curve:

If any point coordinate is P on contour line_m=(x_m,y_m), then the equation of its corresponding parameter space For:

$\stackrel{\&OverBar;}{\mathrm{\&rho;}}=x_m\cos \stackrel{\&OverBar;}{\mathrm{\&theta;}}+y_m\sin \stackrel{\&OverBar;}{\mathrm{\&theta;}}---\left(53\right)$

In formula 53, $\stackrel{\&OverBar;}{\mathrm{\&theta;}}\&Element;[{\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;},{\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;}].$

To formula 53 derivation, then have:

$\frac{d\stackrel{\&OverBar;}{\mathrm{\&rho;}}}{d\stackrel{\&OverBar;}{\mathrm{\&theta;}}}=\sqrt{x_m^2+y_m^2}\sin (\stackrel{\&OverBar;}{\mathrm{\&theta;}}+\arctan (-\frac{y_m}{x_m}))---\left(54\right)$

When $\stackrel{\&OverBar;}{\mathrm{\&theta;}}\&Element;[{\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;},{\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;}]$ Time, $\frac{d\stackrel{\&OverBar;}{\text{\&rho;}}}{d\stackrel{\&OverBar;}{\text{\&theta;}}}>0,$ Then have:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\min }=x_m\cos ({\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;})+y_m\sin ({\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;})---\left(55\right)$

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\max }=x_m\cos ({\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;})+y_m\sin ({\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;})---\left(56\right)$

When $\stackrel{\&OverBar;}{\mathrm{\&theta;}}\&Element;[{\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;},{\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;}]$ Time, $\frac{d\stackrel{\&OverBar;}{\mathrm{\&rho;}}}{d\stackrel{\&OverBar;}{\text{\&theta;}}}<0,$ Then have:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\max }=x_m\cos ({\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;})+y_m\sin ({\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;})---\left(57\right)$

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\max }=x_m\cos ({\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;})+y_m\sin ({\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;})---\left(58\right)$

When $\stackrel{\&OverBar;}{\mathrm{\&theta;}}\&Element;[{\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;},{\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;}]$ Time, exist $\frac{d^2\stackrel{\&OverBar;}{\mathrm{\&rho;}}}{d{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}^2}=0,$ Now $\stackrel{\&OverBar;}{\mathrm{\&theta;}}=\mathrm{\&beta;},$ The second order of formula 53 Derivative is:

$\frac{d^2\stackrel{\&OverBar;}{\mathrm{\&rho;}}}{d{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}^2}=\sqrt{x_m^2+y_m^2}\sin (\stackrel{\&OverBar;}{\mathrm{\&theta;}}+\arctan \left(\frac{x_m}{y_m}\right))---\left(59\right)$

Judged by formula 53,59 $\stackrel{\&OverBar;}{\mathrm{\&theta;}}\&Element;[{\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;},{\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;}]$ On intervalExtreme value.

(1) when $\stackrel{\&OverBar;}{\mathrm{\&theta;}}\&Element;[{\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;},{\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;}]$ Time, $\frac{d^2\stackrel{\&OverBar;}{\mathrm{\&rho;}}}{d{\stackrel{\&OverBar;}{\mathrm{\&rho;}}}^2}<0,$ Then have:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\min }=x_m\cos \mathrm{\&beta;}+y_m\sin \mathrm{\&beta;}---\left(60\right)$

As [x_mcos(θ_j+Δθ)+y_msin(θ_j+Δθ)]-[x_mcos(θ_j-Δθ)+y_msin(θ_j-Δ θ)] ＞ 0 time:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\max }=x_m\cos ({\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;})+y_m\sin ({\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;})---\left(61\right)$

As [x_mcos(θ_j+Δθ)+y_msin(θ_j+Δθ)]-[x_mcos(θ_j-Δθ)+y_msin(θ_j-Δ θ)] ＜ 0 time:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\max }=x_m\cos ({\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;})+y_m\sin ({\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;})---\left(62\right)$

(2) when $\stackrel{\&OverBar;}{\mathrm{\&theta;}}\&Element;[{\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;},{\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;}]$ Time, $\frac{d^2\stackrel{\&OverBar;}{\mathrm{\&rho;}}}{d{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}^2}>0,$ Then have:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\max }=x_m\cos \mathrm{\&beta;}+y_m\sin \mathrm{\&beta;}---\left(63\right)$

As [x_mcos(θ_j+Δθ)+y_msin(θ_j+Δθ)]-[x_mcos(θ_j-Δθ)+y_mSin (θ j-Δ θ)] ＞ 0 time:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\min }=x_m\cos ({\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;})+y_m\mathrm{sim}({\mathrm{\&theta;}}_j-\mathrm{\&Delta;\&theta;})---\left(64\right)$

As [x_mcos(θ_j+Δθ)+y_msin(θ_j+Δθ)]-[x_mcos(θ_j-Δθ)+y_msin(θ_j-Δ θ)] ＜ 0 time:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\min }=x_m\cos ({\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;})+y_m\sin ({\mathrm{\&theta;}}_j+\mathrm{\&Delta;\&theta;})---\left(65\right)$

Can obtain to formula 65 according to formula 53Interval be:

$\stackrel{\&OverBar;}{\mathrm{\&rho;}}\&Element;[{\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\min },{\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\max }]---\left(32\right)$

When meeting formula 67, then some pm is required point.

$[{\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\min },{\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\max }]\&cap;[{\mathrm{\&rho;}}_j-\mathrm{\&Delta;\&rho;},{\mathrm{\&rho;}}_j+\mathrm{\&Delta;\&rho;}]\&NotEqual;\mathrm{\&phi;}---\left(67\right)$

Step 4: ask for through (θ_j,ρ_j) set of neighborhood profile point.

When pm value in neighborhood meets formula 67, point (θ_j,ρ_j) value of corresponding accumulator elementFor:

$\stackrel{\&OverBar;}{A}({\mathrm{\&theta;}}_j,{\mathrm{\&rho;}}_j)=\stackrel{\&OverBar;}{A}({\mathrm{\&theta;}}_j,{\mathrm{\&rho;}}_j)+1---\left(68\right)$

Calculate all as A (θ_j,ρ_jDuring)=α >=1, with (θ_j,ρ_jCentered by) defined adjacent under study for action In territory, meet the point of above-mentioned requirements.TakeTime, with (θ_j,ρ_j) it is The set of the profile point tried to achieve at center is to meet and simplifies the point required.

Step 5: ask for the approximation line segment of profile point.Calculate profile point and concentrate the seat apart from big 2 Mark, these 2 Origin And Destinations being approximation line segment.

Step 6: take the starting point that terminal is lower bar line segment of required line segment, and will obtain in line segment The point set comprised is concentrated from profile point and is removed.Perform step one and arrive step 5, when calculating i-th line When the point of section concentrates the starting point comprising Article 1 straight line, the starting point of connection i-th line section and the 1st article of line segment Terminal, gained closed area be target simplify after contour line.

The present invention is not only limited to above-mentioned specific embodiment, and persons skilled in the art are according to this Disclosure of invention, can use other multiple detailed description of the invention to implement the present invention, therefore, all It is design structure and the thinking using the present invention, does some simply change or designs of change, all fall Enter protection scope of the present invention.

Claims (2)

1. the image outline method for simplifying using nearest neighbor method Hough transform, it is characterised in that: concrete Step is as follows:

(1) bianry image carries out morphologic filtering process smoothed image edge, fill up vacancy；Institute The morphological image stated is filtered into:

If bianry image is that (u, v), the size of the filter window of definition Morphologic filters is B (2p_m+1)×(2p_m+ 1), order:

In formula 1, (i, j) be bianry image B (u, v) in a point；

Step one: in order to remove the weak thin straight line in image target area, isolated miscellaneous point, the completeest The morphological erosion becoming image operates:

In formula 2, i ∈ [u-p_m,u+p_m],j∈[v-p_m,v+p_m]；(u is v) rotten through morphology to E Image after erosion process；

Step 2: after morphological erosion processes, the pixel in object edge is also corroded； In order to reduce target area deformation, fill up hole present in target area, it is ensured that object edge complete Whole property, will carry out morphological dilations process to target area；

In formula 3, condition I be E (u, v)=1；Condition II be B (u, v)=1, and exist a bit (p,q)(p∈[u-p_m,u+p_m],q∈[v-p_m,v+p_m]) meet E (p, q)=1；(u v) is warp to D The image that morphological dilations processes；

After morphological dilations processes, object edge and shape are recovered, in target area Weak thin straight line, isolated miscellaneous point is removed, hole is filled up；

(2) calculate the moment characteristics of target area, obtain center-of-mass coordinate and the main shaft equation of target, by meter Calculate the distance relation of profile point and target main shaft, use minimum distance method to obtain the datum mark that profile simplifies； Algorithm is as follows:

Step one: calculate target centroid

Assume that on image, target has homogenous properties, and its distributed function f (x, y), then the matter of target The heart (X₀,Y₀) coordinate representation is:

$m_{00}=\underset{i}{\&Sigma;}\underset{j}{\&Sigma;}f(i,j)---\left(4\right)$

$m_{10}=\underset{i}{\&Sigma;}\underset{j}{\&Sigma;}if(i,j)---\left(5\right)$

$m_{01}=\underset{i}{\&Sigma;}\underset{j}{\&Sigma;}jf(i,j)---\left(6\right)$

$X_0=\frac{m_{10}}{m_{00}}---\left(7\right)$

$Y_0=\frac{m_{01}}{m_{00}}---\left(8\right)$

Wherein, m₀₀It it is the zeroth order square of target；m₁₀、m₀₁It it is the first moment of target；(X₀,Y₀) it is mesh Target barycenter；

Step 2: calculate target main shaft equation

After obtaining target centroid, calculate target main shaft equation；The central moment of target is tried to achieve in normalization:

$U_{xx}=\frac{\underset{i=1}{\overset{N}{\&Sigma;}}{(x_i-X_0)}^2}{U_{00}}---\left(9\right)$

$U_{yy}=\frac{\underset{i=1}{\overset{N}{\&Sigma;}}{(y_i-Y_0)}^2}{U_{00}}---\left(10\right)$

$U_{xy}=\frac{\underset{i=1}{\overset{N}{\&Sigma;}}(x_i-X_0)(y_i-Y_0)}{U_{00}}---\left(11\right)$

Wherein, N represents the pixel number of given image-region, x_iAnd y_iIt is respectively ith pixel point Abscissa and vertical coordinate；U₀₀=m₀₀；X₀And Y₀It is respectively abscissa and the vertical coordinate of target centroid； U_xx、U_yy、U_xyFor target area central moment；

The target slice major axes orientation that can be calculated is:

$\mathrm{\&phi;}=\left\{\begin{array}{cc}\arctan (2*U_{xy},U_{xx}-U_{yy}+\sqrt{{(U_{xx}-U_{yy})}^2+4*U_{xy}^2})& U_{xx}\&GreaterEqual;U_{yy}\\ \arctan (U_{yy}-U_{xx}+\sqrt{{(U_{yy}-U_{xx})}^2+4*U_{xy}^2},2*U_{xy})& U_{xx}\&le;U_{yy}\end{array}\right.---\left(12\right)$

In formula 12, φ is the angle of target main shaft and coordinate axes,When image wraps During containing geography information, φ can also represent the course of Ship Target；

The main shaft equation calculating target according to formula 7,8,12 is:

Y=x tan φ+Y₀-X₀tanφ (13)

Step 3: ask and simplify the datum mark calculated

Extract single pixel profile of target in image with Robert algorithm, then the point set of objective contour is {(x₁,y₁),(x₂,y₂),...,(x_i,y_i) (i=1,2...., n), wherein n be in profile number a little；

Objective contour point to the distance of target main shaft is:

${\mathrm{\&Delta;h}}_i=\frac{|x_{i}tan\mathrm{\&phi;}-y_i+(Y_0-X_{0}tan\mathrm{\&phi;})|}{\sqrt{1+{\left(tan\mathrm{\&phi;}\right)}^2}}---\left(14\right)$

WhenTime corresponding contour line on the intersection point that point is profile and target main shaft (x_j,y_j), by point all of on contour line with (x_j,y_j) it is that starting point sorts in clockwise direction, Obtain point set:

P={P₁,P₂,...,P_i(i=1,2...., n), P_i=(x_i,y_i) (15)；

(3) use nearest neighbor method Hough transform to ask for profile and simplify line segment, parameter space introduces neighborhood Concept, the curved line number being verified in defined neighborhood, so that it is determined that can approximate in image space For the coordinate of point on straight line, ask for the Origin And Destination of line segment according to ultimate range method, and will ask The point taken is connected in turn and forms a closed curve being made up of line segment, reaches to simplify image outline Purpose；The profile of described target is reduced to:

Defining a two-dimensional array P and deposit the coordinate figure of profile point, first unit of array is profile base P on schedule₁Coordinate, second unit deposits P₂Coordinate, follow-up location contents the like, until P_n；P₁To P_nRankine-Hugoniot relations with P₁For starting point, by profile point at the clock-wise order of image space；

Define a two-dimensional array ThetaR (θ_n-1,ρ_n-1) deposit C₁With C_jThe θ of intersection point_j、ρ_jValue；C₁、 C_jRepresented curve it is mapped in parameter space for image space midpoint；θ_j、ρ_jFor C₁With C_j? The horizontal stroke of the aerial intersection point of parameter, vertical coordinate；

Define an one-dimension array A (θ_n-1,ρ_n-1) as accumulator, to intersection point phase same in parameter space Hand over number of times to count, initialize A (θ_n-1,ρ_n-1) each unit is zero；

Define an one-dimension arrayAs accumulator, deposit satisfied (θ_j,ρ_j) neighborhood pass The coordinate of the profile point of system, initializesEach unit is zero；

Step one: calculate C₁, C_jIntersection point (θ in parameter space_j,ρ_j), and be stored in ThetaR(θ_j,ρ_j)；

$\left\{\begin{array}{c}{\mathrm{\&rho;}}_j=x_1{\mathrm{cos\&theta;}}_j+y_1{\mathrm{sin\&theta;}}_j\\ {\mathrm{\&rho;}}_j=x_j{\mathrm{cos\&theta;}}_j+y_j{\mathrm{sin\&theta;}}_j\end{array}\right.---\left(16\right)$

$\left\{\begin{array}{c}{\mathrm{\&theta;}}_j=\arctan \left(\frac{x_1-x_j}{y_j-y_1}\right)\\ {\mathrm{\&rho;}}_j=x_1\cos \&lsqb;\arctan \left(\frac{x_1-x_j}{y_j-y_1}\right)\&rsqb;+y_1\sin \&lsqb;\arctan \left(\frac{x_1-x_j}{y_j-y_1}\right)\&rsqb;\end{array}\right.---\left(17\right)$

Step 2: calculate A (θ_n-1,ρ_n-1) each unit numerical value, travel through on contour line by the sequence number in formula 15 Except P₁Outer all of point, calculates curve and curve C that they are videoed in parameter space₁Intersection point；

When there is θ_j=θ_q=γ, ρ_j=ρ_q=s and j ＜ q (j, q=1,2 ..., n-1) time, will (θ_j,ρ_j) the accumulator element value at place adds 1, such as formula 18:

$\left\{\begin{array}{c}A({\mathrm{\&theta;}}_j,{\mathrm{\&rho;}}_j)=A({\mathrm{\&theta;}}_j,{\mathrm{\&rho;}}_j)+1\\ A({\mathrm{\&theta;}}_q,{\mathrm{\&rho;}}_q)=0\end{array}\right.---\left(18\right)$

As curve corresponding to k objective contour point and curve C₁(γ, time s), and j is for occurring to meet at same point It is identical that (γ, s) corresponding smallest sequence number profile point, then have A (θ_j,ρ_j)=k, other k-1 some correspondence is cumulative The value of device is zero；

Step 3: whether the curve calculating the corresponding parameter space of the point on contour line passes through (θ_j,ρ_j) place Neighborhood；

Calculate as accumulator A (θ_j,ρ_jDuring)=α >=1, with point (θ in parameter space_j,ρ_jNeighbour centered by) Territory comprises the bar number of curve:

If any point coordinate is P on contour line_m=(x_m,y_m), then the equation of its corresponding parameter space For:

$\stackrel{\&OverBar;}{\mathrm{\&rho;}}=x_{m}cos\stackrel{\&OverBar;}{\mathrm{\&theta;}}+y_{m}sin\stackrel{\&OverBar;}{\mathrm{\&theta;}}---\left(19\right)$

In formula 19,

To formula 19 derivation, then have:

$\frac{d\stackrel{\&OverBar;}{\mathrm{\&rho;}}}{d\stackrel{\&OverBar;}{\mathrm{\&theta;}}}=\sqrt{x_m^2+y_m^2}sin(\stackrel{\&OverBar;}{\mathrm{\&theta;}}+\arctan (-\frac{y_m}{x_m}\left)\right)---\left(20\right)$

WhenTime,Then have:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{min}=x_{m}cos({\mathrm{\&theta;}}_j-\mathrm{\&Delta;}\mathrm{\&theta;})+y_{m}sin({\mathrm{\&theta;}}_j-\mathrm{\&Delta;}\mathrm{\&theta;})---\left(21\right)$

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{max}=x_{m}cos({\mathrm{\&theta;}}_j+\mathrm{\&Delta;}\mathrm{\&theta;})+y_{m}sin({\mathrm{\&theta;}}_j+\mathrm{\&Delta;}\mathrm{\&theta;})---\left(22\right)$

WhenTime,Then have:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{min}=x_{m}cos({\mathrm{\&theta;}}_j+\mathrm{\&Delta;}\mathrm{\&theta;})+y_{m}sin({\mathrm{\&theta;}}_j+\mathrm{\&Delta;}\mathrm{\&theta;})---\left(23\right)$

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{max}=x_{m}cos({\mathrm{\&theta;}}_j-\mathrm{\&Delta;}\mathrm{\&theta;})+y_{m}sin({\mathrm{\&theta;}}_j-\mathrm{\&Delta;}\mathrm{\&theta;})---\left(24\right)$

WhenTime, existNowThe second order of formula 19 Derivative is:

$\frac{d^2\stackrel{\&OverBar;}{\mathrm{\&rho;}}}{d{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}^2}=\sqrt{x_m^2+y_m^2}sin(\stackrel{\&OverBar;}{\mathrm{\&theta;}}+\arctan (\frac{x_m}{y_m}\left)\right)---\left(25\right)$

Judged by formula 19,25On intervalExtreme value；

(1) whenTime,Then have:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{min}=x_{m}cos\mathrm{\&beta;}+y_{m}sin\mathrm{\&beta;}---\left(26\right)$

As [x_mcos(θ_j+Δθ)+y_msin(θ_j+Δθ)]-[x_mcos(θ_j-Δθ)+y_msin(θ_j-Δ θ)] ＞ 0 time:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{max}=x_{m}cos({\mathrm{\&theta;}}_j+\mathrm{\&Delta;}\mathrm{\&theta;})+y_{m}sin({\mathrm{\&theta;}}_j+\mathrm{\&Delta;}\mathrm{\&theta;})---\left(27\right)$

As [x_mcos(θ_j+Δθ)+y_msin(θ_j+Δθ)]-[x_mcos(θ_j-Δθ)+y_msin(θ_j-Δ θ)] ＜ 0 time:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{max}=x_{m}cos({\mathrm{\&theta;}}_j-\mathrm{\&Delta;}\mathrm{\&theta;})+y_{m}sin({\mathrm{\&theta;}}_j-\mathrm{\&Delta;}\mathrm{\&theta;})---\left(28\right)$

(2) whenTime,Then have:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{max}=x_{m}cos\mathrm{\&beta;}+y_{m}sin\mathrm{\&beta;}---\left(29\right)$

As [x_mcos(θ_j+Δθ)+y_msin(θ_j+Δθ)]-[x_mcos(θ_j-Δθ)+y_msin(θ_j-Δ θ)] ＞ 0 time:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{min}=x_{m}cos({\mathrm{\&theta;}}_j-\mathrm{\&Delta;}\mathrm{\&theta;})+y_{m}sin({\mathrm{\&theta;}}_j-\mathrm{\&Delta;}\mathrm{\&theta;})---\left(30\right)$

As [x_mcos(θ_j+Δθ)+y_msin(θ_j+Δθ)]-[x_mcos(θ_j-Δθ)+y_msin(θ_j-Δ θ)] ＜ 0 time:

${\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{\min }=x_{m}cos({\mathrm{\&theta;}}_j+\mathrm{\&Delta;}\mathrm{\&theta;})+y_{m}sin({\mathrm{\&theta;}}_j+\mathrm{\&Delta;}\mathrm{\&theta;})---\left(31\right)$

Can obtain to formula 31 according to formula 19Interval be:

$\stackrel{\&OverBar;}{\mathrm{\&rho;}}\&Element;\&lsqb;{\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{min},{\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{max}\&rsqb;---\left(32\right)$

When meeting formula 33, then put p_mIt is required point；

$\&lsqb;{\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{min},{\stackrel{\&OverBar;}{\mathrm{\&rho;}}}_{max}\&rsqb;\&cap;\&lsqb;{\mathrm{\&rho;}}_j-\mathrm{\&Delta;}\mathrm{\&rho;},{\mathrm{\&rho;}}_j+\mathrm{\&Delta;}\mathrm{\&rho;}\&rsqb;\&NotEqual;\mathrm{\&phi;}---\left(33\right)$

Step 4: ask for through (θ_j,ρ_j) set of neighborhood profile point；

Work as p_mValue in neighborhood meets formula 33, point (θ_j,ρ_j) value of corresponding accumulator elementFor:

$\stackrel{\&OverBar;}{A}({\mathrm{\&theta;}}_j,{\mathrm{\&rho;}}_j)=\stackrel{\&OverBar;}{A}({\mathrm{\&theta;}}_j,{\mathrm{\&rho;}}_j)+1---\left(34\right)$

According to formula 19-34, calculate A (θ_j,ρ_jAll profile point of)=α >=1；TakeTime, with (θ_j,ρ_jThe set of the profile point tried to achieve centered by) It is to meet and simplifies the point required.

2. the image outline method for simplifying of the employing nearest neighbor method Hough transform piece described in Ju claim 1, It is characterized in that: described neighborhood concepts is with image space cathetus L_jThe point being mapped in parameter space (θ_j,ρ_jCentered by), define a neighborhood U₁(θ_j,Δθ)*U₂(ρ_j, Δ ρ), by by this neighborhood The point on the objective contour on image space corresponding to all curves all regards straight line L as_jOn point； Straight line L is determined by the distribution of point set_jOrigin And Destination, and connect into approximate target profile Line segment, finally gives the simplification profile of image through loop computation.