CN101639355A - Three dimensional plane extraction method - Google Patents

Three dimensional plane extraction method Download PDF

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CN101639355A
CN101639355A CN200910044192A CN200910044192A CN101639355A CN 101639355 A CN101639355 A CN 101639355A CN 200910044192 A CN200910044192 A CN 200910044192A CN 200910044192 A CN200910044192 A CN 200910044192A CN 101639355 A CN101639355 A CN 101639355A
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文贡坚
王继阳
回丙伟
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National University of Defense Technology
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Abstract

本发明提供一种三维平面提取方法,技术方案的输入是以任意方法得到的目标场景中的三维直线特征集和稀疏数字高程模型数据、从两个不同视角获取的目标场景的立体图像及获取它们的立体相机的所有内、外参数、对立体图像的直线提取结果,技术方案的输出是定位精度高、边界完整性和准确性好的三维平面。技术方案的处理过程中,将三维平面的定位与平面区域边界的提取分离成两个前后衔接的阶段,分别独立完成这两个任务,减小它们之间的误差影响,使两个阶段的结果分别达到最优,从而实现三维平面提取的最优。

Figure 200910044192

The invention provides a three-dimensional plane extraction method. The input of the technical solution is the three-dimensional straight line feature set and sparse digital elevation model data in the target scene obtained by any method, the three-dimensional images of the target scene obtained from two different perspectives, and the acquisition of them. All internal and external parameters of the stereo camera, and the straight line extraction results of the stereo image, the output of the technical solution is a three-dimensional plane with high positioning accuracy, boundary integrity and accuracy. During the processing of the technical solution, the positioning of the three-dimensional plane and the extraction of the boundary of the plane area are separated into two consecutive stages, and these two tasks are independently completed to reduce the influence of errors between them, so that the results of the two stages Respectively achieve the optimal, so as to realize the optimal extraction of three-dimensional plane.

Figure 200910044192

Description

一种三维平面提取方法 A 3D Plane Extraction Method

技术领域 technical field

本发明涉及摄影测量与遥感、计算机视觉领域,尤其涉及一种三维平面的提取方法。The invention relates to the fields of photogrammetry, remote sensing and computer vision, in particular to a method for extracting a three-dimensional plane.

背景技术 Background technique

人造目标三维重建在城市规划、灾害监测、通信设施建设等领域具有非常广泛的应用,长期以来,一直是摄影测量与遥感、计算机视觉领域长期研究的重要课题。在各种多面体结构的目标三维重建方案中,三维平面提取是其中的一个关键步骤。The 3D reconstruction of man-made objects has a wide range of applications in the fields of urban planning, disaster monitoring, and communication facility construction. It has long been an important research topic in the fields of photogrammetry, remote sensing, and computer vision. In the target 3D reconstruction scheme of various polyhedral structures, 3D plane extraction is a key step.

目前,提取三维平面的方法主要有四种:第一种,利用直接获得的密集数字高程模型(Digital Elevation Model,简称DEM)数据,通过拟合获得三维平面;第二种,通过获取不同光照条件下的目标区域影像,结合目标表面的反射特性模型,建立多个反射图,利用光度学立体视觉法计算每个目标表面点的表面方向,实现对目标表面的重建;第三种,利用直线特征感知编组的方法,假定已知目标表面平面区域的形状,对已提取的直线特征进行特定规则的编组,如针对矩形平面区域的轮廓编组,生成多个平面假设,进而依据影像或其它先验知识进行平面假设的验证;第四种,对提取的直线特征进行匹配,通过前方交会计算对应的三维直线,对三维直线进行共面分析,生成所有可能的平面假设并构造平面边界以形成闭合平面区域,利用各种证据对这些闭合区域进行证实。相比较而言,第一种方法的缺点是随着高程点数据密度的增加,获取成本会同时增加,而且对于结构复杂的目标,如果缺少平面数量或目标结构的大致类型作为先验知识,提取目标表面的所有平面是很困难的。第二种方法的缺点是需要精确已知多个光照条件,因而该方法多见于单纯光照环境下的小体积目标的表面重建,对于光照条件复杂、背景多样或目标表面反射特性存在差异等各种非合作情况,该方法难以取得满意的结果。第三种方法的缺点在于可提取平面的几何形状受到限制。较前三种方法而言,第四种方法适应性更强,而且对高程数据的依赖小,甚至可以不需要高程数据的支持,同时,该方法采用一种自底向上的提取过程,可以解决各种多边形形状的三维平面提取。At present, there are four main methods for extracting 3D planes: the first method is to use the directly obtained dense digital elevation model (Digital Elevation Model, DEM) data to obtain a 3D plane by fitting; The image of the target area below, combined with the reflection characteristic model of the target surface, establishes multiple reflection maps, uses the photometric stereo vision method to calculate the surface direction of each target surface point, and realizes the reconstruction of the target surface; the third method uses the linear feature The method of perceptual grouping, assuming that the shape of the target surface plane area is known, grouping the extracted straight line features with specific rules, such as grouping the outline of a rectangular plane area, generating multiple plane hypotheses, and then based on images or other prior knowledge Carry out plane hypothesis verification; fourth, match the extracted straight line features, calculate the corresponding 3D straight line through forward intersection, perform coplanar analysis on the 3D straight line, generate all possible plane hypotheses and construct the plane boundary to form a closed plane area , using various evidences to confirm these closed regions. In comparison, the disadvantage of the first method is that as the data density of elevation points increases, the acquisition cost will increase at the same time, and for targets with complex structures, if there is no prior knowledge of the number of planes or the approximate type of target structure, the extraction All planes of the target surface are difficult. The disadvantage of the second method is that multiple lighting conditions need to be known accurately, so this method is mostly used in the surface reconstruction of small-volume targets in a simple lighting environment. In the cooperation situation, this method is difficult to obtain satisfactory results. The disadvantage of the third method is that the geometry of the planes that can be extracted is limited. Compared with the first three methods, the fourth method is more adaptable and less dependent on elevation data, and may not even need the support of elevation data. At the same time, this method adopts a bottom-up extraction process, which can solve 3D plane extraction of various polygonal shapes.

在三维平面提取过程中,需要得到的三维平面信息包含两个方面:一是三维平面定位信息,表现为它所在开放平面的平面方程;二是三维平面边界信息,表现为闭合平面的完整和准确的边界线。现有的第四种三维平面提取技术中,将提取这两种信息作为一个任务同时完成,平面的定位精度会受到边界定位误差的影响,而由平面相交得到的边界定位精度又会因为平面的定位参数误差而降低。In the process of 3D plane extraction, the 3D plane information that needs to be obtained includes two aspects: one is the positioning information of the 3D plane, which is expressed as the plane equation of the open plane where it is located; the other is the boundary information of the 3D plane, which is expressed as the completeness and accuracy of the closed plane borderline. In the fourth existing 3D plane extraction technology, the two kinds of information are extracted as one task at the same time, the positioning accuracy of the plane will be affected by the boundary positioning error, and the boundary positioning accuracy obtained by the intersection of the planes will be affected by the boundary positioning error of the plane. Positioning parameter errors are reduced.

发明内容 Contents of the invention

本发明要解决的技术问题是在三维平面提取过程中,将三维平面的定位与平面区域边界的提取分离成两个前后衔接的阶段,分别独立完成这两个任务,减小它们之间的误差影响,使两个阶段的结果分别达到最优,从而实现三维平面提取的最优。The technical problem to be solved by the present invention is to separate the positioning of the three-dimensional plane and the extraction of the boundary of the plane area into two consecutive stages in the process of extracting the three-dimensional plane, to complete these two tasks independently, and to reduce the error between them Influence, so that the results of the two stages are optimal respectively, so as to realize the optimal extraction of the three-dimensional plane.

本发明技术方案的输入是以任意方法得到的目标场景中的三维直线特征集和稀疏数字高程模型数据、从两个不同视角获取的目标场景的立体图像及获取它们的立体相机的所有内、外参数、对立体图像的直线提取结果,技术方案的输出是定位精度高、边界完整性和准确性好的三维平面,实施技术方案的目的是为后续多面体结构目标的表面重建工作打下基础。The input of the technical solution of the present invention is the three-dimensional straight line feature set and sparse digital elevation model data in the target scene obtained by any method, the stereo images of the target scene obtained from two different perspectives, and all the interior and exterior images of the stereo camera that obtained them. The output of the technical solution is a three-dimensional plane with high positioning accuracy, boundary integrity and accuracy. The purpose of implementing the technical solution is to lay a foundation for the subsequent surface reconstruction of polyhedral structural targets.

本发明的思路是:把三维平面提取过程分成前后衔接的两个优化步骤,第一个步骤是获取目标表面所在开放三维平面的高精度定位信息,实现方法是一种分割假设证实方法,具体流程是利用三维直线特征生成开放的三维平面假设→利用三维平面假设的成员直线特征分割开放的三维平面假设形成多个平面块假设→选取最优平面块假设→合并最优的平面块假设,生成高定位精度的开放三维平面;第二个步骤是构造平面的边界,基于生成的开放三维平面,以平面相交获取的平面交线、图像中提取的直线特征以及与目标结构相关的启发性知识作为生成平面边界的依据,为开放平面构造完整的边界,具体流程是利用开放平面的成员直线特征和平面间交线构造平面的边界→利用立体图像中提取的直线特征和启发式规则补充缺失的平面边界→由平面边界生成闭合平面假设,利用优化方法选择最优的闭合平面假设。The idea of the present invention is: the three-dimensional plane extraction process is divided into two optimization steps connected before and after. The first step is to obtain the high-precision positioning information of the open three-dimensional plane where the target surface is located. The realization method is a segmentation hypothesis verification method. The specific process It uses 3D line features to generate open 3D planar hypotheses → utilizes member line features of 3D planar hypotheses to split the open 3D planar hypotheses to form multiple planar block hypotheses → selects the optimal planar block hypothesis → merges the optimal planar block hypotheses to generate high The open 3D plane of positioning accuracy; the second step is to construct the boundary of the plane, based on the generated open 3D plane, the plane intersection line obtained by plane intersection, the straight line features extracted from the image, and the heuristic knowledge related to the target structure are used as the generated The basis of the plane boundary is to construct a complete boundary for the open plane. The specific process is to use the member straight line features of the open plane and the intersection lines between the planes to construct the plane boundary → use the straight line features extracted from the stereo image and heuristic rules to supplement the missing plane boundaries → Generate a closed plane hypothesis from the plane boundary, and use an optimization method to select the optimal closed plane hypothesis.

本发明的技术方案是一种三维平面提取方法,具体包括下述步骤:The technical solution of the present invention is a three-dimensional plane extraction method, specifically comprising the following steps:

已知某一场景的不同视角的两幅图像,分别称为左、右图像,以及摄取这两幅图像的左、右相机的所有内、外参数,其中左、右相机摄影中心在世界坐标系中的坐标分别记为(XO1,YO1,ZO1)和(XO2,YO2,ZO2);由这两幅图像中提取的直线特征集,分别记为H1和H2;已知该场景中的NS条三维直线特征,记为S={Li|i=1,2,…,NS},该集合中的任意元素Li表示一条三维直线段,它的两个端点分别为(Xi1,Yi1,Zi1)和(Xi2,Yi2,Zi2);已知该场景中的稀疏DEM数据,记为高程点集合

Figure G2009100441925D00031
该集合中元素的平均误差为λ。Two images of different viewing angles of a certain scene are known, called left and right images respectively, and all internal and external parameters of the left and right cameras that capture these two images, where the photography centers of the left and right cameras are in the world coordinate system The coordinates in are denoted as (X O1 , YO1 , Z O1 ) and (X O2 , YO2 , Z O2 ) respectively; the linear feature sets extracted from these two images are denoted as H 1 and H 2 respectively; The features of N S three-dimensional straight lines in the scene are known, recorded as S={L i |i=1, 2,..., N S }, any element L i in this set represents a three-dimensional straight line segment, and its two The endpoints are (X i1 , Y i1 , Z i1 ) and (X i2 , Y i2 , Z i2 ); the sparse DEM data in this scene are known, recorded as a collection of elevation points
Figure G2009100441925D00031
The average error of the elements in this set is λ.

第一步,求解目标表面所在的开放三维平面The first step is to solve the open 3D plane where the target surface is located

第(一)步,三维平面假设生成Step (1), three-dimensional plane hypothesis generation

设定角度门限φ,如果两条三维直线的夹角小于φ,则判定它们的方向矢量近似一致;设定距离门限ε,如果一个三维空间点到一条三维直线或一个三维平面的距离小于ε,则判定该三维空间点在这条三维直线上或这个三维平面上。两个门限的取值根据已知三维直线特征的定位精度而定。Set the angle threshold φ, if the angle between two 3D straight lines is smaller than φ, it is determined that their direction vectors are approximately the same; set the distance threshold ε, if the distance between a 3D space point and a 3D line or a 3D plane is less than ε, Then it is determined that the point in the three-dimensional space is on the three-dimensional straight line or on the three-dimensional plane. The values of the two thresholds are determined according to the positioning accuracy of the known 3D straight line features.

设定目标表面的任意一个平面的两条相邻边界所在直线的夹角范围为

Figure G2009100441925D00041
目标表面的任意两个相邻平面间的最小夹角为
Figure G2009100441925D00042
它们的取值与目标表面的结构特点有关。Set the range of the included angle between the straight lines where two adjacent boundaries of any plane of the target surface are located is
Figure G2009100441925D00041
The minimum angle between any two adjacent planes on the target surface is
Figure G2009100441925D00042
Their values are related to the structural characteristics of the target surface.

第1步,计算三维平面假设Step 1, calculate the 3D plane hypothesis

对集合S中的三维直线特征两两组合,生成三维平面假设,具体步骤为:Combine the 3D straight line features in the set S in pairs to generate a 3D plane hypothesis. The specific steps are:

任选两个三维直线特征Lu∈S和Lv∈S,Lu的端点为(Xu1,Yu1,Zu1)和(Xu2,Yu2,Zu2),Lv的端点为(Xv1,Yv1,Zv1)和(Xv2,Yv2,Zv2),利用它们定义目标函数f(a,b,c,d):Choose two 3D linear features L u ∈ S and L v ∈ S, the endpoints of L u are (X u1 , Y u1 , Z u1 ) and (X u2 , Y u2 , Z u2 ), the endpoints of L v are ( X v1 , Y v1 , Z v1 ) and (X v2 , Y v2 , Z v2 ), use them to define the objective function f(a, b, c, d):

f(a,b,c,d)=(a·Xu1+b·Yu1+c·Zu1+d)2+(a·Xu2+b·Yu2+c·Zu2+d)2+(a·Xv1+b·Yv1+c·Zv1+d)2+(a·Xv2+b·Yv2+c·Zv2+d)2 f(a, b, c, d) = (a X u1 + b Y u1 + c Z u1 + d) 2 + (a X u2 + b Y u2 + c Z u2 + d) 2 +(a X v1 +b Y v1 +c Z v1 +d) 2 +(a X v2 +b Y v2 +c Z v2 +d) 2

采用梯度下降法计算f(a,b,c,d)取最小值时的(a,b,c,d),得到由Lu和Lv指定的三维平面假设方程为a·X+b·Y+c·Z+d=0。Using the gradient descent method to calculate (a, b, c, d) when f (a, b, c, d) takes the minimum value, the hypothetical equation of the three-dimensional plane specified by Lu and L v is a·X+b· Y+c·Z+d=0.

按照上述步骤得到的所有三维平面假设集合Ω={Pi|i=1,2,…,N},对于其中任意一个三维平面假设Pi,平面方程为ai·X+bi·Y+ci·Z+di=0,用于生成它的两个三维直线特征称为Pi的成员直线特征。All three-dimensional plane hypothetical sets Ω={P i |i=1, 2, ..., N} obtained according to the above steps, for any one of the three-dimensional plane hypotheses P i , the plane equation is a i ·X+bi · Y+ c i ·Z+d i =0, the two three-dimensional line features used to generate it are called the member line features of P i .

第2步,合并共面的三维平面假设Step 2, incorporate coplanar 3D plane assumptions

对于任意一个三维平面假设Pi∈Ω,计算它的两个成员直线特征中点连线的中点在平面Pi上的投影,记为(XM i,YM i,ZM i)。For any three-dimensional plane hypothesis P i ∈ Ω, calculate the projection of the midpoint of the line connecting the midpoints of its two member straight features on the plane P i , denoted as (X M i , Y M i , Z M i ).

第1)步,判定三维平面假设的共面关系Step 1) Determine the coplanar relationship of the three-dimensional plane hypothesis

对于任意两个三维平面假设Pi∈Ω和Pj∈Ω,如果满足下面的不等式组,则判定Pi和Pj共面:For any two three-dimensional plane assumptions P i ∈ Ω and P j ∈ Ω, if the following inequalities are satisfied, it is determined that P i and P j are coplanar:

|| aa ii &CenterDot;&Center Dot; aa jj ++ bb ii &CenterDot;&Center Dot; bb jj ++ cc ii &CenterDot;&CenterDot; cc jj || aa ii 22 ++ bb ii 22 ++ cc ii 22 &CenterDot;&Center Dot; aa jj 22 ++ bb jj 22 ++ cc jj 22 >> coscos &phi;&phi; || aa ii &CenterDot;&Center Dot; Xx Mm jj ++ bb ii &CenterDot;&Center Dot; YY Mm jj ++ cc ii &CenterDot;&Center Dot; ZZ Mm jj ++ dd ii || aa ii 22 ++ bb ii 22 ++ cc ii 22 << &epsiv;&epsiv; || aa jj &CenterDot;&Center Dot; Xx Mm ii ++ bb jj &CenterDot;&Center Dot; YY Mm ii ++ cc jj &CenterDot;&Center Dot; ZZ Mm ii ++ dd jj || aa jj 22 ++ bb jj 22 ++ cc jj 22 << &epsiv;&epsiv;

第2)步,搜索共面的三维平面假设Step 2), search for coplanar 3D plane hypotheses

生成一个无向图G,将每个三维平面假设Pi∈Ω作为一个节点,如果任意两个三维平面假设是共面的,则它们对应的节点之间保持连接,否则不连接。Generate an undirected graph G, and take each three-dimensional plane hypothesis P i ∈ Ω as a node. If any two three-dimensional plane hypotheses are coplanar, their corresponding nodes are connected, otherwise they are not connected.

计算无向图G的所有极大团,记为{QW|w=1,2,…,NQ},任意一个极大团Qw是Ω的一个子集,记为 { P &Gamma; w ( &kappa; ) | &kappa; = 1,2 , &CenterDot; &CenterDot; &CenterDot; m w } , 子集中任意两个三维平面假设都是共面的。Calculate all the maximal cliques of the undirected graph G, denoted as {Q W |w=1, 2, ..., N Q }, any maximal clique Q w is a subset of Ω, denoted as { P &Gamma; w ( &kappa; ) | &kappa; = 1,2 , &CenterDot; &CenterDot; &CenterDot; m w } , Any two 3D planes in the subset are assumed to be coplanar.

第3)步,合并共面的三维平面假设Step 3) Merging coplanar 3D plane assumptions

对于任意一个极大团Qw,合并它包含的所有三维平面假设,得到一个新的三维平面假设Pw,它的平面方程记为aw·X+bw·Y+cw·Z+dw=0,其中:For any maximal clique Q w , combine all the three-dimensional plane hypotheses it contains to get a new three-dimensional plane hypothesis P w , whose plane equation is recorded as a w ·X+b w ·Y+c w ·Z+d w = 0, where:

aa &OverBar;&OverBar; ww == 11 mm ww &Sigma;&Sigma; &kappa;&kappa; == 11 mm ww aa &Gamma;&Gamma; ww (( &kappa;&kappa; )) aa &Gamma;&Gamma; ww (( &kappa;&kappa; )) 22 ++ bb &Gamma;&Gamma; ww (( &kappa;&kappa; )) 22 ++ cc &Gamma;&Gamma; ww (( &kappa;&kappa; )) 22

bb &OverBar;&OverBar; ww == 11 mm ww &Sigma;&Sigma; &kappa;&kappa; == 11 mm ww bb &Gamma;&Gamma; ww (( &kappa;&kappa; )) aa &Gamma;&Gamma; ww (( &kappa;&kappa; )) 22 ++ bb &Gamma;&Gamma; ww (( &kappa;&kappa; )) 22 ++ cc &Gamma;&Gamma; ww (( &kappa;&kappa; )) 22

cc &OverBar;&OverBar; ww == 11 mm ww &Sigma;&Sigma; &kappa;&kappa; == 11 mm ww cc &Gamma;&Gamma; ww (( &kappa;&kappa; )) aa &Gamma;&Gamma; ww (( &kappa;&kappa; )) 22 ++ bb &Gamma;&Gamma; ww (( &kappa;&kappa; )) 22 ++ cc &Gamma;&Gamma; ww (( &kappa;&kappa; )) 22

dd &OverBar;&OverBar; ww == -- 11 mm ww (( aa &OverBar;&OverBar; ww &CenterDot;&Center Dot; &Sigma;&Sigma; &kappa;&kappa; == 11 mm ww Xx Mm &Gamma;&Gamma; ww (( &kappa;&kappa; )) ++ bb &OverBar;&OverBar; ww &CenterDot;&Center Dot; &Sigma;&Sigma; &kappa;&kappa; == 11 mm ww YY Mm &Gamma;&Gamma; ww (( &kappa;&kappa; )) ++ cc &OverBar;&OverBar; ww &CenterDot;&Center Dot; &Sigma;&Sigma; &kappa;&kappa; == 11 mm ww ZZ Mm &Gamma;&Gamma; ww (( &kappa;&kappa; )) ))

{ P &Gamma; w ( &kappa; ) | &kappa; = 1,2 , &CenterDot; &CenterDot; &CenterDot; m w } 中每个三维平面假设的所有成员直线特征投影到Pw上,得到的所有投影直线段成为Pw的成员直线特征。Will { P &Gamma; w ( &kappa; ) | &kappa; = 1,2 , &CenterDot; &CenterDot; &CenterDot; m w } All the member straight line features of each three-dimensional plane hypothesized in are projected onto P w , and all the projected straight line segments obtained become the member straight line features of P w .

由所有极大团得到的三维平面假设集合记为Ω={Pw|w=1,2,…,N},其中,N=NQ,任意三维平面假设Pw的成员直线特征集合记为 M &OverBar; w = { L &OverBar; j w | j = 1,2 , &CenterDot; &CenterDot; &CenterDot; , n &OverBar; w } , 任意一个成员直线特征Lj w的端点记为(Xj,1 w,Yj,1 w,Zj,1 w)和(Zj,2 w,Yj,2 w,Zj,2 w)。The set of three-dimensional plane hypotheses obtained from all maximal cliques is denoted as Ω={P w |w=1, 2,...,N}, where N=N Q , and the feature set of member straight lines of any three-dimensional plane hypothesis P w is denoted as m &OverBar; w = { L &OverBar; j w | j = 1,2 , &CenterDot; &CenterDot; &Center Dot; , no &OverBar; w } , The endpoints of any member line feature L j w are recorded as (X j, 1 w , Y j, 1 w , Z j, 1 w ) and (Z j, 2 w , Y j, 2 w , Z j, 2 w ).

第(二)步,三维平面假设证实In the second step, the three-dimensional plane hypothesis is confirmed

第1步,分割三维平面假设为平面块假设Step 1, segment the 3D planar hypothesis into planar block hypotheses

对于任意一个三维平面假设Pi∈Ω,利用其成员直线特征Mi对它进行分割,得到若干平面块假设。具体方法包括以下步骤:For any three-dimensional planar hypothesis P i ∈ Ω, use its member line feature M i to segment it, and obtain several planar block hypotheses. The specific method includes the following steps:

第1)步,合并共线的成员直线特征Step 1) Merge the collinear member line features

对于Pi的任意两个成员直线特征Lj i L &OverBar; k i &Element; M &OverBar; i , 计算它们所在直线的夹角,记为μ,以及Lj i的中点到Lk i所在直线的距离和Lk i的中点到Lj i所在直线的距离,分别记为η1和η2For any two member line features L j i of P i , L &OverBar; k i &Element; m &OverBar; i , Calculate the angle between the straight lines where they are located, denoted as μ, and the distance from the midpoint of L j i to the straight line of L k i and the distance from the midpoint of L k i to the straight line of L j i , respectively denoted as η 1 and η 2 ;

如果μ<φ并且η1<ε和μ2<ε,则判定Lj i和Lk i共线,用它们生成一个新的三维直线特征,该三维直线特征的两个端点为Lj i和Lk i的相距最远的两个端点,将新的三维直线特征添加到集合Mi中,并从集合Mi中删除Lj i和Lk iIf μ<φ and η 1 <ε and μ 2 <ε, it is judged that L j i and L k i are collinear, and they are used to generate a new three-dimensional straight line feature, the two endpoints of this three-dimensional straight line feature are L j i and The two farthest endpoints of L k i add new 3D straight line features to set M i , and delete L j i and L k i from set M i .

第2)步,计算Pi的所有成员直线特征的交点Step 2) Calculate the intersection of all member straight line features of P i

对于Pi的任意两个成员直线特征Lα i L &OverBar; &beta; i &Element; M &OverBar; i , 如果它们所在直线的夹角θ满足则计算Lα i和Lβ i的交点或者它们的延长线的交点。For any two member line features L α i of P i , L &OverBar; &beta; i &Element; m &OverBar; i , If the included angle θ of their straight lines satisfies Then calculate the intersection point of L α i and L β i or the intersection point of their extension lines.

所有计算的交点组成的集合记为RiThe set of all calculated intersection points is denoted as R i .

第3)步,生成用于分割平面假设的无向图Step 3) Generate an undirected graph for splitting plane hypotheses

生成一个无向图,记为Gi,它的节点由两类点生成:一是交点集合Ri中的每个元素,二是对于Mi中任意一个成员直线特征,如果在它的一个端点外的延长线上没有与Mi中其它成员直线特征的交点,并且该端点不是集合Ri中的元素,则该端点成为图Gi的一个节点。Generate an undirected graph, denoted as G i , whose nodes are generated by two types of points: one is each element in the intersection point set R i , and the other is a straight line feature for any member in M i , if at one of its endpoints If there is no intersection point on the extension line outside the line with other members of M i , and this end point is not an element in set R i , then this end point becomes a node of graph G i .

无向图Gi中任意两个节点间的连接关系包括两类:一是如果它们对应的点在同一个成员直线特征或其延长线上,并且这两个点的连接线段上没有与图Gi中其它节点相对应的点,则它们保持连接;二是构造的连接关系,构造方法如下:The connection relationship between any two nodes in the undirected graph G i includes two types: one is if their corresponding points are on the same member straight line feature or its extension line, and there is no connection between the two points on the line segment of the graph G Points corresponding to other nodes in i , they remain connected; the second is the connection relationship constructed, and the construction method is as follows:

对于任意一个成员直线特征Lα i,如果它的一个或两个端点是图Gi的节点,则分两种情况进行处理:For any member straight line feature L α i , if one or both of its endpoints are nodes of graph G i , it can be processed in two cases:

第一种情况,Lα i与Mi中的任意其它成员直线特征均不相交In the first case, L α i does not intersect with any other member line features in M i

选择任意成员直线特征 L &OverBar; &beta; i &Element; M &OverBar; i , 且β≠α,指定它所在直线上与图Gi中两个节点相对应的两个点,这两个点位于Lβ i的中点两侧并且它们之间的连接线段上没有与图Gi中节点相对应的点,将它们记为(X1,Y1,Z1)和(X2,Y2,Z2),并计算Select any member line feature L &OverBar; &beta; i &Element; m &OverBar; i , And β≠α, specify two points corresponding to the two nodes in the graph G i on the straight line where it is located, these two points are located on both sides of the midpoint of L β i and there is no connection line segment between them and the graph G Points corresponding to nodes in i , record them as (X 1 , Y 1 , Z 1 ) and (X 2 , Y 2 , Z 2 ), and calculate

ee 11 == (( Xx &OverBar;&OverBar; &alpha;&alpha; ,, 11 ii -- Xx 11 )) &CenterDot;&Center Dot; YY 22 -- YY 11 Xx 22 -- Xx 11 ++ YY 11 -- YY &OverBar;&OverBar; &alpha;&alpha; ,, 11 ii

ee 22 == (( Xx &OverBar;&OverBar; &alpha;&alpha; ,, 22 ii -- Xx 11 )) &CenterDot;&CenterDot; YY 22 -- YY 11 Xx 22 -- Xx 11 ++ YY 11 -- YY &OverBar;&OverBar; &alpha;&alpha; ,, 22 ii

其中,(Xα,1 i,Yα,1 i,Zα,1 i)和(Xα,2 i,Yα,2 i,Zα,2 i)是Lα i的端点。对于任意 L &OverBar; l i &Element; M &OverBar; i , 且l≠α、l≠β,计算Among them, (X α,1 i , Y α,1 i , Z α,1 i ) and (X α,2 i , Y α,2 i , Z α,2 i ) are endpoints of L α i . for any L &OverBar; l i &Element; m &OverBar; i , And l≠α, l≠β, calculate

ff 11 == (( Xx &OverBar;&OverBar; ll ,, 11 ii -- Xx 11 )) &CenterDot;&Center Dot; YY 22 -- YY 11 Xx 22 -- Xx 11 ++ YY 11 -- YY &OverBar;&OverBar; ll ,, 11 ii

ff 22 == (( Xx &OverBar;&OverBar; ll ,, 22 ii -- Xx 11 )) &CenterDot;&Center Dot; YY 22 -- YY 11 Xx 22 -- Xx 11 ++ YY 11 -- YY &OverBar;&OverBar; ll ,, 22 ii

ff 33 == (( Xx &OverBar;&OverBar; ll ,, 11 ii -- Xx &OverBar;&OverBar; &alpha;&alpha; ,, 11 ii )) &CenterDot;&Center Dot; YY &OverBar;&OverBar; &alpha;&alpha; ,, 22 ii -- YY &OverBar;&OverBar; &alpha;&alpha; ,, 11 ii Xx &OverBar;&OverBar; &alpha;&alpha; ,, 22 ii -- Xx &OverBar;&OverBar; &alpha;&alpha; ,, 11 ii ++ YY &OverBar;&OverBar; &alpha;&alpha; ,, 11 ii -- YY &OverBar;&OverBar; ll ,, 11 ii

ff 44 == (( Xx &OverBar;&OverBar; ll ,, 22 ii -- Xx &OverBar;&OverBar; &alpha;&alpha; ,, 11 ii )) &CenterDot;&CenterDot; YY &OverBar;&OverBar; &alpha;&alpha; ,, 22 ii -- YY &OverBar;&OverBar; &alpha;&alpha; ,, 11 ii Xx &OverBar;&OverBar; &alpha;&alpha; ,, 22 ii -- Xx &OverBar;&OverBar; &alpha;&alpha; ,, 11 ii ++ YY &OverBar;&OverBar; &alpha;&alpha; ,, 11 ii -- YY &OverBar;&OverBar; ll ,, 22 ii

其中,(Xl,1 i,Yl,1 i,Zl,1 i)和(Xl,2 i,Yl,2 i,Zl,2 i)是Ll i的端点。如果满足Among them, (X l, 1 i , Y l, 1 i , Z l, 1 i ) and (X l, 2 i , Y l, 2 i , Z l, 2 i ) are endpoints of L l i . if satisfied

ee 11 &CenterDot;&CenterDot; ee 22 >> 00 ff 11 &CenterDot;&CenterDot; ff 33 >> 00 ff 22 &CenterDot;&CenterDot; ff 44 >> 00 ff 11 &CenterDot;&Center Dot; ff 22 >> 00

则计算Then calculate

ff 55 == (( Xx &OverBar;&OverBar; ll ,, 11 ii -- Xx 11 )) &CenterDot;&Center Dot; YY &OverBar;&OverBar; ll ,, 22 ii -- YY 11 Xx &OverBar;&OverBar; ll ,, 22 ii -- Xx 11 ++ YY 11 -- YY &OverBar;&OverBar; ll ,, 11 ii

ff 66 == (( Xx 22 -- Xx 11 )) &CenterDot;&Center Dot; YY &OverBar;&OverBar; ll ,, 22 ii -- YY 11 Xx &OverBar;&OverBar; ll ,, 22 ii -- Xx 11 ++ YY 11 -- YY 22

如果f5·f6>0,则连接点(Xα,i 1,Yα,1 i,Zα,1 i)和(X2,Y2,Z2)在图Gi中对应的节点,以及点(Xα,2 i,Yα,2 i,Zα,2 i)和(X1,Y1,Z1)在图Gi中对应的节点。If f 5 ·f 6 > 0, connect the points (X α, i 1 , Y α, 1 i , Z α, 1 i ) and (X 2 , Y 2 , Z 2 ) the corresponding nodes in the graph G i , and the corresponding nodes of points (X α, 2 i , Y α, 2 i , Z α, 2 i ) and (X 1 , Y 1 , Z 1 ) in graph G i .

第二种情况,Lα i与Mi中的至少一个其它成员直线特征相交In the second case, L α i intersects with at least one other member line feature in M i

如果(Xα,1 i,Yα,1 i,Zα,1 i)与图Gi的一个节点相对应,选择成员直线特征Lα i或其延长线上与(Xα,1 i,Yα,1 i,Zα,1 i)距离最近的交点,记为(X3,Y3,Z3),假定该交点是Lα i与成员直线特征Lβ i的交点。选择Lβ i或其延长线上的与图Gi中节点相对应的点,如果在它和点(X3,Y3,Z3)之间的连线上没有与图Gi中节点相对应的点,则将它记为(X4,Y4,Z4),如果这样的点有两个,则分别将它们记为(X4,Y4,Z4)和(X5,Y5,Z5)。If (X α, 1 i , Y α, 1 i , Z α, 1 i ) corresponds to a node of graph G i , select the member line feature L α i or its extended line with (X α, 1 i , Y α, 1 i , Z α, 1 i ) is the nearest intersection point, denoted as (X 3 , Y 3 , Z 3 ), it is assumed that the intersection point is the intersection point of L α i and member straight line feature L β i . Select the point corresponding to the node in graph G i on L β i or its extension line, if there is no node corresponding to the node in graph G i on the line between it and point (X 3 , Y 3 , Z 3 ) If there are two such points, record them as (X 4 , Y 4 , Z 4 ) and (X 5 , Y 5 , Z5 ).

如果(X4,Y4,Z4)是Lβ i与其它成员直线特征的交点,假定该成员直线特征为Ll i,并且Ll i或其延长线上存在与图Gi中节点相对应的点,它满足与点(Xα,1 i,Yα,1 i,Zα,1 i)落在Lβ i同一侧,并且与点(X4,Y4,Z4)之间的连线上没有与图Gi中节点相对应的点,则连接它与点(Xα,1 i,Yα,1 i,Zα,1 i)在图Gi中对应的节点;否则,连接点(Xα,1 i,Yα,1 i,Zα,1 i)和点(X4,Y4,Z4)在图Gi中对应的节点。如果(X5,Y5,Z5)存在,则按照与(X4,Y4,Z4)同样的方法进行处理。If (X 4 , Y 4 , Z 4 ) is the intersection point of L β i and other member line features, it is assumed that the member line feature is L l i , and there is a node on L l i or its extension line that is related to the node in graph G i The corresponding point, which satisfies the point (X α, 1 i , Y α, 1 i , Z α, 1 i ) on the same side of L β i and between the point (X 4 , Y 4 , Z 4 ) There is no point corresponding to the node in the graph G i on the connection line of , then connect it with the node corresponding to the point (X α, 1 i , Y α, 1 i , Z α, 1 i ) in the graph G i ; otherwise , connecting point (X α, 1 i , Y α, 1 i , Z α, 1 i ) and point (X 4 , Y 4 , Z 4 ) corresponding nodes in graph G i . If (X 5 , Y 5 , Z 5 ) exists, it is processed in the same way as (X 4 , Y 4 , Z 4 ).

如果(Xα,2 i,Yα,2 i,Zα,2 i)是图Gi的节点,则构造连接关系的方法与(Xα,1 i,Yα,1 i,Zα,1 i)相同。If (X α, 2 i , Y α, 2 i , Z α, 2 i ) is a node of graph G i , then the method of constructing the connection relationship is the same as (X α, 1 i , Y α, 1 i , Z α, 1i ) same.

第4)步,分割生成平面块假设Step 4) Segmentation generates planar block hypotheses

搜索无向图Gi中的所有环,对于其中任意一个环,如果它包含的属于同一个成员直线特征或其延长线的节点数不超过2个,则用它生成一个平面块假设,它包含的所有节点按照其连接关系组成一个有序点集,表示该平面块假设的顶点集。Search all the rings in the undirected graph G i , for any one of the rings, if it contains no more than 2 nodes belonging to the same member straight line feature or its extension, use it to generate a planar block hypothesis, which contains All the nodes of are formed into an ordered point set according to their connection relationship, which represents the hypothetical vertex set of the plane block.

由Ω中所有平面分割得到的平面块假设的集合记为 &Omega; ^ = { P ^ w | w = 1,2 , &CenterDot; &CenterDot; &CenterDot; , N ^ } , 其中,任意一个平面块假设

Figure G2009100441925D00092
由它的顶点集来表示,记为 { ( X ^ wj , Y ^ wj , Z ^ wj ) | j = 1,2 , &CenterDot; &CenterDot; &CenterDot; , n ^ w } , Ω中分割得到的三维平面假设的方程记为 a ^ w &CenterDot; X + b ^ w &CenterDot; Y + c ^ w &CenterDot; Z + d ^ w = 0 . The set of planar block hypotheses obtained from all planar partitions in Ω is denoted as &Omega; ^ = { P ^ w | w = 1,2 , &CenterDot; &CenterDot; &CenterDot; , N ^ } , Among them, any plane block assumes
Figure G2009100441925D00092
Represented by its vertex set, denoted as { ( x ^ wj , Y ^ wj , Z ^ wj ) | j = 1,2 , &CenterDot; &Center Dot; &Center Dot; , no ^ w } , divided into Ω to get The equation for the three-dimensional plane assumption of is written as a ^ w &Center Dot; x + b ^ w &Center Dot; Y + c ^ w &CenterDot; Z + d ^ w = 0 .

第2步,计算每个平面块假设的可靠性测度Step 2, calculate the reliability measure for each planar block hypothesis

平面块假设可靠性测度的计算依据已知的高程点集合J和左、右图像,计算J中每个点在左、右图像中的投影,得到两个平面点集,分别记为

Figure G2009100441925D00096
Figure G2009100441925D00097
其中,
Figure G2009100441925D00098
Figure G2009100441925D00099
分别是J中的高程点在左、右图像中的投影点。The calculation of the reliability measure of the plane block hypothesis is based on the known elevation point set J and the left and right images, and the projection of each point in J in the left and right images is calculated to obtain two plane point sets, which are denoted as
Figure G2009100441925D00096
and
Figure G2009100441925D00097
in,
Figure G2009100441925D00098
and
Figure G2009100441925D00099
are the elevation points in J Projection points in the left and right images.

对任意平面块假设 P ^ w &Element; &Omega; ^ , 计算其可靠性测度,包括下述步骤:Assume for any planar block P ^ w &Element; &Omega; ^ , Calculate its reliability measure, including the following steps:

第1)步,搜索被

Figure G2009100441925D000912
覆盖的已知高程点In step 1), the search is
Figure G2009100441925D000912
Covered known elevation points

投影平面块假设

Figure G2009100441925D000913
的顶点集到左、右图像,记任意顶点
Figure G2009100441925D000914
在左、右图像中的投影分别为
Figure G2009100441925D000915
Figure G2009100441925D000916
按照顶点集中各定点的顺序分别在左、右图像中连接顶点的投影,形成
Figure G2009100441925D000917
的两个多边形投影区域。Projected planar block assumption
Figure G2009100441925D000913
The vertex set to the left and right images, remember any vertex
Figure G2009100441925D000914
The projections in the left and right images are respectively
Figure G2009100441925D000915
and
Figure G2009100441925D000916
According to the order of each fixed point in the vertex set, connect the projections of the vertices in the left and right images respectively, forming
Figure G2009100441925D000917
The two polygonal projection areas of .

对于任意高程点

Figure G2009100441925D000918
如果其左图像中的投影位于
Figure G2009100441925D000920
在左图像中的投影区域内,并且其右图像中的投影
Figure G2009100441925D000921
位于
Figure G2009100441925D000922
在右图像中的投影区域内,则判定高程点
Figure G2009100441925D000924
覆盖。For any elevation point
Figure G2009100441925D000918
If the projection in its left image lie in
Figure G2009100441925D000920
in the projected area in the left image, and its projected in the right image
Figure G2009100441925D000921
lie in
Figure G2009100441925D000922
In the projected area in the right image, determine the elevation point quilt
Figure G2009100441925D000924
cover.

Figure G2009100441925D000925
覆盖的所有高程点组成的集合记为 &Theta; w &Subset; J , 集合中元素的数量为αw。quilt
Figure G2009100441925D000925
The set of all elevation points covered is denoted as &Theta; w &Subset; J , The number of elements in the set is α w .

第2)步,计算高程一致性测度Step 2) Calculate the elevation consistency measure

如果

Figure G2009100441925D00101
覆盖的高程点个数αw>0,则计算
Figure G2009100441925D00102
的高程一致性测度为if
Figure G2009100441925D00101
If the number of covered elevation points α w >0, calculate
Figure G2009100441925D00102
The elevation consistency measure of is

Figure G2009100441925D00103
(公式一)
Figure G2009100441925D00103
(Formula 1)

其中,in,

Figure G2009100441925D00104
Figure G2009100441925D00104

Figure G2009100441925D00105
Figure G2009100441925D00105

如果αw=0,则 E w D = 0 . If α w =0, then E. w D. = 0 .

第3)步,计算灰度相似性测度Step 3) Calculate the gray similarity measure

Figure G2009100441925D00107
在左图像中的投影区域内所有图像点组成一个点集,记为Kw L,它们中所有点的图像灰度值组成一个数组,记为{hwt|t=1,2,…,ow}。
Figure G2009100441925D00107
All image points in the projection area in the left image form a point set, which is recorded as K w L , and the image gray values of all points in them form an array, which is recorded as {h wt |t=1, 2, ..., o w }.

依据在左、右图像中三对同名点

Figure G2009100441925D00109
Figure G2009100441925D001010
Figure G2009100441925D001011
以及
Figure G2009100441925D001012
Figure G2009100441925D001013
计算集合Kw L中所有元素点在右图像中的同名点,这些同名点组成的集合记为Kw R,它们中所有点的图像灰度值由插值算法得到,这些灰度值组成另一个数组,记为{hwt′|t=1,2,…,ow}。计算
Figure G2009100441925D001014
在左、右图像中的投影区域之间的灰度相似性测度为in accordance with Three pairs of points with the same name in the left and right images
Figure G2009100441925D00109
and
Figure G2009100441925D001010
and
Figure G2009100441925D001011
as well as
Figure G2009100441925D001012
and
Figure G2009100441925D001013
Calculate the same-name points of all element points in the set K w L in the right image. The set of these same-name points is recorded as K w R . The image gray values of all points in them are obtained by interpolation algorithm, and these gray values form another An array, denoted as {h wt ′|t=1, 2, ..., o w }. calculate
Figure G2009100441925D001014
The gray-scale similarity measure between the projected regions in the left and right images is

E w G = &Sigma; t = 1 O w ( h wt - h &OverBar; w ) ( h wt &prime; - h &OverBar; w &prime; ) [ &Sigma; t = 1 O w ( h wt - h &OverBar; w ) 2 ] &CenterDot; [ &Sigma; t = 1 O w ( h wt &prime; - h &OverBar; w &prime; ) 2 ] (公式二) E. w G = &Sigma; t = 1 o w ( h wt - h &OverBar; w ) ( h wt &prime; - h &OverBar; w &prime; ) [ &Sigma; t = 1 o w ( h wt - h &OverBar; w ) 2 ] &Center Dot; [ &Sigma; t = 1 o w ( h wt &prime; - h &OverBar; w &prime; ) 2 ] (Formula 2)

其中, h &OverBar; w = 1 O w &Sigma; t = 1 O w h wt , h &OverBar; w &prime; = 1 O w &Sigma; t = 1 O w h wt &prime; . in, h &OverBar; w = 1 o w &Sigma; t = 1 o w h wt , h &OverBar; w &prime; = 1 o w &Sigma; t = 1 o w h wt &prime; .

第4)步,计算的可靠性测度Step 4) Calculate reliability measure

Figure G2009100441925D00111
的可靠性测度为
Figure G2009100441925D00111
The reliability measure for

E w = E w D &CenterDot; ( E w G + 1 ) / 2 E w D > 0 ( E w G + 1 ) / 2 E w D = 0 (公式三) E. w = E. w D. &CenterDot; ( E. w G + 1 ) / 2 E. w D. > 0 ( E. w G + 1 ) / 2 E. w D. = 0 (Formula 3)

第3步,搜索可靠的平面块假设Step 3, search for reliable planar block hypotheses

分两个步骤完成:This is done in two steps:

第1)步,搜索有已知高程点覆盖的可靠平面块假设Step 1), search for reliable planar block hypotheses covered by known elevation points

建立一个无向图,对于任意一个平面块假设 P ^ w &Element; &Omega; ^ , 如果 E w D > 0 , 则用它生成该无向图的一个节点,节点的属性值为Ew;对于任意两个节点,如果它们对应的平面块假设在左、右图像中的投影区域均不存在重叠部分,则连接它们。Build an undirected graph, for any planar block assume P ^ w &Element; &Omega; ^ , if E. w D. > 0 , Then use it to generate a node of the undirected graph, and the attribute value of the node is E w ; for any two nodes, if their corresponding planar blocks assume that there is no overlap in the projected areas in the left and right images, then the connection they.

计算该无向图的所有极大团,从中选取节点属性值之和最大的极大团包含的所有平面块假设的集合,记为K1,它表示有覆盖的高程点作为衡量依据的可靠平面块假设。Calculate all the maximal cliques of the undirected graph, and select the hypothetical set of all planar blocks contained in the maximal clique with the largest sum of node attribute values, denoted as K 1 , which represents a reliable plane with covered elevation points as the basis for measurement block assumptions.

第2)步,搜索没有已知高程点覆盖的可靠平面块假设Step 2) Search for reliable planar block hypotheses without known elevation point coverage

建立一个无向图,对于任意一个平面块假设 P ^ w &Element; &Omega; ^ , 如果 E w D = 0 并且它与K1中任意一个平面块假设在左、右图像中的投影区域均不存在重叠部分,则用

Figure G2009100441925D00117
生成该无向图的一个节点,节点的属性值为Ew;对于任意两个节点,如果它们对应的平面块假设在左、右图像中的投影区域均不存在重叠部分,则连接它们。Build an undirected graph, for any planar block assume P ^ w &Element; &Omega; ^ , if E. w D. = 0 And it is assumed that there is no overlap with any planar block in K1 in the projection area of the left and right images, then use
Figure G2009100441925D00117
Generate a node of the undirected graph, and the attribute value of the node is E w ; for any two nodes, if their corresponding planar blocks assume that there is no overlap in the projected areas in the left and right images, connect them.

计算该无向图的所有极大团,从中选取节点属性值之和最大的极大团包含的所有平面块假设的集合,记为K2,它表示没有覆盖的高程点作为衡量依据并且与K1中所有平面块假设可共存的可靠平面块假设。Calculate all the maximal cliques of the undirected graph, select the hypothetical set of all plane blocks contained in the maximal clique with the largest sum of node attribute values, and record it as K 2 , which represents the elevation point without coverage as the basis for measurement and is related to K A reliable planar block assumption that all planar block hypotheses in 1 can co-exist.

将K1∪K2包含的所有平面块假设作为可靠平面块假设。Take all planar block hypotheses contained in K 1 ∪K 2 as reliable planar block hypotheses.

第4步,合并共面的可靠平面块假设Step 4, Merge Coplanar Reliable Planar Block Hypotheses

第1)步,判定平面块假设的共面关系Step 1) Determine the coplanar relationship assumed by the planar block

对于任意两个可靠平面块假设

Figure G2009100441925D00121
P ^ v &Element; K 1 &cup; K 2 , 如果下面的不等式组成立,则判定
Figure G2009100441925D00123
Figure G2009100441925D00124
共面:For any two reliable planar blocks assume
Figure G2009100441925D00121
P ^ v &Element; K 1 &cup; K 2 , If the following inequalities are established, then the decision
Figure G2009100441925D00123
and
Figure G2009100441925D00124
Coplanar:

|| aa ^^ uu &CenterDot;&CenterDot; aa ^^ vv ++ bb ^^ uu &CenterDot;&CenterDot; bb ^^ vv ++ cc ^^ uu ++ cc ^^ vv || aa ^^ uu 22 ++ bb ^^ uu 22 ++ cc ^^ uu 22 &CenterDot;&CenterDot; aa ^^ vv 22 ++ bb ^^ vv 22 ++ cc ^^ vv 22 >> coscos (( 33 22 &CenterDot;&Center Dot; &phi;&phi; )) || aa ^^ uu &CenterDot;&Center Dot; Xx ^^ vv 11 ++ bb ^^ uu &CenterDot;&Center Dot; YY ^^ vv 11 ++ cc ^^ uu &CenterDot;&Center Dot; ZZ ^^ vv 11 ++ dd ^^ uu || aa ^^ uu 22 ++ bb ^^ uu 22 ++ cc ^^ uu 22 << 33 22 &CenterDot;&Center Dot; &epsiv;&epsiv; || aa ^^ vv &CenterDot;&Center Dot; Xx ^^ uu 11 ++ bb ^^ vv &CenterDot;&Center Dot; YY ^^ uu 11 ++ cc ^^ vv &CenterDot;&Center Dot; ZZ ^^ uu 11 ++ dd ^^ vv || aa ^^ vv 22 ++ bb ^^ vv 22 ++ cc ^^ vv 22 << 33 22 &CenterDot;&Center Dot; &epsiv;&epsiv;

第2)步,搜索共面的可靠平面块假设Step 2), search for coplanar reliable planar block hypotheses

生成一个无向图,记为

Figure G2009100441925D00126
用K1∪K2中每一个可靠平面块假设生成一个节点,如果任意两个可靠平面块假设是共面的,则连接它们对应的节点。Generate an undirected graph, denoted as
Figure G2009100441925D00126
Use each reliable plane block hypothesis in K 1K 2 to generate a node, if any two reliable plane block hypotheses are coplanar, connect their corresponding nodes.

计算图

Figure G2009100441925D00127
的所有极大团,记为 { Q ^ s | s = 1,2 , &CenterDot; &CenterDot; &CenterDot; , N ^ Q } , 任意一个极大团
Figure G2009100441925D00129
是图的一个子集,记为 { P ^ &Gamma; ^ s ( &kappa; ) | &kappa; = 1,2 , &CenterDot; &CenterDot; &CenterDot; m ^ s } &Subset; K 1 &cup; K 2 , 该子集中任意两个可靠平面块假设都是共面的。Computation graph
Figure G2009100441925D00127
All maximal cliques of , denoted as { Q ^ the s | the s = 1,2 , &Center Dot; &Center Dot; &Center Dot; , N ^ Q } , any maximal group
Figure G2009100441925D00129
is a picture A subset of , denoted as { P ^ &Gamma; ^ the s ( &kappa; ) | &kappa; = 1,2 , &Center Dot; &Center Dot; &Center Dot; m ^ the s } &Subset; K 1 &cup; K 2 , Any two reliable planar blocks in this subset are assumed to be coplanar.

第3)步,合并共面的可靠平面块假设Step 3) Merge coplanar reliable planar block hypotheses

对于任意一个极大团

Figure G2009100441925D001212
合并它包含的所有可靠平面块假设,得到一个开放平面平面方程记为 a &OverBar; &OverBar; s &CenterDot; X + b &OverBar; &OverBar; s &CenterDot; Y + c &OverBar; &OverBar; s &CenterDot; Z + d &OverBar; &OverBar; s = 0 , 其中:For any maximal group
Figure G2009100441925D001212
Merges all reliable planar block hypotheses it contains, resulting in an open planar The plane equation is denoted as a &OverBar; &OverBar; the s &Center Dot; x + b &OverBar; &OverBar; the s &Center Dot; Y + c &OverBar; &OverBar; the s &Center Dot; Z + d &OverBar; &OverBar; the s = 0 , in:

aa &OverBar;&OverBar; &OverBar;&OverBar; sthe s == 11 mm ^^ sthe s &Sigma;&Sigma; &kappa;&kappa; == 11 mm ^^ sthe s aa ^^ &Gamma;&Gamma; ^^ sthe s (( &kappa;&kappa; )) aa ^^ &Gamma;&Gamma; ^^ sthe s (( &kappa;&kappa; )) 22 ++ bb ^^ &Gamma;&Gamma; ^^ sthe s (( &kappa;&kappa; )) 22 ++ cc ^^ &Gamma;&Gamma; ^^ sthe s (( &kappa;&kappa; )) 22

bb &OverBar;&OverBar; &OverBar;&OverBar; sthe s == 11 mm ^^ sthe s &Sigma;&Sigma; &kappa;&kappa; == 11 mm ^^ sthe s bb ^^ &Gamma;&Gamma; ^^ sthe s (( &kappa;&kappa; )) aa ^^ &Gamma;&Gamma; ^^ sthe s (( &kappa;&kappa; )) 22 ++ bb ^^ &Gamma;&Gamma; ^^ sthe s (( &kappa;&kappa; )) 22 ++ cc ^^ &Gamma;&Gamma; ^^ sthe s (( &kappa;&kappa; )) 22

cc &OverBar;&OverBar; &OverBar;&OverBar; sthe s == 11 mm ^^ sthe s &Sigma;&Sigma; &kappa;&kappa; == 11 mm ^^ sthe s cc ^^ &Gamma;&Gamma; ^^ sthe s (( &kappa;&kappa; )) aa ^^ &Gamma;&Gamma; ^^ sthe s (( &kappa;&kappa; )) 22 ++ bb ^^ &Gamma;&Gamma; ^^ sthe s (( &kappa;&kappa; )) 22 ++ cc ^^ &Gamma;&Gamma; ^^ sthe s (( &kappa;&kappa; )) 22

dd &OverBar;&OverBar; &OverBar;&OverBar; sthe s == -- 11 mm ^^ sthe s (( aa &OverBar;&OverBar; &OverBar;&OverBar; sthe s &CenterDot;&CenterDot; &Sigma;&Sigma; &kappa;&kappa; == 11 mm ^^ sthe s Xx ^^ &Gamma;&Gamma; ^^ sthe s (( &kappa;&kappa; )) ,, 11 ++ bb &OverBar;&OverBar; &OverBar;&OverBar; sthe s &CenterDot;&Center Dot; &Sigma;&Sigma; &kappa;&kappa; == 11 mm ^^ sthe s YY ^^ &Gamma;&Gamma; ^^ sthe s (( &kappa;&kappa; )) ,, 11 ++ cc &OverBar;&OverBar; &OverBar;&OverBar; sthe s &CenterDot;&Center Dot; &Sigma;&Sigma; &kappa;&kappa; == 11 mm ^^ sthe s ZZ ^^ &Gamma;&Gamma; ^^ sthe s (( &kappa;&kappa; )) ,, 11 ))

{ P ^ &Gamma; ^ s ( &kappa; ) | &kappa; = 1,2 , &CenterDot; &CenterDot; &CenterDot; m ^ s } 中每个可靠平面块假设的所有顶点投影到

Figure G2009100441925D00136
上,按照平面块假设中顶点间的顺序顺次连接所有投影点,得到的所有投影直线段成为
Figure G2009100441925D00137
的成员直线特征。Will { P ^ &Gamma; ^ the s ( &kappa; ) | &kappa; = 1,2 , &CenterDot; &CenterDot; &CenterDot; m ^ the s } All vertices assumed by each reliable planar block in
Figure G2009100441925D00136
, connect all projected points sequentially according to the order of vertices in the planar block hypothesis, and all projected straight line segments obtained become
Figure G2009100441925D00137
The member line feature of .

合并后得到的所有开放平面记为 &Omega; &OverBar; &OverBar; = { P &OverBar; &OverBar; s | s = 1,2 , &CenterDot; &CenterDot; &CenterDot; , N &OverBar; &OverBar; } , 其中, N &OverBar; &OverBar; = N ^ Q , 它们对应的三维平面真实边界是未知的,任意开放平面

Figure G2009100441925D001310
的平面方程为 a &OverBar; &OverBar; s &CenterDot; X + b &OverBar; &OverBar; s &CenterDot; Y + c &OverBar; &OverBar; s &CenterDot; Z + d &OverBar; &OverBar; s = 0 , 它的所有成员直线特征的集合记为 M &OverBar; &OverBar; s = { L &OverBar; &OverBar; l s | l = 1,2 , &CenterDot; &CenterDot; &CenterDot; , n &OverBar; &OverBar; s } , 任意一个成员直线特征
Figure G2009100441925D001313
的端点记为
Figure G2009100441925D001314
All open planes obtained after merging are denoted as &Omega; &OverBar; &OverBar; = { P &OverBar; &OverBar; the s | the s = 1,2 , &Center Dot; &CenterDot; &Center Dot; , N &OverBar; &OverBar; } , in, N &OverBar; &OverBar; = N ^ Q , The real boundaries of their corresponding three-dimensional planes are unknown, and any open plane
Figure G2009100441925D001310
The plane equation of is a &OverBar; &OverBar; the s &Center Dot; x + b &OverBar; &OverBar; the s &Center Dot; Y + c &OverBar; &OverBar; the s &CenterDot; Z + d &OverBar; &OverBar; the s = 0 , The collection of all its member line features is denoted as m &OverBar; &OverBar; the s = { L &OverBar; &OverBar; l the s | l = 1,2 , &CenterDot; &Center Dot; &CenterDot; , no &OverBar; &OverBar; the s } , Any member line feature
Figure G2009100441925D001313
The endpoint of is denoted as
Figure G2009100441925D001314
and

第二步,构造平面的边界The second step is to construct the boundary of the plane

第(一)步,利用开放平面的成员直线特征和平面交线构造平面边界假设The first step is to use the member line features of the open plane and the plane intersection to construct the plane boundary hypothesis

第1步,生成平面交线集合Step 1, generate a set of plane intersection lines

任意选取两个开放平面

Figure G2009100441925D001316
P &OverBar; &OverBar; j &Element; &Omega; &OverBar; &OverBar; , 如果Pick two open planes arbitrarily
Figure G2009100441925D001316
P &OverBar; &OverBar; j &Element; &Omega; &OverBar; &OverBar; , if

Figure G2009100441925D001318
Figure G2009100441925D001318

判定它们可相交,计算它们的交线方程,记为 X - X &OverBar; &OverBar; 0 p &OverBar; &OverBar; ij = Y - Y &OverBar; &OverBar; 0 q &OverBar; &OverBar; ij = Z - Z &OverBar; &OverBar; 0 r &OverBar; &OverBar; ij . Determine that they can intersect, calculate their intersection line equation, denoted as x - x &OverBar; &OverBar; 0 p &OverBar; &OverBar; ij = Y - Y &OverBar; &OverBar; 0 q &OverBar; &OverBar; ij = Z - Z &OverBar; &OverBar; 0 r &OverBar; &OverBar; ij .

由所有的平面交线组成的集合记为U。The set consisting of all plane intersections is denoted as U.

第2步,判定平面交线与平面成员直线特征间的共线关系并修改开放平面的成员直线特征集合The second step is to determine the collinear relationship between the plane intersection line and the member line features of the plane and modify the member line feature set of the open plane

对于任意开放平面

Figure G2009100441925D00141
在U中选择它与
Figure G2009100441925D00142
的任意开放平面的交线,如果它的任意成员直线特征 L &OverBar; &OverBar; l i &Element; M &OverBar; &OverBar; i 所在直线与该平面交线的夹角小于φ,并且
Figure G2009100441925D00144
的中点到平面交线的距离小于ε,则判定
Figure G2009100441925D00145
与该平面交线共线,并从开放平面
Figure G2009100441925D00146
的成员直线特征集合中删除
Figure G2009100441925D00148
For any open plane
Figure G2009100441925D00141
Select it in U with
Figure G2009100441925D00142
An intersection line of any open plane of , if any of its member line features L &OverBar; &OverBar; l i &Element; m &OverBar; &OverBar; i The angle between the straight line and the intersection of the plane is less than φ, and
Figure G2009100441925D00144
The distance from the midpoint of the plane to the intersection line of the plane is less than ε, then it is judged
Figure G2009100441925D00145
collinear with the plane intersection and from the open plane
Figure G2009100441925D00146
The member line feature set of delete in
Figure G2009100441925D00148

由所有与

Figure G2009100441925D00149
的某个成员直线特征共线的平面交线组成的集合,记为 by all with
Figure G2009100441925D00149
A set of collinear plane intersection lines of a member line feature, denoted as

第3步,补充缺失的开放平面Step 3, Supplement the missing open plane

对于任意一个开放平面的任意一个成员直线特征 L &OverBar; &OverBar; l i &Element; M &OverBar; &OverBar; i , 应用一种由单条三维直线生成平面假设的方法,由它生成一个新的开放平面,

Figure G2009100441925D001413
成为它的成员直线特征,添加该平面到集合中。For any open plane Any member line feature of L &OverBar; &OverBar; l i &Element; m &OverBar; &OverBar; i , apply a method of generating a plane hypothesis from a single 3D line, from which a new open plane is generated,
Figure G2009100441925D001413
Make it a member line feature, add the plane to the collection middle.

第4步,针对补充的开放平面计算平面交线Step 4, Compute plane intersections for supplementary open planes

对于任意一个开放平面 P &OverBar; &OverBar; u &Element; &Omega; &OverBar; &OverBar; , 如果它是由单条三维直线生成的,则利用与第1步相同的方法,计算它与任意其它开放平面 P &OverBar; &OverBar; v &Element; &Omega; &OverBar; &OverBar; 的交线,并将该平面交线添加到集合U中;如果该平面交线与的任意成员直线特征所在直线的夹角小于φ,并且

Figure G2009100441925D001418
的中点到平面交线的距离小于ε,则将该平面交线添加到直线集合
Figure G2009100441925D001419
中,并从集合
Figure G2009100441925D001420
中删除这个共线的成员直线特征。For any open plane P &OverBar; &OverBar; u &Element; &Omega; &OverBar; &OverBar; , If it is generated by a single 3D line, calculate its relation to any other open plane using the same method as in step 1 P &OverBar; &OverBar; v &Element; &Omega; &OverBar; &OverBar; , and add the plane intersection to the set U; if the plane intersection and The included angle of the straight line of any member of the straight line feature is less than φ, and
Figure G2009100441925D001418
The distance from the midpoint of the plane intersection to the plane intersection line is less than ε, then add the plane intersection line to the straight line set
Figure G2009100441925D001419
in, and from the set of
Figure G2009100441925D001420
Delete this collinear member line feature.

第5步,添加平面交线为开放平面的成员直线特征Step 5, add the plane intersection line as a member line feature of the open plane

对于集合U中的任意一条平面交线,假定它是开放平面

Figure G2009100441925D001422
的交线,则将该平面交线添加到
Figure G2009100441925D001423
的成员直线特征集合
Figure G2009100441925D001425
Figure G2009100441925D001426
中去。For any plane intersection line in the set U, it is assumed to be an open plane and
Figure G2009100441925D001422
, then add the plane intersection to
Figure G2009100441925D001423
and The member line feature set of
Figure G2009100441925D001425
and
Figure G2009100441925D001426
to go.

新增加的成员直线特征的端点此时是未定的。The endpoints of the newly added member line feature are undetermined at this time.

第6步,判定开放平面的成员直线特征集合中任意平面交线与其它成员直线特征是否相交Step 6, determine whether any plane intersection line in the member line feature set of the open plane intersects with other member line features

对于任意开放平面

Figure G2009100441925D00151
的任意一个成员直线特征,如果它属于集合U,并且它与
Figure G2009100441925D00152
的任意其它成员直线特征的夹角属于范围
Figure G2009100441925D00153
则判定这两个成员直线特征相交。For any open plane
Figure G2009100441925D00151
Any member line feature of , if it belongs to the set U, and it is related to
Figure G2009100441925D00152
The included angle of any other member of the line feature belongs to the range
Figure G2009100441925D00153
Then it is determined that the two member line features intersect.

第7步,由开放平面的成员直线特征构造平面边界Step 7, Construct the plane boundary from the member line features of the open plane

方法包括以下步骤:The method includes the following steps:

第1)步,拆分开放平面的所有成员直线特征到两个子集Step 1), split all member line features of the open plane into two subsets

对于任意一个开放平面将它的所有成员直线特征拆分成两个子集,一个子集是 M &OverBar; &OverBar; u &cap; M &OverBar; &OverBar; u &prime; , 记为U1,另一个子集是

Figure G2009100441925D00156
记为U2。For any open plane Split all its member line features into two subsets, one subset is m &OverBar; &OverBar; u &cap; m &OverBar; &OverBar; u &prime; , Denoted as U 1 , another subset is
Figure G2009100441925D00156
Denote it as U 2 .

第2)步,生成无向图集合Step 2) Generate an undirected graph collection

生成U2的所有子集。Generate all subsets of U2 .

选取U2的任意一个子集与U1组成一个新的成员直线特征集,记为U0,生成一个无向图。该无向图以U0中每个成员直线特征作为节点,如果任意两个节点对应的成员直线特征相交,则将它们连接起来。Select any subset of U 2 and U 1 to form a new member line feature set, denoted as U 0 , and generate an undirected graph. The undirected graph takes each member straight line feature in U 0 as a node, and if the member straight line features corresponding to any two nodes intersect, they will be connected.

第3)步,生成平面边界Step 3) Generate plane boundary

在上一步得到的每个无向图中搜索它的所有Hamilton圈,对于每个不重复的圈,按照其中包含的节点间连接顺序,计算相邻节点对应的成员直线特征的交点,把它们作为顶点,生成一个闭合平面假设;如果不存在Hamilton圈,则搜索所有Hamilton路径,对于每个不重复的路径,按照其中包含的节点间连接顺序,计算相邻节点对应的成员直线特征的交点,把它们作为顶点,生成一个半开放平面假设。Search for all Hamilton circles in each undirected graph obtained in the previous step. For each non-repeating circle, according to the order of connections between nodes contained in it, calculate the intersection points of the member straight line features corresponding to adjacent nodes, and use them as vertex, generate a closed plane hypothesis; if there is no Hamilton cycle, then search all Hamilton paths, for each non-repeating path, according to the connection sequence between nodes contained in it, calculate the intersection point of the member straight line features corresponding to adjacent nodes, put They act as vertices, generating a semi-open planar hypothesis.

生成的所有闭合平面假设和半开放平面假设组成的集合记为Ω′={P′k|k=1,2,…,N′},其中任意一个元素P′k用它的有序顶点集合表示,记为 M k &prime; = { T k j | j = 1,2 , &CenterDot; &CenterDot; &CenterDot; , n k &prime; } , 如果 T k 1 = T k n k &prime; , 表示P′k是一个闭合平面假设,否则,表示P′k是一个半开放平面假设。任意一个闭合或半开放平面假设P′k的平面方程与生成它的开放平面是相同的,为了与Ω′对应,该平面方程重写为a′k·X+b′k·Y+c′k·Z+d′k=0。The generated set of all closed plane hypotheses and semi-open plane hypotheses is denoted as Ω′={P′ k |k=1, 2,…, N′}, where any element P′ k uses its ordered vertex set express as m k &prime; = { T k j | j = 1,2 , &Center Dot; &CenterDot; &CenterDot; , no k &prime; } , if T k 1 = T k no k &prime; , Indicates that P′ k is a closed plane assumption, otherwise, indicates that P′ k is a semi-open plane assumption. Any closed or semi-open plane assumes that the plane equation of P′ k is the same as the open plane that generates it. In order to correspond to Ω′, the plane equation is rewritten as a′ k X+b′ k Y+c′ k ·Z+d′ k =0.

第(二)步,利用提取直线特征以及启发式规则补充半开放平面假设的缺失边界In the second step, the missing boundary of the semi-open plan hypothesis is supplemented by extracting straight line features and heuristic rules

对于任意平面假设P′k∈Ω′,如果它是半开放的,则为它补充缺失的边界,首先利用提取直线特征H1和H2,如果能够搜索到可用于补充缺失边界的直线特征,则该步骤结束,生成一个或多个闭合平面假设;否则,利用启发式规则补充缺失的边界,生成一个闭合平面假设。方法如下:For any plane hypothesis P′ k ∈ Ω′, if it is semi-open, then supplement the missing boundary for it, firstly use the straight line features H 1 and H 2 to be extracted, if the straight line features that can be used to supplement the missing boundary can be searched, Then this step ends, and one or more closed plane hypotheses are generated; otherwise, heuristic rules are used to supplement missing boundaries to generate a closed plane hypothesis. Methods as below:

第1步,利用图像中提取的直线特征补充半开放平面假设的缺失边界Step 1, use the straight line features extracted from the image to supplement the missing boundary of the semi-open plane hypothesis

第1)步,计算P′k的缺失边界在左、右图像中投影的所在区域Step 1) Calculate the area where the missing boundary of P′k is projected in the left and right images

投影P′k到左、右图像,分别得到点集 { Tl k j | j = 1,2 , &CenterDot; &CenterDot; &CenterDot; , n k &prime; } { Tr k j | j = 1,2 , &CenterDot; &CenterDot; &CenterDot; , n k &prime; } . Project P′ k to the left and right images to obtain point sets respectively { Tl k j | j = 1,2 , &Center Dot; &Center Dot; &Center Dot; , no k &prime; } and { Tr k j | j = 1,2 , &Center Dot; &CenterDot; &Center Dot; , no k &prime; } .

设定一个表示直线特征完整性的距离门限hd,它表示,在一幅图像中,如果一个直线特征是未提取完整的,那么它的真正端点到它被提取的端点之间的距离小于hdSet a distance threshold h d representing the completeness of the straight line feature, which means that in an image, if a straight line feature is not fully extracted, then the distance between its real end point and its extracted end point is less than h d .

指定P′k的缺失边界在左图像中投影所在的区域为一个四边形,记为Rk L,它的顶点分别记为g1、g2、g3和g4,其中, g 1 = Tl k 1 , g 2 = Tl k n k &prime; , g3是线段

Figure G2009100441925D00167
的延长线上到点g2的距离为hd的点,g4是线段Tlk 2Tlk 1的延长线上到点g1的距离为hd的点,连接它们成为Rk L的四条边g1g2、g2g3、g3g4和g4g1。The region where the missing boundary of P′ k is projected in the left image is a quadrilateral, denoted as R k L , and its vertices are denoted as g 1 , g 2 , g 3 and g 4 , where, g 1 = Tl k 1 , g 2 = Tl k no k &prime; , g 3 is the line segment
Figure G2009100441925D00167
The point on the extension line of Tl k 2 Tl k 1 is the point at the distance h d from the point g 2, and g 4 is the point on the extension line of the line segment Tl k 2 Tl k 1 at the distance h d from the point g 1 , connecting them to become the four lines of R k L Edges g 1 g 2 , g 2 g 3 , g 3 g 4 and g 4 g 1 .

利用类似的方法,计算P′k的缺失边界在右图像中投影所在的区域Rk R,它的四个顶点分别记为g′1、g′2、g′3和g′4Using a similar method, calculate the region R k R where the missing boundary of P′ k is projected in the right image, and its four vertices are denoted as g′ 1 , g′ 2 , g′ 3 and g′ 4 .

第2)步,利用区域Rk L和Rk R内的H1和H2中的直线特征补充半开放平面假设的缺失边界Step 2), using the features of straight lines in H1 and H2 within the regions RkL and RkR to supplement the missing boundaries of the semi - open plane assumption

对于H1中任意一个直线特征,如果它的两个端点都落在Rk L内,并且它的延长线与线段g2g3和g4g1相交,则连接两个交点得到一条直线段,根据P′k的平面方程以及摄影测量理论中的共线方程,计算该直线段的两个端点对应的三维空间点,与P′k的顶点集中除Tk 1

Figure G2009100441925D00171
之外的所有顶点一起,生成一个闭合平面假设。对于H2中任意一个直线特征,如果它的两个端点都落在Rk R内,并且它的延长线与线段g′2g′3和g′4g′1相交,计算它的端点在左图像中的同名点,同名点的计算方法与本发明的第一步中第(二)步的第2步的第3)步的同名点计算方法相同,连接两个同名点为一条直线段,如果这条直线段与左图像中搜索到的任意一条缺失边界都不共线,则计算在右图像中该直线特征的延长线与线段g′2g′3和g′4g′1的交点,连接这两个交点得到一条直线段,根据P′k的平面方程以及摄影测量理论中的共线方程,计算该直线段的两个端点对应的三维空间点,与P′k的顶点集中除Tk 1
Figure G2009100441925D00172
之外的所有顶点一起,生成一个闭合平面假设。For any straight line feature in H 1 , if its two endpoints fall within R k L , and its extension line intersects the line segments g 2 g 3 and g 4 g 1 , then connect the two intersection points to get a straight line segment , according to the plane equation of P′ k and the collinear equation in photogrammetry theory, calculate the three-dimensional space points corresponding to the two endpoints of the straight line segment, and divide T k 1 and
Figure G2009100441925D00171
All vertices other than , together, generate a closed plane hypothesis. For any straight line feature in H 2 , if its two endpoints fall within R k R , and its extension line intersects the line segments g′ 2 g′ 3 and g′ 4 g′ 1 , calculate its endpoint in The same-named point in the left image, the calculation method of the same-named point is identical with the same-named point calculation method of the 3rd) step of the 2nd step of the (2) step in the first step of the present invention, connects two homonymous points to be a straight line segment , if this line segment is not collinear with any of the missing boundaries searched in the left image, then calculate the extension of the line feature in the right image and the line segment g′ 2 g′ 3 and g′ 4 g′ 1 Intersection point, connect these two intersection points to obtain a straight line segment, according to the plane equation of P′ k and the collinear equation in photogrammetry theory, calculate the three-dimensional space point corresponding to the two endpoints of the straight line segment, and concentrate with the vertex of P′ k except T k 1 and
Figure G2009100441925D00172
All vertices other than , together, generate a closed plane hypothesis.

第2步,依据启发式规则补充半开放平面假设的缺失边界Step 2. Supplement the missing boundary of the semi-open plane assumption according to the heuristic rules

如果分别以Tk 2Tk 1

Figure G2009100441925D00173
为共用边界、与Pk′相邻的两个平面假设均已得到完整的边界,依据相邻多边形平面共用边等长的规则,在Tk 2Tk 1或其延长线上指定一点Φ1,使得线段Tk 2Φ1和以Tk 2Tk 1为共用边界的相邻平面假设的共用边界等长,在
Figure G2009100441925D00174
或其延长线上指定一点Φ2,使得线段
Figure G2009100441925D00175
和以
Figure G2009100441925D00176
为共用边界的相邻平面假设的共用边界等长,线段Φ1Φ2成为补充的缺失边界。If T k 2 T k 1 and
Figure G2009100441925D00173
Assuming that the two planes adjacent to P k ′ share the boundary and have obtained a complete boundary, according to the rule that adjacent polygonal planes share sides of equal length, specify a point Φ 1 on T k 2 T k 1 or its extension , so that the line segment T k 2 Φ 1 has the same length as the hypothetical common boundary of the adjacent plane with T k 2 T k 1 as the common boundary, in
Figure G2009100441925D00174
Specify a point Φ 2 on or its extension such that the line segment
Figure G2009100441925D00175
and to
Figure G2009100441925D00176
The shared boundaries assumed to be of the same length for the adjacent planes that share the boundary, the line segment Φ 1 Φ 2 becomes the supplementary missing boundary.

如果分别以Tk 2Tk 1

Figure G2009100441925D00177
为共用边界与P′k相邻的两个平面假设中只有一个平面假设已得到完整的边界,假定这个相邻平面假设为P′l,并且Tk 2Tk 1是P′l与P′k的共用边界,则依据相邻多边形平面共用边界等长的规则,在Tk 2Tk 1或其延长线上指定一点,使得线段Tk 2Φ1与P′l的共用边界等长,在
Figure G2009100441925D00181
或其延长线上指定一点Φ2,使得线段与Tk 2Φ1等长,线段Φ1Φ2成为补充的缺失边界。If T k 2 T k 1 and
Figure G2009100441925D00177
Assume that only one of the two planes adjacent to P′ k for the common boundary assumes that a complete boundary has been obtained. Suppose that this adjacent plane is assumed to be P′ l , and T k 2 T k 1 is P′ l and P′ For the shared boundary of k , according to the rule that the shared boundary of adjacent polygon planes is equal in length, specify a point on T k 2 T k 1 or its extension line, so that the shared boundary of the line segment T k 2 Φ 1 and P′ l is equal in length, exist
Figure G2009100441925D00181
Specify a point Φ 2 on or its extension such that the line segment Isometric to T k 2 Φ 1 , the line segment Φ 1 Φ 2 becomes the supplementary missing boundary.

根据P′k的平面方程以及摄影测量理论中的共线方程,计算Φ1和Φ2对应的三维空间点,与P′k的顶点集中除Tk 1

Figure G2009100441925D00183
之外的所有顶点一起,生成一个闭合平面假设。According to the plane equation of P′ k and the collinear equation in photogrammetry theory, calculate the three-dimensional space points corresponding to Φ 1 and Φ 2 , and divide T k 1 and
Figure G2009100441925D00183
All vertices other than , together, generate a closed plane hypothesis.

这里的共用边界是指相邻两个多边形平面共同拥有的边界线。The shared boundary here refers to the boundary line shared by two adjacent polygonal planes.

将Ω′中的闭合平面假设与由Ω′中的半开放平面假设生成的闭合平面假设合并成为闭合平面假设集合,记为Ω″={P″i|i=1,2,…,N″},其中,任意闭合平面假设P″i由n″i条边界线组成。Combine the closed plane hypothesis in Ω′ and the closed plane hypothesis generated by the semi-open planar hypothesis in Ω′ into a closed plane hypothesis set, which is denoted as Ω″={P″ i |i=1, 2,…, N″ }, where any closed plane assumes that P″ i is composed of n″ i boundary lines.

第(三)步,选取全局最优的闭合平面假设Step (3), select the globally optimal closed plane hypothesis

第1步,计算闭合平面假设的可靠性测度Step 1, calculate the reliability measure of the closed plane hypothesis

对于任意一个闭合平面假设P″i∈Ω″,利用公式一计算P″i的高程一致性测度,如果P″i覆盖的已知高程点数为0,则指定它的高程一致性测度为1/(2·λ),利用公式三,计算它的可靠性测度,记为E″i;在P″i所在平面上,指定它的每条边界线外紧邻的、以该边界线作为一条长边的一个矩形平面区域,长等于边界线的长度,宽等于长的一半,利用公式一计算该矩形平面区域的高程一致性测度,如果它覆盖的已知高程点数为0,则指定它的高程一致性测度为1/(2·λ),利用公式三,计算它的可靠性测度,记由第j条边界线对应的矩形平面区域的可靠性测度为E″i,j,依据下面的公式计算闭合平面假设P″i的边界对应真实平面边界的可靠性测度为For any closed plane hypothesis P″ i ∈ Ω″, use formula 1 to calculate the elevation consistency measure of P″ i , if the number of known elevation points covered by P″ i is 0, specify its elevation consistency measure as 1/ (2·λ), use formula three to calculate its reliability measure, which is denoted as E″ i ; on the plane where P″ i is located, specify the border line adjacent to each of its border lines as a long side A rectangular planar area, whose length is equal to the length of the boundary line, and whose width is equal to half of the length, use formula 1 to calculate the elevation consistency measure of this rectangular planar area, if the number of known elevation points it covers is 0, specify its elevation consistency The reliability measure is 1/(2·λ), use formula 3 to calculate its reliability measure, record the reliability measure of the rectangular plane area corresponding to the jth boundary line as E″ i, j , and calculate according to the following formula The reliability measure of the boundary of the closed plane hypothesis P″ i corresponding to the real plane boundary is

&gamma; i = E i &prime; &prime; &Sigma; j = 1 n i &prime; &prime; E i , j &prime; &prime; (公式四) &gamma; i = E. i &prime; &prime; &Sigma; j = 1 no i &prime; &prime; E. i , j &prime; &prime; (Formula 4)

第2步,求解最优闭合平面假设Step 2, solve the optimal closed plane assumption

生成一个无向图,记为G″,以集合Ω″中每个闭合平面假设为一个节点,利用公式四计算的可靠性测度作为与闭合平面假设相对应的节点的属性,如果任意两个节点对应的闭合平面假设在左图像和右图像中的投影区域均不存在重叠部分,则连接这两个节点,否则不连接。计算图G″的所有极大团,从中选取节点属性值之和最大的极大团包含的所有闭合平面假设,作为最优闭合平面假设,即三维平面提取结果。Generate an undirected graph, denoted as G″, with each closed plane hypothesis in the set Ω″ as a node, and use the reliability measure calculated by formula 4 as the attribute of the node corresponding to the closed plane hypothesis, if any two nodes The corresponding closed plane assumes that there is no overlap between the projected areas in the left image and the right image, then connect the two nodes, otherwise do not connect. Calculate all the maximal cliques of the graph G″, and select all closed plane hypotheses contained in the maximal clique with the largest sum of node attribute values as the optimal closed plane hypothesis, that is, the three-dimensional plane extraction result.

附图说明 Description of drawings

图1是本发明所述的三维平面提取方法流程的示意图;Fig. 1 is a schematic diagram of the process flow of the three-dimensional plane extraction method described in the present invention;

图2是计算平面块假设的高程一致性测度的示意图;Figure 2 is a schematic diagram of calculating the elevation consistency measure assumed by the planar block;

图3是依据启发式规则补充半开放平面假设的缺失边界的示意图。Fig. 3 is a schematic diagram of supplementing the missing boundary of the semi-open plane assumption according to the heuristic rules.

具体实施方式 Detailed ways

下面结合附图对本发明作进一步解释。The present invention will be further explained below in conjunction with the accompanying drawings.

图1是本发明方法所述的三维平面提取方法流程的示意图;技术方案包括:第一步,求解目标表面所在的开放三维平面,本步骤又分为两步,第(一)步,三维平面假设生成,本步骤又分为两个子步骤,第1步,计算三维平面假设,第2步,合并共面的三维平面假设,第(二)步,三维平面假设证实,本步骤又分为四个子步骤,第1步,分割三维平面假设为平面块假设,第2步,计算每个平面块假设的可靠性测度,第3步,搜索最可靠的平面块假设,第4步,合并共面的可靠平面块假设;第二步,构造平面的边界,本步骤又分为三步,第(一)步,利用开放平面的成员直线特征和平面交线构造平面边界假设,本步骤又分为七个子步骤,第1步,生成平面交线集合,第2步,判定平面交线与平面成员直线特征间的共线关系并修改开放平面的成员直线特征集合,第3步,补充缺失的开放平面,第4步,针对补充的开放平面计算平面交线,第5步,添加平面交线为开放平面的成员直线特征,第6步,判定开放平面的成员直线特征集合中任意平面交线与其它成员直线特征是否相交,第7步,由开放平面的成员直线特征构造平面边界,第(二)步,利用提取直线特征以及启发式规则补充半开放平面假设的缺失边界,本步骤又分为两个子步骤,第1步,利用图像中提取的直线特征补充半开放平面假设的缺失边界,第2步,依据启发式规则补充半开放平面假设的缺失边界,第(三)步,选取全局最优的闭合平面假设,本步骤又分为两个子步骤,第1步,计算闭合平面假设的可靠性测度,第2步,求解最优闭合平面假设。Fig. 1 is the schematic diagram of the process flow of the three-dimensional plane extraction method described in the method of the present invention; The technical scheme comprises: the first step, solve the open three-dimensional plane where the target surface is located, this step is divided into two steps again, the first (1) step, the three-dimensional plane Hypothesis generation, this step is divided into two sub-steps, the first step is to calculate the three-dimensional plane hypothesis, the second step is to merge the coplanar three-dimensional plane hypothesis, the second step is to confirm the three-dimensional plane hypothesis, this step is divided into four Step 1, split the 3D planar hypothesis into planar block hypotheses, Step 2, calculate the reliability measure of each planar block hypothesis, Step 3, search for the most reliable planar block hypothesis, Step 4, merge coplanar The second step is to construct the boundary of the plane. This step is divided into three steps. The first step is to use the member straight line features of the open plane and the plane intersection to construct the plane boundary assumption. This step is divided into three steps: Seven sub-steps, the first step is to generate the plane intersection line set, the second step is to determine the collinear relationship between the plane intersection line and the plane member line features and modify the member line feature set of the open plane, the third step is to supplement the missing open Plane, the fourth step is to calculate the plane intersection line for the supplementary open plane, the fifth step is to add the plane intersection line as the member line feature of the open plane, the sixth step is to determine the intersection line of any plane in the member line feature set of the open plane and Whether other member straight line features intersect, the seventh step is to construct the plane boundary from the member straight line features of the open plane, the second step is to use the extracted straight line features and heuristic rules to supplement the missing boundary of the semi-open plane hypothesis, this step is divided into Two sub-steps, the first step is to use the straight line feature extracted from the image to supplement the missing boundary of the semi-open plane hypothesis, the second step is to supplement the missing boundary of the semi-open plane hypothesis according to the heuristic rules, the third step is to select the global most The optimal closed plane hypothesis, this step is divided into two sub-steps, the first step is to calculate the reliability measure of the closed plane hypothesis, and the second step is to solve the optimal closed plane hypothesis.

图2是本发明方法所述的第一步的第(二)步中第2步计算平面块假设的高程一致性测度的示意图:O1和O2是左右相机中心,高程点

Figure G2009100441925D00201
与O1和O2的连线分别交平面块于点F1和F2。该方法分别计算高程点
Figure G2009100441925D00202
与点F1、F2的距离,如果这两个距离越小,表示该平面块假设与高程点
Figure G2009100441925D00203
的一致性越好。Fig. 2 is the schematic diagram of the height consistency measure that the 2nd step calculates the assumption of plane block in the 2nd (2) step of the first step described in the method of the present invention: O 1 and O 2 are the left and right camera centers, the elevation points
Figure G2009100441925D00201
The lines connecting O 1 and O 2 intersect the plane blocks at points F 1 and F 2 respectively. This method calculates elevation points separately
Figure G2009100441925D00202
The distance from points F 1 and F 2 , if the two distances are smaller, it means that the plane block is assumed to be the same as the elevation point
Figure G2009100441925D00203
The better the consistency.

图3是本发明方法所述的第二步的第(二)步中第2步利用启发式规则补充半开放平面假设的缺失边界的示意图:假定半开放平面假设具有两条已知的边界,图3a)中,A和B都是具有完整边界的闭合平面假设,C是一个半开放平面假设,Tk 2Tk 1是A和C的共用边界线,Tk 2Tk 3是B和C的共用边界线,虚线是依据共用边界线等长的启发性规则为平面C补充的缺失边界线Φ1Φ2;图3b)中,与图3a)中不同的地方是,B也是一个半开放平面假设,首先,利用A和C的共用边界线等长的规则,指定C的缺失边界线的一个端点Φ1,再根据B和C的共用边界线与A和C的共用边界线等长的规则,指定缺失边界线的另一个端点Φ2,连接它们,生成半开放平面假设C的缺失边界线。Fig. 3 is the sketch map that the 2nd step utilizes the heuristic rule to supplement the missing boundary of the half-open plane assumption in the (2) step of the second step described in the method of the present invention: assume that the semi-open plane assumption has two known boundaries, In Fig. 3a), A and B are both closed plane assumptions with complete boundaries, C is a semi-open plane assumption, T k 2 T k 1 is the common boundary line of A and C, T k 2 T k 3 is B and The shared boundary line of C, the dotted line is the missing boundary line Φ 1 Φ 2 supplemented for plane C according to the heuristic rule of equal length of the shared boundary line; in Fig. 3b), the difference from Fig. 3a) is that B is also a half Assumption of open plane, firstly, using the rule that the common boundary line of A and C is equal length, designate an end point Φ 1 of the missing boundary line of C, and then according to the common boundary line of B and C and the common boundary line of A and C The rule of , specify the other end point Φ 2 of the missing boundary line, connect them, and generate the missing boundary line of the semi-open plane hypothesis C.

下面详细说明本发明中的有关细节。The relevant details in the present invention will be described in detail below.

第一点,计算三维空间点在三维平面上的投影The first point, calculate the projection of the three-dimensional space point on the three-dimensional plane

本发明中所述的投影三维空间点到三维平面的步骤均采用下面的方法,具体内容为:The step of projecting the three-dimensional space point to the three-dimensional plane described in the present invention all adopts the following method, and the specific content is:

假定三维空间点为(X0,Y0,Z0),三维平面方程为a·X+b·Y+c·Z+d=0,则该三维空间点在这个三维平面上的投影(X0′,Y0′,Z0′),其中Assuming that the point in the three-dimensional space is (X 0 , Y 0 , Z 0 ), and the equation of the three-dimensional plane is a·X+b·Y+c·Z+d=0, then the projection of the point in the three-dimensional space on the three-dimensional plane (X 0 ′, Y 0 ′, Z 0 ′), where

Xx 00 &prime;&prime; == Xx 00 -- aa &CenterDot;&Center Dot; (( aa &CenterDot;&Center Dot; Xx 00 ++ bb &CenterDot;&Center Dot; YY 00 ++ cc &CenterDot;&Center Dot; ZZ 00 ++ dd )) aa 22 ++ bb 22 ++ cc 22 YY 00 &prime;&prime; == YY 00 -- bb &CenterDot;&Center Dot; (( aa &CenterDot;&Center Dot; Xx 00 ++ bb &CenterDot;&Center Dot; YY 00 ++ cc &CenterDot;&Center Dot; ZZ 00 ++ dd )) aa 22 ++ bb 22 ++ cc 22 ZZ 00 &prime;&prime; == ZZ 00 -- cc &CenterDot;&CenterDot; (( aa &CenterDot;&CenterDot; Xx 00 ++ bb &CenterDot;&Center Dot; YY 00 ++ cc &CenterDot;&Center Dot; ZZ 00 ++ dd )) aa 22 ++ bb 22 ++ cc 22

第二点,计算三维空间点在左、右图像上的投影The second point is to calculate the projection of the three-dimensional space point on the left and right images

本发明中所述的投影三维空间点到已知的左、右图像的步骤均采用下面的方法,具体内容为:The steps of projecting three-dimensional space points to known left and right images described in the present invention all adopt the following methods, and the specific contents are:

根据摄影测量理论,利用已知的左、右图像成像参数,可以得到每幅图像相关的共线方程组,依据共线方程可以计算三维空间点在左、右图像中的投影。According to the theory of photogrammetry, using the known imaging parameters of the left and right images, the collinear equations related to each image can be obtained, and the projection of three-dimensional space points in the left and right images can be calculated according to the collinear equations.

由图像成像参数获得共线方程组的方法具体参见Wang Zhizhuo,Principlesof photogrammetry with remote sensing,Press of Wuhan Technical University ofSurveying and Mapping,1990,18-26。The method of obtaining collinear equations from image imaging parameters can be found in Wang Zhizhuo, Principles of photogrammetry with remote sensing, Press of Wuhan Technical University of Surveying and Mapping, 1990, 18-26.

第三点,计算三维直线段在三维平面上的投影The third point is to calculate the projection of the three-dimensional straight line segment on the three-dimensional plane

本发明中所述的投影三维直线段到三维平面的步骤均采用下面的方法,具体内容为:The steps of projecting the three-dimensional straight line segment to the three-dimensional plane described in the present invention all adopt the following method, and the specific content is:

按照投影三维空间点到三维平面的方法,分别投影直线段的两个端点到三维平面上,连接得到的两个投影点为一条直线段,这条直线段就是所求的投影。According to the method of projecting the three-dimensional space point to the three-dimensional plane, respectively project the two endpoints of the straight line segment onto the three-dimensional plane, and connect the two projected points to form a straight line segment, which is the desired projection.

第四点,计算三维直线段在左、右图像上的投影The fourth point, calculate the projection of the 3D straight line segment on the left and right images

本发明中所述的投影三维直线段到已知的左、右图像的步骤均采用下面的方法,具体内容为:The steps of projecting the three-dimensional straight line segment to the known left and right images described in the present invention all adopt the following method, the specific content is:

按照投影三维直线段到左、右图像的方法,分别投影直线段的两个端点到左、右图像上,连接每幅图像中得到的两个投影点为一条直线段,这两条直线段就是所求的投影。According to the method of projecting a three-dimensional straight line segment to the left and right images, respectively project the two endpoints of the straight line segment to the left and right images, and connect the two projected points obtained in each image to form a straight line segment. These two straight line segments are The requested projection.

第五点,梯度下降法计算平面方程The fifth point, the gradient descent method calculates the plane equation

在本发明第一步的第(一)步中第2步中,采用梯度下降法计算由Lu和Lv指定的平面方程的方法,具体内容为:In the 2nd step in the first (1) step of the first step of the present invention, adopt gradient descending method to calculate the method for the plane equation specified by Lu and Lv , concrete content is:

已知用于生成平面的两个三维直线特征Lu∈S和Lv∈S,端点分别为(Xu1,Yu1,Zu1)和(Xu2,Yu2,Zu2)以及(Xv1,Yv1,Zv1)和(Xv2,Yv2,Zv2),计算由它们确定平面方程的方法由以下三个步骤组成:Two 3D line features L u ∈ S and L v ∈ S are known to be used to generate the plane, and the endpoints are (X u1 , Y u1 , Z u1 ) and (X u2 , Y u2 , Z u2 ) and (X v1 , Y v1 , Z v1 ) and (X v2 , Y v2 , Z v2 ), the method to calculate the plane equation determined by them consists of the following three steps:

第1)步,求解平面方程系数的初值Step 1), solve the initial value of the plane equation coefficient

计算点(X0,Y0,Z0)和两个量值Du、Dv,其中Calculation point (X 0 , Y 0 , Z 0 ) and two quantities D u , D v , where

Xx 00 == (( Xx uu 11 ++ Xx uu 22 ++ Xx vv 11 ++ Xx vv 22 )) // 44 YY 00 == (( YY uu 11 ++ YY uu 22 ++ YY vv 11 ++ YY vv 22 )) // 44 ZZ 00 == (( ZZ uu 11 ++ ZZ uu 22 ++ ZZ vv 11 ++ ZZ vv 22 )) // 44

DD. uu == (( Xx uu 11 -- Xx uu 22 )) 22 ++ (( YY uu 11 -- YY uu 22 )) 22 ++ (( ZZ uu 11 -- ZZ uu 22 )) 22

DD. vv == (( Xx vv 11 -- Xx vv 22 )) 22 ++ (( YY vv 11 -- YY vv 22 )) 22 ++ (( ZZ vv 11 -- ZZ vv 22 )) 22

得到Lu和Lv的归一化方向矢量为 v u = X u 1 - X u 2 D u Y u 1 - Y u 2 D u Z u 1 - Z u 2 D u v v = X v 1 - X v 2 D v Y v 1 - Y v 2 D v Z v 1 - Z v 2 D v , 以及它们的叉积vu×vv=(αβγ)。The normalized direction vectors of Lu u and L v are obtained as v u = x u 1 - x u 2 D. u Y u 1 - Y u 2 D. u Z u 1 - Z u 2 D. u and v v = x v 1 - x v 2 D. v Y v 1 - Y v 2 D. v Z v 1 - Z v 2 D. v , And their cross product v u × v v = (αβγ).

计算χ=-(α·X0+β·Y0+λ·Z0),得到一个四维矢量(αβγχ);计算三维矢量(α′β′γ′)=(vu+vv)×(αβγ),进一步计算χ′=-(α′·X0+β′·Y0+γ′·Z0)得到另一个四维矢量(α′β′γ′χ′)。Calculate χ=-(α·X 0 +β·Y 0 +λ·Z 0 ) to get a four-dimensional vector (αβγχ); calculate the three-dimensional vector (α′β′γ′)=(v u +v v )×( αβγ), further calculate χ′=-(α′·X 0 +β′·Y 0 +γ′·Z 0 ) to obtain another four-dimensional vector (α′β′γ′χ′).

依据目标函数According to the objective function

f(a,b,c,d)=(a·Xu1+b·Yu1+c·Zu1+d)2+(a·Xu2+b·Yu2+c·Zu2+d)2+(a·Xv1+b·Yv1+c·Zv1+d)2+(a·Xv2+b·Yv2+c·Zv2+d)2 f(a, b, c, d) = (a X u1 + b Y u1 + c Z u1 + d) 2 + (a X u2 + b Y u2 + c Z u2 + d) 2 +(a X v1 +b Y v1 +c Z v1 +d) 2 +(a X v2 +b Y v2 +c Z v2 +d) 2

如果f(αβγχ)>f(α′β′γ′χ′),平面方程系数的初值为(α′β′γ′χ′),否则为(αβγχ)。If f(αβγχ)>f(α′β′γ′χ′), the initial value of the plane equation coefficient is (α′β′γ′χ′), otherwise (αβγχ).

第2)步,利用梯度下降法计算平面方程系数Step 2) Use the gradient descent method to calculate the coefficients of the plane equation

目标函数为f(a,b,c,d),平面方程系数初值为

Figure G2009100441925D00233
迭代收敛条件为
Figure G2009100441925D00234
利用梯度下降法计算最优的平面方程系数。The objective function is f(a, b, c, d), and the initial coefficient of the plane equation is
Figure G2009100441925D00233
The iteration convergence condition is
Figure G2009100441925D00234
Calculate the optimal plane equation coefficients using gradient descent.

梯度下降法的计算过程参见阳明盛、罗长童著的《最优化原理、方法及求解软件》,北京:科学出版社,2006年,第23-31页。For the calculation process of the gradient descent method, refer to "Optimization Principles, Methods and Solution Software" by Yang Mingsheng and Luo Changtong, Beijing: Science Press, 2006, pp. 23-31.

第3)步,得到平面假设Step 3) Get the plane hypothesis

计算I=f(a(k),b(k),c(k),d(k)),如果I>4ε2,则不生成平面假设;否则得到平面假设方程为a(k)·X+b(k)·Y+c(k)·Z+d(k)=0。Calculate I=f(a (k) , b (k) , c (k) , d (k) ), if I>4ε 2 , no plane hypothesis is generated; otherwise, the plane hypothesis equation is a (k) X +b (k) ·Y+c (k) ·Z+d (k) =0.

第六点,极大团快速求解方法Sixth point, the fast solution method of the maximal clique

本发明中所有的极大团求解均采用一种快速的极大团求解方法,具体参见Tomita E,Tanaka A,Takahashia H.The worst-case time complexity for generating allmaximal cliques and computational experiments.Theoretical Computer Science,2006,363:28-42。All maximal clique solutions in the present invention adopt a fast maximal clique solution method, specifically refer to Tomita E, Tanaka A, Takahashia H. The worst-case time complexity for generating all maximal cliques and computational experiments. Theoretical Computer Science, 2006, 363: 28-42.

第七点,依据

Figure G2009100441925D00241
在左、右图像中三对同名点
Figure G2009100441925D00242
Figure G2009100441925D00244
以及
Figure G2009100441925D00246
Figure G2009100441925D00247
计算集合Kw L中任意一点在右图像中的同名点Seventh point, according to
Figure G2009100441925D00241
Three pairs of points with the same name in the left and right images
Figure G2009100441925D00242
and
Figure G2009100441925D00244
and as well as
Figure G2009100441925D00246
and
Figure G2009100441925D00247
Calculate the homonym of any point in the set K w L in the right image

在本发明第一步的第(二)步中的第2步的第3)步中,利用

Figure G2009100441925D00248
在左、右图像中三对同名点
Figure G2009100441925D00249
Figure G2009100441925D002410
Figure G2009100441925D002411
以及
Figure G2009100441925D002412
Figure G2009100441925D002413
计算左投影多边形内任意点在右图像中的同名点,具体内容为:In the 3) step of the 2nd step in the (2) step of the first step of the present invention, utilize
Figure G2009100441925D00248
Three pairs of points with the same name in the left and right images
Figure G2009100441925D00249
and
Figure G2009100441925D002410
and
Figure G2009100441925D002411
as well as
Figure G2009100441925D002412
and
Figure G2009100441925D002413
Calculate the point with the same name of any point in the left projection polygon in the right image, the specific content is:

生成矢量V和矩阵M分别为The generated vector V and matrix M are respectively

VV == xx ^^ rr ww 11 xx ^^ rr ww 22 xx ^^ rr ww 33 ythe y ^^ rr ww 11 ythe y ^^ rr ww 22 ythe y ^^ rr ww 33 ,, Mm == xx ^^ ll ww 11 ythe y ^^ ll ww 11 11 00 00 00 xx ^^ ll ww 22 ythe y ^^ ll ww 22 11 00 00 00 xx ^^ ll ww 22 ythe y ^^ ll ww 33 11 00 00 00 00 00 00 xx ^^ ll ww 11 ythe y ^^ ll ww 11 11 00 00 00 xx ^^ ll ww 22 ythe y ^^ ll ww 22 11 00 00 00 xx ^^ ll ww 22 ythe y ^^ ll ww 33 11

计算

Figure G2009100441925D002416
平面上同名点之间的变换式系数calculate
Figure G2009100441925D002416
Transformation coefficients between points of the same name on the plane

[a11 a12 a13 a21 a22 a23]T=M-1V[a 11 a 12 a 13 a 21 a 22 a 23 ] T = M -1 V

Figure G2009100441925D002417
在左图像中投影区域内任意点
Figure G2009100441925D002418
在右图像中的同名点为 ( a 11 &CenterDot; x ^ 1 + a 12 &CenterDot; y ^ 1 + a 13 , a 21 &CenterDot; x ^ 1 + a 22 &CenterDot; y ^ 1 + a 23 ) . but
Figure G2009100441925D002417
Any point within the projected area in the left image
Figure G2009100441925D002418
The point of the same name in the right image is ( a 11 &Center Dot; x ^ 1 + a 12 &CenterDot; the y ^ 1 + a 13 , a twenty one &Center Dot; x ^ 1 + a twenty two &CenterDot; the y ^ 1 + a twenty three ) .

第八点,本发明第一步的第(二)步的第2步的第3)步中计算Kw R中任意图像点灰度值的插值算法采用双线性插值算法。The eighth point, the interpolation algorithm for calculating the gray value of any image point in KwR in the step (2 ) of the first step of the present invention, step 2, step 3) adopts a bilinear interpolation algorithm.

第九点,由单条三维直线生成平面假设的方法The ninth point, the method of generating a plane hypothesis from a single three-dimensional straight line

在本发明第二步的第(一)步中的第3步中,利用单条三维直线特征生成三维平面假设的方法参见一种利用多视图像提取三维平面并重建建筑物的方法,即Baillard C,Zisserman A.A plane-sweep strategy for the 3D reconstruction ofbuildings from multiple images.ISPRS Journal of Photogrammetry and RemoteSensing,2000,33(B2):56-62。In the 3rd step in the (1) step of the second step of the present invention, the method for generating a 3D plane assumption by using a single 3D straight line feature refers to a method for extracting a 3D plane and reconstructing a building using a multi-view image, i.e. Baillard C , Zisserman A.A plane-sweep strategy for the 3D reconstruction of buildings from multiple images. ISPRS Journal of Photogrammetry and RemoteSensing, 2000, 33(B2): 56-62.

第十点,在无向图中搜索Hamilton圈和Hamilton路径的方法The tenth point, the method of searching Hamilton circle and Hamilton path in undirected graph

在本发明第二步的第(一)步中的第7步中,在无向图中搜索Hamilton圈和Hamilton路径的方法参考范益政,汪毅,龚世才等译的《图论导引》,北京:人民邮电出版社,Gary Chartrand,Ping Zhang著.2007年,第122-136页。In the 7th step in the (1) step of the second step of the present invention, the method for searching Hamilton circle and Hamilton path in the undirected graph refers to Fan Yizheng, Wang Yi, "Graph Theory Guide" translated by Gong Shicai, etc., Beijing : People's Posts and Telecommunications Press, Gary Chartrand, Ping Zhang. 2007, pp. 122-136.

Claims (11)

1. A three-dimensional plane extraction method is characterized by comprising the following steps of knowing a three-dimensional linear feature set and sparse digital elevation model data in a target scene, three-dimensional images of the target scene acquired from two different visual angles, all internal and external parameters of a three-dimensional camera for acquiring the three-dimensional images and linear extraction results of the three-dimensional images:
firstly, acquiring high-precision positioning information of an open three-dimensional plane where a target surface is located
Generating an open three-dimensional plane hypothesis by using the three-dimensional linear characteristics, dividing the open three-dimensional plane hypothesis by using member linear characteristics of the three-dimensional plane hypothesis to form a plurality of plane block hypotheses, selecting an optimal plane block hypothesis by using the plurality of plane block hypotheses, and finally combining the optimal plane block hypotheses to generate an open three-dimensional plane;
second, constructing the boundaries of the planes
Based on the generated open three-dimensional plane, constructing a plane boundary by using member linear features of the open three-dimensional plane and an intersection line between planes, supplementing a missing plane boundary by using linear features extracted from a stereo image and a heuristic rule, generating a closed plane hypothesis from the plane boundary, and selecting an optimal closed plane hypothesis by using an optimization method.
2. The three-dimensional plane extraction method according to claim 1, characterized in that:
two images of a scene from different perspectives are known, and are respectively called as a left image and a right image, and all internal and external parameters of a left camera and a right camera which take the two images are known, wherein the coordinates of the shooting centers of the left camera and the right camera in a world coordinate system are respectively marked as (X)O1,YO1,ZO1) And (X)O2,YO2,ZO2) (ii) a The linear feature sets extracted from the two images are respectively marked as H1And H2(ii) a Knowing N in the sceneSA three-dimensional line feature, denoted as S ═ Li|i=1,2,…,NSH, any element L in the setiRepresents a three-dimensional straight line segment, two end points of which are respectively (X)i1,Yi1,Zi1) And (X)i2,Yi2,Zi2) (ii) a The sparse DEM data in the scene is known and is recorded as an elevation point set
Figure A2009100441920002C1
The average error of the elements in the set is λ;
firstly, solving an open three-dimensional plane where a target surface is located
The method comprises the following steps that firstly, a set angle threshold phi is generated on the assumption of a three-dimensional plane; setting a distance threshold epsilon;
setting the included angle range of straight lines of two adjacent boundaries of any plane of the target surface as
Figure A2009100441920003C1
The minimum included angle between any two adjacent planes of the target surface is
Step 1, calculating three-dimensional plane hypothesis
Combining the three-dimensional straight line features in the set S pairwise to generate a three-dimensional plane hypothesis set omega ═ Pi1, 2, …, N, for any of which three-dimensional planes P is assumediThe plane equation is ai·X+bi·Y+ci·Z+diTwo three-dimensional straight line features used to generate it are called P0iA member line feature of (a);
step 2, merging coplanar three-dimensional plane assumptions
All coplanar three-dimensional plane assumptions in the set omega are merged, and the obtained three-dimensional plane assumption set is marked as omega { P ═ Pw1, 2, …, N, arbitrary three-dimensional plane hypothesis PwIs recorded as a member straight line feature set <math> <mrow> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>w</mi> </msub> <mo>=</mo> <mo>{</mo> <msubsup> <mover> <mi>L</mi> <mo>&OverBar;</mo> </mover> <mi>j</mi> <mi>w</mi> </msubsup> <mo>|</mo> <mi>j</mi> <mo>=</mo> <mn>1,2</mn> <mo>,</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>,</mo> <msub> <mover> <mi>n</mi> <mo>&OverBar;</mo> </mover> <mi>w</mi> </msub> <mo>}</mo> <mo>,</mo> </mrow> </math> Any one member straight line feature Lj wIs noted as (X)j,1 w,Yj,1 w,Zj,1 w) And (X)j,2 w,Yj,2 w,Zj,2 w);
Second, three-dimensional plane hypothesis validation
Step 1, dividing a three-dimensional plane into plane block hypotheses
Assuming P for any one three-dimensional planeiBelongs to omega, and utilizes the member straight line characteristic MiDividing the image to obtain a plurality of plane block hypotheses;
the set of plane block hypotheses resulting from the splitting of all planes in Ω is denoted as <math> <mrow> <mover> <mi>&Omega;</mi> <mo>^</mo> </mover> <mo>=</mo> <mo>{</mo> <msub> <mover> <mi>P</mi> <mo>^</mo> </mover> <mi>w</mi> </msub> <mo>|</mo> <mi>w</mi> <mo>=</mo> <mn>1,2</mn> <mo>,</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>,</mo> <mover> <mi>N</mi> <mo>^</mo> </mover> <mo>}</mo> <mo>,</mo> </mrow> </math> Wherein any one of the plane blocks is assumed to be
Figure A2009100441920003C5
Is represented by its set of vertices, denoted as { ( X ^ wj , Y ^ wj , Z ^ wj ) | j = 1,2 , . . . , n ^ w } , Is obtained by division in omegaIs expressed as the equation of the three-dimensional plane hypothesis <math> <mrow> <msub> <mover> <mi>a</mi> <mo>^</mo> </mover> <mi>w</mi> </msub> <mo>&CenterDot;</mo> <mi>X</mi> <mo>+</mo> <msub> <mover> <mi>b</mi> <mo>^</mo> </mover> <mi>w</mi> </msub> <mo>&CenterDot;</mo> <mi>Y</mi> <mo>+</mo> <msub> <mover> <mi>c</mi> <mo>^</mo> </mover> <mi>w</mi> </msub> <mo>&CenterDot;</mo> <mi>Z</mi> <mo>+</mo> <msub> <mover> <mi>d</mi> <mo>^</mo> </mover> <mi>w</mi> </msub> <mo>=</mo> <mn>0</mn> <mo>;</mo> </mrow> </math>
Step 2, calculating the reliability measure of each plane block hypothesis
Calculating the hypothesis reliability measure of the plane block according to the known height point set J and the left and right images, calculating the projection of each point in the J in the left and right images to obtain two plane point sets which are respectively marked as
Figure A2009100441920004C1
And
Figure A2009100441920004C2
wherein,
Figure A2009100441920004C3
and
Figure A2009100441920004C4
respectively, is the elevation point in J
Figure A2009100441920004C5
Projection points in the left and right images;
for arbitrary plane block hypothesis <math> <mrow> <msub> <mover> <mi>P</mi> <mo>^</mo> </mover> <mi>w</mi> </msub> <mo>&Element;</mo> <mover> <mrow> <mi>&Omega;</mi> <mo>,</mo> </mrow> <mo>^</mo> </mover> </mrow> </math> Calculating the reliability measure;
step 3, searching a reliable plane block hypothesis;
step 4, combining the reliable plane block hypothesis of the coplanarity;
all open planes obtained after merging are recorded as <math> <mrow> <mover> <mover> <mi>&Omega;</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mo>=</mo> <mo>{</mo> <msub> <mover> <mover> <mi>P</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>s</mi> </msub> <mo>|</mo> <mi>s</mi> <mo>=</mo> <mn>1,2</mn> <mo>,</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>,</mo> <mover> <mover> <mi>N</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mo>}</mo> <mo>,</mo> </mrow> </math> The real boundary of the corresponding three-dimensional plane is unknown, and any open plane is
Figure A2009100441920004C8
Has the plane equation of <math> <mrow> <msub> <mover> <mover> <mi>a</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>s</mi> </msub> <mo>&CenterDot;</mo> <mi>X</mi> <mo>+</mo> <msub> <mover> <mover> <mi>b</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>s</mi> </msub> <mo>&CenterDot;</mo> <mi>Y</mi> <mo>+</mo> <msub> <mover> <mover> <mi>c</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>s</mi> </msub> <mo>&CenterDot;</mo> <mi>Z</mi> <mo>+</mo> <msub> <mover> <mover> <mi>d</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>s</mi> </msub> <mo>=</mo> <mn>0</mn> <mo>,</mo> </mrow> </math> Its set of all member linear features is denoted as <math> <mrow> <msub> <mover> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>s</mi> </msub> <mo>=</mo> <mo>{</mo> <msubsup> <mover> <mover> <mi>L</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>l</mi> <mi>s</mi> </msubsup> <mo>|</mo> <mi>l</mi> <mo>=</mo> <mn>1,2</mn> <mo>,</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>,</mo> <msub> <mover> <mover> <mi>n</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>s</mi> </msub> <mo>}</mo> <mo>,</mo> </mrow> </math> Any one member line feature
Figure A2009100441920004C11
Is marked asAnd
Figure A2009100441920004C13
second, constructing the boundaries of the planes
The first step is to use the member straight line characteristic of the open plane and the plane intersection line to construct the plane boundary hypothesis
Step 1, generating a plane intersection set
Arbitrarily selecting two open planes <math> <mrow> <msub> <mover> <mover> <mi>P</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>i</mi> </msub> <mo>,</mo> <msub> <mover> <mover> <mi>P</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>j</mi> </msub> <mo>&Element;</mo> <mover> <mover> <mi>&Omega;</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mo>,</mo> </mrow> </math> If it is not
Figure A2009100441920004C15
Judging that they can be crossed, calculating their cross line equation and recording it as <math> <mrow> <mfrac> <mrow> <mi>X</mi> <mo>-</mo> <msub> <mover> <mover> <mi>X</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mn>0</mn> </msub> </mrow> <msub> <mover> <mover> <mi>p</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>ij</mi> </msub> </mfrac> <mo>=</mo> <mfrac> <mrow> <mi>Y</mi> <mo>-</mo> <msub> <mover> <mover> <mi>Y</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mn>0</mn> </msub> </mrow> <msub> <mover> <mover> <mi>q</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>ij</mi> </msub> </mfrac> <mo>=</mo> <mfrac> <mrow> <mi>Z</mi> <mo>-</mo> <msub> <mover> <mover> <mi>Z</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mn>0</mn> </msub> </mrow> <msub> <mover> <mover> <mi>r</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>ij</mi> </msub> </mfrac> <mo>;</mo> </mrow> </math>
The set formed by all the intersecting lines of the planes is marked as U;
step 2, judging the collinear relation between the plane intersection line and the plane member linear feature and modifying the member linear feature set of the open plane
For any open plane
Figure A2009100441920004C17
Selecting it from U and
Figure A2009100441920004C18
any open plane of intersection of its if any member is a straight line feature <math> <mrow> <msubsup> <mover> <mover> <mi>L</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>l</mi> <mi>i</mi> </msubsup> <mo>&Element;</mo> <msub> <mover> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>i</mi> </msub> </mrow> </math> The included angle between the straight line and the intersection line of the plane is less than phi, and
Figure A2009100441920004C20
is less than epsilon, it is determined
Figure A2009100441920005C1
Collinear with the plane and from the open plane
Figure A2009100441920005C2
Member straight line feature set of
Figure A2009100441920005C3
Deletion in
Figure A2009100441920005C4
By all of
Figure A2009100441920005C5
Is a set of plane intersections of collinear linear features of a member, and is recorded as
Figure A2009100441920005C6
Step 3, supplementing missing open planes
For any one open plane
Figure A2009100441920005C7
Any one member of the straight line characteristics <math> <mrow> <msubsup> <mover> <mover> <mi>L</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>l</mi> <mi>i</mi> </msubsup> <mo>&Element;</mo> <msub> <mover> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>i</mi> </msub> <mo>,</mo> </mrow> </math> Applying a method of generating a plane hypothesis from a single three-dimensional straight line, generating a new open plane,
Figure A2009100441920005C9
adding the plane to the set as its member line features
Figure A2009100441920005C10
Performing the following steps;
step 4, calculating plane intersection lines aiming at the supplemented open planes
For any one open plane <math> <mrow> <msub> <mover> <mover> <mi>P</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>u</mi> </msub> <mo>&Element;</mo> <mover> <mover> <mi>&Omega;</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mo>,</mo> </mrow> </math> If it is generated from a single three-dimensional straight line, calculate it from any other open planes <math> <mrow> <msub> <mover> <mover> <mi>P</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>v</mi> </msub> <mo>&Element;</mo> <mover> <mover> <mi>&Omega;</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> </mrow> </math> And adding the plane intersection to the set U; if the plane intersects with
Figure A2009100441920005C13
Is smaller than phi, and
Figure A2009100441920005C14
is less than epsilon, the plane intersection is added to the set of lines
Figure A2009100441920005C15
And from the set
Figure A2009100441920005C16
Deleting the collinear member straight line feature;
step 5, adding member straight line characteristics of which the plane intersection line is an open plane
For any plane intersection in the set U, it is assumed to be an open plane
Figure A2009100441920005C17
And
Figure A2009100441920005C18
the intersection of the planes is added to
Figure A2009100441920005C19
And
Figure A2009100441920005C20
member straight line feature set of
Figure A2009100441920005C21
Andremoving;
step 6, judging whether any plane intersection line in the member linear feature set of the open plane intersects with other member linear features or not
For any open plane
Figure A2009100441920005C23
If it belongs to the set U, and it is compared with
Figure A2009100441920005C24
Fall within the range of any other member linear feature
Figure A2009100441920005C25
Judging that the two member straight line features are intersected;
step 7, constructing plane boundary by member straight line characteristic of open plane
The generated set composed of all the closed plane hypotheses and the semi-open plane hypotheses is denoted as Ω '═ P'k1, 2, …, N ', any one of which is P'kExpressed in its ordered set of vertices, denoted <math> <mrow> <msup> <msub> <mi>M</mi> <mi>k</mi> </msub> <mo>&prime;</mo> </msup> <mo>=</mo> <mo>{</mo> <msubsup> <mi>T</mi> <mi>k</mi> <mi>j</mi> </msubsup> <mo>|</mo> <mi>j</mi> <mo>=</mo> <mn>1,2</mn> <mo>,</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>,</mo> <msup> <msub> <mi>n</mi> <mi>k</mi> </msub> <mo>&prime;</mo> </msup> <mo>}</mo> <msup> <msub> <mi>P</mi> <mi>k</mi> </msub> <mo>&prime;</mo> </msup> </mrow> </math> Is a'k·X+b′k·Y+c′k·Z+d′k=0;
The second step, the missing boundary of the semi-open plane hypothesis is supplemented by using the extracted straight line characteristics and the heuristic rule
Let P 'for any plane'kE.omega', if it is semi-open, thenTo supplement it with missing boundaries, first use the extracted straight line feature H1And H2If the straight line feature which can be used for supplementing the missing boundary can be searched, the step is ended, and one or more closed plane hypotheses are generated; otherwise, supplementing the missing boundary by using a heuristic rule to generate a closed plane hypothesis;
step three, selecting a global optimal closed plane hypothesis
Step 1, calculating reliability measure of the closed plane hypothesis;
step 2, solving the optimal closed plane hypothesis
Generating an undirected graph, marked as G ', assuming each closed plane in the set omega' as a node, taking the reliability measure of the node as the attribute of the node corresponding to the closed plane assumption, if the closed plane assumptions corresponding to any two nodes do not have overlapping parts in the projection areas of the left image and the right image, connecting the two nodes, otherwise, not connecting the two nodes; and calculating all the maximum cliques of the graph G' and selecting all closed plane hypotheses contained in the maximum cliques with the maximum sum of the node attribute values as a three-dimensional plane extraction result.
3. The three-dimensional plane extraction method according to claim 2, wherein the three-dimensional straight line features in the set S are combined pairwise to generate a three-dimensional plane hypothesis, and the specific steps are as follows:
optionally two three-dimensional straight line features LuE.g. S and Lv∈S,LuHas an endpoint of (X)u1,Yu1,Zu1) And (X)u2,Yu2,Zu2),LvHas an endpoint of (X)v1,Yv1,Zv1) And (X)v2,Yv2,Zv2) With which the objective function f (a, b, c, d) is defined:
f(a,b,c,d)=(a·Xu1+b·Yu1+c·Zu1+d)2+(a·Xu2+b·Yu2+c·Zu2+d)2
+(a·Xv1+b·Yv1+c·Zv1+d)2+(a·Xv2+b·Yv2+c·Zv2+d)2
calculating (a, b, c, d) when f (a, b, c, d) is minimum by gradient descent method to obtain LuAnd LvThe given three-dimensional plane assumes the equation a · X + b · Y + c · Z + d to be 0.
4. The three-dimensional plane extraction method according to claim 3, wherein the method of merging coplanar three-dimensional plane hypotheses is:
assuming P for any one three-dimensional planeiE omega, calculating the midpoint of the connecting line of the midpoints of the two member straight line features in the plane PiProjection onto, denoted as (X)M i,YM i,ZM i);
Step 1), judging the assumed coplanarity relationship of three-dimensional planes
Assuming P for any two three-dimensional planesiE.g. omega and PjBelongs to omega, if the following inequality group is satisfied, P is determinediAnd PjCoplanarity:
<math> <mfenced open='{' close=''> <mtable> <mtr> <mtd> <mfrac> <mrow> <mo>|</mo> <msub> <mi>a</mi> <mi>i</mi> </msub> <mo>&CenterDot;</mo> <msub> <mi>a</mi> <mi>j</mi> </msub> <mo>+</mo> <msub> <mi>b</mi> <mi>i</mi> </msub> <mo>&CenterDot;</mo> <msub> <mi>b</mi> <mi>j</mi> </msub> <mo>+</mo> <msub> <mi>c</mi> <mi>i</mi> </msub> <mo>&CenterDot;</mo> <msub> <mi>c</mi> <mi>j</mi> </msub> <mo>|</mo> </mrow> <mrow> <msqrt> <msup> <msub> <mi>a</mi> <mi>i</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>b</mi> <mi>i</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>c</mi> <mi>i</mi> </msub> <mn>2</mn> </msup> </msqrt> <mo>&CenterDot;</mo> <msqrt> <msup> <msub> <mi>a</mi> <mi>j</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>b</mi> <mi>j</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>c</mi> <mi>j</mi> </msub> <mn>2</mn> </msup> </msqrt> </mrow> </mfrac> <mo>></mo> <mi>cos</mi> <mi>&phi;</mi> </mtd> </mtr> <mtr> <mtd> <mfrac> <mrow> <mo>|</mo> <msub> <mi>a</mi> <mi>i</mi> </msub> <mo>&CenterDot;</mo> <msubsup> <mi>X</mi> <mi>M</mi> <mi>j</mi> </msubsup> <mo>+</mo> <msub> <mi>b</mi> <mi>i</mi> </msub> <mo>&CenterDot;</mo> <msubsup> <mi>Y</mi> <mi>M</mi> <mi>j</mi> </msubsup> <mo>+</mo> <msub> <mi>c</mi> <mi>i</mi> </msub> <mo>&CenterDot;</mo> <msubsup> <mi>Z</mi> <mi>M</mi> <mi>j</mi> </msubsup> <mo>+</mo> <msub> <mi>d</mi> <mi>i</mi> </msub> <mo>|</mo> </mrow> <msqrt> <msup> <msub> <mi>a</mi> <mi>i</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>b</mi> <mi>i</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>c</mi> <mi>i</mi> </msub> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>&lt;</mo> <mi>&epsiv;</mi> </mtd> </mtr> <mtr> <mtd> <mfrac> <mrow> <mo>|</mo> <msub> <mi>a</mi> <mi>j</mi> </msub> <mo>&CenterDot;</mo> <msubsup> <mi>X</mi> <mi>M</mi> <mi>i</mi> </msubsup> <mo>+</mo> <msub> <mi>b</mi> <mi>j</mi> </msub> <mo>&CenterDot;</mo> <msubsup> <mi>Y</mi> <mi>M</mi> <mi>i</mi> </msubsup> <mo>+</mo> <msub> <mi>c</mi> <mi>j</mi> </msub> <mo>&CenterDot;</mo> <msubsup> <mi>Z</mi> <mi>M</mi> <mi>i</mi> </msubsup> <mo>+</mo> <msub> <mi>d</mi> <mi>j</mi> </msub> <mo>|</mo> </mrow> <msqrt> <msup> <msub> <mi>a</mi> <mi>j</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>b</mi> <mi>j</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>c</mi> <mi>j</mi> </msub> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>&lt;</mo> <mi>&epsiv;</mi> </mtd> </mtr> </mtable> </mfenced> </math>
step 2), searching coplanar three-dimensional plane hypothesis
Generating an undirected graph G, and assuming each three-dimensional plane as PiE omega is used as a node, if any two three-dimensional planes are assumed to be coplanar, the corresponding nodes are connected, otherwise, the nodes are not connected;
calculate all the very big cliques of undirected graph G, noted as { Qw|w=1,2,…,NQAny one of the very big clusters QwIs a subset of Ω, denoted <math> <mrow> <mo>{</mo> <msub> <mi>P</mi> <mrow> <msub> <mi>&Gamma;</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <mo>|</mo> <mi>&kappa;</mi> <mo>=</mo> <mn>1,2</mn> <mo>,</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <msub> <mi>m</mi> <mi>w</mi> </msub> <mo>}</mo> <mo>;</mo> </mrow> </math>
Step 3) merging coplanar three-dimensional plane assumptions
For any one extremely large group QwCombining all three-dimensional plane hypotheses contained in the three-dimensional plane hypothesis to obtain a new three-dimensional plane hypothesis PwIts plane equation is denoted as aw·X+bw·Y+cw·Z+dw0, wherein:
<math> <mrow> <msub> <mover> <mi>a</mi> <mo>&OverBar;</mo> </mover> <mi>w</mi> </msub> <mo>=</mo> <mfrac> <mn>1</mn> <msub> <mi>m</mi> <mi>w</mi> </msub> </mfrac> <munderover> <mi>&Sigma;</mi> <mrow> <mi>&kappa;</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mi>m</mi> <mi>w</mi> </msub> </munderover> <mfrac> <msub> <mi>a</mi> <mrow> <msub> <mi>&Gamma;</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <msqrt> <msup> <msub> <mi>a</mi> <mrow> <msub> <mi>&Gamma;</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>b</mi> <mrow> <msub> <mi>&Gamma;</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>c</mi> <mrow> <msub> <mi>&Gamma;</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> </msqrt> </mfrac> </mrow> </math>
<math> <mrow> <msub> <mover> <mi>b</mi> <mo>&OverBar;</mo> </mover> <mi>w</mi> </msub> <mo>=</mo> <mfrac> <mn>1</mn> <msub> <mi>m</mi> <mi>w</mi> </msub> </mfrac> <munderover> <mi>&Sigma;</mi> <mrow> <mi>&kappa;</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mi>m</mi> <mi>w</mi> </msub> </munderover> <mfrac> <msub> <mi>b</mi> <mrow> <msub> <mi>&Gamma;</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <msqrt> <msup> <msub> <mi>a</mi> <mrow> <msub> <mi>&Gamma;</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>b</mi> <mrow> <msub> <mi>&Gamma;</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>c</mi> <mrow> <msub> <mi>&Gamma;</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> </msqrt> </mfrac> </mrow> </math>
<math> <mrow> <msub> <mover> <mi>c</mi> <mo>&OverBar;</mo> </mover> <mi>w</mi> </msub> <mo>=</mo> <mfrac> <mn>1</mn> <msub> <mi>m</mi> <mi>w</mi> </msub> </mfrac> <munderover> <mi>&Sigma;</mi> <mrow> <mi>&kappa;</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mi>m</mi> <mi>w</mi> </msub> </munderover> <mfrac> <msub> <mi>c</mi> <mrow> <msub> <mi>&Gamma;</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <msqrt> <msup> <msub> <mi>a</mi> <mrow> <msub> <mi>&Gamma;</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>b</mi> <mrow> <msub> <mi>&Gamma;</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>c</mi> <mrow> <msub> <mi>&Gamma;</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> </msqrt> </mfrac> </mrow> </math>
<math> <mrow> <msub> <mover> <mi>d</mi> <mo>&OverBar;</mo> </mover> <mi>w</mi> </msub> <mo>=</mo> <mo>-</mo> <mfrac> <mn>1</mn> <msub> <mi>m</mi> <mi>w</mi> </msub> </mfrac> <mrow> <mo>(</mo> <msub> <mover> <mi>a</mi> <mo>&OverBar;</mo> </mover> <mi>w</mi> </msub> <mo>&CenterDot;</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>&kappa;</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mi>m</mi> <mi>w</mi> </msub> </munderover> <msubsup> <mi>X</mi> <mi>M</mi> <mrow> <msub> <mi>&Gamma;</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msubsup> <mo>+</mo> <msub> <mover> <mi>b</mi> <mo>&OverBar;</mo> </mover> <mi>w</mi> </msub> <mo>&CenterDot;</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>&kappa;</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mi>m</mi> <mi>w</mi> </msub> </munderover> <msubsup> <mi>Y</mi> <mi>M</mi> <mrow> <msub> <mi>&Gamma;</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msubsup> <mo>+</mo> <msub> <mover> <mi>c</mi> <mo>&OverBar;</mo> </mover> <mi>w</mi> </msub> <mo>&CenterDot;</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>&kappa;</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mi>m</mi> <mi>w</mi> </msub> </munderover> <msubsup> <mi>Z</mi> <mi>M</mi> <mrow> <msub> <mi>&Gamma;</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msubsup> <mo>)</mo> </mrow> </mrow> </math>
will be provided with <math> <mrow> <mo>{</mo> <msub> <mi>P</mi> <mrow> <msub> <mi>&Gamma;</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <mo>|</mo> <mi>&kappa;</mi> <mo>=</mo> <mn>1,2</mn> <mo>,</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>,</mo> <msub> <mi>m</mi> <mi>w</mi> </msub> <mo>}</mo> </mrow> </math> All member linear features of each three-dimensional plane hypothesis are projected to PwAll the resulting projected straight line segments become PwIs a member line feature of (1).
5. The three-dimensional plane extraction method according to claim 4, wherein P is assumed for any one three-dimensional planeiBelongs to omega, and utilizes the member straight line characteristic MiThe method comprises the following steps of:
step 1) merging collinear member straight line features
For PiOf any two member straight line features Lj i <math> <mrow> <msubsup> <mover> <mi>L</mi> <mo>&OverBar;</mo> </mover> <mi>k</mi> <mi>i</mi> </msubsup> <mo>&Element;</mo> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>i</mi> </msub> <mo>,</mo> </mrow> </math> Calculate the angle between the straight lines, denoted as μ, and Lj iFrom midpoint to Lk iDistance sum L of straight linesk iFrom midpoint to Lj iThe distances of the straight lines are respectively marked as eta1And η2
If μ < φ and η1< ε and μ2If < epsilon, then L is determinedj iAnd Lk iCollineates, and uses them to generate a new three-dimensional linear feature with two endpoints Lj iAnd Lk iThe two end points which are farthest away, and a new three-dimensional straight line feature is added to the set MiAnd from the set MiDeletion in Lj iAnd Lk i
Step 2) calculating PiOf all member straight line features of
For PiOf any two member straight line features Lα i <math> <mrow> <msubsup> <mover> <mi>L</mi> <mo>&OverBar;</mo> </mover> <mi>&beta;</mi> <mi>i</mi> </msubsup> <mo>&Element;</mo> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>i</mi> </msub> <mo>,</mo> </mrow> </math> If the included angle theta of the straight lines satisfies
Figure A2009100441920009C1
Then calculate Lα iAnd Lβ iOr the intersection of their extensions;
the set of all calculated intersections is denoted as Ri
Step 3), generating an undirected graph for the segmentation of the planar hypotheses
Generating an undirected graph marked as GiIts nodes are generated by two types of points: one is an intersection set RiFor each element in M, IIiAny one member of the linear features, if it is not connected to M on an extension outside one of its endpointsiThe other members of the set, and the end point is not the set RiThe end point becomes the graph GiA node of (2);
undirected graph GiThe connection relationship between any two nodes includes two types: one is if their corresponding points are on the same member straight line feature or its extension and the connecting line segment of these two points is not identical to that of graph GiIf the nodes correspond to the other nodes, the nodes are kept connected; secondly, the connection relationship of the structure, the construction method is as follows:
for any one member line feature Lα iIf one or both of its endpoints is a graphGiThe node (b) is processed in two cases:
first case, Lα iAnd MiIs not intersected by any other member straight line feature
Selecting any member line feature <math> <mrow> <msubsup> <mover> <mi>L</mi> <mo>&OverBar;</mo> </mover> <mi>&beta;</mi> <mi>i</mi> </msubsup> <mo>&Element;</mo> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>i</mi> </msub> <mo>,</mo> </mrow> </math> And β ≠ α, designating it as on the straight line with graph GiTwo points corresponding to the two nodes, the two points are positioned at Lβ iOn both sides of the midpoint and on the connecting line segment between them, not connected to the graph GiThe points corresponding to the middle node are marked as (X)1,Y1,Z1) And (X)2,Y2,Z2) And calculate
<math> <mrow> <msub> <mi>e</mi> <mn>1</mn> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msubsup> <mover> <mi>X</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>&alpha;</mi> <mo>,</mo> <mn>1</mn> </mrow> <mi>i</mi> </msubsup> <mo>-</mo> <msub> <mi>X</mi> <mn>1</mn> </msub> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <mfrac> <mrow> <msub> <mi>Y</mi> <mn>2</mn> </msub> <mo>-</mo> <msub> <mi>Y</mi> <mn>1</mn> </msub> </mrow> <mrow> <msub> <mi>X</mi> <mn>2</mn> </msub> <mo>-</mo> <msub> <mi>X</mi> <mn>1</mn> </msub> </mrow> </mfrac> <mo>+</mo> <msub> <mi>Y</mi> <mn>1</mn> </msub> <mo>-</mo> <msubsup> <mover> <mi>Y</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>&alpha;</mi> <mo>,</mo> <mn>1</mn> </mrow> <mi>i</mi> </msubsup> </mrow> </math>
<math> <mrow> <msub> <mi>e</mi> <mn>2</mn> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msubsup> <mover> <mi>X</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>&alpha;</mi> <mo>,</mo> <mn>2</mn> </mrow> <mi>i</mi> </msubsup> <mo>-</mo> <msub> <mi>X</mi> <mn>1</mn> </msub> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <mfrac> <mrow> <msub> <mi>Y</mi> <mn>2</mn> </msub> <mo>-</mo> <msub> <mi>Y</mi> <mn>1</mn> </msub> </mrow> <mrow> <msub> <mi>X</mi> <mn>2</mn> </msub> <mo>-</mo> <msub> <mi>X</mi> <mn>1</mn> </msub> </mrow> </mfrac> <mo>+</mo> <msub> <mi>Y</mi> <mn>1</mn> </msub> <mo>-</mo> <msubsup> <mover> <mi>Y</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>&alpha;</mi> <mo>,</mo> <mn>2</mn> </mrow> <mi>i</mi> </msubsup> </mrow> </math>
Wherein (X)α,1 i,Yα,1 i,Zα,1 i) And (X)α,2 i,Yα,2 i,Zα,2 i) Is Lα iThe endpoint of (1); for any one <math> <mrow> <msubsup> <mover> <mi>L</mi> <mo>&OverBar;</mo> </mover> <mi>l</mi> <mi>i</mi> </msubsup> <mo>&Element;</mo> <msub> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mi>i</mi> </msub> <mo>,</mo> </mrow> </math> And l ≠ α, l ≠ β, calculating
<math> <mrow> <msub> <mi>f</mi> <mn>1</mn> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msubsup> <mover> <mi>X</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>l</mi> <mo>,</mo> <mn>1</mn> </mrow> <mi>i</mi> </msubsup> <mo>-</mo> <msub> <mi>X</mi> <mn>1</mn> </msub> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <mfrac> <mrow> <msub> <mi>Y</mi> <mn>2</mn> </msub> <mo>-</mo> <msub> <mi>Y</mi> <mn>1</mn> </msub> </mrow> <mrow> <msub> <mi>X</mi> <mn>2</mn> </msub> <mo>-</mo> <msub> <mi>X</mi> <mn>1</mn> </msub> </mrow> </mfrac> <mo>+</mo> <msub> <mi>Y</mi> <mn>1</mn> </msub> <mo>-</mo> <msubsup> <mover> <mi>Y</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>l</mi> <mo>,</mo> <mn>1</mn> </mrow> <mi>i</mi> </msubsup> </mrow> </math>
<math> <mrow> <msub> <mi>f</mi> <mn>2</mn> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msubsup> <mover> <mi>X</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>l</mi> <mo>,</mo> <mn>2</mn> </mrow> <mi>i</mi> </msubsup> <mo>-</mo> <msub> <mi>X</mi> <mn>1</mn> </msub> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <mfrac> <mrow> <msub> <mi>Y</mi> <mn>2</mn> </msub> <mo>-</mo> <msub> <mi>Y</mi> <mn>1</mn> </msub> </mrow> <mrow> <msub> <mi>X</mi> <mn>2</mn> </msub> <mo>-</mo> <msub> <mi>X</mi> <mn>1</mn> </msub> </mrow> </mfrac> <mo>+</mo> <msub> <mi>Y</mi> <mn>1</mn> </msub> <mo>-</mo> <msubsup> <mover> <mi>Y</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>l</mi> <mo>,</mo> <mn>2</mn> </mrow> <mi>i</mi> </msubsup> </mrow> </math>
<math> <mrow> <msub> <mi>f</mi> <mn>3</mn> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msubsup> <mover> <mi>X</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>l</mi> <mo>,</mo> <mn>1</mn> </mrow> <mi>i</mi> </msubsup> <mo>-</mo> <msubsup> <mover> <mi>X</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>&alpha;</mi> <mo>,</mo> <mn>1</mn> </mrow> <mi>i</mi> </msubsup> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <mfrac> <mrow> <msubsup> <mover> <mi>Y</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>&alpha;</mi> <mo>,</mo> <mn>2</mn> </mrow> <mi>i</mi> </msubsup> <mo>-</mo> <msubsup> <mover> <mi>Y</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>&alpha;</mi> <mo>,</mo> <mn>1</mn> </mrow> <mi>i</mi> </msubsup> </mrow> <mrow> <msubsup> <mover> <mi>X</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>&alpha;</mi> <mo>,</mo> <mn>2</mn> </mrow> <mi>i</mi> </msubsup> <mo>-</mo> <msubsup> <mover> <mi>X</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>&alpha;</mi> <mo>,</mo> <mn>1</mn> </mrow> <mi>i</mi> </msubsup> </mrow> </mfrac> <mo>+</mo> <msubsup> <mover> <mi>Y</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>&alpha;</mi> <mo>,</mo> <mn>1</mn> </mrow> <mi>i</mi> </msubsup> <mo>-</mo> <msubsup> <mover> <mi>Y</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>l</mi> <mo>,</mo> <mn>1</mn> </mrow> <mi>i</mi> </msubsup> </mrow> </math>
<math> <mrow> <msub> <mi>f</mi> <mn>4</mn> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msubsup> <mover> <mi>X</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>l</mi> <mo>,</mo> <mn>2</mn> </mrow> <mi>i</mi> </msubsup> <mo>-</mo> <msubsup> <mover> <mi>X</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>&alpha;</mi> <mo>,</mo> <mn>1</mn> </mrow> <mi>i</mi> </msubsup> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <mfrac> <mrow> <msubsup> <mover> <mi>Y</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>&alpha;</mi> <mo>,</mo> <mn>2</mn> </mrow> <mi>i</mi> </msubsup> <mo>-</mo> <msubsup> <mover> <mi>Y</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>&alpha;</mi> <mo>,</mo> <mn>1</mn> </mrow> <mi>i</mi> </msubsup> </mrow> <mrow> <msubsup> <mover> <mi>X</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>&alpha;</mi> <mo>,</mo> <mn>2</mn> </mrow> <mi>i</mi> </msubsup> <mo>-</mo> <msubsup> <mover> <mi>X</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>&alpha;</mi> <mo>,</mo> <mn>1</mn> </mrow> <mi>i</mi> </msubsup> </mrow> </mfrac> <mo>+</mo> <msubsup> <mover> <mi>Y</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>&alpha;</mi> <mo>,</mo> <mn>1</mn> </mrow> <mi>i</mi> </msubsup> <mo>-</mo> <msubsup> <mover> <mi>Y</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>l</mi> <mo>,</mo> <mn>2</mn> </mrow> <mi>i</mi> </msubsup> </mrow> </math>
Wherein (X)l,1 i,Yl,1 i,Zl,1 i) And(Xl,2 i,Yl,2 i,Zl,2 i) Is Ll iThe endpoint of (1); if it is satisfied with
<math> <mfenced open='{' close=''> <mtable> <mtr> <mtd> <msub> <mi>e</mi> <mn>1</mn> </msub> <mo>&CenterDot;</mo> <msub> <mi>e</mi> <mn>2</mn> </msub> <mo>></mo> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <msub> <mi>f</mi> <mn>1</mn> </msub> <mo>&CenterDot;</mo> <msub> <mi>f</mi> <mn>3</mn> </msub> <mo>></mo> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <msub> <mi>f</mi> <mn>2</mn> </msub> <mo>&CenterDot;</mo> <msub> <mi>f</mi> <mn>4</mn> </msub> <mo>></mo> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <msub> <mi>f</mi> <mn>1</mn> </msub> <mo>&CenterDot;</mo> <msub> <mi>f</mi> <mn>2</mn> </msub> <mo>></mo> <mn>0</mn> </mtd> </mtr> </mtable> </mfenced> </math>
Then calculate
<math> <mrow> <msub> <mi>f</mi> <mn>5</mn> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msubsup> <mover> <mi>X</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>l</mi> <mo>,</mo> <mn>1</mn> </mrow> <mi>i</mi> </msubsup> <mo>-</mo> <msub> <mi>X</mi> <mrow> <mi></mi> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <mfrac> <mrow> <msubsup> <mover> <mi>Y</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>l</mi> <mo>,</mo> <mn>2</mn> </mrow> <mi>i</mi> </msubsup> <mo>-</mo> <msub> <mi>Y</mi> <mrow> <mi></mi> <mn>1</mn> </mrow> </msub> </mrow> <mrow> <msubsup> <mover> <mi>X</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>l</mi> <mo>,</mo> <mn>2</mn> </mrow> <mi>i</mi> </msubsup> <mo>-</mo> <msub> <mi>X</mi> <mn>1</mn> </msub> </mrow> </mfrac> <mo>+</mo> <msub> <mi>Y</mi> <mn>1</mn> </msub> <mo>-</mo> <msubsup> <mover> <mi>Y</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>l</mi> <mo>,</mo> <mn>1</mn> </mrow> <mi>i</mi> </msubsup> </mrow> </math>
<math> <mrow> <msub> <mi>f</mi> <mn>6</mn> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msub> <mi>X</mi> <mn>2</mn> </msub> <mo>-</mo> <msub> <mi>X</mi> <mrow> <mi></mi> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <mfrac> <mrow> <msubsup> <mover> <mi>Y</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>l</mi> <mo>,</mo> <mn>2</mn> </mrow> <mi>i</mi> </msubsup> <mo>-</mo> <msub> <mi>Y</mi> <mrow> <mi></mi> <mn>1</mn> </mrow> </msub> </mrow> <mrow> <msubsup> <mover> <mi>X</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>l</mi> <mo>,</mo> <mn>2</mn> </mrow> <mi>i</mi> </msubsup> <mo>-</mo> <msub> <mi>X</mi> <mn>1</mn> </msub> </mrow> </mfrac> <mo>+</mo> <msub> <mi>Y</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>Y</mi> <mn>2</mn> </msub> </mrow> </math>
If f is5·f6> 0, then the point of attachment (X)α,1 i,Yα,1 i,Zα,1 i) And (X)2,Y2,Z2) In the figure GiAnd a corresponding node in (X), and a point (X)α,2 i,Yα,2 i,Xα,2 i) And (X)1,Y1,Z1) In the figure GiThe corresponding node in (1);
second case, Lα iAnd MiAt least one other member line feature of
If (X)α,1 i,Yα,1 i,Zα,1 i) And graph GiCorresponding to one node, and selecting the member straight line characteristic Lα iOr on an extension thereof with (X)α,1 i,Yα,1 i,Zα,1 i) The closest intersection point, denoted as (X)3,Y3,Z3) Assuming that the intersection point is Lα iLinear feature of member Lβ iThe intersection point of (a); selection of Lβ iOr on an extension thereof with the diagram GiThe point corresponding to the middle node if it is at point (X)3,Y3,Z3) Do not have a connection with the graph GiIf the middle node corresponds to a point, it is marked as (X)4,Y4,Z4) If there are two such points, they are respectively denoted as (X)4,Y4,Z4) And (X)5,Y5,Z5);
If (X)4,Y4,Z4) Is Lβ iThe intersection point with other member straight line features is assumed to be Ll iAnd L isl iOr on the extension thereof with the graph GiA point corresponding to the middle node, which satisfies the relation with point (X)α,1 i,Yα,1 i,Zα,1 i) Falls on Lβ iSame side, and with point (X)4,Y4,Z4) Do not have a connection with the graph GiThe point corresponding to the middle node is connected with the point (X)α,1 i,Yα,1 i,Zα,1 i) In the figure GiThe corresponding node in (1); otherwise, the point of attachment (X)α,1 i,Yα,1 i,Zα,1 i) And point (X)4,Y4,Z4) In the figure GiThe corresponding node in (1); if (X)5,Y5,Z5) If present, then according to (X)4,Y4,Z4) Treating by the same method;
if (X)α,2 i,Yα,2 i,Zα,2 i) Is a drawing GiNode (2), method for constructing connection relation and (X)α,1 i,Yα,1 i,Zα,1 i) The same;
step 4), dividing to generate the plane block hypothesis
Search undirected graph GiFor any one of the rings, if it contains a straight feature or extension thereof belonging to the same memberIf the number of the nodes is not more than 2, generating a plane block hypothesis by using the nodes, wherein all the nodes contained in the plane block hypothesis form an ordered point set according to the connection relation and represent the vertex set of the plane block hypothesis.
6. The three-dimensional plane extraction method according to claim 5, wherein an arbitrary plane block is assumed <math> <mrow> <msub> <mover> <mi>P</mi> <mo>^</mo> </mover> <mi>w</mi> </msub> <mo>&Element;</mo> <mover> <mi>&Omega;</mi> <mo>^</mo> </mover> <mo>,</mo> </mrow> </math> Calculating the reliability measure thereof, comprising the steps of:
step 1) searching for
Figure A2009100441920011C2
Known elevation points of coverage
Block hypothesis of projection planeThe vertex of the image is collected to the left and right images, and any vertex is recorded
Figure A2009100441920011C4
The projections in the left and right images are respectively
Figure A2009100441920011C5
And
Figure A2009100441920011C6
connecting the projection of the vertex in the left and right images according to the order of the fixed points in the vertex set to form
Figure A2009100441920011C7
Two polygonal projection areas of (a);
for any elevation point
Figure A2009100441920011C8
If projection in its left image
Figure A2009100441920011C9
Is located at
Figure A2009100441920011C10
Within the projection area in the left image, and the projection in the right image thereof
Figure A2009100441920011C11
Is located at
Figure A2009100441920011C12
Within the projection area in the right image, the elevation point is determined
Figure A2009100441920011C13
Quilt
Figure A2009100441920011C14
Covering;
quilt
Figure A2009100441920011C15
The set of all the elevation points covered is recorded as <math> <mrow> <msub> <mi>&Theta;</mi> <mi>w</mi> </msub> <mo>&Subset;</mo> <mi>J</mi> <mo>,</mo> </mrow> </math> The number of elements in the set being alphaw
Step 2), calculating the elevation consistency measure
If it is not
Figure A2009100441920012C1
If the number of covered elevation points alpha w is more than 0, calculating
Figure A2009100441920012C2
Elevation ofMeasure of consistency of
(formula one)
Wherein,
Figure A2009100441920012C4
Figure A2009100441920012C5
if α iswWhen the value is equal to 0, then E w D = 0 ;
Step 3), calculating the gray level similarity measure
Figure A2009100441920012C7
All image points in the projection region in the left image form a point set, denoted as Kw LThe gray values of the images of all the points form an array, which is marked as { h }wt|t=1,2,…,ow};
According to
Figure A2009100441920012C8
Three pairs of homonymous points in left and right images
Figure A2009100441920012C9
And
Figure A2009100441920012C11
and
Figure A2009100441920012C12
and
Figure A2009100441920012C13
and
Figure A2009100441920012C14
computing a set Kw LAll element points in the image are homonymous points in the right image, and the set of the homonymous points is marked as Kw RThe gray values of the image of all the points are obtained by interpolation algorithm, and the gray values form another array, which is marked as { h }wt′|t=1,2,…,ow}; computing
Figure A2009100441920012C15
The gray level similarity measure between the projected regions in the left and right images is
<math> <mrow> <msubsup> <mi>E</mi> <mi>w</mi> <mi>G</mi> </msubsup> <mo>=</mo> <mfrac> <mrow> <munderover> <mi>&Sigma;</mi> <mrow> <mi>t</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mi>o</mi> <mi>w</mi> </msub> </munderover> <mrow> <mo>(</mo> <msub> <mi>h</mi> <mi>wt</mi> </msub> <mo>-</mo> <msub> <mover> <mi>h</mi> <mo>&OverBar;</mo> </mover> <mi>w</mi> </msub> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <msup> <msub> <mi>h</mi> <mi>wt</mi> </msub> <mo>&prime;</mo> </msup> <mo>-</mo> <msup> <msub> <mover> <mi>h</mi> <mo>&OverBar;</mo> </mover> <mi>w</mi> </msub> <mo>&prime;</mo> </msup> <mo>)</mo> </mrow> </mrow> <msqrt> <mo>[</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>t</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mi>o</mi> <mi>w</mi> </msub> </munderover> <msup> <mrow> <mo>(</mo> <msub> <mi>h</mi> <mi>wt</mi> </msub> <mo>-</mo> <msub> <mover> <mi>h</mi> <mo>&OverBar;</mo> </mover> <mi>w</mi> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>]</mo> <mo>&CenterDot;</mo> <mo>[</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>t</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mi>o</mi> <mi>w</mi> </msub> </munderover> <msup> <mrow> <mo>(</mo> <msup> <msub> <mi>h</mi> <mi>wt</mi> </msub> <mo>&prime;</mo> </msup> <mo>-</mo> <msup> <msub> <mover> <mi>h</mi> <mo>&OverBar;</mo> </mover> <mi>w</mi> </msub> <mo>&prime;</mo> </msup> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>]</mo> </msqrt> </mfrac> </mrow> </math> (formula two)
Wherein, <math> <mrow> <msub> <mover> <mi>h</mi> <mo>&OverBar;</mo> </mover> <mi>w</mi> </msub> <mo>=</mo> <mfrac> <mn>1</mn> <msub> <mi>o</mi> <mi>w</mi> </msub> </mfrac> <munderover> <mi>&Sigma;</mi> <mrow> <mi>t</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mi>o</mi> <mi>w</mi> </msub> </munderover> <msub> <mi>h</mi> <mi>wt</mi> </msub> <mo>,</mo> </mrow> </math> <math> <mrow> <msup> <msub> <mover> <mi>h</mi> <mo>&OverBar;</mo> </mover> <mi>w</mi> </msub> <mo>&prime;</mo> </msup> <mo>=</mo> <mfrac> <mn>1</mn> <msub> <mi>o</mi> <mi>w</mi> </msub> </mfrac> <munderover> <mi>&Sigma;</mi> <mrow> <mi>t</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mi>o</mi> <mi>w</mi> </msub> </munderover> <msup> <msub> <mi>h</mi> <mi>wt</mi> </msub> <mo>&prime;</mo> </msup> <mo>;</mo> </mrow> </math>
step 4) calculating
Figure A2009100441920012C19
Measure of reliability of
Figure A2009100441920012C20
Measure of reliability of
<math> <mrow> <msub> <mi>E</mi> <mi>w</mi> </msub> <mo>=</mo> <mfenced open='{' close=''> <mtable> <mtr> <mtd> <msubsup> <mi>E</mi> <mi>w</mi> <mi>D</mi> </msubsup> <mo>&CenterDot;</mo> <mrow> <mo>(</mo> <msubsup> <mi>E</mi> <mi>w</mi> <mi>G</mi> </msubsup> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>/</mo> <mn>2</mn> </mtd> <mtd> <msubsup> <mi>E</mi> <mi>w</mi> <mi>D</mi> </msubsup> <mo>></mo> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>(</mo> <msubsup> <mi>E</mi> <mi>w</mi> <mi>G</mi> </msubsup> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>/</mo> <mn>2</mn> </mtd> <mtd> <msubsup> <mi>E</mi> <mi>w</mi> <mi>D</mi> </msubsup> <mo>=</mo> <mn>0</mn> </mtd> </mtr> </mtable> </mfenced> </mrow> </math> (formula three).
7. The three-dimensional plane extraction method according to claim 6, wherein an arbitrary plane block is assumed <math> <mrow> <msub> <mover> <mi>P</mi> <mo>^</mo> </mover> <mi>w</mi> </msub> <mo>&Element;</mo> <mover> <mi>&Omega;</mi> <mo>^</mo> </mover> <mo>,</mo> </mrow> </math> The steps for searching the most reliable plane block hypothesis are:
step 1), searching for reliable plane block hypothesis with known elevation point coverage
Establishing an undirected graph, assuming for any one plane block <math> <mrow> <msub> <mover> <mi>P</mi> <mo>^</mo> </mover> <mi>w</mi> </msub> <mo>&Element;</mo> <mover> <mi>&Omega;</mi> <mo>^</mo> </mover> <mo>,</mo> </mrow> </math> If it is not E w D > 0 , It is used to generate a node of the undirected graph having an attribute value of Ew(ii) a For any two nodes, if the projection areas of the plane blocks corresponding to the two nodes do not have overlapping parts in the left image and the right image, connecting the two nodes;
calculating all the huge cliques of the undirected graph, and selecting a set of all plane block hypotheses contained in the huge cliques with the maximum sum of node attribute values from the huge cliques, wherein the set is marked as K1It represents the reliable flat block hypothesis with the covered elevation points as the basis for the measurement;
step 2), searching for reliable plane block hypothesis without known elevation point coverage
Establishing an undirected graph, assuming for any one plane block <math> <mrow> <msub> <mover> <mi>P</mi> <mo>^</mo> </mover> <mi>w</mi> </msub> <mo>&Element;</mo> <mover> <mi>&Omega;</mi> <mo>^</mo> </mover> <mo>,</mo> </mrow> </math> If it is not E w D > 0 And it is reacted with K1In which any plane block assumes that there is no overlap in the projection areas in the left and right imagesIn part, use
Figure A2009100441920013C7
Generating a node of the undirected graph, wherein the attribute value of the node is Ew(ii) a For any two nodes, if the projection areas of the plane blocks corresponding to the two nodes do not have overlapping parts in the left image and the right image, connecting the two nodes;
calculating all the huge cliques of the undirected graph, and selecting a set of all plane block hypotheses contained in the huge cliques with the maximum sum of node attribute values from the huge cliques, wherein the set is marked as K2It represents the point of elevation without coverage as a measure and is related to K1All plane block hypotheses may coexist with a reliable plane block hypothesis;
will K1∪K2All the plane block hypotheses are included as reliable plane block hypotheses.
8. The three-dimensional plane extraction method according to claim 7, wherein the method of merging the coplanar reliable plane block hypotheses is;
step 1), judging the assumed coplanarity relation of the plane blocks
For any two reliable plane block hypotheses
Figure A2009100441920014C1
<math> <mrow> <msub> <mover> <mi>P</mi> <mo>^</mo> </mover> <mi>v</mi> </msub> <mo>&Element;</mo> <msub> <mi>K</mi> <mn>1</mn> </msub> <mo>&cup;</mo> <msub> <mi>K</mi> <mn>2</mn> </msub> <mo>,</mo> </mrow> </math> If the following inequality set is true, it is determined
Figure A2009100441920014C3
And
Figure A2009100441920014C4
coplanarity:
<math> <mfenced open='{' close=''> <mtable> <mtr> <mtd> <mfrac> <mrow> <mo>|</mo> <msub> <mover> <mi>a</mi> <mo>^</mo> </mover> <mi>u</mi> </msub> <mo>&CenterDot;</mo> <msub> <mover> <mi>a</mi> <mo>^</mo> </mover> <mi>v</mi> </msub> <mo>+</mo> <msub> <mover> <mi>b</mi> <mo>^</mo> </mover> <mi>u</mi> </msub> <mo>&CenterDot;</mo> <msub> <mover> <mi>b</mi> <mo>^</mo> </mover> <mi>v</mi> </msub> <mo>+</mo> <msub> <mover> <mi>c</mi> <mo>^</mo> </mover> <mi>u</mi> </msub> <mo>&CenterDot;</mo> <msub> <mover> <mi>c</mi> <mo>^</mo> </mover> <mi>v</mi> </msub> <mo>|</mo> </mrow> <mrow> <msqrt> <msup> <msub> <mover> <mi>a</mi> <mo>^</mo> </mover> <mi>u</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mover> <mi>b</mi> <mo>^</mo> </mover> <mi>u</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mover> <mi>c</mi> <mo>^</mo> </mover> <mi>u</mi> </msub> <mn>2</mn> </msup> </msqrt> <mo>&CenterDot;</mo> <msqrt> <msup> <msub> <mover> <mi>a</mi> <mo>^</mo> </mover> <mi>v</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mover> <mi>b</mi> <mo>^</mo> </mover> <mi>v</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mover> <mi>c</mi> <mo>^</mo> </mover> <mi>v</mi> </msub> <mn>2</mn> </msup> </msqrt> </mrow> </mfrac> <mo>></mo> <mi>cos</mi> <mrow> <mo>(</mo> <mfrac> <mn>3</mn> <mn>2</mn> </mfrac> <mo>&CenterDot;</mo> <mi>&phi;</mi> <mo>)</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mfrac> <mrow> <mo>|</mo> <msub> <mover> <mi>a</mi> <mo>^</mo> </mover> <mi>u</mi> </msub> <mo>&CenterDot;</mo> <msub> <mover> <mi>X</mi> <mo>^</mo> </mover> <mrow> <mi>v</mi> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mover> <mi>b</mi> <mo>^</mo> </mover> <mi>u</mi> </msub> <mo>&CenterDot;</mo> <msub> <mover> <mi>Y</mi> <mo>^</mo> </mover> <mrow> <mi>v</mi> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mover> <mi>c</mi> <mo>^</mo> </mover> <mi>u</mi> </msub> <mo>&CenterDot;</mo> <msub> <mover> <mi>Z</mi> <mo>^</mo> </mover> <mrow> <mi>v</mi> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mover> <mi>d</mi> <mo>^</mo> </mover> <mi>u</mi> </msub> <mo>|</mo> </mrow> <msqrt> <msup> <msub> <mover> <mi>a</mi> <mo>^</mo> </mover> <mi>u</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mover> <mi>b</mi> <mo>^</mo> </mover> <mi>u</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mover> <mi>c</mi> <mo>^</mo> </mover> <mi>u</mi> </msub> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>&lt;</mo> <mfrac> <mn>3</mn> <mn>2</mn> </mfrac> <mo>&CenterDot;</mo> <mi>&epsiv;</mi> </mtd> </mtr> <mtr> <mtd> <mfrac> <mrow> <mo>|</mo> <msub> <mover> <mi>a</mi> <mo>^</mo> </mover> <mi>v</mi> </msub> <mo>&CenterDot;</mo> <msub> <mover> <mi>X</mi> <mo>^</mo> </mover> <mrow> <mi>u</mi> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mover> <mi>b</mi> <mo>^</mo> </mover> <mi>v</mi> </msub> <mo>&CenterDot;</mo> <msub> <mover> <mi>Y</mi> <mo>^</mo> </mover> <mrow> <mi>u</mi> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mover> <mi>c</mi> <mo>^</mo> </mover> <mi>v</mi> </msub> <mo>&CenterDot;</mo> <msub> <mover> <mi>Z</mi> <mo>^</mo> </mover> <mrow> <mi>u</mi> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mover> <mi>d</mi> <mo>^</mo> </mover> <mi>v</mi> </msub> <mo>|</mo> </mrow> <msqrt> <msup> <msub> <mover> <mi>a</mi> <mo>^</mo> </mover> <mi>v</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mover> <mi>b</mi> <mo>^</mo> </mover> <mi>v</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mover> <mi>c</mi> <mo>^</mo> </mover> <mi>v</mi> </msub> <mn>2</mn> </msup> </msqrt> </mfrac> <mo>&lt;</mo> <mfrac> <mn>3</mn> <mn>2</mn> </mfrac> <mo>&CenterDot;</mo> <mi>&epsiv;</mi> </mtd> </mtr> </mtable> </mfenced> </math>
step 2), searching the hypothesis of the coplanar reliable plane block to generate an undirected graph which is recorded as
Figure A2009100441920014C6
By K1∪K2Generating a node by each reliable plane block hypothesis, and if any two reliable plane blocks are assumed to be coplanar, connecting the corresponding nodes;
calculation chart
Figure A2009100441920014C7
All very big groups of (1), noted { Q ^ s | s = 1,2 , . . . , N ^ Q } , Any one of the extremely large groups
Figure A2009100441920014C9
Is shown in the figure
Figure A2009100441920014C10
A subset of <math> <mrow> <mo>{</mo> <msub> <mover> <mi>P</mi> <mo>^</mo> </mover> <mrow> <msub> <mover> <mi>&Gamma;</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <mo>|</mo> <mi>&kappa;</mi> <mo>=</mo> <mn>1,2</mn> <mo>,</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <msub> <mover> <mi>m</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mo>}</mo> <mo>&Subset;</mo> <msub> <mi>K</mi> <mn>1</mn> </msub> <mo>&cup;</mo> <msub> <mi>K</mi> <mn>2</mn> </msub> <mo>;</mo> </mrow> </math>
Step 3) merging the coplanar reliable plane block hypotheses
For any one maximal clique
Figure A2009100441920014C12
All reliable plane block hypotheses contained by the plane block are combined to obtain an open planeThe plane equation is expressed as <math> <mrow> <msub> <mover> <mover> <mi>a</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>s</mi> </msub> <mo>&CenterDot;</mo> <mi>X</mi> <mo>+</mo> <msub> <mover> <mover> <mi>b</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>s</mi> </msub> <mo>&CenterDot;</mo> <mi>Y</mi> <mo>+</mo> <msub> <mover> <mover> <mi>c</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>s</mi> </msub> <mo>&CenterDot;</mo> <mi>Z</mi> <mo>+</mo> <msub> <mover> <mover> <mi>d</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>s</mi> </msub> <mo>=</mo> <mn>0</mn> <mo>,</mo> </mrow> </math> Wherein:
<math> <mrow> <msub> <mover> <mover> <mi>a</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>s</mi> </msub> <mo>=</mo> <mfrac> <mn>1</mn> <msub> <mover> <mi>m</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> </mfrac> <munderover> <mi>&Sigma;</mi> <mrow> <mi>&kappa;</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mover> <mi>m</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> </munderover> <mfrac> <msub> <mover> <mi>a</mi> <mo>^</mo> </mover> <mrow> <msub> <mover> <mi>&Gamma;</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <msqrt> <msup> <msub> <mover> <mi>a</mi> <mo>^</mo> </mover> <mrow> <msub> <mover> <mi>&Gamma;</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mover> <mi>b</mi> <mo>^</mo> </mover> <mrow> <msub> <mover> <mi>&Gamma;</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mover> <mi>c</mi> <mo>^</mo> </mover> <mrow> <msub> <mover> <mi>&Gamma;</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> </msqrt> </mfrac> </mrow> </math>
<math> <mrow> <msub> <mover> <mover> <mi>b</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>s</mi> </msub> <mo>=</mo> <mfrac> <mn>1</mn> <msub> <mover> <mi>m</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> </mfrac> <munderover> <mi>&Sigma;</mi> <mrow> <mi>&kappa;</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mover> <mi>m</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> </munderover> <mfrac> <msub> <mover> <mi>b</mi> <mo>^</mo> </mover> <mrow> <msub> <mover> <mi>&Gamma;</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <msqrt> <msup> <msub> <mover> <mi>a</mi> <mo>^</mo> </mover> <mrow> <msub> <mover> <mi>&Gamma;</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mover> <mi>b</mi> <mo>^</mo> </mover> <mrow> <msub> <mover> <mi>&Gamma;</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mover> <mi>c</mi> <mo>^</mo> </mover> <mrow> <msub> <mover> <mi>&Gamma;</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> </msqrt> </mfrac> </mrow> </math>
<math> <mrow> <msub> <mover> <mover> <mi>c</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>s</mi> </msub> <mo>=</mo> <mfrac> <mn>1</mn> <msub> <mover> <mi>m</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> </mfrac> <munderover> <mi>&Sigma;</mi> <mrow> <mi>&kappa;</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mover> <mi>m</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> </munderover> <mfrac> <msub> <mover> <mi>c</mi> <mo>^</mo> </mover> <mrow> <msub> <mover> <mi>&Gamma;</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <msqrt> <msup> <msub> <mover> <mi>a</mi> <mo>^</mo> </mover> <mrow> <msub> <mover> <mi>&Gamma;</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mover> <mi>b</mi> <mo>^</mo> </mover> <mrow> <msub> <mover> <mi>&Gamma;</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mover> <mi>c</mi> <mo>^</mo> </mover> <mrow> <msub> <mover> <mi>&Gamma;</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> </msqrt> </mfrac> </mrow> </math>
<math> <mrow> <msub> <mover> <mover> <mi>d</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>s</mi> </msub> <mo>=</mo> <mo>-</mo> <mfrac> <mn>1</mn> <msub> <mover> <mi>m</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> </mfrac> <mrow> <mo>(</mo> <msub> <mover> <mover> <mi>a</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>s</mi> </msub> <mo>&CenterDot;</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>&kappa;</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mover> <mi>m</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> </munderover> <msub> <mover> <mi>X</mi> <mo>^</mo> </mover> <mrow> <msub> <mover> <mi>&Gamma;</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> <mo>,</mo> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mover> <mover> <mi>b</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>s</mi> </msub> <mo>&CenterDot;</mo> <munderover> <mi>&Sigma;</mi> <mrow> <mi>&kappa;</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mover> <mi>m</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> </munderover> <msub> <mover> <mi>Y</mi> <mo>^</mo> </mover> <mrow> <msub> <mover> <mi>&Gamma;</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> <mo>,</mo> <mn>1</mn> </mrow> </msub> <msub> <mrow> <mo>+</mo> <mover> <mover> <mi>c</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> </mrow> <mi>s</mi> </msub> <munderover> <mi>&Sigma;</mi> <mrow> <mi>&kappa;</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mover> <mi>m</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> </munderover> <msub> <mover> <mi>Z</mi> <mo>^</mo> </mover> <mrow> <msub> <mover> <mi>&Gamma;</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> <mo>,</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </math>
will be provided with <math> <mrow> <mo>{</mo> <msub> <mover> <mi>P</mi> <mo>^</mo> </mover> <mrow> <msub> <mover> <mi>&Gamma;</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>&kappa;</mi> <mo>)</mo> </mrow> </mrow> </msub> <mo>|</mo> <mi>&kappa;</mi> <mo>=</mo> <mn>1,2</mn> <mo>,</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <msub> <mover> <mi>m</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mo>}</mo> </mrow> </math> All vertices of each reliable plane block hypothesis are projected to
Figure A2009100441920015C2
In the above, all the projection points are connected in order of the vertices in the plane block hypothesis, and all the obtained projection straight-line segments become
Figure A2009100441920015C3
Is a member line feature of (1).
9. The three-dimensional plane extraction method according to claim 8, wherein the method of constructing the plane boundary from the member straight line features of the open plane comprises the steps of:
step 1), splitting all member linear characteristics of the open plane into two subsets
For any one open plane
Figure A2009100441920015C4
Splitting all member line features into two subsets, one subset being <math> <mrow> <msub> <mover> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>u</mi> </msub> <mo>&cap;</mo> <msup> <msub> <mover> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>u</mi> </msub> <mo>&prime;</mo> </msup> <mo>,</mo> </mrow> </math> Is marked as U1Another subset is
Figure A2009100441920015C6
Is marked as U2
Step 2) generating an undirected graph set
Generating U2All subsets of (a);
selecting U2Any subset of (2) and U1Form a new member linear feature set, which is marked as U0Generating an undirected graph; the undirected graph is represented by U0Taking each member straight line feature as a node, and connecting the member straight line features if the member straight line features corresponding to any two nodes are intersected;
step 3) generating plane boundary
Searching all Hamilton rings in each undirected graph obtained in the previous step, calculating intersection points of member linear features corresponding to adjacent nodes for each nonrepeating ring according to the connection sequence among the nodes contained in the undirected graph, and taking the intersection points as vertexes to generate a closed plane hypothesis; and if no Hamilton circle exists, searching all Hamilton paths, calculating the intersection points of member linear features corresponding to adjacent nodes according to the connection sequence among the nodes contained in each nonrepeating path, and taking the intersection points as vertexes to generate a semi-open plane hypothesis.
10. The method of claim 9, wherein the missing boundary of the semi-open plane hypothesis is supplemented by using the extracted straight line features and heuristic rules, as follows:
step 1, supplementing the assumed missing boundary of the semi-open plane by using the linear features extracted from the image
Step 1), calculating P'kIs projected in the region where the missing boundary of (1) is projected in the left and right images'kTo the left and right images, respectively obtaining point sets <math> <mrow> <mo>{</mo> <msubsup> <mi>Tl</mi> <mi>k</mi> <mi>j</mi> </msubsup> <mo>|</mo> <mi>j</mi> <mo>=</mo> <mn>1,2</mn> <mo>,</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>,</mo> <msup> <msub> <mi>n</mi> <mi>k</mi> </msub> <mo>&prime;</mo> </msup> <mo>}</mo> </mrow> </math> And <math> <mrow> <mrow> <mo>{</mo> <msubsup> <mi>Tr</mi> <mi>k</mi> <mi>j</mi> </msubsup> <mo>|</mo> <mi>j</mi> <mo>=</mo> <mn>1,2</mn> <mo>,</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>,</mo> <msup> <msub> <mi>n</mi> <mi>k</mi> </msub> <mo>&prime;</mo> </msup> <mo>}</mo> </mrow> <mo>;</mo> </mrow> </math>
setting a distance threshold h representing the integrity of the linear featured
Designated P'kThe area where the missing boundary projected in the left image is a quadrangle, which is marked as Pk LIts vertexes are respectively marked as g1、g2、g3And g4Wherein g 1 = Tl k 1 , <math> <mrow> <msub> <mi>g</mi> <mn>2</mn> </msub> <mo>=</mo> <msubsup> <mi>Tl</mi> <mi>k</mi> <msup> <msub> <mi>n</mi> <mi>k</mi> </msub> <mo>&prime;</mo> </msup> </msubsup> <mo>,</mo> </mrow> </math> g3is a line segment
Figure A2009100441920016C5
To point g on the extension line of2Is a distance hdPoint of (a), g4Is line segment Tlk 2Tlk 1Is not limited toOn the long line to point g1Is a distance hdConnecting them to Rk LFour sides g of1g2、g2g3、g3g4And g4g1
P 'was calculated using a similar method'kIs projected in the right image in the region Rk RIts four vertexes are respectively marked as g1′、g2′、g3' and g4′;
Step 2) using the region Rk LAnd Rk RInner H1And H2The straight line feature in (1) complements the missing boundary of the semi-open plane hypothesis
For H1Any straight line feature if its two end points fall on Rk LInside, and its extension line and line segment g2g3And g4g1Intersecting, connecting two intersections to obtain a straight line segment according to P'kAnd a collinear equation in the photogrammetry theory, calculating three-dimensional space points corresponding to two end points of the straight line segment, and P'kThe vertex of (1) divides T intensivelyk 1And
Figure A2009100441920016C6
all the other vertexes are taken together to generate a closed plane hypothesis; for H2Any straight line feature if its two end points fall on Rk RInside, and its extension line and line segment g2′g3' and g4′g1Intersecting, calculating the homonymous point of the end point in the left image, connecting two homonymous points to form a straight line segment, and calculating the extension line of the straight line feature and a line segment g in the right image if the straight line segment is not collinear with any missing boundary searched in the left image2′g3' and g4′g1'intersection point connecting the two intersection points to obtain a straight line segment, according to P'kAnd collinearity in photogrammetry theoryCalculating three-dimensional space points corresponding to two endpoints of the straight line segment, and P'kThe vertex of (1) divides T intensivelyk 1And
Figure A2009100441920017C1
all the other vertexes are taken together to generate a closed plane hypothesis;
step 2, supplementing the assumed missing boundary of the semi-open plane according to a heuristic rule
If respectively with Tk 2Tk 1And
Figure A2009100441920017C2
is a common boundary, and P'kAssuming that the two adjacent planes have all obtained complete boundaries, and according to the rule that the shared edges of the adjacent polygonal planes are equal in length, at Tk 2Tk 1Or a point phi designated on the extension thereof1So that the line segment Tk 2Φ1And with Tk 2Tk 1The common boundaries assumed for adjacent planes of the common boundary are of equal length, in
Figure A2009100441920017C3
Or a point phi designated on the extension thereof2Make the line segment
Figure A2009100441920017C4
And with
Figure A2009100441920017C5
The common boundaries assumed for adjacent planes of the common boundary are of equal length, the line segment phi1Φ2A missing boundary that becomes a complement;
if respectively with Tk 2Tk 1And
Figure A2009100441920017C6
is a common boundary and P'kOnly one of the two adjacent plane hypotheses has been assumedComplete boundaries are obtained, assuming this neighboring plane is assumed to be P'lAnd T isk 2Tk 1Is P'lAnd P'kAccording to the rule that the shared boundaries of adjacent polygon planes are equal in length, at Tk 2Tk 1Or a point specified on the extension thereof so that the line segment Tk 2Φ1And P'lThe common boundary of (B) is equal in length in
Figure A2009100441920017C7
Or a point phi designated on the extension thereof2Make the line segment
Figure A2009100441920017C8
And Tk 2Φ1Equal length, line phi1Φ2A missing boundary that becomes a complement;
according to P'kAnd collinearity equation in photogrammetry theory, calculating phi1And phi2Corresponding three-dimensional space point, and P'kThe vertex of (1) divides T intensivelyk 1And
Figure A2009100441920017C9
all the other vertexes are taken together to generate a closed plane hypothesis;
the occlusion plane hypotheses in Ω ' and the occlusion plane hypotheses generated from the semi-open plane hypotheses in Ω ' are merged into a set of occlusion plane hypotheses, which is denoted by Ω ″ = { P 'iI | (1, 2, …, N ″), where any closed plane assumes P ″iFrom n ″)iThe boundary lines of the edges.
11. The three-dimensional plane extraction method according to claim 10, wherein the reliability measure of the closed plane hypothesis is calculated by:
assume P "for any one closed planeiBelongs to omega', and utilizes a formula I to calculate P ″)iA measure of elevation coherence, e.g.Fruit PiIf the number of covered known elevation points is 0, the elevation consistency measure of the known elevation points is designated as 1/(2. lambda.), and the reliability measure is calculated by using a formula III and is marked as E ″i(ii) a At P ″)iOn the plane, a rectangular plane area which is adjacent to the outside of each boundary line of the rectangular plane area and takes the boundary line as a long edge is designated, the length is equal to the length of the boundary line, the width is equal to half of the length, the height consistency measure of the rectangular plane area is calculated by using a formula I, if the number of covered known elevation points is 0, the height consistency measure of the rectangular plane area is designated as 1/(2 & lambda), the reliability measure of the rectangular plane area is calculated by using a formula III, and the reliability measure of the rectangular plane area corresponding to the boundary line of the jth edge is recorded as Ei,j", the closed plane hypothesis P", is calculated according to the following formulaiThe reliability measure of the boundary corresponding to the real plane boundary is
<math> <mrow> <msub> <mi>&gamma;</mi> <mi>i</mi> </msub> <mo>=</mo> <mfrac> <msup> <msub> <mi>E</mi> <mi>i</mi> </msub> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> <mrow> <munderover> <mi>&Sigma;</mi> <mrow> <mi>j</mi> <mo>=</mo> <mn>1</mn> </mrow> <msubsup> <mi>n</mi> <mi>i</mi> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msubsup> </munderover> <msup> <msub> <mi>E</mi> <mrow> <mi>i</mi> <mo>,</mo> <mi>j</mi> </mrow> </msub> <mrow> <mo>&prime;</mo> <mo>&prime;</mo> </mrow> </msup> </mrow> </mfrac> </mrow> </math> (equation four).
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CN103411590A (en) * 2013-09-02 2013-11-27 武汉大学 Method and system for automatically determining installation position of camera in photogrammetric survey
CN107464258A (en) * 2017-07-28 2017-12-12 南京华捷艾米软件科技有限公司 A kind of method that three-dimensional planar is extracted from image
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CN103411590A (en) * 2013-09-02 2013-11-27 武汉大学 Method and system for automatically determining installation position of camera in photogrammetric survey
CN103411590B (en) * 2013-09-02 2015-09-02 武汉大学 Automatically the method and system of photogrammetric middle video camera installation position are determined
CN107464258A (en) * 2017-07-28 2017-12-12 南京华捷艾米软件科技有限公司 A kind of method that three-dimensional planar is extracted from image
CN109859314A (en) * 2019-03-12 2019-06-07 上海曼恒数字技术股份有限公司 Three-dimensional rebuilding method, device, electronic equipment and storage medium
CN112747734A (en) * 2019-10-31 2021-05-04 深圳拓邦股份有限公司 Environment map direction adjusting method, system and device
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