CN101639355A

CN101639355A - Three dimensional plane extraction method

Info

Publication number: CN101639355A
Application number: CN200910044192A
Authority: CN
Inventors: 文贡坚; 王继阳; 回丙伟
Original assignee: National University of Defense Technology
Current assignee: National University of Defense Technology
Priority date: 2009-08-26
Filing date: 2009-08-26
Publication date: 2010-02-03
Anticipated expiration: 2029-08-26
Also published as: CN101639355B

Abstract

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.

Description

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(X_O1，Y_O1，Z_O1)和(X_O2，Y_O2，Z_O2)；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N_S条三维直线特征，记为S＝{L_i|i＝1，2，…，N_S}，该集合中的任意元素L_i表示一条三维直线段，它的两个端点分别为(X_i1，Y_i1，Z_i1)和(X_i2，Y_i2，Z_i2)；已知该场景中的稀疏DEM数据，记为高程点集合

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

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

目标表面的任意两个相邻平面间的最小夹角为

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

The minimum angle between any two adjacent planes on the target surface is

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L_u∈S和L_v∈S，L_u的端点为(X_u1，Y_u1，Z_u1)和(X_u2，Y_u2，Z_u2)，L_v的端点为(X_v1，Y_v1，Z_v1)和(X_v2，Y_v2，Z_v2)，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·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 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L_u和L_v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＝{P_i|i＝1，2，…，N}，对于其中任意一个三维平面假设P_i，平面方程为a_i·X+b_i·Y+c_i·Z+d_i＝0，用于生成它的两个三维直线特征称为P_i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P_i∈Ω，计算它的两个成员直线特征中点连线的中点在平面P_i上的投影，记为(X_M ⁱ，Y_M ⁱ，Z_M 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P_i∈Ω和P_j∈Ω，如果满足下面的不等式组，则判定P_i和P_j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:

$\{\begin{matrix} \frac{| | {a a}_{i i} \cdot &Center Dot; {a a}_{j j} + + {b b}_{i i} \cdot &Center Dot; {b b}_{j j} + + {c c}_{i i} \cdot \cdot {c c}_{j j} | |}{\sqrt{{a a}_{i i}^{22} + + {b b}_{i i}^{22} + + {c c}_{i i}^{22}} \cdot &Center Dot; \sqrt{{a a}_{j j}^{22} + + {b b}_{}^{j j} + + {c c}_{}^{j j}}} > > cos cos φ φ \\ \frac{| | {a a}_{i i} \cdot &Center Dot; {X x}_{M m}^{j j} + + {b b}_{i i} \cdot &Center Dot; {Y Y}_{M m}^{j j} + + {c c}_{i i} \cdot &Center Dot; {Z Z}_{M m}^{j j} + + {d d}_{i i} | |}{\sqrt{{a a}_{i i}^{22} + + {b b}_{i i}^{22} + + {c c}_{i i}^{22}}} < < ϵ ϵ \\ \frac{| | {a a}_{j j} \cdot &Center Dot; {X x}_{M m}^{i i} + + {b b}_{j j} \cdot &Center Dot; {Y Y}_{M m}^{i i} + + {c c}_{j j} \cdot &Center Dot; {Z Z}_{M m}^{i i} + + {d d}_{j j} | |}{\sqrt{{a a}_{j j}^{22} + + {b b}_{}^{j j} + + {c c}_{}^{j j}}} < < ϵ ϵ \end{matrix}$

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P_i∈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{Q_W|w＝1，2，…，N_Q}，任意一个极大团Q_w是Ω的一个子集，记为 ${P_{Γ_{w} (κ)} | κ = 1,2, \cdot \cdot \cdot m_{w}},$ 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 ${P_{Γ_{w} (κ)} | κ = 1,2, \cdot \cdot \cdot m_{w}},$ 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Q_w，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P_w，它的平面方程记为a_w·X+b_w·Y+c_w·Z+d_w＝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:

${\overset{&OverBar; &OverBar;}{a a}}_{w w} = = \frac{11}{{m m}_{w w}} {Σ Σ}_{κ κ = = 11}^{{m m}_{w w}} \frac{{a a}_{{Γ Γ}_{w w} ((κ κ))}}{\sqrt{{a a}_{{Γ Γ}_{w w} ((κ κ))}^{22} + + {b b}_{{Γ Γ}_{w w} ((κ κ))}^{22} + + {c c}_{{Γ Γ}_{w w} ((κ κ))}^{22}}}$

${\overset{&OverBar; &OverBar;}{b b}}_{w w} = = \frac{11}{{m m}_{w w}} {Σ Σ}_{κ κ = = 11}^{{m m}_{w w}} \frac{{b b}_{{Γ Γ}_{w w} ((κ κ))}}{\sqrt{{a a}_{{Γ Γ}_{w w} ((κ κ))}^{22} + + {b b}_{{Γ Γ}_{w w} ((κ κ))}^{22} + + {c c}_{{Γ Γ}_{w w} ((κ κ))}^{22}}}$

${\overset{&OverBar; &OverBar;}{c c}}_{w w} = = \frac{11}{{m m}_{w w}} {Σ Σ}_{κ κ = = 11}^{{m m}_{w w}} \frac{{c c}_{{Γ Γ}_{w w} ((κ κ))}}{\sqrt{{a a}_{{Γ Γ}_{w w} ((κ κ))}^{22} + + {b b}_{{Γ Γ}_{w w} ((κ κ))}^{22} + + {c c}_{{Γ Γ}_{w w} ((κ κ))}^{22}}}$

${\overset{&OverBar; &OverBar;}{d d}}_{w w} = = - - \frac{11}{{m m}_{w w}} (({\overset{&OverBar; &OverBar;}{a a}}_{w w} \cdot &Center Dot; {Σ Σ}_{κ κ = = 11}^{{m m}_{w w}} {X x}_{M m}^{{Γ Γ}_{w w} ((κ κ))} + + {\overset{&OverBar; &OverBar;}{b b}}_{w w} \cdot &Center Dot; {Σ Σ}_{κ κ = = 11}^{{m m}_{w w}} {Y Y}_{M m}^{{Γ Γ}_{w w} ((κ κ))} + + {\overset{&OverBar; &OverBar;}{c c}}_{w w} \cdot &Center Dot; {Σ Σ}_{κ κ = = 11}^{{m m}_{w w}} {Z Z}_{M m}^{{Γ Γ}_{w w} ((κ κ))}))$

将 ${P_{Γ_{w} (κ)} | κ = 1,2, \cdot \cdot \cdot m_{w}}$ 中每个三维平面假设的所有成员直线特征投影到P_w上，得到的所有投影直线段成为P_w的成员直线特征。Will ${P_{Γ_{w} (κ)} | κ = 1,2, \cdot \cdot \cdot m_{w}}$ 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＝{P_w|w＝1，2，…，N}，其中，N＝N_Q，任意三维平面假设P_w的成员直线特征集合记为 ${\overset{&OverBar;}{M}}_{w} = {{\overset{&OverBar;}{L}}_{j}^{w} | j = 1,2, \cdot \cdot \cdot, {\overset{&OverBar;}{n}}_{w}},$ 任意一个成员直线特征L_j ^w的端点记为(X_j，1 ^w，Y_j，1 ^w，Z_j，1 ^w)和(Z_j，2 ^w，Y_j，2 ^w，Z_j，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 ${\overset{&OverBar;}{m}}_{w} = {{\overset{&OverBar;}{L}}_{j}^{w} | j = 1,2, \cdot \cdot &Center Dot;, {\overset{&OverBar;}{no}}_{w}},$ 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P_i∈Ω，利用其成员直线特征M_i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P_i的任意两个成员直线特征L_j ⁱ， ${\overset{&OverBar;}{L}}_{k}^{i} &Element; {\overset{&OverBar;}{M}}_{i},$ 计算它们所在直线的夹角，记为μ，以及L_j ⁱ的中点到L_k ⁱ所在直线的距离和L_k ⁱ的中点到L_j ⁱ所在直线的距离，分别记为η₁和η₂；For any two member line features L _j ⁱ of P _i , ${\overset{&OverBar;}{L}}_{k}^{i} &Element; {\overset{&OverBar;}{m}}_{i},$ 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L_j ⁱ和L_k 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L_j ⁱ和L_k 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M_i中，并从集合M_i中删除L_j ⁱ和L_k ⁱ。If 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P_i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P_i的任意两个成员直线特征L_α ⁱ， ${\overset{&OverBar;}{L}}_{β}^{i} &Element; {\overset{&OverBar;}{M}}_{i},$ 如果它们所在直线的夹角θ满足则计算L_α ⁱ和L_β 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 , ${\overset{&OverBar;}{L}}_{β}^{i} &Element; {\overset{&OverBar;}{m}}_{i},$ 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R_i。The 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G_i，它的节点由两类点生成：一是交点集合R_i中的每个元素，二是对于M_i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M_i中其它成员直线特征的交点，并且该端点不是集合R_i中的元素，则该端点成为图G_i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G_i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G_i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L_α ⁱ，如果它的一个或两个端点是图G_i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，L_α ⁱ与M_i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 ${\overset{&OverBar;}{L}}_{β}^{i} &Element; {\overset{&OverBar;}{M}}_{i},$ 且β≠α，指定它所在直线上与图G_i中两个节点相对应的两个点，这两个点位于L_β ⁱ的中点两侧并且它们之间的连接线段上没有与图G_i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 ${\overset{&OverBar;}{L}}_{β}^{i} &Element; {\overset{&OverBar;}{m}}_{i},$ 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

${e e}_{11} = = (({\overset{&OverBar; &OverBar;}{X x}}_{α α,, 11}^{i i} - - {X x}_{11})) \cdot &Center Dot; \frac{{Y Y}_{22} - - {Y Y}_{11}}{{X x}_{22} - - {X x}_{11}} + + {Y Y}_{11} - - {\overset{&OverBar; &OverBar;}{Y Y}}_{α α,, 11}^{i i}$

${e e}_{22} = = (({\overset{&OverBar; &OverBar;}{X x}}_{α α,, 22}^{i i} - - {X x}_{11})) \cdot \cdot \frac{{Y Y}_{22} - - {Y Y}_{11}}{{X x}_{22} - - {X x}_{11}} + + {Y Y}_{11} - - {\overset{&OverBar; &OverBar;}{Y Y}}_{α α,, 22}^{i i}$

其中，(X_α，1 ⁱ，Y_α，1 ⁱ，Z_α，1 ⁱ)和(X_α，2 ⁱ，Y_α，2 ⁱ，Z_α，2 ⁱ)是L_α ⁱ的端点。对于任意 ${\overset{&OverBar;}{L}}_{l}^{i} &Element; {\overset{&OverBar;}{M}}_{i},$ 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 ${\overset{&OverBar;}{L}}_{l}^{i} &Element; {\overset{&OverBar;}{m}}_{i},$ And l≠α, l≠β, calculate

${f f}_{11} = = (({\overset{&OverBar; &OverBar;}{X x}}_{l l,, 11}^{i i} - - {X x}_{11})) \cdot &Center Dot; \frac{{Y Y}_{22} - - {Y Y}_{11}}{{X x}_{22} - - {X x}_{11}} + + {Y Y}_{11} - - {\overset{&OverBar; &OverBar;}{Y Y}}_{l l,, 11}^{i i}$

${f f}_{22} = = (({\overset{&OverBar; &OverBar;}{X x}}_{l l,, 22}^{i i} - - {X x}_{11})) \cdot &Center Dot; \frac{{Y Y}_{22} - - {Y Y}_{11}}{{X x}_{22} - - {X x}_{11}} + + {Y Y}_{11} - - {\overset{&OverBar; &OverBar;}{Y Y}}_{l l,, 22}^{i i}$

${f f}_{33} = = (({\overset{&OverBar; &OverBar;}{X x}}_{l l,, 11}^{i i} - - {\overset{&OverBar; &OverBar;}{X x}}_{α α,, 11}^{i i})) \cdot &Center Dot; \frac{{\overset{&OverBar; &OverBar;}{Y Y}}_{α α,, 22}^{i i} - - {\overset{&OverBar; &OverBar;}{Y Y}}_{α α,, 11}^{i i}}{{\overset{&OverBar; &OverBar;}{X x}}_{α α,, 22}^{i i} - - {\overset{&OverBar; &OverBar;}{X x}}_{α α,, 11}^{i i}} + + {\overset{&OverBar; &OverBar;}{Y Y}}_{α α,, 11}^{i i} - - {\overset{&OverBar; &OverBar;}{Y Y}}_{l l,, 11}^{i i}$

${f f}_{44} = = (({\overset{&OverBar; &OverBar;}{X x}}_{l l,, 22}^{i i} - - {\overset{&OverBar; &OverBar;}{X x}}_{α α,, 11}^{i i})) \cdot \cdot \frac{{\overset{&OverBar; &OverBar;}{Y Y}}_{α α,, 22}^{i i} - - {\overset{&OverBar; &OverBar;}{Y Y}}_{α α,, 11}^{i i}}{{\overset{&OverBar; &OverBar;}{X x}}_{α α,, 22}^{i i} - - {\overset{&OverBar; &OverBar;}{X x}}_{α α,, 11}^{i i}} + + {\overset{&OverBar; &OverBar;}{Y Y}}_{α α,, 11}^{i i} - - {\overset{&OverBar; &OverBar;}{Y Y}}_{l l,, 22}^{i i}$

其中，(X_l，1 ⁱ，Y_l，1 ⁱ，Z_l，1 ⁱ)和(X_l，2 ⁱ，Y_l，2 ⁱ，Z_l，2 ⁱ)是L_l 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

$\{\begin{matrix} {e e}_{11} \cdot \cdot {e e}_{22} > > 00 \\ {f f}_{11} \cdot \cdot {f f}_{33} > > 00 \\ {f f}_{22} \cdot \cdot {f f}_{44} > > 00 \\ {f f}_{11} \cdot &Center Dot; {f f}_{22} > > 00 \end{matrix}$

则计算Then calculate

${f f}_{55} = = (({\overset{&OverBar; &OverBar;}{X x}}_{l l,, 11}^{i i} - - {X x}_{11})) \cdot &Center Dot; \frac{{\overset{&OverBar; &OverBar;}{Y Y}}_{l l,, 22}^{i i} - - {Y Y}_{11}}{{\overset{&OverBar; &OverBar;}{X x}}_{l l,, 22}^{i i} - - {X x}_{11}} + + {Y Y}_{11} - - {\overset{&OverBar; &OverBar;}{Y Y}}_{l l,, 11}^{i i}$

${f f}_{66} = = (({X x}_{22} - - {X x}_{11})) \cdot &Center Dot; \frac{{\overset{&OverBar; &OverBar;}{Y Y}}_{l l,, 22}^{i i} - - {Y Y}_{11}}{{\overset{&OverBar; &OverBar;}{X x}}_{l l,, 22}^{i i} - - {X x}_{11}} + + {Y Y}_{11} - - {Y Y}_{22}$

如果f₅·f₆＞0，则连接点(X_α，i ¹，Y_α，1 ⁱ，Z_α，1 ⁱ)和(X₂，Y₂，Z₂)在图G_i中对应的节点，以及点(X_α，2 ⁱ，Y_α，2 ⁱ，Z_α，2 ⁱ)和(X₁，Y₁，Z₁)在图G_i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，L_α ⁱ与M_i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(X_α，1 ⁱ，Y_α，1 ⁱ，Z_α，1 ⁱ)与图G_i的一个节点相对应，选择成员直线特征L_α ⁱ或其延长线上与(X_α，1 ⁱ，Y_α，1 ⁱ，Z_α，1 ⁱ)距离最近的交点，记为(X₃，Y₃，Z₃)，假定该交点是L_α ⁱ与成员直线特征L_β ⁱ的交点。选择L_β ⁱ或其延长线上的与图G_i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G_i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L_β ⁱ与其它成员直线特征的交点，假定该成员直线特征为L_l ⁱ，并且L_l ⁱ或其延长线上存在与图G_i中节点相对应的点，它满足与点(X_α，1 ⁱ，Y_α，1 ⁱ，Z_α，1 ⁱ)落在L_β ⁱ同一侧，并且与点(X₄，Y₄，Z₄)之间的连线上没有与图G_i中节点相对应的点，则连接它与点(X_α，1 ⁱ，Y_α，1 ⁱ，Z_α，1 ⁱ)在图G_i中对应的节点；否则，连接点(X_α，1 ⁱ，Y_α，1 ⁱ，Z_α，1 ⁱ)和点(X₄，Y₄，Z₄)在图G_i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(X_α，2 ⁱ，Y_α，2 ⁱ，Z_α，2 ⁱ)是图G_i的节点，则构造连接关系的方法与(X_α，1 ⁱ，Y_α，1 ⁱ，Z_α，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G_i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 $\hat{Ω} = {{\hat{P}}_{w} | w = 1,2, \cdot \cdot \cdot, \hat{N}},$ 其中，任意一个平面块假设

由它的顶点集来表示，记为

{({\hat{X}}_{wj}, {\hat{Y}}_{wj}, {\hat{Z}}_{wj}) | j = 1,2, \cdot \cdot \cdot, {\hat{n}}_{w}},

Ω中分割得到的三维平面假设的方程记为

{\hat{a}}_{w} \cdot X + {\hat{b}}_{w} \cdot Y + {\hat{c}}_{w} \cdot Z + {\hat{d}}_{w} = 0 .

The set of planar block hypotheses obtained from all planar partitions in Ω is denoted as

\hat{Ω} = {{\hat{P}}_{w} | w = 1,2, \cdot \cdot \cdot, \hat{N}},

Among them, any plane block assumes

Represented by its vertex set, denoted as

{({\hat{x}}_{wj}, {\hat{Y}}_{wj}, {\hat{Z}}_{wj}) | j = 1,2, \cdot &Center Dot; &Center Dot;, {\hat{no}}_{w}},

divided into Ω to get The equation for the three-dimensional plane assumption of is written as

{\hat{a}}_{w} &Center Dot; x + {\hat{b}}_{w} &Center Dot; Y + {\hat{c}}_{w} \cdot Z + {\hat{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

和

其中，

与

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

and

in,

and

are the elevation points in J Projection points in the left and right images.

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

第1)步，搜索被

覆盖的已知高程点In step 1), the search is

Covered known elevation points

投影平面块假设

的顶点集到左、右图像，记任意顶点

在左、右图像中的投影分别为

和

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

的两个多边形投影区域。Projected planar block assumption

The vertex set to the left and right images, remember any vertex

The projections in the left and right images are respectively

and

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

The two polygonal projection areas of .

对于任意高程点

如果其左图像中的投影位于

在左图像中的投影区域内，并且其右图像中的投影

位于

在右图像中的投影区域内，则判定高程点被

覆盖。For any elevation point

If the projection in its left image lie in

in the projected area in the left image, and its projected in the right image

lie in

In the projected area in the right image, determine the elevation point quilt

cover.

被

覆盖的所有高程点组成的集合记为

Θ_{w} &Subset; J,

集合中元素的数量为α_w。quilt

The set of all elevation points covered is denoted as

Θ_{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

覆盖的高程点个数α_w＞0，则计算

的高程一致性测度为if

If the number of covered elevation points α _w >0, calculate

The elevation consistency measure of is

(公式一)

(Formula 1)

其中，in,

如果α_w＝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

在左图像中的投影区域内所有图像点组成一个点集，记为K_w ^L，它们中所有点的图像灰度值组成一个数组，记为{h_wt|t＝1，2，…，o_w}。

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

与

与

以及

与

计算集合K_w ^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K_w ^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{h_wt′|t＝1，2，…，o_w}。计算

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

and

as well as

and

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

The gray-scale similarity measure between the projected regions in the left and right images is

$E_{w}^{G} = \frac{Σ_{t = 1}^{O_{w}} (h_{wt} - {\overset{&OverBar;}{h}}_{w}) ({h_{wt}}^{'} - {\overset{&OverBar;}{h}}_{w}^{'})}{\sqrt{[Σ_{t = 1}^{O_{w}} {(h_{wt} - {\overset{&OverBar;}{h}}_{w})}^{2}] \cdot [Σ_{t = 1}^{O_{w}} {({h_{wt}}^{'} - {\overset{&OverBar;}{h}}_{w}^{'})}^{2}]}}$ (公式二) ${E.}_{w}^{G} = \frac{Σ_{t = 1}^{o_{w}} (h_{wt} - {\overset{&OverBar;}{h}}_{w}) ({h_{wt}}^{'} - {\overset{&OverBar;}{h}}_{w}^{'})}{\sqrt{[Σ_{t = 1}^{o_{w}} {(h_{wt} - {\overset{&OverBar;}{h}}_{w})}^{2}] &Center Dot; [Σ_{t = 1}^{o_{w}} {({h_{wt}}^{'} - {\overset{&OverBar;}{h}}_{w}^{'})}^{2}]}}$ (Formula 2)

其中， ${\overset{&OverBar;}{h}}_{w} = \frac{1}{O_{w}} Σ_{t = 1}^{O_{w}} h_{wt},$ ${\overset{&OverBar;}{h}}_{w}^{'} = \frac{1}{O_{w}} Σ_{t = 1}^{O_{w}} {h_{wt}}^{'} .$ in, ${\overset{&OverBar;}{h}}_{w} = \frac{1}{o_{w}} Σ_{t = 1}^{o_{w}} h_{wt},$ ${\overset{&OverBar;}{h}}_{w}^{'} = \frac{1}{o_{w}} Σ_{t = 1}^{o_{w}} {h_{wt}}^{'} .$

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

的可靠性测度为

The reliability measure for

$E_{w} = \{\begin{matrix} E_{w}^{D} \cdot (E_{w}^{G} + 1) / 2 & E_{w}^{D} > 0 \\ (E_{w}^{G} + 1) / 2 & E_{w}^{D} = 0 \end{matrix}$ (公式三) ${E.}_{w} = \{\begin{matrix} {E.}_{w}^{D.} \cdot ({E.}_{w}^{G} + 1) / 2 & {E.}_{w}^{D.} > 0 \\ ({E.}_{w}^{G} + 1) / 2 & {E.}_{w}^{D.} = 0 \end{matrix}$ (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 ${\hat{P}}_{w} &Element; \hat{Ω},$ 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E_w；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 ${\hat{P}}_{w} &Element; \hat{Ω},$ 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 ${\hat{P}}_{w} &Element; \hat{Ω},$ 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

生成该无向图的一个节点，节点的属性值为E_w；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

{\hat{P}}_{w} &Element; \hat{Ω},

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

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

{\hat{P}}_{v} &Element; K_{1} \cup K_{2},

如果下面的不等式组成立，则判定

和

共面：For any two reliable planar blocks assume

{\hat{P}}_{v} &Element; K_{1} \cup K_{2},

If the following inequalities are established, then the decision

and

Coplanar:

$\{\begin{matrix} \frac{| | {\overset{^^}{a a}}_{u u} \cdot \cdot {\overset{^^}{a a}}_{v v} + + {\overset{^^}{b b}}_{u u} \cdot \cdot {\overset{^^}{b b}}_{v v} + + {\overset{^^}{c c}}_{u u} + + {\overset{^^}{c c}}_{v v} | |}{\sqrt{{\overset{^^}{a a}}_{u u}^{22} + + {\overset{^^}{b b}}_{u u}^{22} + + {\overset{^^}{c c}}_{u u}^{22}} \cdot \cdot \sqrt{{\overset{^^}{a a}}_{v v}^{22} + + {\overset{^^}{b b}}_{v v}^{22} + + {\overset{^^}{c c}}_{v v}^{22}}} > > cos cos ((\frac{33}{22} \cdot &Center Dot; φ φ)) \\ \frac{| | {\overset{^^}{a a}}_{u u} \cdot &Center Dot; {\overset{^^}{X x}}_{v v 11} + + {\overset{^^}{b b}}_{u u} \cdot &Center Dot; {\overset{^^}{Y Y}}_{v v 11} + + {\overset{^^}{c c}}_{u u} \cdot &Center Dot; {\overset{^^}{Z Z}}_{v v 11} + + {\overset{^^}{d d}}_{u u} | |}{\sqrt{{\overset{^^}{a a}}_{u u}^{22} + + {\overset{^^}{b b}}_{u u}^{22} + + {\overset{^^}{c c}}_{u u}^{22}}} < < \frac{33}{22} \cdot &Center Dot; ϵ ϵ \\ \frac{| | {\overset{^^}{a a}}_{v v} \cdot &Center Dot; {\overset{^^}{X x}}_{u u 11} + + {\overset{^^}{b b}}_{v v} \cdot &Center Dot; {\overset{^^}{Y Y}}_{u u 11} + + {\overset{^^}{c c}}_{v v} \cdot &Center Dot; {\overset{^^}{Z Z}}_{u u 11} + + {\overset{^^}{d d}}_{v v} | |}{\sqrt{{\overset{^^}{a a}}_{v v}^{22} + + {\overset{^^}{b b}}_{v v}^{22} + + {\overset{^^}{c c}}_{v v}^{22}}} < < \frac{33}{22} \cdot &Center Dot; ϵ ϵ \end{matrix}$

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

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

Use each reliable plane block hypothesis in K ₁ ∪ _{K 2} to generate a node, if any two reliable plane block hypotheses are coplanar, connect their corresponding nodes.

计算图

的所有极大团，记为

{{\hat{Q}}_{s} | s = 1,2, \cdot \cdot \cdot, {\hat{N}}_{Q}},

任意一个极大团

是图的一个子集，记为

{{\hat{P}}_{{\hat{Γ}}_{s} (κ)} | κ = 1,2, \cdot \cdot \cdot {\hat{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

All maximal cliques of , denoted as

{{\hat{Q}}_{the s} | the s = 1,2, &Center Dot; &Center Dot; &Center Dot;, {\hat{N}}_{Q}},

any maximal group

is a picture A subset of , denoted as

{{\hat{P}}_{{\hat{Γ}}_{the s} (κ)} | κ = 1,2, &Center Dot; &Center Dot; &Center Dot; {\hat{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

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

{\overset{&OverBar;}{\overset{&OverBar;}{a}}}_{s} \cdot X + {\overset{&OverBar;}{\overset{&OverBar;}{b}}}_{s} \cdot Y + {\overset{&OverBar;}{\overset{&OverBar;}{c}}}_{s} \cdot Z + {\overset{&OverBar;}{\overset{&OverBar;}{d}}}_{s} = 0,

其中：For any maximal group

Merges all reliable planar block hypotheses it contains, resulting in an open planar The plane equation is denoted as

{\overset{&OverBar;}{\overset{&OverBar;}{a}}}_{the s} &Center Dot; x + {\overset{&OverBar;}{\overset{&OverBar;}{b}}}_{the s} &Center Dot; Y + {\overset{&OverBar;}{\overset{&OverBar;}{c}}}_{the s} &Center Dot; Z + {\overset{&OverBar;}{\overset{&OverBar;}{d}}}_{the s} = 0,

in:

${\overset{&OverBar; &OverBar;}{\overset{&OverBar; &OverBar;}{a a}}}_{s the s} = = \frac{11}{{\overset{^^}{m m}}_{s the s}} {Σ Σ}_{κ κ = = 11}^{{\overset{^^}{m m}}_{s the s}} \frac{{\overset{^^}{a a}}_{{\overset{^^}{Γ Γ}}_{s the s} ((κ κ))}}{\sqrt{{\overset{^^}{a a}}_{{\overset{^^}{Γ Γ}}_{s the s} ((κ κ))}^{22} + + {\overset{^^}{b b}}_{{\overset{^^}{Γ Γ}}_{s the s} ((κ κ))}^{22} + + {\overset{^^}{c c}}_{{\overset{^^}{Γ Γ}}_{s the s} ((κ κ))}^{22}}}$

${\overset{&OverBar; &OverBar;}{\overset{&OverBar; &OverBar;}{b b}}}_{s the s} = = \frac{11}{{\overset{^^}{m m}}_{s the s}} {Σ Σ}_{κ κ = = 11}^{{\overset{^^}{m m}}_{s the s}} \frac{{\overset{^^}{b b}}_{{\overset{^^}{Γ Γ}}_{s the s} ((κ κ))}}{\sqrt{{\overset{^^}{a a}}_{{\overset{^^}{Γ Γ}}_{s the s} ((κ κ))}^{22} + + {\overset{^^}{b b}}_{{\overset{^^}{Γ Γ}}_{s the s} ((κ κ))}^{22} + + {\overset{^^}{c c}}_{{\overset{^^}{Γ Γ}}_{s the s} ((κ κ))}^{22}}}$

${\overset{&OverBar; &OverBar;}{\overset{&OverBar; &OverBar;}{c c}}}_{s the s} = = \frac{11}{{\overset{^^}{m m}}_{s the s}} {Σ Σ}_{κ κ = = 11}^{{\overset{^^}{m m}}_{s the s}} \frac{{\overset{^^}{c c}}_{{\overset{^^}{Γ Γ}}_{s the s} ((κ κ))}}{\sqrt{{\overset{^^}{a a}}_{{\overset{^^}{Γ Γ}}_{s the s} ((κ κ))}^{22} + + {\overset{^^}{b b}}_{{\overset{^^}{Γ Γ}}_{s the s} ((κ κ))}^{22} + + {\overset{^^}{c c}}_{{\overset{^^}{Γ Γ}}_{s the s} ((κ κ))}^{22}}}$

${\overset{&OverBar; &OverBar;}{\overset{&OverBar; &OverBar;}{d d}}}_{s the s} = = - - \frac{11}{{\overset{^^}{m m}}_{s the s}} (({\overset{&OverBar; &OverBar;}{\overset{&OverBar; &OverBar;}{a a}}}_{s the s} \cdot \cdot {Σ Σ}_{κ κ = = 11}^{{\overset{^^}{m m}}_{s the s}} {\overset{^^}{X x}}_{{\overset{^^}{Γ Γ}}_{s the s} ((κ κ)),, 11} + + {\overset{&OverBar; &OverBar;}{\overset{&OverBar; &OverBar;}{b b}}}_{s the s} \cdot &Center Dot; {Σ Σ}_{κ κ = = 11}^{{\overset{^^}{m m}}_{s the s}} {\overset{^^}{Y Y}}_{{\overset{^^}{Γ Γ}}_{s the s} ((κ κ)),, 11} + + {\overset{&OverBar; &OverBar;}{\overset{&OverBar; &OverBar;}{c c}}}_{s the s} \cdot &Center Dot; {Σ Σ}_{κ κ = = 11}^{{\overset{^^}{m m}}_{s the s}} {\overset{^^}{Z Z}}_{{\overset{^^}{Γ Γ}}_{s the s} ((κ κ)),, 11}))$

将 ${{\hat{P}}_{{\hat{Γ}}_{s} (κ)} | κ = 1,2, \cdot \cdot \cdot {\hat{m}}_{s}}$ 中每个可靠平面块假设的所有顶点投影到

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

的成员直线特征。Will

{{\hat{P}}_{{\hat{Γ}}_{the s} (κ)} | κ = 1,2, \cdot \cdot \cdot {\hat{m}}_{the s}}

All vertices assumed by each reliable planar block in

, connect all projected points sequentially according to the order of vertices in the planar block hypothesis, and all projected straight line segments obtained become

The member line feature of .

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

的平面方程为

{\overset{&OverBar;}{\overset{&OverBar;}{a}}}_{s} \cdot X + {\overset{&OverBar;}{\overset{&OverBar;}{b}}}_{s} \cdot Y + {\overset{&OverBar;}{\overset{&OverBar;}{c}}}_{s} \cdot Z + {\overset{&OverBar;}{\overset{&OverBar;}{d}}}_{s} = 0,

它的所有成员直线特征的集合记为

{\overset{&OverBar;}{\overset{&OverBar;}{M}}}_{s} = {{\overset{&OverBar;}{\overset{&OverBar;}{L}}}_{l}^{s} | l = 1,2, \cdot \cdot \cdot, {\overset{&OverBar;}{\overset{&OverBar;}{n}}}_{s}},

任意一个成员直线特征

的端点记为

和 All open planes obtained after merging are denoted as

\overset{&OverBar;}{\overset{&OverBar;}{Ω}} = {{\overset{&OverBar;}{\overset{&OverBar;}{P}}}_{the s} | the s = 1,2, &Center Dot; \cdot &Center Dot;, \overset{&OverBar;}{\overset{&OverBar;}{N}}},

in,

\overset{&OverBar;}{\overset{&OverBar;}{N}} = {\hat{N}}_{Q},

The real boundaries of their corresponding three-dimensional planes are unknown, and any open plane

The plane equation of is

{\overset{&OverBar;}{\overset{&OverBar;}{a}}}_{the s} &Center Dot; x + {\overset{&OverBar;}{\overset{&OverBar;}{b}}}_{the s} &Center Dot; Y + {\overset{&OverBar;}{\overset{&OverBar;}{c}}}_{the s} \cdot Z + {\overset{&OverBar;}{\overset{&OverBar;}{d}}}_{the s} = 0,

The collection of all its member line features is denoted as

{\overset{&OverBar;}{\overset{&OverBar;}{m}}}_{the s} = {{\overset{&OverBar;}{\overset{&OverBar;}{L}}}_{l}^{the s} | l = 1,2, \cdot &Center Dot; \cdot, {\overset{&OverBar;}{\overset{&OverBar;}{no}}}_{the s}},

Any member line feature

The endpoint of is denoted as

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

{\overset{&OverBar;}{\overset{&OverBar;}{P}}}_{j} &Element; \overset{&OverBar;}{\overset{&OverBar;}{Ω}},

如果Pick two open planes arbitrarily

{\overset{&OverBar;}{\overset{&OverBar;}{P}}}_{j} &Element; \overset{&OverBar;}{\overset{&OverBar;}{Ω}},

if

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

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

在U中选择它与

的任意开放平面的交线，如果它的任意成员直线特征

{\overset{&OverBar;}{\overset{&OverBar;}{L}}}_{l}^{i} &Element; {\overset{&OverBar;}{\overset{&OverBar;}{M}}}_{i}

所在直线与该平面交线的夹角小于φ，并且

的中点到平面交线的距离小于ε，则判定

与该平面交线共线，并从开放平面

的成员直线特征集合中删除

For any open plane

Select it in U with

An intersection line of any open plane of , if any of its member line features

{\overset{&OverBar;}{\overset{&OverBar;}{L}}}_{l}^{i} &Element; {\overset{&OverBar;}{\overset{&OverBar;}{m}}}_{i}

The angle between the straight line and the intersection of the plane is less than φ, and

The distance from the midpoint of the plane to the intersection line of the plane is less than ε, then it is judged

collinear with the plane intersection and from the open plane

The member line feature set of delete in

由所有与

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

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 ${\overset{&OverBar;}{\overset{&OverBar;}{L}}}_{l}^{i} &Element; {\overset{&OverBar;}{\overset{&OverBar;}{M}}}_{i},$ 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，

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

{\overset{&OverBar;}{\overset{&OverBar;}{L}}}_{l}^{i} &Element; {\overset{&OverBar;}{\overset{&OverBar;}{m}}}_{i},

apply a method of generating a plane hypothesis from a single 3D line, from which a new open plane is generated,

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 ${\overset{&OverBar;}{\overset{&OverBar;}{P}}}_{u} &Element; \overset{&OverBar;}{\overset{&OverBar;}{Ω}},$ 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 ${\overset{&OverBar;}{\overset{&OverBar;}{P}}}_{v} &Element; \overset{&OverBar;}{\overset{&OverBar;}{Ω}}$ 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

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

中，并从集合

中删除这个共线的成员直线特征。For any open plane

{\overset{&OverBar;}{\overset{&OverBar;}{P}}}_{u} &Element; \overset{&OverBar;}{\overset{&OverBar;}{Ω}},

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

{\overset{&OverBar;}{\overset{&OverBar;}{P}}}_{v} &Element; \overset{&OverBar;}{\overset{&OverBar;}{Ω}}

, 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

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

in, and from the set of

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，假定它是开放平面和

的交线，则将该平面交线添加到

和的成员直线特征集合

和

中去。For any plane intersection line in the set U, it is assumed to be an open plane and

, then add the plane intersection to

and The member line feature set of

and

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

的任意一个成员直线特征，如果它属于集合U，并且它与

的任意其它成员直线特征的夹角属于范围

则判定这两个成员直线特征相交。For any open plane

Any member line feature of , if it belongs to the set U, and it is related to

The included angle of any other member of the line feature belongs to the range

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 ${\overset{&OverBar;}{\overset{&OverBar;}{M}}}_{u} \cap {\overset{&OverBar;}{\overset{&OverBar;}{M}}}_{u}^{'},$ 记为U₁，另一个子集是

记为U₂。For any open plane Split all its member line features into two subsets, one subset is

{\overset{&OverBar;}{\overset{&OverBar;}{m}}}_{u} \cap {\overset{&OverBar;}{\overset{&OverBar;}{m}}}_{u}^{'},

Denoted as U ₁ , another subset is

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 $M_{k}^{'} = {T_{k}^{j} | j = 1,2, \cdot \cdot \cdot, n_{k}^{'}},$ 如果 $T_{k}^{1} = T_{k}^{n_{k}^{'}},$ 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 $m_{k}^{'} = {T_{k}^{j} | j = 1,2, &Center Dot; \cdot \cdot, {no}_{k}^{'}},$ if $T_{k}^{} = T_{k}^{{no}_{k}^{'}},$ 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 ${{Tl}_{k}^{j} | j = 1,2, \cdot \cdot \cdot, n_{k}^{'}}$ 和 ${{Tr}_{k}^{j} | j = 1,2, \cdot \cdot \cdot, n_{k}^{'}} .$ Project P′ _k to the left and right images to obtain point sets respectively ${{Tl}_{k}^{j} | j = 1,2, &Center Dot; &Center Dot; &Center Dot;, {no}_{k}^{'}}$ and ${{Tr}_{k}^{j} | j = 1,2, &Center Dot; \cdot &Center Dot;, {no}_{k}^{'}} .$

设定一个表示直线特征完整性的距离门限h_d，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h_d。Set 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R_k ^L，它的顶点分别记为g₁、g₂、g₃和g₄，其中， $g_{1} = {Tl}_{k}^{1},$ $g_{2} = {Tl}_{k}^{n_{k}^{'}},$ g₃是线段

的延长线上到点g₂的距离为h_d的点，g₄是线段Tl_k ²Tl_k ¹的延长线上到点g₁的距离为h_d的点，连接它们成为R_k ^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}^{},

g_{2} = {Tl}_{k}^{{no}_{k}^{'}},

g ₃ is the line segment

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R_k ^R，它的四个顶点分别记为g′₁、g′₂、g′₃和g′₄。Using 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R_k ^L和R_k ^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R_k ^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T_k ¹和

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R_k ^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T_k ¹和

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

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

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T_k ²T_k ¹和

为共用边界、与P_k′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T_k ²T_k ¹或其延长线上指定一点Φ₁，使得线段T_k ²Φ₁和以T_k ²T_k ¹为共用边界的相邻平面假设的共用边界等长，在

或其延长线上指定一点Φ₂，使得线段

和以

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

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

Specify a point Φ ₂ on or its extension such that the line segment

and to

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T_k ²T_k ¹和

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T_k ²T_k 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T_k ²T_k ¹或其延长线上指定一点，使得线段T_k ²Φ₁与P′_l的共用边界等长，在

或其延长线上指定一点Φ₂，使得线段与T_k ²Φ₁等长，线段Φ₁Φ₂成为补充的缺失边界。If T _k ² T _k ¹ and

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

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T_k ¹和

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

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

$γ_{i} = \frac{E_{i}^{''}}{Σ_{j = 1}^{n_{i}^{''}} {E_{i, j}}^{''}}$ (公式四) $γ_{i} = \frac{{E.}_{i}^{''}}{Σ_{j = 1}^{{no}_{i}^{''}} {E.}_{i, j}^{''}}$ (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

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

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

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

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

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

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，T_k ²T_k ¹是A和C的共用边界线，T_k ²T_k 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

$\{\begin{matrix} {X x}_{00}^{' '} = = {X x}_{00} - - \frac{a a \cdot &Center Dot; ((a a \cdot &Center Dot; {X x}_{00} + + b b \cdot &Center Dot; {Y Y}_{00} + + c c \cdot &Center Dot; {Z Z}_{00} + + d d))}{{a a}^{22} + + {b b}^{22} + + {c c}^{22}} \\ {Y Y}_{00}^{' '} = = {Y Y}_{00} - - \frac{b b \cdot &Center Dot; ((a a \cdot &Center Dot; {X x}_{00} + + b b \cdot &Center Dot; {Y Y}_{00} + + c c \cdot &Center Dot; {Z Z}_{00} + + d d))}{{a a}^{22} + + {b b}^{22} + + {c c}^{22}} \\ {Z Z}_{00}^{' '} = = {Z Z}_{00} - - \frac{c c \cdot \cdot ((a a \cdot \cdot {X x}_{00} + + b b \cdot &Center Dot; {Y Y}_{00} + + c c \cdot &Center Dot; {Z Z}_{00} + + d d))}{{a a}^{22} + + {b b}^{22} + + {c c}^{22}} \end{matrix}$

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L_u和L_v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L_u∈S和L_v∈S，端点分别为(X_u1，Y_u1，Z_u1)和(X_u2，Y_u2，Z_u2)以及(X_v1，Y_v1，Z_v1)和(X_v2，Y_v2，Z_v2)，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D_u、D_v，其中Calculation point (X ₀ , Y ₀ , Z ₀ ) and two quantities D _u , D _v , where

$\{\begin{matrix} {X x}_{00} = = (({X x}_{u u 11} + + {X x}_{u u 22} + + {X x}_{v v 11} + + {X x}_{v v 22})) / / 44 \\ {Y Y}_{00} = = (({Y Y}_{u u 11} + + {Y Y}_{u u 22} + + {Y Y}_{v v 11} + + {Y Y}_{v v 22})) / / 44 \\ {Z Z}_{00} = = (({Z Z}_{u u 11} + + {Z Z}_{u u 22} + + {Z Z}_{v v 11} + + {Z Z}_{v v 22})) / / 44 \end{matrix}$

${D D.}_{u u} = = \sqrt{{(({X x}_{u u 11} - - {X x}_{u u 22}))}^{22} + + {(({Y Y}_{u u 11} - - {Y Y}_{u u 22}))}^{22} + + {(({Z Z}_{u u 11} - - {Z Z}_{u u 22}))}^{22}}$

${D D.}_{v v} = = \sqrt{{(({X x}_{v v 11} - - {X x}_{v v 22}))}^{22} + + {(({Y Y}_{v v 11} - - {Y Y}_{v v 22}))}^{22} + + {(({Z Z}_{v v 11} - - {Z Z}_{v v 22}))}^{22}}$

得到L_u和L_v的归一化方向矢量为 $v_{u} = (\begin{matrix} \frac{X_{u 1} - X_{u 2}}{D_{u}} & \frac{Y_{u 1} - Y_{u 2}}{D_{u}} & \frac{Z_{u 1} - Z_{u 2}}{D_{u}} \end{matrix})$ 和 $v_{v} = (\begin{matrix} \frac{X_{v 1} - X_{v 2}}{D_{v}} & \frac{Y_{v 1} - Y_{v 2}}{D_{v}} & \frac{Z_{v 1} - Z_{v 2}}{D_{v}} \end{matrix}),$ 以及它们的叉积v_u×v_v＝(αβγ)。The normalized direction vectors of Lu _u and L _v are obtained as $v_{u} = (\begin{matrix} \frac{x_{u 1} - x_{u 2}}{{D.}_{u}} & \frac{Y_{u 1} - Y_{u 2}}{{D.}_{u}} & \frac{Z_{u 1} - Z_{u 2}}{{D.}_{u}} \end{matrix})$ and $v_{v} = (\begin{matrix} \frac{x_{v 1} - x_{v 2}}{{D.}_{v}} & \frac{Y_{v 1} - Y_{v 2}}{{D.}_{v}} & \frac{Z_{v 1} - Z_{v 2}}{{D.}_{v}} \end{matrix}),$ 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′)＝(v_u+v_v)×(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

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

迭代收敛条件为

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

The iteration convergence condition is

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

在左、右图像中三对同名点

与

与以及

与

计算集合K_w ^L中任意一点在右图像中的同名点Seventh point, according to

Three pairs of points with the same name in the left and right images

and

and as well as

and

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

在左、右图像中三对同名点

与

与

以及

与

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

Three pairs of points with the same name in the left and right images

and

as well as

and

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

$V V = = [\begin{matrix} \overset{^^}{x x} {r r}_{w w 11} \\ \overset{^^}{x x} {r r}_{w w 22} \\ \overset{^^}{x x} {r r}_{w w 33} \\ \overset{^^}{y the y} {r r}_{w w 11} \\ \overset{^^}{y the y} {r r}_{w w 22} \\ \overset{^^}{y the y} {r r}_{w w 33} \end{matrix}],,$ $M m = = [\begin{matrix} \overset{^^}{x x} {l l}_{w w 11} & \overset{^^}{y the y} {l l}_{w w 11} & 11 & 00 & 00 & 00 \\ \overset{^^}{x x} {l l}_{w w 22} & \overset{^^}{y the y} {l l}_{w w 22} & 11 & 00 & 00 & 00 \\ \overset{^^}{x x} {l l}_{w w 22} & \overset{^^}{y the y} {l l}_{w w 33} & 11 & 00 & 00 & 00 \\ 00 & 00 & 00 & \overset{^^}{x x} {l l}_{w w 11} & \overset{^^}{y the y} {l l}_{w w 11} & 11 \\ 00 & 00 & 00 & \overset{^^}{x x} {l l}_{w w 22} & \overset{^^}{y the y} {l l}_{w w 22} & 11 \\ 00 & 00 & 00 & \overset{^^}{x x} {l l}_{w w 22} & \overset{^^}{y the y} {l l}_{w w 33} & 11 \end{matrix}]$

计算

平面上同名点之间的变换式系数calculate

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

则

在左图像中投影区域内任意点

在右图像中的同名点为

(a_{11} \cdot {\hat{x}}_{1} + a_{12} \cdot {\hat{y}}_{1} + a_{13}, a_{21} \cdot {\hat{x}}_{1} + a_{22} \cdot {\hat{y}}_{1} + a_{23}) .

but

Any point within the projected area in the left image

The point of the same name in the right image is

(a_{11} &Center Dot; {\hat{x}}_{1} + a_{12} \cdot {\hat{the y}}_{1} + a_{13}, a_{twenty one} &Center Dot; {\hat{x}}_{1} + a_{twenty two} \cdot {\hat{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K_w ^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

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，Y_O1，Z_O1) And (X)_O2，Y_O2，Z_O2) (ii) a The linear feature sets extracted from the two images are respectively marked as H₁And H₂(ii) a Knowing N in the scene_SA three-dimensional line feature, denoted as S ═ L_i|i＝1，2，…，N_SH, any element L in the set_iRepresents a three-dimensional straight line segment, two end points of which are respectively (X)_i1，Y_i1，Z_i1) And (X)_i2，Y_i2，Z_i2) (ii) a The sparse DEM data in the scene is known and is recorded as an elevation point set

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

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 ═ P_i1, 2, …, N, for any of which three-dimensional planes P is assumed_iThe plane equation is a_i·X+b_i·Y+c_i·Z+d_iTwo three-dimensional straight line features used to generate it are called P0_iA 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 ═ P_w1, 2, …, N, arbitrary three-dimensional plane hypothesis P_wIs 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 L_j ^wIs noted as (X)_j，1 ^w，Y_j，1 ^w，Z_j，1 ^w) And (X)_j，2 ^w，Y_j，2 ^w，Z_j，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 plane_iBelongs to omega, and utilizes the member straight line characteristic M_iDividing 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

Wherein any one of the plane blocks is assumed to be

Is represented by its set of vertices, denoted as

{({\hat{X}}_{wj}, {\hat{Y}}_{wj}, {\hat{Z}}_{wj}) | j = 1,2, . . ., {\hat{n}}_{w}},

Is obtained by division in omegaIs expressed as the equation of the three-dimensional plane hypothesis

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

And

wherein,

and

respectively, is the elevation point in J

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>Ω</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>Ω</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

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>·</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>·</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>·</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

Is marked asAnd

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>Ω</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mo>,</mo> </mrow> </math>

If it is not

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

Selecting it from U and

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

is less than epsilon, it is determined

Collinear with the plane and from the open plane

Member straight line feature set of

Deletion in

By all of

Is a set of plane intersections of collinear linear features of a member, and is recorded as

Step 3, supplementing missing open planes

For any one open plane

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,

adding the plane to the set as its member line features

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>Ω</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>Ω</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

Is smaller than phi, and

is less than epsilon, the plane intersection is added to the set of lines

And from the set

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

And

the intersection of the planes is added to

And

member straight line feature set of

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

If it belongs to the set U, and it is compared with

Fall within the range of any other member linear feature

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

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 L_uE.g. S and L_v∈S，L_uHas an endpoint of (X)_u1，Y_u1，Z_u1) And (X)_u2，Y_u2，Z_u2)，L_vHas an endpoint of (X)_v1，Y_v1，Z_v1) And (X)_v2，Y_v2，Z_v2) With which the objective function f (a, b, c, d) is defined:

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

calculating (a, b, c, d) when f (a, b, c, d) is minimum by gradient descent method to obtain L_uAnd L_vThe 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 plane_iE omega, calculating the midpoint of the connecting line of the midpoints of the two member straight line features in the plane P_iProjection onto, denoted as (X)_M ⁱ，Y_M ⁱ，Z_M ⁱ)；

Step 1), judging the assumed coplanarity relationship of three-dimensional planes

Assuming P for any two three-dimensional planes_iE.g. omega and P_jBelongs to omega, if the following inequality group is satisfied, P is determined_iAnd P_jCoplanarity:

step 2), searching coplanar three-dimensional plane hypothesis

Generating an undirected graph G, and assuming each three-dimensional plane as P_iE 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 { Q_w|w＝1，2，…，N_QAny one of the very big clusters Q_wIs a subset of Ω, denoted

Step 3) merging coplanar three-dimensional plane assumptions

For any one extremely large group Q_wCombining all three-dimensional plane hypotheses contained in the three-dimensional plane hypothesis to obtain a new three-dimensional plane hypothesis P_wIts plane equation is denoted as a_w·X+b_w·Y+c_w·Z+d_w0, 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>Σ</mi> <mrow> <mi>κ</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mi>m</mi> <mi>w</mi> </msub> </munderover> <mfrac> <msub> <mi>a</mi> <mrow> <msub> <mi>Γ</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>κ</mi> <mo>)</mo> </mrow> </mrow> </msub> <msqrt> <msup> <msub> <mi>a</mi> <mrow> <msub> <mi>Γ</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>κ</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>b</mi> <mrow> <msub> <mi>Γ</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>κ</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>c</mi> <mrow> <msub> <mi>Γ</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>κ</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>Σ</mi> <mrow> <mi>κ</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mi>m</mi> <mi>w</mi> </msub> </munderover> <mfrac> <msub> <mi>b</mi> <mrow> <msub> <mi>Γ</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>κ</mi> <mo>)</mo> </mrow> </mrow> </msub> <msqrt> <msup> <msub> <mi>a</mi> <mrow> <msub> <mi>Γ</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>κ</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>b</mi> <mrow> <msub> <mi>Γ</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>κ</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>c</mi> <mrow> <msub> <mi>Γ</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>κ</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>Σ</mi> <mrow> <mi>κ</mi> <mo>=</mo> <mn>1</mn> </mrow> <msub> <mi>m</mi> <mi>w</mi> </msub> </munderover> <mfrac> <msub> <mi>c</mi> <mrow> <msub> <mi>Γ</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>κ</mi> <mo>)</mo> </mrow> </mrow> </msub> <msqrt> <msup> <msub> <mi>a</mi> <mrow> <msub> <mi>Γ</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>κ</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>b</mi> <mrow> <msub> <mi>Γ</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>κ</mi> <mo>)</mo> </mrow> </mrow> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>c</mi> <mrow> <msub> <mi>Γ</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>κ</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>·</mo> <munderover> <mi>Σ</mi> <mrow> <mi>κ</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>Γ</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>κ</mi> <mo>)</mo> </mrow> </mrow> </msubsup> <mo>+</mo> <msub> <mover> <mi>b</mi> <mo>&OverBar;</mo> </mover> <mi>w</mi> </msub> <mo>·</mo> <munderover> <mi>Σ</mi> <mrow> <mi>κ</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>Γ</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>κ</mi> <mo>)</mo> </mrow> </mrow> </msubsup> <mo>+</mo> <msub> <mover> <mi>c</mi> <mo>&OverBar;</mo> </mover> <mi>w</mi> </msub> <mo>·</mo> <munderover> <mi>Σ</mi> <mrow> <mi>κ</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>Γ</mi> <mi>w</mi> </msub> <mrow> <mo>(</mo> <mi>κ</mi> <mo>)</mo> </mrow> </mrow> </msubsup> <mo>)</mo> </mrow> </mrow> </math>

will be provided with

All member linear features of each three-dimensional plane hypothesis are projected to P_wAll the resulting projected straight line segments become P_wIs 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 plane_iBelongs to omega, and utilizes the member straight line characteristic M_iThe method comprises the following steps of:

step 1) merging collinear member straight line features

For P_iOf any two member straight line features L_j 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 L_j ⁱFrom midpoint to L_k ⁱDistance sum L of straight lines_k ⁱFrom midpoint to L_j ⁱThe distances of the straight lines are respectively marked as eta₁And η₂；

If μ < φ and η₁< ε and μ₂If < epsilon, then L is determined_j ⁱAnd L_k ⁱCollineates, and uses them to generate a new three-dimensional linear feature with two endpoints L_j ⁱAnd L_k ⁱThe two end points which are farthest away, and a new three-dimensional straight line feature is added to the set M_iAnd from the set M_iDeletion in L_j ⁱAnd L_k ⁱ；

Step 2) calculating P_iOf all member straight line features of

For P_iOf any two member straight line features L_α ⁱ，

<math> <mrow> <msubsup> <mover> <mi>L</mi> <mo>&OverBar;</mo> </mover> <mi>β</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

Then calculate L_α ⁱAnd L_β ⁱOr the intersection of their extensions;

the set of all calculated intersections is denoted as R_i；

Step 3), generating an undirected graph for the segmentation of the planar hypotheses

Generating an undirected graph marked as G_iIts nodes are generated by two types of points: one is an intersection set R_iFor each element in M, II_iAny one member of the linear features, if it is not connected to M on an extension outside one of its endpoints_iThe other members of the set, and the end point is not the set R_iThe end point becomes the graph G_iA node of (2);

undirected graph G_iThe 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 G_iIf 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 L_α ⁱIf one or both of its endpoints is a graphG_iThe node (b) is processed in two cases:

first case, L_α ⁱAnd M_iIs not intersected by any other member straight line feature

Selecting any member line feature

And β ≠ α, designating it as on the straight line with graph G_iTwo points corresponding to the two nodes, the two points are positioned at L_β ⁱOn both sides of the midpoint and on the connecting line segment between them, not connected to the graph G_iThe 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>α</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>·</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>α</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>α</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>·</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>α</mi> <mo>,</mo> <mn>2</mn> </mrow> <mi>i</mi> </msubsup> </mrow> </math>

Wherein (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) Is L_α 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>·</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>·</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>α</mi> <mo>,</mo> <mn>1</mn> </mrow> <mi>i</mi> </msubsup> <mo>)</mo> </mrow> <mo>·</mo> <mfrac> <mrow> <msubsup> <mover> <mi>Y</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>α</mi> <mo>,</mo> <mn>2</mn> </mrow> <mi>i</mi> </msubsup> <mo>-</mo> <msubsup> <mover> <mi>Y</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>α</mi> <mo>,</mo> <mn>1</mn> </mrow> <mi>i</mi> </msubsup> </mrow> <mrow> <msubsup> <mover> <mi>X</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>α</mi> <mo>,</mo> <mn>2</mn> </mrow> <mi>i</mi> </msubsup> <mo>-</mo> <msubsup> <mover> <mi>X</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>α</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>α</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>α</mi> <mo>,</mo> <mn>1</mn> </mrow> <mi>i</mi> </msubsup> <mo>)</mo> </mrow> <mo>·</mo> <mfrac> <mrow> <msubsup> <mover> <mi>Y</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>α</mi> <mo>,</mo> <mn>2</mn> </mrow> <mi>i</mi> </msubsup> <mo>-</mo> <msubsup> <mover> <mi>Y</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>α</mi> <mo>,</mo> <mn>1</mn> </mrow> <mi>i</mi> </msubsup> </mrow> <mrow> <msubsup> <mover> <mi>X</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>α</mi> <mo>,</mo> <mn>2</mn> </mrow> <mi>i</mi> </msubsup> <mo>-</mo> <msubsup> <mover> <mi>X</mi> <mo>&OverBar;</mo> </mover> <mrow> <mi>α</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>α</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，Y_l，1 ⁱ，Z_l，1 ⁱ) And(X_l，2 ⁱ，Y_l，2 ⁱ，Z_l，2 ⁱ) Is L_l ⁱThe endpoint of (1); if it is satisfied with

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>·</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>·</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，Y_α，1 ⁱ，Z_α，1 ⁱ) And (X)₂，Y₂，Z₂) In the figure G_iAnd a corresponding node in (X), and a point (X)_α，2 ⁱ，Y_α，2 ⁱ，X_α，2 ⁱ) And (X)₁，Y₁，Z₁) In the figure G_iThe corresponding node in (1);

second case, L_α ⁱAnd M_iAt least one other member line feature of

If (X)_α，1 ⁱ，Y_α，1 ⁱ，Z_α，1 ⁱ) And graph G_iCorresponding to one node, and selecting the member straight line characteristic L_α ⁱOr on an extension thereof with (X)_α，1 ⁱ，Y_α，1 ⁱ，Z_α，1 ⁱ) The closest intersection point, denoted as (X)₃，Y₃，Z₃) Assuming that the intersection point is L_α ⁱLinear feature of member L_β ⁱThe intersection point of (a); selection of L_β ⁱOr on an extension thereof with the diagram G_iThe point corresponding to the middle node if it is at point (X)₃，Y₃，Z₃) Do not have a connection with the graph G_iIf 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 L_β ⁱThe intersection point with other member straight line features is assumed to be L_l ⁱAnd L is_l ⁱOr on the extension thereof with the graph G_iA point corresponding to the middle node, which satisfies the relation with point (X)_α，1 ⁱ，Y_α，1 ⁱ，Z_α，1 ⁱ) Falls on L_β ⁱSame side, and with point (X)₄，Y₄，Z₄) Do not have a connection with the graph G_iThe point corresponding to the middle node is connected with the point (X)_α，1 ⁱ，Y_α，1 ⁱ，Z_α，1 ⁱ) In the figure G_iThe corresponding node in (1); otherwise, the point of attachment (X)_α，1 ⁱ，Y_α，1 ⁱ，Z_α，1 ⁱ) And point (X)₄，Y₄，Z₄) In the figure G_iThe 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，Y_α，2 ⁱ，Z_α，2 ⁱ) Is a drawing G_iNode (2), method for constructing connection relation and (X)_α，1 ⁱ，Y_α，1 ⁱ，Z_α，1 ⁱ) The same;

step 4), dividing to generate the plane block hypothesis

Search undirected graph G_iFor 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>Ω</mi> <mo>^</mo> </mover> <mo>,</mo> </mrow> </math>

Calculating the reliability measure thereof, comprising the steps of:

step 1) searching for

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

The projections in the left and right images are respectively

And

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

Two polygonal projection areas of (a);

for any elevation point

If projection in its left image

Is located at

Within the projection area in the left image, and the projection in the right image thereof

Is located at

Within the projection area in the right image, the elevation point is determined

Quilt

Covering;

quilt

The set of all the elevation points covered is recorded as

<math> <mrow> <msub> <mi>Θ</mi> <mi>w</mi> </msub> <mo>&Subset;</mo> <mi>J</mi> <mo>,</mo> </mrow> </math>

The number of elements in the set being alpha_w；

Step 2), calculating the elevation consistency measure

If it is not

If the number of covered elevation points alpha w is more than 0, calculating

Elevation ofMeasure of consistency of

(formula one)

Wherein,

if α is_wWhen the value is equal to 0, then

E_{w}^{D} = 0;

Step 3), calculating the gray level similarity measure

All image points in the projection region in the left image form a point set, denoted as K_w ^LThe gray values of the images of all the points form an array, which is marked as { h }_wt|t＝1，2，…，o_w}；

According to

Three pairs of homonymous points in left and right images

And

and

and

and

computing a set K_w ^LAll element points in the image are homonymous points in the right image, and the set of the homonymous points is marked as K_w ^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，…，o_w}; computing

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>Σ</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>′</mo> </msup> <mo>-</mo> <msup> <msub> <mover> <mi>h</mi> <mo>&OverBar;</mo> </mover> <mi>w</mi> </msub> <mo>′</mo> </msup> <mo>)</mo> </mrow> </mrow> <msqrt> <mo>[</mo> <munderover> <mi>Σ</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>·</mo> <mo>[</mo> <munderover> <mi>Σ</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>′</mo> </msup> <mo>-</mo> <msup> <msub> <mover> <mi>h</mi> <mo>&OverBar;</mo> </mover> <mi>w</mi> </msub> <mo>′</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>Σ</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>′</mo> </msup> <mo>=</mo> <mfrac> <mn>1</mn> <msub> <mi>o</mi> <mi>w</mi> </msub> </mfrac> <munderover> <mi>Σ</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>′</mo> </msup> <mo>;</mo> </mrow> </math>

step 4) calculating

Measure of reliability of

Measure of reliability of

(formula three).

7. The three-dimensional plane extraction method according to claim 6, wherein an arbitrary plane block is assumed

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

If it is not

E_{w}^{D} > 0,

It is used to generate a node of the undirected graph having an attribute value of E_w(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

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

Generating a node of the undirected graph, wherein the attribute value of the node is E_w(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

<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>∪</mo> <msub> <mi>K</mi> <mn>2</mn> </msub> <mo>,</mo> </mrow> </math>

If the following inequality set is true, it is determined

And

coplanarity:

step 2), searching the hypothesis of the coplanar reliable plane block to generate an undirected graph which is recorded as

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

All very big groups of (1), noted

{{\hat{Q}}_{s} | s = 1,2, . . ., {\hat{N}}_{Q}},

Any one of the extremely large groups

Is shown in the figure

A subset of

<math> <mrow> <mo>{</mo> <msub> <mover> <mi>P</mi> <mo>^</mo> </mover> <mrow> <msub> <mover> <mi>Γ</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>κ</mi> <mo>)</mo> </mrow> </mrow> </msub> <mo>|</mo> <mi>κ</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>∪</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

All reliable plane block hypotheses contained by the plane block are combined to obtain an open planeThe plane equation is expressed as

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>Σ</mi> <mrow> <mi>κ</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>Γ</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>κ</mi> <mo>)</mo> </mrow> </mrow> </msub> <msqrt> <msup> <msub> <mover> <mi>a</mi> <mo>^</mo> </mover> <mrow> <msub> <mover> <mi>Γ</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>κ</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>Γ</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>κ</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>Γ</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>κ</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>Σ</mi> <mrow> <mi>κ</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>Γ</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>κ</mi> <mo>)</mo> </mrow> </mrow> </msub> <msqrt> <msup> <msub> <mover> <mi>a</mi> <mo>^</mo> </mover> <mrow> <msub> <mover> <mi>Γ</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>κ</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>Γ</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>κ</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>Γ</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>κ</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>Σ</mi> <mrow> <mi>κ</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>Γ</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>κ</mi> <mo>)</mo> </mrow> </mrow> </msub> <msqrt> <msup> <msub> <mover> <mi>a</mi> <mo>^</mo> </mover> <mrow> <msub> <mover> <mi>Γ</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>κ</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>Γ</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>κ</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>Γ</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>κ</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>·</mo> <munderover> <mi>Σ</mi> <mrow> <mi>κ</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>Γ</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>κ</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>·</mo> <munderover> <mi>Σ</mi> <mrow> <mi>κ</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>Γ</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>κ</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>Σ</mi> <mrow> <mi>κ</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>Γ</mi> <mo>^</mo> </mover> <mi>s</mi> </msub> <mrow> <mo>(</mo> <mi>κ</mi> <mo>)</mo> </mrow> <mo>,</mo> <mn>1</mn> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </math>

will be provided with

All vertices of each reliable plane block hypothesis are projected to

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

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

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>∩</mo> <msup> <msub> <mover> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>u</mi> </msub> <mo>′</mo> </msup> <mo>,</mo> </mrow> </math>

Is marked as U₁Another subset is

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

And

setting a distance threshold h representing the integrity of the linear feature_d；

Designated P'_kThe area where the missing boundary projected in the left image is a quadrangle, which is marked as P_k ^LIts vertexes are respectively marked as g₁、g₂、g₃And g₄Wherein

g_{1} = {Tl}_{k}^{1},

g₃is a line segment

To point g on the extension line of₂Is a distance h_dPoint of (a), g₄Is line segment Tl_k ²Tl_k ¹Is not limited toOn the long line to point g₁Is a distance h_dConnecting them to R_k ^LFour sides g of₁g₂、g₂g₃、g₃g₄And g₄g₁；

P 'was calculated using a similar method'_kIs projected in the right image in the region R_k ^RIts four vertexes are respectively marked as g₁′、g₂′、g₃' and g₄′；

Step 2) using the region R_k ^LAnd R_k ^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 R_k ^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 intensively_k ¹And

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 R_k ^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 intensively_k ¹And

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 T_k ²T_k ¹And

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 T_k ²T_k ¹Or a point phi designated on the extension thereof₁So that the line segment T_k ²Φ₁And with T_k ²T_k ¹The common boundaries assumed for adjacent planes of the common boundary are of equal length, in

Or a point phi designated on the extension thereof₂Make the line segment

And with

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 T_k ²T_k ¹And

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 is_k ²T_k ¹Is P'_lAnd P'_kAccording to the rule that the shared boundaries of adjacent polygon planes are equal in length, at T_k ²T_k ¹Or a point specified on the extension thereof so that the line segment T_k ²Φ₁And P'_lThe common boundary of (B) is equal in length in

Or a point phi designated on the extension thereof₂Make the line segment

And T_k 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 intensively_k ¹And

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 plane_iBelongs to omega', and utilizes a formula I to calculate P ″)_iA measure of elevation coherence, e.g.Fruit P_iIf 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 E_i，j", the closed plane hypothesis P", is calculated according to the following formula_iThe reliability measure of the boundary corresponding to the real plane boundary is

(equation four).