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 scheme is that any method is used for obtaining a three dimensional line feature set and a sparse digitalelevation model data in a target scene, obtaining stereo image of a target scene from two different visual angles and obtaining all the inner parameters and the outer parameters of the stereoscopic camera as well as the line extracting result of the stereo image; the output of the technical scheme is the three dimensional plane with high positioning precision and favorable boundary integrity andaccuracy. In the treating process of the technical scheme, the positioning of the three dimensional plane and the extraction and separation of the plane zone boundary are divided into two stages whichare connected for independently finishing two tasks and reducing error impact between the two tasks, which causes the results of the two stages to be respectively optimal, so that the three dimensional plane extraction is optimal.

Description

Three-dimensional plane extraction method

Technical Field

The invention relates to the field of photogrammetry, remote sensing and computer vision, in particular to a method for extracting a three-dimensional plane.

Background

The three-dimensional reconstruction of the artificial target has very wide application in the fields of urban planning, disaster monitoring, communication facility construction and the like, and has been an important subject of long-term research in the fields of photogrammetry, remote sensing and computer vision for a long time. In various target three-dimensional reconstruction schemes with polyhedral structures, three-dimensional plane extraction is a key step.

At present, there are four main methods for extracting three-dimensional planes: firstly, a three-dimensional plane is obtained by fitting by using directly obtained Dense Elevation Model (DEM) data; secondly, establishing a plurality of reflection maps by acquiring images of a target area under different illumination conditions and combining a reflection characteristic model of a target surface, and calculating the surface direction of each target surface point by using a photometric stereo vision method to realize the reconstruction of the target surface; thirdly, by using a straight line feature sensing grouping method, assuming the shape of a known target surface plane region, carrying out specific regular grouping on the extracted straight line features, such as contour grouping aiming at a rectangular plane region, generating a plurality of plane hypotheses, and further carrying out verification on the plane hypotheses according to images or other prior knowledge; fourthly, matching the extracted linear features, calculating corresponding three-dimensional linear through forward intersection, performing coplanar analysis on the three-dimensional linear to generate all possible plane hypotheses and construct plane boundaries to form closed plane areas, and verifying the closed areas by using various evidences. In contrast, the first method has the disadvantage that as the density of the elevation data increases, the acquisition cost increases simultaneously, and for a structurally complex object, it is difficult to extract all planes of the object surface if the number of planes or the approximate type of the object structure is missing as a priori knowledge. The second method has the disadvantages that a plurality of illumination conditions need to be accurately known, so that the method is mostly used for surface reconstruction of a small-volume target in a pure illumination environment, and the method cannot achieve satisfactory results for various uncooperative conditions such as complicated illumination conditions, various backgrounds or differences of reflection characteristics of the target surface. The third method has the disadvantage that the geometry of the extractable plane is limited. Compared with the first three methods, the fourth method has stronger adaptability and small dependence on elevation data, even does not need the support of the elevation data, and simultaneously adopts a bottom-up extraction process to solve the problem of three-dimensional plane extraction of various polygonal shapes.

In the three-dimensional plane extraction process, three-dimensional plane information required to be obtained comprises two aspects: the three-dimensional plane positioning information is expressed as a plane equation of an open plane where the three-dimensional plane positioning information is located; and the second is three-dimensional plane boundary information which is expressed as a complete and accurate boundary line of the closed plane. In the existing fourth three-dimensional plane extraction technology, the two kinds of information are extracted as one task to be completed simultaneously, the positioning accuracy of the plane is affected by the boundary positioning error, and the boundary positioning accuracy obtained by the intersection of the plane is reduced due to the positioning parameter error of the plane.

Disclosure of Invention

The invention aims to solve the technical problem that in the process of extracting the three-dimensional plane, the positioning of the three-dimensional plane and the extraction of the plane area boundary are separated into two stages which are connected in front and back, the two tasks are respectively and independently completed, the error influence between the two tasks is reduced, and the results of the two stages are respectively optimal, so that the optimization of the three-dimensional plane extraction is realized.

The input of the technical scheme of the invention is three-dimensional linear feature set and sparse digital elevation model data in a target scene obtained by any method, three-dimensional images of the target scene obtained from two different visual angles, all internal and external parameters of a three-dimensional camera for obtaining the three-dimensional images and linear extraction results of the three-dimensional images, the output of the technical scheme is a three-dimensional plane with high positioning precision and good boundary integrity and accuracy, and the implementation technical scheme aims to lay a foundation for the subsequent surface reconstruction work of a polyhedral structure target.

The idea of the invention is as follows: dividing a three-dimensional plane extraction process into two optimization steps which are connected in sequence, wherein the first step is to acquire high-precision positioning information of an open three-dimensional plane where a target surface is located, the implementation method is a segmentation hypothesis confirmation method, and the specific process is to generate an open three-dimensional plane hypothesis by using three-dimensional linear characteristics → use member linear characteristics of the three-dimensional plane hypothesis to segment the open three-dimensional plane hypothesis to form a plurality of plane block hypotheses → select an optimal plane block hypothesis → combine the optimal plane block hypotheses to generate the open three-dimensional plane with high positioning precision; the second step is to construct the boundary of a plane, construct a complete boundary for the open plane by taking the plane intersecting line acquired by plane intersection, the linear feature extracted from the image and the heuristic knowledge related to the target structure as the basis for generating the plane boundary based on the generated open three-dimensional plane, and the specific process is to construct the boundary of the plane by using the member linear feature of the open plane and the intersecting line between the planes → to supplement the missing plane boundary by using the linear feature extracted from the stereo image and the heuristic rule → to generate the closed plane hypothesis from the plane boundary, and select the optimal closed plane hypothesis by using an optimization method.

The technical scheme of the invention is a three-dimensional plane extraction method, which specifically comprises the following steps:

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 From the two imagesThe linear feature set extracted in (1) is 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

First step, three-dimensional plane hypothesis generation

Setting an angle threshold phi, and if the included angle of the two three-dimensional straight lines is smaller than phi, judging that the direction vectors of the two three-dimensional straight lines are approximately consistent; and setting a distance threshold epsilon, and if the distance from a three-dimensional space point to a three-dimensional straight line or a three-dimensional plane is less than epsilon, judging that the three-dimensional space point is on the three-dimensional straight line or the three-dimensional plane. The values of the two thresholds are determined according to the positioning accuracy of the known three-dimensional straight line characteristics.

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

Their values are related to the structural characteristics of the target surface.

Step 1, calculating three-dimensional plane hypothesis

Combining the three-dimensional straight line characteristics in the set S pairwise to generate a three-dimensional plane hypothesis, and specifically comprising the following steps of:

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.

All three-dimensional plane assumption sets Ω ═ P obtained according to the

above steps

_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_iIs a member line feature of (1).

Step 2, merging coplanar three-dimensional plane assumptions

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 Ω is taken 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

Any two three-dimensional planes in the subset are assumed to be coplanar.

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 sideThe distance is marked 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).

The set of three-dimensional plane assumptions from all maxima cliques is denoted as Ω ═ P_w|w＝1，2，…, N }, wherein N ═ N_QAny 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 (Z)_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_iIt is segmented to get several plane block hypotheses. The method comprises the following steps:

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 satisfiesThen 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_iMiddle and other members of the straight lineIntersection of features, 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 graph G_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 ≠Beta, calculation of

<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> <mn>1</mn> </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> <mn>1</mn> </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> <mn>1</mn> </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> <mn>1</mn> </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，i ¹，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，Z_α，2 ⁱ) And (X)₁，Y₁，Z₁) In the figure G_iTo the 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_iTo the corresponding node in (1). If (X)₅，Y₅，Z₅) If present, then according to (X)₄，Y₄，Z₄) The same method is used for processing.

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 is true.

Step 4), dividing to generate the plane block hypothesis

Search undirected graph G_iIf the number of the nodes which belong to the same member straight line feature or the extension line thereof in any one ring is not more than 2, the ring is used for generating a plane block hypothesis, and all the nodes contained in the ring form an ordered point set according to the connection relationship of the nodes, and the ordered point set represents the vertex set of the plane block hypothesis.

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

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 JProjected 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> <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 plane

The 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.

For any elevation point

If projection in its left imageIs 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 determinedQuilt

And (6) 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

Number of covered elevation points alpha_wIf > 0, then calculate

Measure of elevation coherence 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 toThree 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) calculatingMeasure of reliability of

Measure of reliability of

(formula three)

Step 3, search for reliable plane block hypothesis

The method is completed in two steps:

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,

Then use itGenerating 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 their corresponding planar blocks in the left and right images do not have overlapping parts, they are connected.

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 metric.

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 K₁If any of the plane blocks does not have overlapping parts in the projection areas of the left and right images, the method is used

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 their corresponding planar blocks in the left and right images do not have overlapping parts, they are connected.

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 of the 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.

Step 4, merging the coplanar reliable plane block hypotheses

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) search for coplanar reliable plane block hypotheses

Generating an undirected graph

By K₁∪K₂Each reliable plane block hypothesis generates a node, and if any two reliable plane block hypotheses are coplanar, their corresponding nodes are connected.

Calculation chart

All very big groups of (1), noted

Any one of the extremely large groups

Is shown in the figureA 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>

Any two reliable flat blocks in the subset are assumed to be coplanar.

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

<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>

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> <mo>+</mo> <msub> <mover> <mover> <mi>c</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>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).

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>

Wherein,

<math> <mrow> <mover> <mover> <mi>N</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mo>=</mo> <msub> <mover> <mi>N</mi> <mo>^</mo> </mover> <mi>Q</mi> </msub> <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

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 as

And

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>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 of all plane intersections is denoted 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 ofDeletion 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 planeAny 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 featuresIn (1).

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, it is computed from any other open planes using the same method as step 1

<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 withIs smaller than phi, and

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

And from the set

The collinear member line feature is deleted.

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 planeAnd

the intersection of the planes is added to

Andmember straight line feature set of

And

and (4) removing.

The end points of the newly added member line feature are now undetermined.

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 openingNoodle

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

Fall within the range of any other member linear feature

The two membership linear features are determined to intersect.

Step 7, constructing plane boundary by member straight line characteristic of open plane

The method comprises the following steps:

step 1), splitting all member linear characteristics of the open plane into two subsets

For any one open planeSplitting 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> <msubsup> <mover> <mover> <mi>M</mi> <mo>&OverBar;</mo> </mover> <mo>&OverBar;</mo> </mover> <mi>u</mi> <mo>′</mo> </msubsup> <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 U₀And generating an undirected graph. The undirected graph is represented by U₀Each member straight line feature is used as a node, and if the member straight line features corresponding to any two nodes are intersected, the member straight line features are connected.

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.

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

If it is not

Represents P'_kIs a closed plane assumption, noThen represents P'_kIs a semi-open plane assumption. Any one closed or semi-open plane is assumed to be P'_kIs the same as the open plane from which it was generated, and is rewritten to a 'to correspond to Ω'_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 Ω', if it is semi-open, it is supplemented with missing boundaries, first by extracting the 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. The method comprises the following steps:

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'_kIn the area where the missing boundary of (2) is projected in the left and right images

Projection P'_kTo the left and right images, respectively obtaining point sets

And

setting a distance threshold h representing the integrity of the linear feature_dIt means that if a straight line feature is not extracted completely in an image, the distance from its true end point to the end point from which it was extracted is less than h_d。

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

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

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 ¹To point g on the extension line of₁Is 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 ^RAnd its four vertexes are respectively marked as g'₁、g′₂、g′₃And g'₄。

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 vertices except for one generate one closed plane hypothesis. For H₂Any straight line feature if its two end points fall on R_k ^RAnd its extension line and line segment g'₂g′₃And g'₄g′₁Intersecting, calculating the homonymy point of the end point in the left image, wherein the homonymy point is calculated by the same method as the method for calculating the homonymy point in the step 3) of the step 2 of the second step in the first step, connecting the two homonymy points into a straight line segment, and calculating the extension line of the straight line feature and the line segment g 'in the right image if the straight line segment is not collinear with any missing boundary searched in the left image'₂g′₃And g'₄g′₁Is connected to form 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 vertices except for one generate one 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, with P_k' two adjacent planes are assumed to have 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 phi₁Φ₂Becoming a supplemental missing boundary.

If respectively with T_k ²T_k ¹And

is a common boundary and P'_kOnly one of the two adjacent plane hypotheses, assumed to be P ', has obtained a complete boundary'_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 segmentAnd T_k ²Φ₁Equal length, line phi₁Φ₂Becoming a supplemental missing boundary.

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

all vertices except for one generate one closed plane hypothesis.

The common boundary here refers to a boundary line that two adjacent polygon planes share.

The closed plane hypotheses in Ω 'and the closed plane hypotheses generated by the semi-open plane hypotheses in Ω' are merged into a closed plane hypothesis set, which is denoted as Ω ″ { P ″_iI | (1, 2, …, N ″), where any closed plane assumes P ″_iFrom n ″)_iThe boundary lines of the edges.

Step three, selecting a global optimal closed plane hypothesis

Step 1, calculating reliability measure of closed plane hypothesis

Assume P "for any one closed plane_iBelongs to omega', and utilizes a formula I to calculate P ″)_iIf P ″ ", is the measure of elevation coherence_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 recorded asE″_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 side is specified, 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 specified to be 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 jth boundary line is recorded as E ″_i，jThe closed plane hypothesis P "is calculated according to the following formula_iThe reliability measure of the boundary corresponding to the real plane boundary is

(formula four)

Step 2, solving the optimal closed plane hypothesis

And generating an undirected graph, marked as G ', setting each closed plane hypothesis in the set omega' as a node, using the reliability measure calculated by the formula IV as the attribute of the node corresponding to the closed plane hypothesis, if the projection areas of the closed plane hypothesis corresponding to any two nodes in the left image and the right image do not have overlapping parts, connecting the two nodes, otherwise, not connecting the two nodes. And calculating all the maximum cliques of the graph G' and selecting all the closed plane hypotheses contained in the maximum clique with the maximum sum of the node attribute values as the optimal closed plane hypothesis, namely the three-dimensional plane extraction result.

Drawings

FIG. 1 is a schematic diagram of a flow of a three-dimensional plane extraction method according to the present invention;

FIG. 2 is a schematic illustration of computing an elevation conformity measure for a planar patch hypothesis;

fig. 3 is a schematic diagram of supplementing the missing boundary of the semi-open plane hypothesis according to a heuristic rule.

Detailed Description

The invention is further explained below with reference to the drawings.

FIG. 1 is a schematic diagram of a flow of a three-dimensional plane extraction method according to the method of the present invention; the technical scheme comprises the following steps: the method comprises the following steps of firstly, solving an open three-dimensional plane where a target surface is located, wherein the step is divided into two steps, namely, the step (I) and the step (II) of generating a three-dimensional plane hypothesis, the step is divided into two sub-steps, namely, the step 1 of calculating the three-dimensional plane hypothesis, the step 2 of combining coplanar three-dimensional plane hypotheses, the step (II) of verifying the three-dimensional plane hypothesis, the step is divided into four sub-steps, namely, the step 1 of dividing the three-dimensional plane hypothesis into plane block hypotheses, the step 2 of calculating the reliability measure of each plane block hypothesis, the step 3 of searching the most reliable plane block hypothesis, the step 4 of combining the coplanar reliable plane block hypotheses; second step, construct the boundary of the plane, this step is divided into three steps again, the first step, utilize member's straight line characteristic and plane intersect line structure plane boundary hypothesis of open plane, this step is divided into seven substeps again, the step 1, produce the set of plane intersect line, the step 2, judge the collinear relation and revise the member's straight line characteristic set of open plane between member's straight line characteristic of plane intersect line and plane, the step 3, supplement the open plane missing, the step 4, calculate the plane intersect line to the open plane supplemented, the step 5, add the member's straight line characteristic of the plane intersect line characteristic of open plane, the step 6, judge whether any plane intersect line and other member's straight line characteristic in the member's straight line characteristic set of open plane, the step 7, construct the plane boundary by the member's straight line characteristic of open plane, the step (second) utilize and withdraw the missing boundary of the semi-open plane hypothesis of heuristic rule, the method comprises the following steps of 1, supplementing the missing boundary of the semi-open plane hypothesis by using the linear features extracted from the image, 2, supplementing the missing boundary of the semi-open plane hypothesis according to heuristic rules, and (third), selecting a globally optimal closed plane hypothesis, wherein the method comprises the following steps of 1, calculating the reliability measure of the closed plane hypothesis, and 2, solving the optimal closed plane hypothesis.

FIG. 2 is a schematic diagram of the calculation of the elevation consistency measure of the planar patch hypothesis in step 2 of the second step of the first step of the method of the present invention: o is₁And O₂Is the center and elevation point of the left and right cameras

And O₁And O₂The connecting lines of (A) respectively intersect the plane blocks at points F₁And F₂. The method calculates elevation points respectively

And point F₁、F₂If the two distances are smaller, the distance between the hypothesis and the elevation point of the plane block is represented

The better the consistency.

Fig. 3 is a schematic diagram of step 2 of the second step of the method of the present invention, which uses heuristic rules to supplement the missing boundary of the semi-open plane hypothesis: assuming a semi-open plane hypothesis with two known boundaries, in FIG. 3a), A and B are both closed plane hypotheses with complete boundaries, C is a semi-open plane hypothesis, T_k ²T_k ¹Is a common boundary line of A and C, T_k ²T_k ³Is a common boundary line of B and C, and the dotted line is a missing boundary line phi which supplements the plane C according to an heuristic rule that the common boundary line is equal in length₁Φ₂(ii) a In fig. 3B), unlike in fig. 3a), B is also a semi-open plane, and first, one end Φ of the missing boundary line of C is specified by using a rule that the common boundary line of a and C is equal in length₁Then, the other end phi of the missing boundary line is specified according to the rule that the common boundary line of B and C is equal to the common boundary line of A and C₂And connecting them to generate a missing boundary line of the semi-open plane hypothesis C.

Details of the present invention will be described below.

First, calculating the projection of three-dimensional space points on a three-dimensional plane

The steps of projecting three-dimensional space points to a three-dimensional plane all adopt the following methods, and the specific contents are as follows:

assume a three-dimensional space point of (X)₀，Y₀，Z₀) If the equation of the three-dimensional plane is a · X + b · Y + c · Z + d is 0, then the projection of the three-dimensional space point on this three-dimensional plane (X + b · Y + c · Z + d) is defined as₀′，Y₀′，Z₀') wherein

Second point, calculating the projection of three-dimensional space point on left and right images

The steps of projecting three-dimensional space points to known left and right images in the invention all adopt the following methods, and the specific contents are as follows:

according to the photogrammetry theory, a collinear equation set related to each image can be obtained by using known imaging parameters of the left image and the right image, and projections of three-dimensional space points in the left image and the right image can be calculated according to the collinear equations.

Methods for deriving the set of collinearity equations from the image imaging parameters are described in detail in Wang Zhizhuo, principles photo grammetry with remote sensing, Press of Wuhan Technical University of Surveing and Mapping, 1990, 18-26.

A third point, calculating the projection of the three-dimensional straight line segment on the three-dimensional plane

The step of projecting the three-dimensional straight line segment to the three-dimensional plane adopts the following method, and the specific content is as follows:

according to the method for projecting the three-dimensional space point to the three-dimensional plane, two endpoints of the straight line segment are respectively projected to the three-dimensional plane, the two projection points obtained by connection are a straight line segment, and the straight line segment is the projection required.

The fourth point, calculating the projection of the three-dimensional straight line segment on the left and right images

The steps of projecting three-dimensional straight-line segments to known left and right images in the invention all adopt the following methods, and the specific contents are as follows:

according to the method for projecting the three-dimensional straight-line segment to the left image and the right image, two endpoints of the straight-line segment are respectively projected to the left image and the right image, two projection points obtained by connecting each image are a straight-line segment, and the two straight-line segments are the required projections.

Fifthly, calculating a plane equation by a gradient descent method

In the 2 nd step of the first step of the present invention, the calculation of L is performed by the gradient descent method_uAnd L_vThe specific content of the method of the designated plane equation is as follows:

two three-dimensional straight-line features L for generating planes are known_uE.g. S and L_vBelongs to S, and the endpoints are respectively (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) Calculate from themThe method for determining the plane equation comprises the following three steps:

step 1) solving initial values of plane equation coefficients

Calculation point (X)₀，Y₀，Z₀) And two magnitudes D_u、D_vWherein

\{\begin{matrix} X_{0} = (X_{u 1} + X_{u 2} + X_{v 1} + X_{v 2}) / 4 \\ Y_{0} = (Y_{u 1} + Y_{u 2} + Y_{v 1} + Y_{v 2}) / 4 \\ Z_{0} = (Z_{u 1} + Z_{u 2} + Z_{v 1} + Z_{v 2}) / 4 \end{matrix}

D_{u} = \sqrt{{(X_{u 1} - X_{u 2})}^{2} + {(Y_{u 1} - Y_{u 2})}^{2} + {(Z_{u 1} - Z_{u 2})}^{2}}

D_{v} = \sqrt{{(X_{v 1} - X_{v 2})}^{2} + {(Y_{v 1} - Y_{v 2})}^{2} + {(Z_{v 1} - Z_{v 2})}^{2}}

To obtain L_uAnd L_vNormalized direction vector of

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)。

Calculating χ ═ - (α · X)₀+β·Y₀+λ·Z₀) Obtaining a four-dimensional vector (alpha beta gamma chi); calculating a three-dimensional vector (α ' β ' γ ') (v ═ v-_u+v_v) X (α β γ), and χ '(. α'. X) was calculated₀+β′·Y₀+γ′·Z₀) Another four-dimensional vector (α 'β' γ 'χ') is obtained.

According to an objective function

If f (alpha beta gamma chi) > f (alpha ' beta ' gamma chi '), the initial value of the plane equation coefficientIs (α 'β' γ 'χ'), otherwise is (α β γ χ).

Step 2), calculating plane equation coefficients by using a gradient descent method

The objective function is f (a, b, c, d), and the initial value of the plane equation coefficient is

The iterative convergence condition is

And calculating the optimal plane equation coefficient by using a gradient descent method.

The calculation process of the gradient descent method is described in optimization principles, methods and solving software written by the great boy of the plain, Beijing: scientific Press, 2006, pages 23-31.

Step 3), obtaining a plane hypothesis

Calculating I ═ f (a)^(k)，b^(k)，c^(k)，d^(k)) If I > 4 ε²If so, generating no plane hypothesis; otherwise, the plane hypothesis equation is obtained as a^(k)·X+b^(k)·Y+c^(k)·Z+d^(k)＝0。

Sixth, fast solving method for maximum clique

All maximal cliques in the invention adopt a rapid maximal clique solving method, which is specifically described in Tomita E, Tanaka a, Takahashia h, the work-time complex for generating an aggregate clique and a functional entity, the theoretical Computer Science, 2006, 363: 28-42.

Seventh point, according to

Three pairs of homonymous points in left and right images

And

andand

and

computing a set K_w ^LAny point in the same name point in the right image

In step 3) of step 2 of the second (second) step of the first step of the present invention, the use of

Three pairs of homonymous points in left and right images

And

and

and

and

calculating the homonymous point of any point in the left projection polygon in the right image, wherein the specific content is as follows:

generating vector V and matrix M as

V = [\begin{matrix} \hat{x} r_{w 1} \\ \hat{x} r_{w 2} \\ \hat{x} r_{w 3} \\ \hat{y} r_{w 1} \\ \hat{y} r_{w 2} \\ \hat{y} r_{w 3} \end{matrix}],

M = [\begin{matrix} \hat{x} l_{w 1} & \hat{y} l_{w 1} & 1 & 0 & 0 & 0 \\ \hat{x} l_{w 2} & \hat{y} l_{w 2} & 1 & 0 & 0 & 0 \\ \hat{x} l_{w 2} & \hat{y} l_{w 3} & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & \hat{x} l_{w 1} & \hat{y} l_{w 1} & 1 \\ 0 & 0 & 0 & \hat{x} l_{w 2} & \hat{y} l_{w 2} & 1 \\ 0 & 0 & 0 & \hat{x} l_{w 2} & \hat{y} l_{w 3} & 1 \end{matrix}]

Computing

Transform coefficients between homonymous points on a plane

[a₁₁ a₁₂ a₁₃ a₂₁ a₂₂ a₂₃]^T＝M^-1V

Then

At any point within the projection region in the left image

The same name point in the right image is

Eighth, K is calculated in step 3) of step 2 of the second (second) step of the first step of the present invention_w ^RThe interpolation algorithm of the gray value of the image point in the image adopts a bilinear interpolation algorithm.

Ninth point, method for generating plane hypothesis from single three-dimensional straight line

In step 3 of the second (first) step of the present invention, the method for generating a three-dimensional plane hypothesis using a single three-dimensional straight line feature refers to a method for extracting a three-dimensional plane and reconstructing a building using a multi-view image, namely Baillard C, Zisserman A.A plane-sweep strategy for the 3D reconstruction of buildings from multiple images, isps Journal of photographic mapping and RemoteSensing, 2000, 33 (B2): 56-62.

Tenth, method for searching Hamilton circle and Hamilton path in undirected graph

In step 7 of the second step (first) of the present invention, the method for searching Hamilton circle and Hamilton path in undirected graph refers to "graph theory guidance", beijing: people post-post press, Gary Chartrand, Ping Zhang 2007, page 122-.

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;

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;

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;

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}},

Step 2, calculating the reliability measure of each plane block hypothesis

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>

Has the plane equation of

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

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

<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;

For any open plane

Selecting it from U and

any open plane of intersection of its if any member is a straight line feature

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

Step 3, supplementing missing open planes

For any one open plane

Any one member of the straight line characteristics

adding the plane to the set as its member line features

Performing the following steps;

For any one open plane

If it is generated from a single three-dimensional straight line, calculate it from any other open planes

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;

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;

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;

Is a'_k·X+b′_k·Y+c′_k·Z+d′_k＝0；

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:

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

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 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;

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:

will be provided with

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，

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，

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；

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);

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:

Selecting any member line feature

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

And l ≠ α, l ≠ β, calculating

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

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

Two polygonal projection areas of (a);

for any elevation point

If projection in its left image

Is located at

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

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

(formula two)

Wherein,

<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:

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;

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;

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

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

Wherein:

<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

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:

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

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), 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′；

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;

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;

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).