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A Generalized Projective Reconstruction Theorem and Depth Constraints for Projective Factorization

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Abstract

This paper presents a generalized version of the classic projective reconstruction theorem which helps to choose or assess depth constraints for projective depth estimation algorithms. The theorem shows that projective reconstruction is possible under a much weaker constraint than requiring all estimated projective depths to be nonzero. This result enables us to present classes of depth constraints under which any reconstruction of cameras and points projecting into given image points is projectively equivalent to the true camera-point configuration. It also completely specifies the possible wrong configurations allowed by other constraints. We demonstrate the application of the theorem by analysing several constraints used in the literature, as well as presenting new constraints with desirable properties. We mention some of the implications of our results on iterative depth estimation algorithms and projective reconstruction via rank minimization. Our theory is verified by running experiments on both synthetic and real data.

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Notes

  1. Actually, in Dai et al. (2010, 2013) the constraint is given as imposing strictly positive depths: \(\hat{\lambda }_{ij} > 0\), giving a non-closed constraint space. However, what can be implemented in practice using semi-definite programming or other iterative methods is non-strict inequalities \(\hat{\lambda }_{ij} \ge 0\).

  2. Throughout the paper we follow the convention that estimated depths, camera matrices and points are denoted using the hatted quantities such as \(\hat{\lambda }_{ij}\), \(\hat{\mathtt {P}}_i\) and \(\hat{\mathbf {X}}_j\), while the ground truth is shown using unhatted symbols like \(\lambda _{ij}\), \(\mathtt {P}_i\) and \(\mathbf {X}_j\).

  3. This term has been borrowed from Sinkhorn (1967).

  4. For simplicity of notation, we are being a bit sloppy here about the projective entities like projective lines, quadric surfaces and twisted cubics. The reader must understand that when talking about a point \(\mathbf {X}\in \mathbb {R}^4\) lying on a projective entity, what we really mean is that the projective point in \(\mathbb {P}^3\) represented by \(\mathbf {X}\) in homogeneous coordinates lies on them.

  5. According to (G3) the \(n-1\) points \(\mathbf {X}_j\) corresponding to nonzero zero vectors \((\hat{\lambda }_{kj}, \hat{\lambda }_{lj})\) and the camera centre of \(\mathtt {P}_l\) do not all lie on a twisted cubic. This is a generic property as \(n-1 \ge 6\) (see Sect. 4.1). Notice that here the matrices \(\mathtt {P}_l\) and \(\mathtt {H}\mathtt {P}_k\) respectively act as \(\mathtt {Q}\) and \(\hat{\mathtt {Q}}\) in Lemma 19. The genericity conditions (G1–G3) provide the conditions (C1, C2) in Lemma 19.

  6. This can be done similarly to (Hartley and Zisserman (2004), Theorem 10.1), so we do not repeat it here.

  7. In fact, the term exclusive might not be a precise term here, as (D1–D3) holding for all \(\hat{\mathtt {\Lambda }}\in C\) is just a sufficient condition for a constraint to exclude false solutions. While, according to Lemma 16, (D3) holding for all \(\hat{\mathtt {\Lambda }}\in C\) is necessary for ruling out false solutions, (D1) and (D2) holding for all members of \(C\) is not necessary for this purpose. This is because there might exist some \(\hat{\mathtt {\Lambda }}\in C\) for which (D1) or (D2) do not hold, but it is excluded by \(\hat{\mathtt {\Lambda }}\odot [\mathbf {x}_{ij}] = \hat{\mathtt {P}}\hat{\mathtt {X}}\). This is why in Sect. 5 we said that (D1–D3) are minimal in a certain sense.

  8. namely \(\{\hat{\mathtt {\Lambda }}\,|\,\hat{\mathtt {\Lambda }}= \mathrm {diag}({\varvec{\tau }}) \,\mathtt {\Lambda }\mathrm {diag}({\varvec{\nu }}),~\mathtt {M}\circ \hat{\mathtt {\Lambda }}= \mathtt {M}\}\).

  9. http://www.robots.ox.ac.uk/~vgg/data/data-mview.html.

  10. Notice that the scale of the vertical axis in Fig. 15a is different from that of Fig. 14a.

  11. The proof is quite simple: The column space of \(\mathtt {Q}\), \(\hat{\mathtt {P}}\) and \(\mathtt {A}\) must be equal and therefore we have \(\hat{\mathtt {P}}= \mathtt {Q}\mathtt {H}\) for some invertible \(4\times 4\) matrix \(\mathtt {H}\). Similarly, we can argue that \(\hat{\mathtt {X}}= \mathtt {G}\mathtt {Y}\) for some invertible \(\mathtt {G}\). Therefore, we have \( \mathtt {Q}\,\mathtt {Y}= \mathtt {Q}\,\mathtt {H}\,\mathtt {G}\, \mathtt {Y}\). As \(\mathtt {Q}\) has full column rank and \(\mathtt {Y}\) has full row rank, the above implies \(\mathtt {H}\,\mathtt {G}= \mathtt {I}\) and hence, \(\mathtt {G}= \mathtt {H}^{-1}\).

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Authors

Corresponding author

Correspondence to Behrooz Nasihatkon.

Additional information

Communicated by Long Quan.

NICTA is funded by the Australian Government as represented by the Department of Broadband, Communications and the Digital Economy and the Australian Research Council through the ICT Centre of Excellence program.

Appendices

Appendix 1: The Triangulation Problem

Triangulation is the process of determining the location of a 3D point given its images in two or more cameras with known camera matrices. The following lemma states that the solution to triangulation is unique in generic cases:

Lemma 18

(Triangulation) Consider two full-row-rank camera matrices \(\mathtt {P}_1,\mathtt {P}_2 \in \mathbb {R}^{3\times 4}\), two points \(\mathbf {X},\mathbf {Y}\in \mathbb {R}^4\), and scalars \(\hat{\lambda }_1\) and \(\hat{\lambda }_2\) such that the vector \((\hat{\lambda }_1,\, \hat{\lambda }_2)\) is nonzero, for which the relations

$$\begin{aligned} \mathtt {P}_1\mathbf {Y}&= \hat{\lambda }_1 \mathtt {P}_1 \mathbf {X}\end{aligned}$$
(58)
$$\begin{aligned} \mathtt {P}_2\mathbf {Y}&= \hat{\lambda }_2 \mathtt {P}_2 \mathbf {X}\end{aligned}$$
(59)

hold. Take nonzero vectors \(\mathbf {C}_1 \in \fancyscript{N}(\mathtt {P}_1)\) and \(\mathbf {C}_2 \in \fancyscript{N}(\mathtt {P}_2)\). If the three vectors \(\mathbf {C}_1\), \(\mathbf {C}_2\) and \(\mathbf {X}\) are linearly independent, then \(\mathbf {Y}\) is equal to \(\mathbf {X}\) up to a nonzero scaling factor.

Notice that the condition of \(\mathbf {C}_1\), \(\mathbf {C}_2\) and \(\mathbf {X}\) being linearly independent means that the two camera centres are distinct and \(\mathbf {X}\) does not lie on the projective line joining them (see footnote 4). A geometric proof of this is given in (Hartley and Zisserman (2004), Theorem 10.1). Here, we give an algebraic proof as one might argue that Hartley and Zisserman (2004) has used projective equality relations which cannot be fully translated to our affine space equations since we do not assume that \(\hat{\lambda }_1\) and \(\hat{\lambda }_2\) are both nonzero in (58) and (59).

Proof

Since \(\mathtt {P}_1\) and \(\mathtt {P}_2\) have full row rank they have a 1D null space. Thus, relations (58) and (59) respectively imply

$$\begin{aligned} \mathbf {Y}&= \alpha _1 \mathbf {C}_1 + \hat{\lambda }_1\mathbf {X}, \end{aligned}$$
(60)
$$\begin{aligned} \mathbf {Y}&= \alpha _2 \mathbf {C}_2 + \hat{\lambda }_2\mathbf {X}, \end{aligned}$$
(61)

for some scalars \(\alpha _1\) and \(\alpha _2\). These give \( \alpha _1 \mathbf {C}_1 + \hat{\lambda }_1\mathbf {X}= \alpha _2 \mathbf {C}_2 + \hat{\lambda }_2\mathbf {X}\) or

$$\begin{aligned} \alpha _1 \mathbf {C}_1 - \alpha _2 \mathbf {C}_2 + (\hat{\lambda }_1 - \hat{\lambda }_2) \mathbf {X}=0 \end{aligned}$$
(62)

As the three vectors \(\mathbf {C}_1\), \(\mathbf {C}_2\) and \(\mathbf {X}\) are linearly independent, (62) implies that \(\alpha _1 = 0\), \(\alpha _2 = 0\) and \(\hat{\lambda }_1 = \hat{\lambda }_2\). Define \(\nu \overset{\underset{\mathrm {def}}{}}{=}\hat{\lambda }_1 = \hat{\lambda }_2\). Then, from (60) we have \(\mathbf {Y}= \nu \mathbf {X}\). Moreover, \(\nu \) must be nonzero as the lemma assumes that the vector \((\hat{\lambda }_1,\hat{\lambda }_2) = (\nu ,\nu )\) is nonzero. \(\square \)

Appendix 2: The Camera Resectioning Problem

Camera resectioning is the task of computing camera parameters given the 3D points and their images. It can be shown that with sufficient 3D points in general locations, the camera matrix can be uniquely determined up to scale (Hartley and Zisserman 2004). Here, we consider a slightly revised version of this problem, which fits our case where the estimated projective depths are not assumed to be all nonzero and the second (estimated) set of camera matrices need not be assumed to have full rank.

Lemma 19

(Resectioning) Consider a \(3\times 4\) matrix \(\mathtt {Q}\) of rank 3 and a set of points \(\mathbf {X}_1, \mathbf {X}_2, \ldots , \mathbf {X}_p\) such that for a nonzero vector \(\mathbf {C}\in \fancyscript{N}(\mathtt {Q})\) we have

(C1):

Any four vectors among \(\mathbf {C}, \mathbf {X}_1, \mathbf {X}_2, \ldots , \mathbf {X}_p\) are linearly independent, and

(C2):

the set of points \(\{\mathbf {C}, \mathbf {X}_1, \mathbf {X}_2, \ldots , \mathbf {X}_n\}\) do not lie on a twisted cubic (see footnote 4) or any of the degenerate critical sets resulting in a resection ambiguity (set out in Hartley and Zisserman 2004, Sect.22.1).

Now, for any \(\hat{\mathtt {Q}}\in \mathbb {R}^{3\times 4}\) if we have

$$\begin{aligned} \alpha _j \mathtt {Q}\mathbf {X}_j = \beta _j \hat{\mathtt {Q}}\mathbf {X}_j \end{aligned}$$
(63)

for all \(j = 1,2,\ldots , p\) where scalars \(\alpha _j\) and \(\beta _j\) are such that the vector \((\alpha _j, \beta _j)\) is nonzero for all \(j\), then \(\hat{\mathtt {Q}}= a \mathtt {Q}\) for some scalar \(a\).

Proof

First, since 6 points in general position completely specify a twisted cubic (Semple and Kneebone 1952), (C2) implies that \(p+1 \ge 7\), or \(p \ge 6\).

If \(\hat{\mathtt {Q}}= \mathtt {0}\), then \(\hat{\mathtt {Q}}= a \mathtt {Q}\) with \(a=0\), proving the claim of the lemma. Thus, in what follows we only consider the case of \(\hat{\mathtt {Q}}\ne \mathtt {0}\).

By (C1), for all \(j\) we have \(\mathtt {Q}\mathbf {X}_j \ne \mathbf {0}\). Therefore, \(\beta _j \ne 0\), as otherwise if \(\beta _j = 0\) from \((\alpha _j, \beta _j)^T \ne \mathbf {0}\) we would have \(\alpha \ne 0\) and therefore \(\mathbf {0}= \beta _j \hat{\mathtt {Q}}\mathbf {X}_j = \alpha _j \mathtt {Q}\mathbf {X}_j \ne \mathbf {0}\), which is a contradiction. From \(\beta _j \ne 0\) and (63) it follows that if \(\alpha _j = 0\) for some \(j\), then \(\mathbf {X}_j \in \fancyscript{N}(\hat{\mathtt {Q}})\). Now, if for 4 indices \(j\) we have \(\alpha _j = 0\), from (C1) it follows that \(\hat{\mathtt {Q}}\) has a 4D null space, or equivalently \(\hat{\mathtt {Q}}= \mathtt {0}\). Since we excluded this case, we conclude that there are less than 4 zero-valued \(\alpha _j\)-s. As \(p \ge 6\), it follows that there are at least three nonzero \(\alpha _j\)-s, namely \(\alpha _{j_1}\), \(\alpha _{j_2}\) and \(\alpha _{j_3}\). Since \(\beta _j\)-s are all nonzero, \(\alpha _j \ne 0\) along with (63) implies that \(\mathtt {Q}\mathbf {X}_j\) is in \(\fancyscript{C}(\hat{\mathtt {Q}})\), the column space of \(\hat{\mathtt {Q}}\). Therefore, we have \(\mathrm {span}(\mathtt {Q}\mathbf {X}_{j_1}, \mathtt {Q}\mathbf {X}_{j_2}, \mathtt {Q}\mathbf {X}_{j_3}) \subseteq \fancyscript{C}(\hat{\mathtt {Q}})\). From (C1) we know that \(\mathrm {span}( \mathbf {X}_{j_1}, \mathbf {X}_{j_2}, \mathbf {X}_{j_3})\) is 3-dimensional and does not contain the null space of \(\mathtt {Q}\). Therefore, \(\mathrm {span}(\mathtt {Q}\mathbf {X}_{j_1}, \mathtt {Q}\mathbf {X}_{j_2}, \mathtt {Q}\mathbf {X}_{j_3})\) is also 3-dimensional. From \(\mathrm {span}(\mathtt {Q}\mathbf {X}_{j_1}, \mathtt {Q}\mathbf {X}_{j_2}, \mathtt {Q}\mathbf {X}_{j_3}) \subseteq \fancyscript{C}(\hat{\mathtt {Q}})\) then we conclude that \(\hat{\mathtt {Q}}\) has full row rank.

As \(\mathrm {rank}(\hat{\mathtt {Q}}) = 3\), we can consider it as a proper camera matrix in multiple view geometry, talking about its camera centre represented by its null space. Therefore, for two camera matrices \(\mathtt {Q}\) and \(\hat{\mathtt {Q}}\) and all the points \(\mathbf {X}_j\) for which \(\alpha _j \ne 0\) we can apply the results of the classic camera resectioning problem: It is known that for two (up to scale) distinct camera matrices \(\mathtt {Q}\) and \(\hat{\mathtt {Q}}\) to see the points \(\mathbf {X}_j\) equally up to a possible nonzero scaling factor, the points \(\mathbf {X}_j\) and the camera centres must lie on a common twisted cubic (or possibly some other specific degenerate sets, see (Hartley and Zisserman 2004; Buchanan 1988)).

Notice that, as \(\mathrm {rank}(\hat{\mathtt {Q}}) = 3\), (C1) implies that among the points \(\mathbf {X}_j\) at most one lies on the null-space of \(\hat{\mathtt {Q}}\) and therefore, by (63) we can say that at most one \(\alpha _j\) can be zero. By possibly relabeling the points we assume that \(\alpha _1,\ldots , \alpha _{p-1}\) are all nonzero.

Now to get a contradiction, assume that there is a resection ambiguity. We consider two cases namely \(\alpha _p \ne 0\) and \(\alpha _p = 0\). If \(\alpha _p \ne 0\) then by \(\alpha _j \mathtt {Q}\mathbf {X}_j = \beta _j \hat{\mathtt {Q}}\mathbf {X}_j\) we know that \(\mathbf {X}_1,\ldots ,\mathbf {X}_p\) are viewed equally up to scale by both \(\mathtt {Q}\) and \(\hat{\mathtt {Q}}\) and thus \(\mathbf {X}_1,\ldots ,\mathbf {X}_6\) along with the camera centre of \(\mathtt {Q}\) must lie on a twisted cubic (or other degenerate sets leading to a resection ambiguity), which is impossible due to (C2). If \(\alpha _6 = 0\), implying \(\mathbf {X}_6 \in \fancyscript{N}(\hat{\mathtt {Q}})\), then again the camera center of \(\mathtt {Q}\), \(\mathbf {X}_1,\ldots ,\mathbf {X}_5\) and \(\mathbf {X}_6\) (this time as the camera centre of \(\hat{\mathtt {Q}}\)) must lie on a twisted cubic (or the degenerate sets), contradicting with (C2). Hence there can be no resection ambiguity and \(\mathtt {Q}\) and \(\hat{\mathtt {Q}}\) must be equal up to a scaling factor. \(\square \)

Appendix 3: Proof of Lemma 1

Proof of Lemma 1

We need to prove that under assumptions of Lemma 1 and the relations

$$\begin{aligned} \lambda _{ij}\mathbf {x}_{ij}&= \mathtt {P}_i \mathbf {X}_j \end{aligned}$$
(64)
$$\begin{aligned} \hat{\lambda }_{ij}\mathbf {x}_{ij}&= \hat{\mathtt {P}}_i \hat{\mathbf {X}}_j \end{aligned}$$
(65)

\((\{\mathtt {P}_i\},\{\mathbf {X}_j\})\) and \((\{\hat{\mathtt {P}}_i\},\{\hat{\mathbf {X}}_j\})\) are projectively equivalent if and only if the matrices \(\mathtt {\Lambda }\) and \(\hat{\mathtt {\Lambda }}\) are diagonally equivalent.

First, assume that \((\{\mathtt {P}_i\},\{\mathbf {X}_j\})\) and \((\{\hat{\mathtt {P}}_i\},\{\hat{\mathbf {X}}_j\})\) are projectively equivalent. Then, there exist nonzero scalars \(\tau _1, \tau _2, \ldots , \tau _m\) and \(\nu _1, \nu _2, \ldots , \nu _n\) and an invertible matrix \(\mathtt {H}\) such that (5) and (6) hold. Therefore we have

$$\begin{aligned} \hat{\lambda }_{ij} \mathtt {P}_i \mathbf {X}_j&= \hat{\lambda }_{ij} \lambda _{ij} \mathbf {x}_{ij} = \lambda _{ij} \hat{\mathtt {P}}_i \hat{\mathbf {X}}_j \\&= \lambda _{ij} \nu _j \tau _i \,\mathtt {P}_i \,\mathtt {H}\,\mathtt {H}^{-1} \,\mathbf {X}_j = \lambda _{ij} \nu _j \tau _i \,\mathtt {P}_i \,\mathbf {X}_j. \end{aligned}$$

where the first, second and third equations above hold respectively from (64), (65) and (5) and (6) together. By condition (i) in the lemma, that is \(\mathtt {P}_i \,\mathbf {X}_j \ne \mathbf {0}\), we have \(\hat{\lambda }_{ij} = \lambda _{ij} \nu _j \tau _i\) for all \(i\) and \(j\). This is equivalent to (7) and hence \(\mathtt {\Lambda }\) and \(\hat{\mathtt {\Lambda }}\) are diagonally equivalent.

To prove the other direction, assume that \(\mathtt {\Lambda }\) and \(\hat{\mathtt {\Lambda }}\) are diagonally equivalent. Then from (7) we have \(\hat{\lambda }_{ij} = \lambda _{ij} \nu _j \tau _i\). This along with (64) and (65) gives

$$\begin{aligned} \hat{\mathtt {P}}_i \hat{\mathbf {X}}_j = \hat{\lambda }_{ij} \mathbf {x}_{ij} = \lambda _{ij} \nu _j \tau _i \mathbf {x}_{ij} = \tau _i \nu _j \mathtt {P}_i \mathbf {X}_j = (\tau _i\mathtt {P}_i) (\nu _j \mathbf {X}_j) \end{aligned}$$
(66)

for \(i=1,\ldots ,m\) and \(j=1,\ldots ,n\). Let \(\mathtt {Q}_i = \tau _i\mathtt {P}_i\) and \(\mathbf {Y}_j = \nu _j \mathbf {X}_j\), so we have \(\hat{\mathtt {P}}_i \hat{\mathbf {X}}_j = \mathtt {Q}_i \mathbf {Y}_j\). Denote by \(\mathtt {Q}\) and \(\hat{\mathtt {P}}\) the vertical concatenations of \(\mathtt {Q}_i\)-s and \(\hat{\mathtt {P}}_i\)-s respectively and denote by \(\mathtt {Y}\) and \(\hat{\mathtt {X}}\) respectively the horizontal concatenations of \(\mathbf {Y}_j\)-s and \(\hat{\mathbf {X}}_j\)-s. From \(\hat{\mathtt {P}}_i \hat{\mathbf {X}}_j = \mathtt {Q}_i \mathbf {Y}_j\) we have

$$\begin{aligned} \hat{\mathtt {P}}\hat{\mathtt {X}}= \mathtt {Q}\mathtt {Y}\overset{\underset{\mathrm {def}}{}}{=}\mathtt {A}. \end{aligned}$$
(67)

From conditions (ii) and (iii) in the lemma along with the fact that \(\tau _i\) and \(\nu _j\) are nonzero, we can conclude that \(\mathtt {Q}\) has full column rank and \(\mathtt {Y}\) has full row rank. Therefore, \(\mathtt {A}\overset{\underset{\mathrm {def}}{}}{=}\mathtt {Q}\mathtt {Y}\) has rank \(4\) and the \( 3m \times 4\) and \(4 \times n\) matrices \(\hat{\mathtt {P}}\) and \(\hat{\mathtt {X}}\) must be full-column- and full-row-rank matrices respectively. As \(\mathtt {Q}\mathtt {Y}\) and \(\hat{\mathtt {P}}\hat{\mathtt {X}}\) are two rank-\(r\) factorizations of \(\mathtt {A}\), having \(\hat{\mathtt {P}}= \mathtt {Q}\, \mathtt {H}\) and \(\hat{\mathtt {X}}= \mathtt {H}^{-1} \mathtt {Y}\) for some invertible matrix \(\mathtt {H}\) is the only possibilityFootnote 11. This is the same thing as

$$\begin{aligned} \hat{\mathtt {P}}_i&= \mathtt {Q}_i \mathtt {H}= \tau _i\mathtt {P}_i\mathtt {H}\end{aligned}$$
(68)
$$\begin{aligned} \hat{\mathbf {X}}_j&= \mathtt {H}^{-1} \mathbf {Y}_j = \nu _j \mathtt {H}^{-1} \mathbf {X}_j \end{aligned}$$
(69)

Thus, \((\{\mathtt {P}_i\},\{\mathbf {X}_j\})\) and \((\{\hat{\mathtt {P}}_i\},\{\hat{\mathbf {X}}_j\})\) are projectively equivalent. \(\square \)

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Nasihatkon, B., Hartley, R. & Trumpf, J. A Generalized Projective Reconstruction Theorem and Depth Constraints for Projective Factorization. Int J Comput Vis 115, 87–114 (2015). https://doi.org/10.1007/s11263-015-0803-3

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