Pose estimation from point and line correspondences

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1 Pose estimation from point and line correspondences Giorgio Panin October 17, Problem formulation Estimate (in a LSE sense) the pose of an object from N correspondences between known object points (3D or D) X i and their noisy projections on the D image plane, x i. The pose is represented by a (3 3) or (4 4) homogeneous transformation matrix T, belonging to a given sub-group G of object-space transformations. G is a subset of all possible transforms, closed under matrix product: T 1, T G (T 1 T ) G The intrinsic camera matrix K ((3 3) or (3 4), respectively) is supposed to be known in advance. Moreover, a reference (constant) transform matrix T may be present, which is pre-multiplied by T (e.g. a fixed displacement in articulated structures, with Denavit-Hartenberg parameters), and this matrix may also not belong to the same transformation group G. Overall, this is a projective geometry problem, that can be formulated in basically two ways: using homogeneous or non-homogeneous coordinates. They raise two different LSE error measures to be optimized: the algebraic and geometric error, respectively. The latter is the real target of our estimation (Maximum-Likelihood solution), but usually non-linear, while the former gives redundant and sub-optimal equations, but usually linear. Therefore, we can use the algebraic solution as a starting-point to minimize the geometric error. 1.1 Geometric error We have the following problem: Given N exact model points X and corresponding noisy image points x in homogeneous coordinates X = (X, Y, 1) or X = (X, Y, Z, 1) x = (x, y, 1) 1

2 find the optimal transformation T belonging to a group G, such that T = min T G N π(k T T Xi ) π(x i ) where π is the nonlinear projection operator, from homogeneous to non-homogeneous coordinates π(x, y, w) = [ x/w y/w ] T and T is a constant matrix, not necessarily belonging to the same group G. Solution: nonlinear LSE optimization (Gauss-Newton), starting from an initial guess T 0, close enough to T in order to ensure convergence. All of the groups we will consider hereafter are smooth manifolds with a Lie group structure, and the local tangent space at T is a Lie algebra, which maps to the whole group through the exponential mapping. Therefore, Gauss-Newton optimization is straightforwardly performed with the Lie generators G i and the compositional update (see e.g. [1][][3]). 1. Algebraic error In homogeneous coordinates, we look for T in G that satifies the equations: T G : i, λ i : ( K T T X i ) = λi x i As we can see, the homogeneous formulation carry the projective ambiguity as coefficients (λ i ) which account for the augmented number of equations (3 instead of ) per point. We remove λ i by writing the problem as a cross-product T G : x i (K T T X i ) = 0, i which provides a redundant set of homogeneous equations in T. That means, one component of each cross product (usually the third) can be discarded from the equations above, which then will be N again. For noisy data, the problem above can be cast into an LSE form T = min T G n xi (K T T X i ) and this problem can be usually put into a linear form min p Ap or min Ap b, p where p is a vector parametrizing the transformation T, and A is a column-rank deficient matrix with 1 solutions in p. By imposing a constraint on p such as a unit-norm condition p = 1, the globally optimal solution can be found in one step, via the SVD algorithm. The resulting algorithm is called DLT (Direct Linear Transform). By resuming, we solve the original problem in two steps: 1. Estimate a T 0 matrix (hopefully close enough to T ), by minimizing the algebraic error (DLT)

3 . Starting from T 0, minimize the geometric error with the Gauss-Newton (or Levenberg-Marquardt) method, in order to obtain T. For this purpose, we use compositional updates T and Lie algebra derivatives In some cases (especially D-D or 3D-3D problems), the geometric error is linear and can be solved at once, without need for T 0. And also some nonlinear cases (e.g. the absolute orientation problem) can still be solved in one step for the geometric error. However, most 3D-D cases have an inherent nonlinearity due to the projection π(), therefore the two-step procedure cannot be generally avoided. In that case, altough step has a common formulation for all poses (apart from different Jacobians, obtained through the respective Lie generators) step 1, instead, must be solved differently for each class. D-D transforms: (3x3) matrix In a D-D problem, both X and x are given by 3 homogeneous coordinates, of which the third is usually set to 1. Moreover, we suppose all (3x3) K matrices to have the simple form: K D = 1 0 r x/ r y / 0 where (r x, r y ) are the horizontal and vertical image resolution, respectively. Therefore, all image data points x i can be pre-processed in order to remove both K and T : x i = T 1 K 1 x i = T 1 ( x i [r x /, r y /, 0] T ) and the two LSE errors are re-formulated as Algebraic: T N = min T G x i (H X i ) Geometric: T N = min T G π(h X i) π( x i ) NOTE: If the last row of T is [0, 0, 1], then the geometric error T = min T G N T X i x i has a linear form, and the problem can be directly solved in non-homogeneous coordinates. 3

4 .1 Additional symbols: I n = (n n) Identity matrix t n = (n 1) Translation vector R n = (n n) Rotation matrix: R T n R n = I n R x,y,z = (3 3) Single-axis rotation matrices s = Uniform scale factor D n = diag(s 1,..., s n ): Non-uniform scale matrix A n = (n n) Linear transformation v n = (n 1) Perspective distortion vector. PoseDTranslation ( dof, min. 1 point) The simplest case is a pure translation I t The geometric error is Figure 1: Pure translation. T = min (t x,t y) N A i t b i with A i = I b i = x i X i I that is, T = min At b with A =, b = t I b 1 b n 4

5 This is a linear LSE, solved by t = A + b; in this case, the LSE solution corresponds to the displacement of point centroids: t = 1 N N ( x i X i ) = µ x µ X.3 PoseD1ScaleTranslation (3 dof, min. point) Next, we add a uniform scale factor si t Figure : Translation and uniform scale. The geometric error results N N p(t X i ) p( x i ) = sx i + t x x i sy i + t y ȳ i = A s t x b t y with A = X Y 1 X N 1 0 Y N, b = This is again a linear LSE problem in (s, t x, t y ). x 1 ȳ 1 x N ȳ N 5

6 .4 PoseDScalesTranslation (4 dof, min. points) A non-uniform scale with translation is given by D t Figure 3: Translation and non-uniform scale. This is similar to the previous problem, but with scales (s 1, s, t x, t y ); the geometric error minimization gives: min (s 1,s,t x,t y) X Y 1 X N 0 0 Y N s 1 s t x t y.5 PoseD1ScaleRotoTranslation (4 dof, min. points) Now we consider transformations involving rotations. These are in principle nonlinear problems (because of the rotation matrix R) but fortunately, due to the nature of the problem, the LSE geometric error can still be globally optimized in one step, by using the SVD decomposition. We start by giving here the solution for the general similarity transform (uniform scale, rotation and translation), and deduce its sub-cases afterwards. sr t From the [Umeyama] paper: the LSE geometric error is optimized by the (s, R, t) parameters obatined from the following steps: x 1 ȳ 1 x N ȳ N 6

7 Figure 4: Similarity (rigid roto-translation and uniform scale). Mean vectors: µ x = 1 n x i, µ X = 1 n X i Variance of the norms: σ x = 1 n x i µ x, σ X = 1 n X i µ X Cross-covariance matrix ( ): Σ xx = 1 n ( x i µ x )(X i µ X ) T SVD of the cross-covariance: Σ xx = UDV T Sign correction for det(r): S = Rotation reconstruction: R = USV T Scale reconstruction: s = 1 σ X tr(ds) { I if det(σ xx ) 0 diag(1, 1, 1,..., 1) if det(σ xx ) < 0 Translation vector: t = µ x srµ X.6 PoseD1ScaleRotation ( dof, min. 1 point) Similar as before, but without translation: sr 0 This implies that the mean vectors are µ x = µ X = 0. Therefore we have: Variance of the norms: σ x = 1 n x i, σ X = 1 n X i Cross-covariance matrix ( ): Σ xx = 1 n x ix T i 7

8 Figure 5: Rotation and uniform scale. SVD of the cross-covariance: Σ xx = UDV T Sign correction for det(r): S = Rotation reconstruction: R = USV T Scale reconstruction: s = 1 σ x tr(ds) { I if det(σ xx ) 0 diag(1, 1, 1,..., 1) if det(σ xx ) < 0.7 PoseDRotoTranslation (3 dof, min. points) By removing the scale (s = 1), we get the Euclidean transform (rigid rototranslation): R t H = where σ x = σ X = 1. The algorithm becomes Mean vectors: µ x = 1 n x i, µ X = 1 n X i Cross-covariance matrix ( ): Σ xx = 1 n ( x i µ x )(X i µ X ) T SVD of the cross-covariance: Σ xx = UDV T Sign correction for det(r): S = Rotation reconstruction: R = USV T { I if det(σ xx ) 0 diag(1, 1, 1,..., 1) if det(σ xx ) < 0 Translation vector: t = µ x Rµ X 8

9 Figure 6: Euclidean transform (rigid roto-translation)..8 PoseDRotation (1 dof, min. 1 point) Finally, if both scale and translation are removed, we obtain the absolute orientation problem: R 0 which is solved by Figure 7: Pure rotation. Cross-covariance matrix ( ): Σ xx = 1 n x ix T i SVD of the cross-covariance: Σ xx = UDV T 9

10 { I if Sign correction for det(r): S = det(σ xx ) 0 diag(1, 1, 1,..., 1) if det(σ xx ) < 0 Rotation reconstruction: R = USV T.9 PoseDScalesRotoTranslation (5 dof, min. 3 points) The problems with non-uniform scale and rotation cannot be solved like the uniform scale cases. Since the number of degrees of freedom is close to that of an affinity (6 vs. 5 dof), we prefer to solve first for an affinity, and then upgrade to the non-uniform scale similarity, by removing 1 degree of freedom. We consider here two sub-cases: with and without translation. The first one is given by R D t Figure 8: Roto-translation with non-uniform scale. and the complete procedure is the following: 1. solve for an affine transform (see below), and find A, t. using the SVD, compute A = R(θ)R( φ)dr(φ) with D = diag(s 1, s ), and s 1 > s 3. set R = R(θ) and remove φ: if φ 0, then keep the order of (s 1, s ) in D; if φ 90 deg, then swap the scales 4. Finally, since this solution is generally not the optimal LSE, it is recommended to run a Gauss-Newton optimization in the geometric error. 10

11 .10 PoseDScalesRotation (3 dof, min. point) Next, we consider the rotation with non-uniform scale: R D H = 0 Figure 9: Rotation with non-uniform scale. This is a special case of the previous one, where t = 0. We can solve it as a purely linear (affine without translation) transform with 4 dof, then upgrading A to the non-uniform scale and rotation matrix as in the previous Section..11 PoseDAffine (6 or 4 dof, respectively 3 or points) A t Figure 10: Affine transform (linear+constant). 11

12 The affine case is again a linear LSE problem in the geometric error [A, t ] = min (A,t) with a the 6 stacked parameters N AX i + t x i = min M i a x i a = [ A 11 A 1 A 1 A t x t y ] T and M i a coefficient matrix function of X i Xi Y M i = i X i Y i By stacking together the M i matrices and the x i vectors, and solving for a, we get the LSE affine parameters. Similar equations can be written for the purely linear case (t = 0) A 0 with 4 parameters only Figure 11: Purely linear transform. a = [ A 11 A 1 A 1 A ] T M i = Xi Y i X i Y i 1

13 .1 PoseDHomography (8 dof, min. 4 points): the DLT algorithm In the most general D-D case, the matrix T can be any linear transform in the homogeneous coordinates. This is defined up to a scale factor, that we remove by setting T (3, 3) = 1: A t v T 1 Figure 1: General D homography. By applying the projection operator p, this leads to a non-linear geometric error; therefore, the pose estimation problem must be formulated in two steps as described in the introduction (algebraic and geometric error minimization). Concerning the first, after pre-processing (removing K and T from the data points x i ) we have an algebraic error of the form min T N x i (T X i ) with X i, x i R 3, T R 3 3. By writing X i = (X i, Y i, Z i ), x i = ( x i, ȳ i, z i ) we take the first two terms of the cross product, and we have, for each point i, two homogeneous equations [ 0 T z i Xi T ȳ i Xi T z i Xi T 0 T x i Xi T ] where h i are the three (transposed) rows of H = h 1 h h 3 h T 1 h T h T 3 = 0. 13

14 If A is the (n 9) stacked matrix of all l.h.s. terms above, we get the DLT equation min Ah with rank-deficient A that, as we expect, has T 1 solutions in h. By imposing h = 1, the solution is obtained by the SVD decomposition A = USV T as the last column of V (corresponding to the minimum singular value in S). In addition, the normalization technique (removing the centroids of X and x, followed by isotropic coordinate scaling) ensures a better numerical stability, and therefore we use it. 1. Removing centroids: For both point sets, we compute the mean values µ x = 1 n x i, µ X = 1 n X i and remove them from each point. Isotropic scaling: Afterwards, we make sure that the average distance from the mass center (which is 0) is equal to, i.e. that the average point of both sets is (1, 1, 1) T These two steps ultimately correspond to multiply the two point sets for two matrices T X, T x. Therefore, after estimating the transformation T the two normalizations are removed by T = T x 1 T T X 14

15 3 3D-D transforms: (4x4), K = (3x4) Here, the calibration matrix K can be more general; in particular, we consider pinhole models, without distortion and skew, and with equal focal lengths: K = [ K 0 ], K = f 0 0 f r x/ r y / 0 Moreover, we have extrinsic transformation matrices the type A3 3 t 3 0 T G 1 Ā3 3 t 3 0 T G 1 belonging to possibly different groups G, G. The base estimation problem becomes: find (A, t ) such that [ A t ] G : i, λ i : ( K [ ĀA Āt + t ] X i ) = λi x i We can pre-process the image data x i x i = (K ) 1 x i and re-write the equations [ ĀA Āt + t ] X i = λ i x i However, unlike the D-D case, in general we cannot remove neither Ā nor t, because of the non-square matrix on the left side 1. Moreover, as already mentioned in the introduction, because of the dimensionality loss (from 3D to D) the projection operator π() always provides a nonlinearity, no matter of the form for A, t. Therefore, the linear approach for the algebraic error (DLT) cannot be avoided, in order to provide an initial guess T 0 for the geometric error optimization. Finally, we need the DLT approach for all transform classes, which may have much less than the maximum number of parameters (1); therefore, we assume that each T G can be parametrized (or at least approximated) by a vector q [ A(q) t(q) ] with dim(q) = d q 1. In what follows, we will call this method generalized, projective DLT (GP-DLT). 1 The only exception is given by t = 0, in which case Ā can also be removed from the data points x i = ( K Ā) 1 xi and the problem becomes a purely projective one [ A t ] X i = P X i = λ i x i 15

16 3.1 Algebraic error for 3D-D projections: the GP-DLT approach We re-write the algebraic error in terms of the reduced parameters q as follows: q = min q R dq n xi ([ ĀA (q) Āt (q) + t ] ) X i that can be written as with and q = min q R dq n F i (q) + f i F i (q) = [x i ] Ā [ A (q) f i = [x i ] [ 03 3 [x i ] = t ] X i 0 z i y i z i 0 x i y i x i 0 t (q) ] X i the cross-product matrix. If we further impose A (q) and t (q) to be linear in (q), then we can show that F i (q) = F i q and the algebraic error becomes a linear LSE problem. This condition seems to be quite restrictive, since most parametrization for transformation groups are nonlinear (particularly if a rotation matrix is involved); nevertheless, we can always find a parametrization q l in a linear group G l that includes G, where dim(g l ) is higher than dim(g), but as close as possible to it. Afterwards, the so obtained transform T l can be upgraded to the actual T by a Procrustes analysis (min T l T T G F ) or by simpler means, such as clamping the affine parameters φ for upgrading to a similarity, etc. (similarly to the D- D cases). In any case, the result of this procedure is needed only as a starting point for the geometric error optimization. 3. Example of linear constraints for 3D pose parameters We mention here a few examples of linear constraints that we can impose to p. Pure translation in 3D This corresponds to impose P = [ I 3 t 3 ] Q = I

17 p 0 = [ ] T q = [ t x t y t z ] T Almost pure rotation around z (without the non-linear constraint c + s = 1) c s 0 0 P = s c This corresponds to Q = p 0 = [ ] T q = [ c 4 Line correspondences s ] T In some cases, the image measurement consists of line segments, that have to be matched to corresponding model segments. For example, when performing a hand detection task (Fig. 13), the Hough transform provides very well-aligned line segments on the fingers, associated to the corresponding model lines. Segment correspondences in principle provide point correspondences (i.e. 4 measurement data). However, as we can see from the fingers in Fig. 13, the end-points of the detected segments are not as well localized as the line itself (direction and distance from the origin), therefore the most reliable matching can be obtained by considering pure line correspondences. This unfortunately provides less equations ( instead of 4) for each feature, but at least assures to use only the most reliable information source for pose estimation. As long as pure lines are concerned, a simpler way to describe correspondences consists of replacing them by point-to-line correspondences (i.e. a segment-to-line correspondence), where the two model points can be arbitrarily chosen onto the respective line (in D or 3D space). Alternatively, the end-points information can be still included, but with a lower weight in the LSE optimization process 17

18 Figure 13: Line detection with the Hough transform for planar hand detection. Therefore, we formulate the problem as follows: given a model segment (L 1, L ) and a corresponding image line l = (l 1, l, l 3 ) T, find a transformation H such that both points HL 1 and HL lie on l. When a noisy measurement l is given, the error can be again formulated in two ways (geometric and algebraic error) which lead to different LSE errors. In order to provide the geometric error, we also assume that the normal direction to the image line n = (l 1, l ) is normalized n = 1; in this way, the third component l 3 = d represents the distance of the line to the origin of image coordinates. In particular, algebraic errors are defined in homogeneous coordinates, and geometric errors in projected (non-homogeneous) coordinates, through the π() function. Algebraic errors: Geometric errors: l T ( K T T L 1) = l T ( K T T L ) = 0 n T π ( K T T L 1) + d = n T π ( K T T L ) + d = 0 which, in a least-squares setting, become Algebraic LSE: T = min n T G ( l T i K T T L 1 i + l T i K T T L ) i 18

19 Geometric LSE: T = min n T G ( n T i π ( K T T L 1 ) i + di + n T i π ( K T T L ) ) i + di 4.1 D-D line correspondences For most D cases (apart from the general homography of Sec..1), the geometric LSE is equivalent to the algebraic one. In fact, if the transformation matrix has the form A t then we have n T π ( K T T L ) + d = l T ( K T T L ) = l T K T ( A L + t 1 where L are the first two (non-homogeneous) coordinates of L. For sake of clarity, in the following we will omit the sign, whenever the context avoids ambiguity of interpretation. The above equations clearly show how the two terms K T can be removed by pre-processing the lines l l l T K ( n T, d) Furthermore, if the parametrization of the group A(q), t(q) is linear in q, then the problem becomes linear, and can be solved in one step via the SVD decomposition. For example, if we consider the general affine transform (Sec..11), parametrized by q = [ A 11 A 1 A 1 A t x t y ] T then we have AL + t = so that the LSE problem becomes q = min q R 6 n [ Lx L y L x L y ] q = ˆLq ( n T i ˆL 1 i q + d i + n T i ˆL i q + d i ) = min q R 6 ) n ˆL i q + d i with [ n T ˆL i = i ˆL1 i n T ˆL i i ] di ; di = d i 19

20 A similar result can be obtained for the similarity case (uniform scale, rotation and translation) with 4 parameters. In order to keep the linearity, we parametrize it as c s t x s c t y 0 with so that in this case q = [ c s t x t y ] T ˆL = [ Lx L y 1 0 L y L x and ˆL i is computed from this expression. However, the pure rotational cases (with c = cos(θ), s = sin(θ)) involve a nonlinearity that, in the point-to-point case, had been dealt with by using the Umeyama approach. In this case, we can simply estimate it as a similarity, and afterwards remove the scale by simply dividing (c, s) by c + s. 5 Point and line correspondences The most general case involves matching points and lines simultaneously. In order to formulate it in an elegant way, we start from the result of the previous Section, and add the point-related terms. Concerning the geometric error term for a linear pose parametrization q, we have K T T (q) X x = K T ˆX q x where the ˆX matrix is defined in the same way as ˆL (for segments). The pre-processing step for points (already been described in the related Section) becomes x = ( K T ) 1 x that we can see as the dual version of the line pre-processing. Therefore, for n l line and n p point correspondences, we have 3 n l q = min ˆL i q + d n p i + ˆX j q x j q R 6 with ˆL i, d i defined in the previous Section. 3 Notice the sign on the second terms. j=1 ] 0

21 A Derivation of the GP-DLT linear equations In order to derive the linear LSE matrix F i, we first consider the internal product Ā [ A (q) t (q) ] X i = W X i with W = Ā [ A (q) t (q) ] a (3 4) matrix. We express it row-wise W = w T 1 w T w T 3 where w j is the j th row (transposed to a column vector). Therefore, we can write it as a matrix-vector product X T i 0 T 0 T w 1 W X i = 0 T X T i 0 T w 0 T 0 T X T i w 3 which, after including the cross-product matrix [x i ], becomes F i = ˆX i w 1 w w 3 and ˆX i = Next, we consider again the W matrix 0 T z i X T i ȳ i X T i z i X T i 0 T x i X T i ȳ i X T i x i X T i 0 T each element of the product is given by W hk = ā T h p k Ā = W = ĀP (q) P (q) = [ A (q) t (q) ] ā T 1 ā T ā T 3, P = [ p 1 p p 3 p 4 ] where Ā has been expressed row-wise, and P column-wise. Therefore, we have ā T h 0 T 0 T 0 T w h = 0 T ā T h 0 T 0 T 0 T 0 T ā T h 0 T p (q) 0 T 0 T 0 T ā T h 1

22 where the vector p contains the 1 entries of P (column-wise), which are parametrized by q. By stacking the matrices M h row-wise, we have where  = W = Âp (q) ā ā ā ā ā ā ā ā ā ā ā ā 3 Our assumption about the linear dependency on q can be expressed by p (q) = Q q + p 0 with suitable values for the 1 d q matrix Q and the vector p 0. Therefore, in the linear case the l.h.s. becomes [x i ] ĀP (q) X i = ˆX i  (Q q + p 0 ) where last term of F i does not depend on q, and therefore can be moved to the right-hand side f i of the LSE. The r.h.s. term can be similarly developed: with [x i ] [ 03 3 t ] X i = ˆX i m m = [ t t t 3 ] T so that, finally, the LSE problem becomes n q = min F i q + f i q R dq where References F i = ˆX i ÂQ f i = ˆX i (m Âp 0 [1] T. Drummond and R. Cipolla, Visual tracking and control using lie algebras, cvpr, vol. 0, p. 65, [], Real-time visual tracking of complex structures, IEEE Trans. Pattern Anal. Mach. Intell., vol. 4, no. 7, pp , 00. [3], Real-time tracking of multiple articulated structures in multiple views, in ECCV 00: Proceedings of the 6th European Conference on Computer Vision-Part II. London, UK: Springer-Verlag, 000, pp ) T