cvpr cvpr2013 cvpr2013-432 knowledge-graph by maker-knowledge-mining
Source: pdf
Author: Ramakrishna Kakarala, Prabhu Kaliamoorthi, Vittal Premachandran
Abstract: We show that bilateral symmetry plane estimation for three-dimensional (3-D) shapes may be carried out accurately, and efficiently, in the spherical harmonic domain. Our methods are valuable for applications where spherical harmonic expansion is already employed, such as 3-D shape registration, morphometry, and retrieval. We show that the presence of bilateral symmetry in the 3-D shape is equivalent to a linear phase structure in the corresponding spherical harmonic coefficients, and provide algorithms for estimating the orientation of the symmetry plane. The benefit of using spherical harmonic phase is that symmetry estimation reduces to matching a compact set of descriptors, without the need to solve a correspondence problem. Our methods work on point clouds as well as large-scale mesh models of 3-D shapes.
Reference: text
sentIndex sentText sentNum sentScore
1 Three-dimensional bilateral symmetry plane estimation in the phase domain Ramakrishna Kakarala, Prabhu Kaliamoorthi, and Vittal Premachandran School of Computer Engineering, Nanyang Technological University, Singapore ramakri shna @ntu . [sent-1, score-0.961]
2 Abstract We show that bilateral symmetry plane estimation for three-dimensional (3-D) shapes may be carried out accurately, and efficiently, in the spherical harmonic domain. [sent-4, score-1.539]
3 Our methods are valuable for applications where spherical harmonic expansion is already employed, such as 3-D shape registration, morphometry, and retrieval. [sent-5, score-0.905]
4 We show that the presence of bilateral symmetry in the 3-D shape is equivalent to a linear phase structure in the corresponding spherical harmonic coefficients, and provide algorithms for estimating the orientation of the symmetry plane. [sent-6, score-2.167]
5 The benefit of using spherical harmonic phase is that symmetry estimation reduces to matching a compact set of descriptors, without the need to solve a correspondence problem. [sent-7, score-1.507]
6 Introduction The motivation to apply spherical harmonic expansion to solve three-dimensional (3-D) computer vision problems stems from at least three well-known properties. [sent-10, score-0.846]
7 First, the expansion summarizes a large number of shape points (ver- tices or surface voxels) in a relatively small number of coefficients. [sent-11, score-0.18]
8 Third, the spherical harmonic coefficients behave predictably under 3-D rotation. [sent-13, score-0.877]
9 Accordingly, spherical harmonics have been successfully employed in computer vision for 3-D shape registration [10], morphometry [3], and recognition [7][9]. [sent-14, score-0.705]
10 In this paper, we show that spherical harmonics also provide an accurate and efficient solution for estimating the bilateral symmetry plane of a 3-D shape. [sent-15, score-1.315]
11 Bilateral symmetry has been studied in numerous works (for example, [12] [6][20]). [sent-16, score-0.491]
12 However, the problem of estimating the symmetry plane through spherical harmonic coefficients alone has not been solved, though there are results for the special case of moment coefficients [11]. [sent-17, score-1.601]
13 Consequently, with current techniques, if the spherical harmonic expansion has already been computed, as would be the case in . [sent-18, score-0.846]
14 sg (a) (b) (c)(d) Figure 1: The image in (a) is reconstructed from only its Fourier phase in (b), illustrating phase’s importance to appearance. [sent-24, score-0.257]
15 The same is true for the sphere: the continental edges in the world map (c) are clearly visible in (d), which is reconstructed from (c) using only the spherical harmonic “phase” as defined in this paper. [sent-25, score-0.759]
16 the applications mentioned above, there is no way to reuse the computationally-expensive expansion to obtain a symmetry plane estimate. [sent-26, score-0.693]
17 And yet, as we show, the information required to obtain the estimate is clearly available in the harmonic coefficients, and a relatively simple estimation algorithm is possible. [sent-27, score-0.322]
18 Our approach makes use of the “phase” of spherical harmonics, which is a term without a widely-accepted definition for spherical harmonics. [sent-28, score-0.874]
19 In contrast, phase is welldefined for the ordinary Fourier transform: on the real line, the phase of transform F is φ in the polar decomposition F = |F|ejφ. [sent-29, score-0.636]
20 The analogous definition for spherical harmFo =nics |F must, as we argue below, consider vectors of coeffi- cients and treat the unit vector direction (vector divided by its length) as the phase. [sent-30, score-0.477]
21 It is well-known that for the ordinary Fourier transform, phase defines the locations of edges and is therefore more important for appearance than magnitude. [sent-31, score-0.29]
22 Figure 1 shows that this is true for phase of spherical 222444999 harmonic coefficients as well. [sent-32, score-1.134]
23 The figure motivates us to examine what must be true for spherical harmonic phase if symmetry exists in the spatial domain. [sent-33, score-1.507]
24 Our main contribution shows that bilateral symmetry in spherical data manifests itself as linear phase structure in the spherical harmonic coefficients. [sent-34, score-2.042]
25 Our results allow symmetry to be determined from any type of spherical harmonic expansion, which is valuable since there are several different expansions in use in the literature. [sent-35, score-1.273]
26 We also propose new algorithms for estimating the orientation of the symmetry plane by maximizing the fit of a linear phase equivalent to the harmonic coefficients. [sent-36, score-1.271]
27 Our methods work for a wide variety of data, including point clouds and polygonal meshes (open as well as watertight), and they do not require that the meshes be aligned to the symmetry or have a star-shaped property. [sent-38, score-0.555]
28 Previous work Given a point cloud in 3-D, one method for finding can- × didates for the bilateral symmetry plane is to evaluate the distance of corresponding points when reflected across each candidate plane. [sent-40, score-0.758]
29 h Ienf tihtse th 3r ×ee eigenvectors determine orthogonal candidates for the symmetry plane. [sent-43, score-0.516]
30 [6] describe a reflective symmetry descriptor that is constructed by measuring the norm of the projection of a voxel set onto the space of bilaterallysymmetric sets. [sent-47, score-0.515]
31 The shape of the descriptor function agrees visually with the perceived symmetries of 3-D shapes, and is valuable for shape registration and classification. [sent-49, score-0.169]
32 However, the problem of estimating the orientation of the symmetry plane is not discussed in [6]. [sent-50, score-0.641]
33 The 3-D Fourier transform is calculated on a psuedo-polar grid to detect symmetry groups in voxel data [1]. [sent-54, score-0.574]
34 The richness of the literature on 3-D symmetry shows that the topic is of considerable interest. [sent-55, score-0.491]
35 While many facets of symmetry have been explored, there is no existing work on determining bilateral symmetry from spherical harmonic expansion alone. [sent-56, score-1.926]
36 [11] show that local minima for the spherical harmonic coefficients of evenorder moments provide candidates for the symmetry axis. [sent-58, score-1.393]
37 To evaluate the suitability of a candidate reflection R, they compute the maximum possible distance between all vertices of S and their nearest points (which are found by minimization) in the reflection of S by R. [sent-59, score-0.281]
38 That calculation is expensive, and does not take advantage of the structure present in the spherical harmonic coefficients. [sent-60, score-0.759]
39 In this paper, we compute a much cheaper and more suitable distance using spherical harmonics coefficients alone, which is a substantial savings since a shape having thousands of vertices may be represented using only a few tens of coefficients. [sent-61, score-0.849]
40 Moreover, the wide-spread use of spherical harmonics in various applications, as mentioned above, makes it attractive to investigate how to estimate the symmetry plane from harmonic coefficients alone. [sent-62, score-1.657]
41 Our methods are compatible with each of the numerous uses of spherical harmonics in the literature, whether applied to the spherical mappings in [6], to the even-order moments in [11], or to the Zernike coefficient mapping in [9]. [sent-63, score-1.138]
42 Notation and background Conventions and notations for spherical harmonics vary across many papers on 3-D vision, and therefore we specify ours in this section, referring to standard works [21] for details. [sent-65, score-0.66]
43 (1) Spherical harmonics form an orthogonal basis for functions on S2, and, for each non-negative integer ? [sent-69, score-0.174]
44 the spherical harmonic coefficients is important to our development. [sent-125, score-0.877]
45 e tRedota btiyo n3 i×n t3he o spherical mhaarmtrioxn Pic domain is determined for each frequency ? [sent-127, score-0.437]
46 Data in 3-D vision tasks are frequently presented as a set of points (xi, yi, zi), or equivalently in spherical coordinates (αi, βi, ρi), for i = 1, . [sent-152, score-0.437]
47 Such data may be approximated with spherical harmonics by choosing coefficients F to minimize the squared error i? [sent-156, score-0.729]
48 The first case minimizes (7) by fitting the vertices from a triangulated mesh model, and is referred to as the vertex mapping in this paper. [sent-172, score-0.332]
49 The EGI mapping of the model is obtained from surface normals, shown superimposed on the model in (d), and approximated in (e) and (f). [sent-176, score-0.173]
50 Symmetry is equivalent to linear phase Taking inspiration from the magnitude-phase decomposition of the real-line Fourier transform value as F = |F|ejφ, we define the magnitude-phase decomposition for t|Fhe| spherical harmonic vector F? [sent-178, score-1.057]
51 unitary), ibnust tnhvea spherical rh raortmaotionnic phase hroetDates m as U? [sent-196, score-0.694]
52 The spherical harmonic phase also transforms in a simple way under reflection. [sent-202, score-1.016]
53 Consequently, p(2ar)a msheotwersi ztahtaiot nth oef spherical eha, αrmo? [sent-205, score-0.437]
54 For each case, the spherical harmonic approximation to the mapping function is necessarily a star-shaped surface, but is still able to capture a variety of surface features, as such as the antennae as illustrated for the vertex mapping in part (c) of Fig. [sent-208, score-1.057]
55 Now suppose that the reflection is across a plane Fwit? [sent-213, score-0.221]
56 (8) Consequently, if f is reflection symmetric across N⊥, then for every rotation P such that PN = Y , we must have that F? [sent-237, score-0.241]
57 (9) This relationship helps to establish conditions for bilateral symmetry in the spherical harmonic domain. [sent-242, score-1.348]
58 If a real-valued function h has symmetry across the origin, i. [sent-255, score-0.514]
59 Such functions are described in signal processing as having linear phase, since the phase component ωx0 is linearly dependent on the shift x0. [sent-261, score-0.257]
60 (10) Note the similarity of (10) to (9), with the translation phase ejωx0 replaced by the rotation phase D? [sent-263, score-0.592]
61 Finding the symmetry plane Since symmetric functions have linear-phase spherical harmonic coefficients, the problem of finding the symmetry plane requires optimizing a linear phase fit to the observed coefficients. [sent-266, score-2.365]
62 n∗, we impose a similar symmetry for the rea=l c (o−e1ffi)cients: R−? [sent-296, score-0.491]
63 opt Inserting back into (12) and simplifying, we obtain our Imnsaienr ithnegoRretical result: the optimal choice for the rotation P that determines the orientation of the symmetry plane is obtained by maximizing the function ? [sent-344, score-0.719]
64 1 Note that (14) is expressed only in terms of the spherical harmonics coefficients F and the rotation P, and the previously unnikcsno cwoenf ireciaeln nctose Fffi acinednt tsh eR r ohtaavtieo bne Pen, aenlidm thineat perdev. [sent-357, score-0.807]
65 Third, if f is linear phase and therefore has coefficients F? [sent-372, score-0.375]
66 1 (15) Setting P = PN, the true symmetry plane, gives the global maximum in this case, as expected. [sent-390, score-0.491]
67 Fourth, if P is a rotation about the y axis (Euler angles α = γ = 0), then D? [sent-391, score-0.16]
68 This is intuitively reasonable, since only two angles are required to specify the orientation of a plane in 3-D. [sent-398, score-0.258]
69 Finally, calculation of (14) is efficient because it depends only on a relatively small number of spherical harmonic coefficients, and not on the much larger number of vertices. [sent-399, score-0.759]
70 Optimum estimates for the symmetry plane The form of Φ in (14? [sent-402, score-0.606]
71 Darker red values indicate improving estimates of the symmetry plane of (a). [sent-460, score-0.606]
72 over three Euler angles α, β, and γ, which is, in principle, unnecessary, as two angles suffice to specify the orientation of the symmetry plane. [sent-462, score-0.716]
73 To perform the search efficiently, we use a nested polar grid search, in which we 222555333 perform initially a coarse grid search by evaluating N values of α, β over [0, π] . [sent-482, score-0.192]
74 As a baseline for both algorithms, we use the three eigenvectors of the covariance matrix to obtain three corresponding candidates for the symmetry plane. [sent-498, score-0.553]
75 Experimental methods and results We estimate spherical harmonic coefficients using the iterated residual fitting (IRF) method [3]. [sent-502, score-0.905]
76 Previous works have tested symmetry estimation on a relatively small number of shapes, and tested robustness by simulating acquisition and topological noise [1][1 1][20]. [sent-509, score-0.515]
77 (a)(b)(c) Figure 5: Uniform densification is illustrated for the plant model in (a), whose 316 vertices are plotted in (b), and whose densified samples are shown in (c). [sent-510, score-0.218]
78 However, the vertices are not uniformly distributed across the shape’s surface, which may affect the symmetry estimation. [sent-515, score-0.598]
79 Therefore, we uniformly sample a dense set of points across the surface of the shape by relying on the triangulated mesh. [sent-516, score-0.176]
80 With such a representation, the points are more uniformly distributed on the shape surface, helping the identification of symmetry planes. [sent-519, score-0.527]
81 In our experiments, every shape is randomly rotated in all three Euler angles prior to symmetry estimation. [sent-521, score-0.609]
82 To provide a shape-independent measure of fit, we define the normalized fitting error as the relative error between the linear phase fit for the optimum rotation Popt and its upper bound: E(Popt) = 100 ×? [sent-522, score-0.46]
83 Figure 6 shows examples from the PSB database of various degrees of symmetry and the associated fit. [sent-534, score-0.491]
84 We found over numerous shapes that the E measure correlated well with our perception of how well the symmetry plane fit the shape. [sent-535, score-0.733]
85 In each case the symmetry plane found by optimizing (14) with SH-ISA is shown superimposed in green. [sent-540, score-0.684]
86 Rows 2-4 show results for the PSB database, and contain respectively the results for the EGI mapping, the vertex mapping using original vertices, denoted (O), and the vertex mapping using the densified vertices (D); similarly for other rows and S 10 and SEG databases. [sent-542, score-0.385]
87 We find that the symmetry plane is signifi- cantly better estimated using the vertex mapping than the EGI mapping, reducing E by at least 30%, and also that densification improves the vertex mapping results substantially, by similar amounts. [sent-545, score-0.975]
88 The EGI mapping is unique only for convex objects, and does not capture shape properties as well as vertex mapping (see Fig. [sent-546, score-0.237]
89 % Figure 7: The symmetry plane from SH-ISA is shown superimposed in green. [sent-556, score-0.655]
90 in the shape, as a benefit of (14) relying on the spherical harmonic expansion alone. [sent-558, score-0.871]
91 The robustness of our linear phase methods are due to the least-squares fit of the spherical harmonic coefficients using IRF, the ability of low-frequency spherical harmonics to smooth fluctuations, and performing optimization of (14) without differentiation. [sent-560, score-1.796]
92 The main limitation in symmetry determination is the spherical mapping. [sent-563, score-0.928]
93 One important case is articulated shapes, which requires an analysis of intrinsic symmetry [17], and would normally fail with the extrinsic methods used here, as shown in the left column of Figure 8. [sent-564, score-0.491]
94 Summary and conclusions Figure 1inspires us to look in the phase domain for information about symmetry. [sent-569, score-0.257]
95 By developing the idea that linear phase in spherical harmonic coefficients is equivalent to bilateral symmetry, our main theoretical contribution shows that optimizing the degree of linear phase fit is equivalent to estimating the symmetry plane. [sent-570, score-2.06]
96 Our method is compatible with those works in that it is able to use any symmetry-preserving mapping to the sphere, including the vertex mapping and EGI mapping. [sent-573, score-0.201]
97 Beyond symmetry, our results suggest that the phase of spherical harmonics is rich in other information about 3-D structure. [sent-574, score-0.868]
98 Furthermore, the structural properties of phase for three-variable spherical harmonics, or SPHARM [16] constitute another interesting extension to explore. [sent-576, score-0.694]
99 3-D symmetry detection and analysis using the pseudo-polar Fourier transform. [sent-581, score-0.491]
100 Rotation invariant spherical harmonic representation of 3D shape descriptors. [sent-635, score-0.795]
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