3D reconstruction for featureless scenes with curvature hints

Andrea Baldacci; Daniele Bernabei; Massimiliano Corsini; Fabio Ganovelli; Roberto Scopigno

doi:10.1007/s00371-015-1144-5

The Visual Computer

December 2016, Volume 32, Issue 12, pp 1605–1620 | Cite as

3D reconstruction for featureless scenes with curvature hints

Authors
Authors and affiliations

Andrea Baldacci
Daniele Bernabei
Massimiliano CorsiniEmail author
Fabio Ganovelli
Roberto Scopigno

Original Article

First Online: 31 July 2015

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Abstract

We present a novel interactive framework for improving 3D reconstruction starting from incomplete or noisy results obtained through image-based reconstruction algorithms. The core idea is to enable the user to provide localized hints on the curvature of the surface, which are turned into constraints during an energy minimization reconstruction. To make this task simple, we propose two algorithms. The first is a multi-view segmentation algorithm that allows the user to propagate the foreground selection of one or more images both to all the images of the input set and to the 3D points, to accurately select the part of the scene to be reconstructed. The second is a fast GPU-based algorithm for the reconstruction of smooth surfaces from multiple views, which incorporates the hints provided by the user. We show that our framework can turn a poor-quality reconstruction produced with state of the art image-based reconstruction methods into a high- quality one.

Keywords

Image-based reconstruction Image-based modeling, surface reconstruction Depth maps fusion Energy minimization on the GPU

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Notes

Acknowledgments

The research leading to these results was funded by EU FP7 project ICT FET Harvest4D (http://www.harvest4d.org/, G.A. no. 323567). The Museum Dataset is courtesy of Chaurasia et al. [45].

Supplementary material

Supplementary material 1 (mp4 28128 KB)

Appendix A: Algebraic derivations of the gradient of the energy terms

Smoothness term

The derivative of the smoothness term with respect to $z_{m,n}$ is simply:

$$\begin{aligned} \frac{\partial }{z_{m,n}} S(\mathbf {z}_h)= & {} \sum _{i,j} 2 \mathbf {z}_{xx}(i,j) \frac{\partial \mathbf {z}_{xx}(i,j)}{\partial z_{m,n}} \nonumber \\&+\,4 \mathbf {z}_{xy}(i,j) \frac{\partial \mathbf {z}_{xy}(i,j)}{\partial z_{m,n}} \nonumber \\&+\, 2 \mathbf {z}_{yy}(i,j) \frac{\partial \mathbf {z}_{yy}(i,j)}{\partial z_{m,n}}\Delta x\Delta y \end{aligned}$$

(14)

The unrolled formula, using central finite differences, is:

$$\begin{aligned} \nabla E_{s}(\mathbf {z})_{m,n}= & {} \frac{\partial }{\partial z_{m,n}}\sum _{i,j}( z_{i+1,j}-2z_{i,j}+z_{i-1,j} )^{2}\nonumber \\&+\,\frac{1}{8} ( z_{i-1,j-1}-z_{i-1,j+1}-z_{i+1,j-1} \nonumber \\&+\, z_{i+1,j+1} )^{2}+\,( z_{i+1,j}-2z_{i,j}+z_{i-1,j} )^{2}\nonumber \\ \end{aligned}$$

(15)

which finally gives:

$$\begin{aligned} \nabla E_{s} (\mathbf {z})_{m,n}= & {} 25\mathbf {z}_{m,n} +\frac{3}{2}(\mathbf {z}_{m,n+2}+\mathbf {z}_{m,n-2}+ \mathbf {z}_{m+2,n}\nonumber \\&+\,\mathbf {z}_{m-2,n}) -\,8(\mathbf {z}_{m+1,n}+\mathbf {z}_{m-1,n}+ \mathbf {z}_{m,n+1} \nonumber \\&+\,\mathbf {z}_{m,n-1})+\,\frac{1}{4}(\mathbf {z}_{m+2,n+2}\nonumber \\&+\,\mathbf {z}_{m+2,n-2} +\mathbf {z}_{m-2,n-2}+\mathbf {z}_{m-2,n+2}) \end{aligned}$$

(16)

Coherence term

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}\mathbf {z}_{i,j}} R(\mathbf {z}_h)= & {} \frac{\mathrm{d}}{\mathrm{d}\mathbf {z}_{i,j}}((g_k(i,j, \mathbf {z}_{i,j})-f_k(i,j,\mathbf {z}_{i,j}))^{2})\nonumber \\= & {} 2(g_k(i,j,z)-f_k(i,j,z))\nonumber \\&\times \bigg (\frac{\mathrm{d}}{\mathrm{d}z}g_k(i,j,z)-\frac{\mathrm{d}}{\mathrm{d}z}f_k(i,j,z)\bigg ) \end{aligned}$$

(17)

So we need the derivatives of $g_k(i,j,z)$ and $f_k(i,j,z)$.

Since the cameras are calibrated, we know the matrix $\mathbf {R}_{h,k}$ that transforms the depth values from camera k to camera h:

$$\begin{aligned} \mathbf {R}_{h,k} = \mathbf {I}_k \mathbf {E}_k \mathbf {I}^{-1}_h \mathbf {E}^{-1}_h \end{aligned}$$

(18)

where $\mathbf {I}$ and $\mathbf {E}$ are the intrinsic and extrinsic matrices of camera h and k.

$g_k(i,j,z)$ is defined as:

$$\begin{aligned} g_k(i,j,z)=s_w \cdot \mathbf {v}(z,i,j) = (\mathbf {r}_{30}i+\mathbf {r}_{31}j+\mathbf {r}_{33})z +\mathbf {r}_{32} \end{aligned}$$

(19)

where $\mathbf {s}_{w}$ is the row vector that selects component w, (i.e., $\mathbf {s}_{w}=[\begin{array}{cccc} 0&0&0&1\end{array}]$) and the $\mathbf {r}_{ij}$ are the components of the $\mathbf {R}$ matrix. The derivative of $g_k$ is then:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}z}g_{i,j}(z)=(\mathbf {r}_{30}i+\mathbf {r}_{31}j +\mathbf {r}_{33}) \end{aligned}$$

(20)

It is no surprise that the derivative does not depend on z, because function $g_k(i,j,z)$ simply returns the distance between a point along a line and a plane, which varies linearly.

For function $f_k(i,j,z)$, things are a little harder, because it describes the depth map $z_k$ along the projection of the line on the image plane of camera k. Let us define the parametric function describing such a projection:

$$\begin{aligned} \mathbf {u}(z):{\mathbb {{R}}}\rightarrow {\mathbb {{R}}}^{2}=\frac{\mathbf {s}_{xy}\cdot \mathbf {v}(z)}{\mathbf {s}_{w}\cdot \mathbf {v}(z)} \end{aligned}$$

(21)

Function f is then the composition of $z(x,y):{\mathbb {{R}}}^{2}\rightarrow {\mathbb {{R}}}$ with $\mathbf {u}$, i.e., $(z_k\cdot \mathbf {u}):{\mathbb {{R}}}\rightarrow {\mathbb {{R}}}$. Therefore, the derivative of the composition is

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}z}f_k(i,j,z)= & {} (z_k\cdot \mathbf {u}(z))'=\left[ \frac{\partial z_k}{\partial x},\frac{\partial z_k}{\partial y}\right] \nonumber \\&\times (\mathbf {u}_{x}(z),\mathbf {u}_{y}(z))\cdot \left[ \frac{\mathrm{d}}{\mathrm{d}z} \mathbf {u}_{x}(z),\frac{\mathrm{d}}{\mathrm{d}z}\mathbf {u}_{y}(z)\right] \nonumber \\ \end{aligned}$$

(22)

We still need to define what $\frac{\mathrm{d}}{\mathrm{d}z}u_{x}(z)$ is (and, by symmetry, this will also yield its y-axis counterpart). This is the derivative

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}z}(\mathbf {s}_{x}\cdot \mathbf {v}(z)/ \mathbf {s}_{w}\cdot \mathbf {v}(z)) \end{aligned}$$

(23)

of a function with the form $\frac{\alpha z+\beta }{\gamma z+\delta }$ whose derivative is $\frac{\alpha \delta -\beta \gamma }{(\gamma z+\delta )^{2}}$. Therefore,

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}z}f_k(i,j,z)= & {} \left[ \frac{\partial z_k}{\partial x},\frac{\partial z_k}{\partial y}\right] \left[ \frac{\mathbf {s}_{xy}\cdot \mathbf {v}(z)}{\mathbf {s}_{w}\cdot \mathbf {v}(z)}\right] \nonumber \\&\times \left[ \frac{\alpha _{x} \mathbf {r}_{32}-\mathbf {r}_{02}\gamma }{(\gamma z+\mathbf {r}_{32})^{2}},\frac{\alpha _{y} \mathbf {r}_{32}-\mathbf {r}_{12}\gamma }{(\gamma z+\mathbf {r}_{32})^{2}}\right] \end{aligned}$$

(24)

where $\alpha _{x} = (\mathbf {r}_{00}i+\mathbf {r}_{01}j+\mathbf {r}_{03}) = \mathrm{d}\mathbf {v}_{x}$, $\alpha _{y} = (\mathbf {r}_{10}i+\mathbf {r}_{11}j$ $+ \mathbf {r}_{13}) = \mathrm{d}\mathbf {v}_{y}$ and $\gamma =(\mathbf {r}_{30}i+\mathbf {r}_{31}j +\mathbf {r}_{33})=\mathrm{d}\mathbf {v}_{w}$. In conclusion, the gradient is:

$$\begin{aligned}&\nabla R(z)_{m,n} = 25\mathbf {z}_{m,n}^{k} \nonumber \\&\quad +\,\frac{3}{2}(\mathbf {z}_{m,n+2}^{k}+\mathbf {z}_{m,n-2}^{k}+ \mathbf {z}_{m+2,n}^{k}+\mathbf {z}_{m-2,n}^{k}) \nonumber \\&\quad -\,8(\mathbf {z}_{m+1,n}^{k} +\mathbf {z}_{m-1,n}^{k}+\mathbf {z}_{m,n+1}^{k}+ \mathbf {z}_{m,n-1}^{k}) \nonumber \\&\quad +\,\frac{1}{4} (\mathbf {z}_{m+2,n+2}^{k}+\mathbf {z}_{m+2,n-2}^{k}+ \mathbf {z}_{m-2,n-2}^{k}+\mathbf {z}_{m-2,n+2}^{k}) \nonumber \\&\quad +\,2(g_{m,n}(\mathbf {z}_{m,n}^{k})-h_{m,n}(\mathbf {z}_{m,n}^{k})) \bigg (\frac{\mathrm{d}}{\mathrm{d}z}g_{m,n}(\mathbf {z}_{m,n}^{k})\nonumber \\&\quad -\frac{\mathrm{d}}{\mathrm{d}z}h_{m,n}(\mathbf {z}_{m,n}^{k})\bigg ) \end{aligned}$$

(25)

Curvature term

Proceeding as for the smoothness term:

$$\begin{aligned}&\nabla C(\mathbf {z})_{m,n} = \frac{\partial }{\partial z_{m,n}}\sum _{i,j} \left( z_{i+1,j}-2z_{i,j}+z_{i-1,j} u^2 \right. \nonumber \\&\quad +\frac{2 \left( z_{i-1,j-1}-z_{i-,j+1}-z_{i+1,j-1}+z_{i+1,j+1}\right) uv}{4}\nonumber \\&\quad +\left. z_{i+1,j}-2z_{i,j}+z_{i-1,j}v^2\right) ^2 \end{aligned}$$

(26)

which, after a trivial but tedious derivation, gives:

$$\begin{aligned}&\nabla C(\mathbf {z})_{m,n} = 2 [ 24 (u^4+v^4)+ 36 u^2 v^2)\ (z_{m,n})\nonumber \\&\quad -\,16(u^4-u^2v^2) (z_{m+1,n}+z_{m-1,n}) \nonumber \\&\quad +\,(4u^4-2u^2v^2) (z_{m+2,n}+z_{m-2,n}) \nonumber \\&\quad -\,16(v^4-u^2v^2) (z_{m+1,n}+z_{m-1,n}) \nonumber \\&\quad +\,(4u^4-2u^2v^2) (z_{m+2,n}+z_{m-2,n} \nonumber \\&\quad +\,(4v^4-2u^2v^2) (z_{m,n+2}+z_{m,n-2} \nonumber \\&\quad +\,8(u^3v+uv^3+u^2v^2) (z_{m+1,n+1}+z_{m-1,n-1}) \nonumber \\&\quad -\,8(u^3v+uv^3+u^2v^2) (z_{m-1,n+1}+z_{m+1,n-1}) \nonumber \\&\quad +\,u^2v^2 (z_{m+2,n+2} +z_{m-2,n-2})] \end{aligned}$$

(27)

Appendix B: An example of handling discontinuities with the LUT table

In this section, we show how the coefficients of the LUT table are derived in a specific case. Let us consider, Eq. (14) for the gradient of the smoothness term, which is a weighted sum of second derivatives, and consider one of the terms of the sum:

$$\begin{aligned} A =\frac{\partial }{\partial z_{m,n}} \mathbf {z}^2_{xx}(n-2,m)= 2 \mathbf {z}_{xx}(n-2,m) \frac{\partial \mathbf {z}_{xx}(n-2,m)}{\partial z_{m,n}} \end{aligned}$$

(28)

If $\mathbf {z}_{xx}(n-2,m)$ is computed by central finite differences, we have:

$$\begin{aligned}&\frac{\partial \mathbf {z}_{xx}(n-2,m)}{\partial z_{m,n}} = \frac{\partial }{\partial z_{m,n}} ( \mathbf {z}(n-3,m) \nonumber \\&\quad - 2 \mathbf {z}(n-2,m) + \mathbf {z}(n-1,m) ) \nonumber \\&\quad = 0\Rightarrow A=0 \end{aligned}$$

(29)

In other words, since $z_{m,n}$ does not appear in the computation of $\mathbf {z}_{xx}(n-2,m)$ the derivative on $z_{m,n}$, and thus A, is zero. Referring to Fig. 7, this is because the entry for this configuration (first row) is null.

On the other hand, if $\mathbf {z}_{xx}(n-2,m)$ is computed by forward finite differences, we have:

$$\begin{aligned} \frac{\partial }{\partial z_{m,n}} \mathbf {z}_{xx}(n-2,m)= & {} \frac{\partial }{\partial z_{m,n}} ( \mathbf {z}(n-2,m)\nonumber \\&- 2 \mathbf {z}(n-1,m) + \mathbf {z}(n,m)) =1\nonumber \\ \end{aligned}$$

(30)

and thus:

$$\begin{aligned} A= & {} 2 \mathbf {z}_{xx}(n-2,m)\nonumber \\= & {} \mathbf {2} \mathbf {z}(n-2,m) \mathbf {-4} \mathbf {z}(n-1,m) + \mathbf {2} \mathbf {z}(n,m) \end{aligned}$$

(31)

This gives the coefficients to apply as just shown in Fig. 7 (second row).

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Authors and Affiliations

Andrea Baldacci
- 1
Daniele Bernabei
- 1
Massimiliano Corsini
- 1
Email author
Fabio Ganovelli
- 1
Roberto Scopigno
- 1

1.ISTI-CNRPisaItaly

3D reconstruction for featureless scenes with curvature hints

Abstract

Keywords

Notes

Acknowledgments

Supplementary material

References

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