In the following I will present a method for deforming three dimensional geometry using a technique relying on *radial basis functions* (RBFs). These are mathematical functions that take a real number as input argument and return a real number. RBFs can be used for creating a smooth interpolation between values known only at a discrete set of positions. The term *radial* is used because the input argument given is typically computed as the distance between a fixed position in 3D space and another position at which we would like to evaluate a certain quantity.

The tutorial will give a short introduction to the linear algebra involved. However the source code contains a working implementation of the technique which may be used as a black box.

Source code with Visual Studio 2005 solution can be found here. The code should also compile on other platforms.

### Interpolation by radial basis functions

Assume that the value of a scalar valued function $$F : mathbb{R}^3 rightarrow mathbb{R}$$ is known in $$M$$ distinct discrete points $$mathbf{x}_i$$ in three dimensional space. Then RBFs provide a means for creating a smooth interpolation function of $$F$$ in the whole domain of $$mathbb{R}^3$$. This function is written as a sum of $$M$$ evaluations of a radial basis function $$g(r_i) : mathbb{R} rightarrow mathbb{R}$$ where $$r_i$$ is the distance between the point $$mathbf{x} = (x, y, z)$$ to be evaluated and $$mathbf{x}_i$$:

$$!F(mathbf{x}) = sum_{i=1}^M a_i g(||mathbf{x} – mathbf{x}_i||) + c_0 + c_1 x + c_2 y + c_3 z, mathbf{x} = (x,y,z) mathbf{(1)}$$

Here $$a_i$$ are scalar coefficients and the last four terms constitute a first degree polynomial with coefficients $$c_0$$ to $$c_3$$. These terms describe an affine transformation which cannot be realised by the radial basis functions alone. From the $$M$$ known function values $$F( x_i, y_i, z_i ) = F_i$$ we can assemble a system of $$M+4$$ linear equations: $$mathbf{G} mathbf{A} = mathbf{F}$$

where $$mathbf{F} = (F_1, F_2, ldots, F_M, 0, 0, 0, 0)$$, $$mathbf{A} = (a_1, a_2, ldots, a_M, c_0, c_1, c_2, c_3)$$ and $$mathbf{G}$$ is an $$(M+4) times (M+4)$$ matrix :

$$! mathbf{G} = left[begin{array}{cccccccccc}g_{11} & g_{12} & bullet & bullet & bullet & g_{1M} & 1 & x_1 & y_1 & z_1 \ g_{21} & g_{22} & bullet & bullet & bullet & g_{2M} & 1 & x_2 & y_2 & z_2 \ bullet & bullet & bullet & bullet & bullet & bullet & bullet & bullet & bullet & bullet \ bullet & bullet & bullet & bullet & bullet & bullet & bullet & bullet & bullet & bullet \ bullet & bullet & bullet & bullet & bullet & bullet & bullet & bullet & bullet & bullet \ g_{M1} & g_{M2} & bullet & bullet & bullet & g_{MM} & 1 & x_M & y_M & z_M \ 1 & 1 & bullet & bullet & bullet & 1 & 0 & 0 & 0 & 0 \ x_1 & x_2 & bullet & bullet & bullet & x_M & 0 & 0 & 0 & 0 \ y_1 & y_2 & bullet & bullet & bullet & y_M & 0 & 0 & 0 & 0 \ z_1 & z_2 & bullet & bullet & bullet & z_M & 0 & 0 & 0 & 0end{array}right] $$

Here $$g_{ij} = g(|| mathbf{x}_i – mathbf{x}_j ||)$$. A number of choices for $$g$$ will result in a unique solution of the system. In this tutorial we use the *shifted log function*:

$$! g(t) = sqrt{log(t^2 + k^2)}, k^2geq 1$$

with $$k = 1$$. Solving the equation system for $$mathbf{A}$$ gives us the coefficients to use in equation $$textbf{(1)}$$ when interpolating between known values.

### Interpolating displacements

How can RBF’s be used for deforming geometry? Well assume that the deformation is known for $$M$$ 3D positions $$mathbf{x}_i$$ and that this information is represented by a vector describing 3D displacement $$mathbf{u}_i$$ of the geometry that was positioned at $$mathbf{x}_i$$ in the original, undeformed state. You can think of the $$mathbf{x}_i$$ positions as control points that have been moved to positions $$mathbf{x}_i+mathbf{u}_i$$. The RBF interpolation method can now be used for interpolating these displacements to other positions.

Using the notation $$mathbf{x}_i = (x_i, y_i, z_i)$$ and $$mathbf{u}_i = (u^x_i, u^y_i, u^z_i)$$ three linear systems are set up as above letting the displacements $$u$$ be the quantity we called $$a$$ in the previous section:

$$!mathbf{G} mathbf{A}_x = (u^x_1, u^x_2, ldots, u^x_M, 0, 0, 0, 0)^T$$

$$!mathbf{G} mathbf{A}_y = (u^y_1, u^y_2, ldots, u^y_M, 0, 0, 0, 0)^T$$

$$!mathbf{G} mathbf{A}_z = (u^z_1, u^z_2, ldots, u^z_M, 0, 0, 0, 0)^T$$

where $$mathbf{G}$$ is assembled as described above. Solving for $$mathbf{A}_x$$, $$mathbf{A}_y$$, and $$mathbf{A}_z$$ involves a single matrix inversion and three matrix-vector multiplications and gives us the coefficients for interpolating displacements in all three directions by the expression $$mathbf{(1)}$$

### The source code

In the source code accompanying this tutorial you will find the class RBFInterpolator which has an interface like this:

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class RBFInterpolator { public: RBFInterpolator(); ~RBFInterpolator(); //create an interpolation function f that obeys F_i = f(x_i, y_i, z_i) RBFInterpolator(vector x, vector y, vector z, vector F); //specify new function values F_i while keeping the same void UpdateFunctionValues(vector F); //evaluate the interpolation function f at the 3D position (x,y,z) real interpolate(float x, float y, float z); private: ... }; |

This class implements the interpolation method described above using the newmat matrix library. It is quite easy to use: just fill stl::vectors with the $$x_i$$, $$y_i$$ and $$z_i$$ components of the positions where the value $$F$$ is known and another stl::vector with the $$F_i$$ values. Then pass these vectors to the RBFInterpolator constructor, and it will be ready to interpolate. The $$F$$ value at any position is then evaluated by calling the ‘interpolate’ function. If some of the $$F_i$$ values change at any time, the interpolator can be quickly updated using the ‘UpdateFunctionValues’ method.

We want to deform a triangle surface mesh. These are stored in a class TriangleMesh, and loaded from OBJ files.

In the source code the allocation of stl::vectors describing the control points and the initialisation of RBFInterpolators looks like this:

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void loadMeshAndSetupControlPoints() { // open an OBJ file to deform string sourceOBJ = "test_dragon.obj"; undeformedMesh = new TriangleMesh(sourceOBJ); deformedMesh = new TriangleMesh(sourceOBJ); // we want 11 control points which we place at different vertex positions const int numControlPoints = 11; const int verticesPerControlPoint = ((int)undeformedMesh->getParticles().size())/numControlPoints; for (int i = 0; i<numControlPoints; i++) { Vector3 pos = undeformedMesh->getParticles()[i*verticesPerControlPoint].getPos(); controlPointPosX.push_back(pos[0]); controlPointPosY.push_back(pos[1]); controlPointPosZ.push_back(pos[2]); } // allocate vectors for storing displacements for (unsigned int i = 0; i<controlPointPosX.size(); i++) { controlPointDisplacementX.push_back(0.0f); controlPointDisplacementY.push_back(0.0f); controlPointDisplacementZ.push_back(0.0f); } // initialize interpolation functions rbfX = RBFInterpolator(controlPointPosX, controlPointPosY, controlPointPosZ, controlPointDisplacementX ); rbfY = RBFInterpolator(controlPointPosX, controlPointPosY, controlPointPosZ, controlPointDisplacementY ); rbfZ = RBFInterpolator(controlPointPosX, controlPointPosY, controlPointPosZ, controlPointDisplacementZ ); } |

Now all displacements are set to zero vectors – not terribly exciting! To make it a bit more fun we can vary the displacements with time:

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// move control points for (unsigned int i = 0; i < controlPointPosX.size(); i++ ) { controlPointDisplacementX[i] = displacementMagnitude*cosf(time+i*timeOffset); controlPointDisplacementY[i] = displacementMagnitude*sinf(2.0f*(time+i*timeOffset)); controlPointDisplacementZ[i] = displacementMagnitude*sinf(4.0f*(time+i*timeOffset)); } // update the control points based on the new control point positions rbfX.UpdateFunctionValues(controlPointDisplacementX); rbfY.UpdateFunctionValues(controlPointDisplacementY); rbfZ.UpdateFunctionValues(controlPointDisplacementZ); // deform the object to render deformObject(deformedMesh, undeformedMesh); |

Here the function ‘deformObject’ looks like this:

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// Code for deforming the mesh 'initialObject' based on the current interpolation functions (global variables). // The deformed vertex positions will be stored in the mesh 'res' // The triangle connectivity is assumed to be already correct in 'res' void deformObject(TriangleMesh* res, TriangleMesh* initialObject) { for (unsigned int i = 0; i < res->getParticles().size(); i++) { Vector3 oldpos = initialObject->getParticles()[i].getPos(); Vector3 newpos; newpos[0] = oldpos[0] + rbfX.interpolate(oldpos[0], oldpos[1], oldpos[2]); newpos[1] = oldpos[1] + rbfY.interpolate(oldpos[0], oldpos[1], oldpos[2]); newpos[2] = oldpos[2] + rbfZ.interpolate(oldpos[0], oldpos[1], oldpos[2]); res->getParticles()[i].setPos(newpos); } } |

That’s it!!! Now I encourage you to download the source code and play with it. Perhaps you can experiment with other radial basis functions? Or make the dragon crawl like a caterpillar? If you code something interesting based on this tutorial send a link to me and we will link to it from this page 🙂

I got a mail from Woo Won Kim from Yonsei University in South Korea who has made a head modeling program that can generate 3D human heads from two pictures of the person using code from this RBF tutorial. Check out a video of this here.

Amr

Feb 23, 2010

thanx for this tutorial .. iam coding a physics engine for games on Xbox so iam using c# xna as my language .. i have problem converting your code to C# wise 🙁

i appreciate your help by any mean

thanx 🙂

yaoyansi

Mar 8, 2010

Thank you very very much for this source code.

Belayouni

Jul 2, 2010

Thanks for this simple and well explained tutorial.

Actually I’m trying to apply this algorithm to deform a small Cartesian grid on MatLab. But since the gii terms of the radial function are zeros and the G matrix is not invertible (det(G)=0) because sqrt(log(0+k^2))=0 k=1).

I wanted to know how did you deal with this? 🙂

Minh Nguyen

Sep 14, 2010

The code is very interesting. However, I doubt that this is practical for relatively large models (wouldn’t say too big) with 100Ks of vertices. Because your matrix G is a dense matrix, most computers won’t even have enough memory to allocate 100k x 100k x sizeof(float) just for the matrix G itself.

admin

Oct 29, 2010

Minh Nguyen: The size of the matrix depends on the number of control points and not the number of vertices to displace. So if you think of the control points you are right, but not if you are thinking about the vertices in the model to deform.

David

May 23, 2011

Thanks for this tutorial. This really helped a lot!

Damien

Jun 9, 2011

Thanks for the tutorial. I would’ve had trouble applying it if it wasn’t this detailed! I’ve been having similar problems and didn’t know how to resolve it until I came across your post. Kudos!

Nina

Oct 25, 2012

Thanks for the great work!

I was wondering if you can post the references of the work too.

thanks

admin

Nov 5, 2012

To Nina

The work we used it for was this:

Surface membrane based bladder registration for evaluation of accumulated dose during brachytherapy in cervical cancer. K. Ø. Noe, K. Tanderup, T.S. Sørensen. IEEE International Symposium on Biomedical Imaging (ISBI) 2011.

We adopted the RBF method from here:

G. K. Matsopoulos, N. A. Mouravliansky, P. A. Asvestas, K. K. Delibasis, and V. Kouloulias. Thoracic non-rigid registration combining selforganizing maps and radial basis functions. Medical Image Analysis, 9(3):237 – 254, 2005.

Nina

Jun 12, 2013

Thank you!

Also can you please explain what are “t” and “k” in the shifted Log Function?

D SURESH BABU

Mar 15, 2017

Hi,

It is excellent work. RBF interpolator is giving the same displacement vector value. So all depenedent points are moving to same magnitude. Am I doing nay mistake.