According to the documentation of System.Windows.Media.Transform.Inverse, the function
Gets the inverse of this transform, if it exists.
But not much further explanation, there. So when would the inverse not exist? Under what circumstances or types of transforms?
I use a TransformGroup that has both a TranslateTransform and RotateTransform which I modify individually. Do I need to worry about it returning null for that?
There are technical ways to describe non-invertible transformation. See e.g. https://en.wikipedia.org/wiki/Affine_transformation#Groups. But intuitively, it's easy to see that, given that a transform can be represented as a matrix, and given that the inverse of a transform matrix is the matrix that multiplied by the original transform matrix gives you the identity matrix, that it's going to come down to the mathematical operations of division and multiplication.
In particular, just as I can provide the multiplicative inverse of any scalar except zero, simply by dividing the multiplicative identity (i.e. 1) by that number, likewise I can provide the multiplicative inverse of any matrix that doesn't require dividing by zero when I "divide" the identity by that matrix.
Geometrically, this fails (i.e. dividing by zero would occur) if your transform somehow causes the transformed geometry to have a non-zero length in some dimension.
From that, we can see that if you use a scaling transform where the scale factor is zero in at least one axis, that transform will be non-invertible.
And indeed:
GeneralTransform t = new ScaleTransform(1, 0).Inverse;
Returns null.
Do you need to worry about it? I don't know. That depends on how you're creating your transforms in the first place. That detail isn't present in your question.
Typically, I don't think it's something you'd need to think about. But if you're in a situation where for whatever reason a transform's scale factor winds up at zero, either through user input (whether entered numerically or through dragging the size of some shape on screen), successively combining fractional scale factors, etc. then sure, it's theoretically possible you could find yourself with a non-invertible transform.
If it were me, unless I could prove with certainty that I'm not doing anything with the transforms that would cause it to be non-invertible, I would go ahead and make sure I handle the null result in some reasonable way. This could either be to change the preconditions to create that certainty, or maybe you allow it and then just not do anything with the inverted transform if it's not invertible (since if it's not, there's probably no reasonable on-screen rendering that would make any sense anyway, that seems like a reasonable approach).
Related
I am now interested in the bundle adjustment in SLAM, where the Rodrigues vectors $R$ of dimension 3 are used as part of variables. Assume, without loss of generality, we use Gauss-Newton method to solve it, then in each step we need to solve the following linear least square problem:
$$J(x_k)\Delta x = -F(x_k),$$
where $J$ is the Jacobi of $F$.
Here I am wondering how to calculate the derivative $\frac{\partial F}{\partial R}$. Is it just like the ordinary Jacobi in mathematic analysis? I have this wondering because when I look for papers, I find many other concepts like exponential map, quaternions, Lie group and Lie algebra. So I suspect if there is any misunderstanding.
This is not an answer, but is too long for a comment.
I think you need to give more information about how the Rodrigues vector appears in your F.
First off, is the vector assumed to be of unit length.? If so that presents some difficulties as now it doesn't have 3 independent components. If you know that the vector will lie in some region (eg that it's z component will always be positive), you can work round this.
If instead the vector is normalised before use, then while you could then compute the derivatives, the resulting Jacobian will be singular.
Another approach is to use the length of the vector as the angle through which you rotate. However this means you need a special case to get a rotation through 0, and the resulting function is not differentiable at 0. Of course if this can never occur, you may be ok.
Good morning, I'm trying to perform a 2D FFT as 2 1-Dimensional FFT.
The problem setup is the following:
There's a matrix of complex numbers generated by an inverse FFT on an array of real numbers, lets call it arr[-nx..+nx][-nz..+nz].
Now, since the original array was made up of real numbers, I exploit the symmetry and reduce my array to be arr[0..nx][-nz..+nz].
My problem starts here, with arr[0..nx][-nz..nz] provided.
Now I should come back in the domain of real numbers.
The question is what kind of transformation I should use in the 2 directions?
In x I use the fftw_plan_r2r_1d( .., .., .., FFTW_HC2R, ..), called Half complex to Real transformation because in that direction I've exploited the symmetry, and that's ok I think.
But in z direction I can't figure out if I should use the same transformation or, the Complex to complex (C2C) transformation?
What is the correct once and why?
In case of needing here, at page 11, the HC2R transformation is briefly described
Thank you
"To easily retrieve a result comparable to that of fftw_plan_dft_r2c_2d(), you can chain a call to fftw_plan_dft_r2c_1d() and a call to the complex-to-complex dft fftw_plan_many_dft(). The arguments howmany and istride can easily be tuned to match the pattern of the output of fftw_plan_dft_r2c_1d(). Contrary to fftw_plan_dft_r2c_1d(), the r2r_1d(...FFTW_HR2C...) separates the real and complex component of each frequency. A second FFTW_HR2C can be applied and would be comparable to fftw_plan_dft_r2c_2d() but not exactly similar.
As quoted on the page 11 of the documentation that you judiciously linked,
'Half of these column transforms, however, are of imaginary parts, and should therefore be multiplied by I and combined with the r2hc transforms of the real columns to produce the 2d DFT amplitudes; ... Thus, ... we recommend using the ordinary r2c/c2r interface.'
Since you have an array of complex numbers, you can either use c2r transforms or unfold real/imaginary parts and try to use HC2R transforms. The former option seems the most practical.Which one might solve your issue?"
-#Francis
The Matlab function bvp4c solves boundary value problems. It takes a differential equation, boundary conditions and an initial guess as input, and returns a structure array containing arrays of x, y and yp (which stands for "y prime", or y').
The length of the output arrays should be the same as that of the initial guess, but I found that it isn't always. I have checked the dimensions of the input (the initial guess, always 1x101 double for x and 16x101 double for y) and the output (sometimes 1x101 double for x and 16x101 double for y and yp as it should be, but often different values, such as 1x91 double and 16x91 double or 1x175 double and 16x175 double).
Looking at the output array x when its length is off, some extra values are squeezed in, or some are taken out. For example, the initial guess has 100 positions between x=0 and x=1, and the x array should be [0 0.01 0.02 ... 1], but sometimes a new position like 0.015 shows up.
Question: Why does this happen, and how can this be solved?
"The length of the output arrays should be the same as that of the initial guess ...." This is incorrect.
As described in the bvp4c documentation, sol.x contains a "[mesh] selected by bvp4c" with an "[approximation] to y(x) at the mesh points of sol.x". In order to evaluate bvp4c's solution on your mesh, use deval.
Why does bvp4c choose a mesh? Quoting from the cited paper1, which you can get in full here if you have a MathWorks account:
Because BVPs can have more than one solution, BVP codes require users to supply a guess for the solution desired. The guess includes a guess for an initial mesh that reveals the behavior of the desired solution. The codes then adapt the mesh so as to obtain an accurate numerical solution with a modest number of mesh points.
Because a steady BVP generally has a global behavior strongly dependent on its boundary values, the spatial mesh between the two boundaries may need to be refined in order to properly approximate the desired solution with the locally chosen basis functions for the method. However, there may also be portions of the mesh that do not need to be refined and can even be coarsened in some cases to maintain a reasonably small residual and accurate approximation. Therefore, for general efficiency, the guess mesh is adaptively refined or coarsened depending on some locally chosen metric (since bvp4c is collocation based, the metric is probably point-based or division-integrated based) such that the mesh returned by bvp4c is, in some sense, adequate enough for generic interpolation within the boundaries.
I'll also note that this is different from numerically solving IVPs since their state is not global across the entire time integration locus and only depends on the current state to the next time-step, and possibly previous time steps if using a multi-step method or solving a delay differential equation, which makes the refinement inherently local. This local behavior of IVPs is what allows functions like ode45 to return a solution at pre-selected time values because it can locally refine the solution at the selected point while performing the time march (this is known as dense output).
1 Shampine, L.F., M.W. Reichelt, and J. Kierzenka, "Solving Boundary Value Problems for Ordinary Differential Equations in MATLAB with bvp4c".
New to python and not sure about efficiency issues here. For vectors x, y, and z that represent the coordinates of n particles I can do the following computation
import numpy as np
X=np.subtract.outer(x,x)
Y=np.subtract.outer(y,y)
Z=np.subtract.outer(z,z)
R=np.sqrt(X**2+Y**2+Z**2)
A=X/R
np.fill_diagonal(A,0)
a=np.sum(A,axis=0)
With this calculation there is about a factor of 2 in redundancy in so far as multiplications and divisions go as the diagonals are not needed and the lower diagonal is just the negative of the upper diagonal. I plan to use this kind of computation inside a function call that is used by odeint - i.e. it would be called a lot and the vectors will be large - as large as my computer will handle. To remove it, naively I would end up doing a for loop which presumably is a stupid thing to do. Can I get rid of this redundancy in a vectorized way or is it even worth the effort?
Update: Based on the suggestions below, the only way I could see to improve was
ut=np.triu_indices(n,1)
X=x[ut[0]]-x[ut[1]]
With similar expressions for Y and Z and using pdist to find R. This construction only calculates the upper triangular part. Looking at the source code for pdist I am not convinced it does anything particularly smart so I think my expression above would be equally good. The use of squareform only produces the symmetric form. For the antisymmetric may as well use
B=np.zeros((n,n),dtype=np.float64)
B(ut[0],ut[1])=A
B=B-B.T
This cannot be slower than square form because this is pretty much exactly what squareform does. Since the function is called often it would seem to me that ut should be made static along with storage for others (X,Y,Z,A,B). However being new to python I'm not sure how that is done.
I have a vector A, represented by an angle and a length. I want to add vector B, updating the original A. B comes from a lookup table, so it can be represented in which ever way makes the computation easier.
Specifically, A is defined thusly:
uint16_t A_angle; // 0-65535 = 0-2π
int16_t A_length;
Approximations are fine. Checking for overflow is not necessary. A fast sin/cos approximation is available.
The fastest way I can think is to have B represented as a component vector, convert A to component, add A and B, convert the result back to angle/length and replace A. (This requires the addition of a fast asin/acos)
I am not especially good at math and wonder if I am missing a more sensible approach?
I am primarily looking for a general approach, but specific answers/comments about useful micro-optimizations in C is also interesting.
If you need to do a lot of additive operations, it would probably be worth considering storing everything in Cartesian coordinates, rather than polar.
Polar is well-suited to rotation operations (and scaling, I guess), but sticking with Cartesian (where a rotation is four multiplies, see below) is probably going to be cheaper than using cos/sin/acos/asin every time you want to do a vector addition. Although, of course, it depends on the distribution of operations in your case.
FYI, a rotation in Cartesian coordinates is as follows (see http://en.wikipedia.org/wiki/Rotation_matrix):
x' = x.cos(a) - y.sin(a)
y' = x.sin(a) + y.cos(a)
If a is known ahead of time, then cos(a) and sin(a) can be precomputed.