hyperbolic | Hyperbolic & Conformal Geometry Visualizations | Graphics library
kandi X-RAY | hyperbolic Summary
kandi X-RAY | hyperbolic Summary
Hyperbolic & Conformal Geometry Visualizations in D3 + React, written in ES6.
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hyperbolic Examples and Code Snippets
Lines of Code : 35License : Non-SPDX (Apache License 2.0)
Copydef log_cosh(y_true, y_pred):
"""Logarithm of the hyperbolic cosine of the prediction error.
`log(cosh(x))` is approximately equal to `(x ** 2) / 2` for small `x` and
to `abs(x) - log(2)` for large `x`. This means that 'logcosh' works mostly
def tanh(x):
"""Hyperbolic tangent activation function.
For example:
>>> a = tf.constant([-3.0,-1.0, 0.0,1.0,3.0], dtype = tf.float32)
>>> b = tf.keras.activations.tanh(a)
>>> b.numpy()
array([-0.9950547, -0.7 Community Discussions
Trending Discussions on hyperbolic
QUESTION
I am writing a 2D simulator and game using the HTML canvas which involves orbital mechanics. One feature of the program is to take the position and velocity vector of a satellite at one point and return the semi-major axis, eccentricity, argument of periapsis, etc of a 2D orbit around one planet. When the eccentricity is less than one, I can easily graph the orbit as an ellipse using ctx.ellipse(). However, for eccentricities greater than one, the correct shape of the orbit is a hyperbola. At the moment, my program just draws nothing if the eccentricity is greater than one, but I would like it to graph the correct hyperbolic orbit. Since there is no built in "hyperbola" function, I need to convert my orbit into a Bézier curve. I am at a bit of a loss as to how to do this. The inputs would be the location of one focus, semi-major axis, eccentricity, and argument of periapsis (basically how far the orbit is rotated) and it should return the correct control points to graph a Bézier curve approximation of a hyperbola. It does not have to be exactly perfect, as long as it is a close enough fit. How can I approach this problem?
...ANSWER
Answered 2021-May-12 at 15:45In terms of conic sections, hyperbola are unfortunately the one class of curves that the Canvas cannot natively render, so you're stuck with approximating the curve you need. There are some options here:
- Flatten your curve, by sampling the hyperbola at one or two points in the distance and lots of points near the extrema so that you can draw a simple polygon that looks like a curve.
- Model the hyperbola with a single "best approximation" quadratic or cubic curve.
- As @fang mentions: sample the curve at a few points and convert the Catmull-Rom spline through those points to Bezier form.
- Combine approaches 1 and 2. using a single Bezier curve to approximate the part of the hyperbola that actually looks curved, and using straight lines for the parts that don't.
- Combine approaches 1 and 3, using a Catmull-Rom spline for the curvy bit, and straight lines for the straight bits.
Curve flattening is basically trivial. Rotate your curve until it's axis-aligned, and then just compute y given x using the standard hyperbolic function, where a is half the distance between the extrema, and b is the semi-minor axis:
QUESTION
I have a hyperbolic function and i need to find the 0 of it. I have tried various classical methods (bisection, newton and so on).
Second derivatives are continuous but not accessible analytically, so i have to exclude methods using them.
For the purpose of my application Newton method is the only one providing sufficient speed but it's relatively unstable if I'm not close enough to the actual zero. Here is a simple screenshot:
The zero is somewhere around 0.05. and since the function diverges at 0, if i take a initial guess value greater then the minimum location of a certain extent, then i obviously have problems with the asymptote.
Is there a more stable method in this case that would eventually offer speeds comparable to Newton?
I also thought of transforming the function in an equivalent better function with the same zero and only then applying Newton but I don't really know which transformations I can do.
Any help would be appreciated.
...ANSWER
Answered 2021-Jan-14 at 08:51How about using log(x) instead of x?
QUESTION
I am interested in the Hyperbolic Tessellations such as those generated by @TilingBot. To narrow it down a bit more, I would like to be able to construct some of the Uniform Tilings on the Hyperbolic Plane, such as this:
The closest answer I have found comes from Math SE and recommends these 3 resources:
- Ajit Datar's master's thesis
- David Joyce's Hyperbolic Tessellations applet
- And David Joyce's corresponding Java source code.
Here I have translated the Java to JavaScript (and preserved the comments), as well as attempted to draw the center shape:
...ANSWER
Answered 2020-Sep-28 at 23:05I would recommend you do a stroke instead of fill, that way you see what that polygon is giving you.
Run the code below so you can see the difference...
Now, comparing that result with your image it looks nothing like what you want
QUESTION
I have two data list (x1, y1), (x2, y2) ...
I do not know the equation between x and y. So, I tried to use neural network to find it.
The hyperbolic.txt file has (x1, y1), (x2, y2) ...
The codes are below, but it does not work.
...ANSWER
Answered 2020-Aug-25 at 13:49You got it right, you need to feed your network a (N,1) tensor and not a (N,) tensor.
The easiest solution might be to add the new dimension on the numpy side, using either np.newaxis (which is None) or the np.reshape function.
So you can apply this in gen_data_batch, replacing yield np.array(x_batch), np.array(y_batch) by yield np.array(x_batch)[:, np.newaxis], np.array(y_batch)[:, np.newaxis] for example.
You can also add this new axis on np_poses_x and np_poses_y :
QUESTION
I use julia 1.4, and running the following code:
...ANSWER
Answered 2020-Aug-19 at 15:55unshift! is existing function in julia 1.4
Where did you see this? It was renamed for Julia 1.0 two years ago to pushfirst!:
QUESTION
I am completely new to docker, and I need to run a code in docker environment.
I get error in building Dockerfile :
I'm running Ubuntu 20.04 via hyper-V, and when I build the Dockerfile, I get the following message:
...ANSWER
Answered 2020-Aug-11 at 22:57The /bin/sh: 1: /opt/conda/bin/conda: not found error is caused by conda not being installed properly. This is because the miniconda.sh file that you download (with curl -o ~/miniconda.sh -O https://repo.continuum.io/miniconda/Miniconda3-latest-Linux-x86_64.sh) is an empty file.
This happens because curl by default doesn't follow redirects, and the response from the url above sends a redirect rather than the miniconda.sh file directly. You can verify this by checking the headers of the URL (e.g. curl -i https://repo.continuum.io/miniconda/Miniconda3-latest-Linux-x86_64.sh shows a HTTP 301 redirect statuscode).
You can fix this by telling curl to follow redirects, by providing the -L flag in your curl command, e.g. :
(I also had to remove sudo)
QUESTION
I've generated a reciprocal ROM using Quartus II and I've developed a circuit that calculates the hyperbolic tangent. I'm facing the following error when I try to simulate a testbench of my circuit. Note that I've used the ROM as an instantiate in my circuit.
...ANSWER
Answered 2020-Jul-29 at 06:08Probably you just need to load the Altera libraries in Modelsim:
- using GUI:
Simulate > Start Simulation > Libraries > Add > altera_mf_ver - using console: add
-L altera_mf_verto your command.
QUESTION
I am trying to convert a C/C++ API package into a Python extension. I have gone through this visual studio tutorial succefully. This is a simple tutorial and maybe not very practical for a real convertion.
Here is my task. I have the include folder, a folder of *.dll and *.lib files, and some demo cpp files.
question 1: where to place *.dll and *.lib folders?
Following the visual studio tutorial, I created a C++ project. Where can I place the *.dll and *.lib files into the project before Release and Build the project?
Question 2: Is there a way to import C++ functions and classed in batch?
To use pybind11, we need to add following code at the bottom of cpp file.
ANSWER
Answered 2020-Jun-09 at 10:321) The dlls will need to be packaged with the python extension (see below) and most likely installed in the same folder.
2) No you need to wrap them individually. This is less effort than it sounds, and even if some sort of batching was available you'd still need to specify them all.
3) This question is about packaging. Your setup.py is probably based on setuptools which can be used to create wheels that can be installed with pip without having to rebuild each time. Anaconda also has its own package management system.
QUESTION
I have a question on a specific implementation of a Nelder-Mead algorithm (1) that handles box contraints in an unusual way. I cannot find in anything about it in any paper (25 papers), textbook (searched 4 of them) or the internet.
I have a typical optimisation problem: min f(x) with a box constraint -0.25 <= x_i <= 250
The expected approach would be using a penalty function and make sure that all instances of f(x) are "unattractive" when x is out of bounds.
The algorithm works differently: the implementation in question does not touch f(x). Instead it distorts the parameter space using an inverse hyperbolic tangens atanh(f). Now the simplex algorithm can freely operate in a space without bounds and pick just any point. Before it gets f(x) in order to assess the solution at x the algorithm switches back into normal space.
At a first glance I found the idea ingenious. This way we avoid the disadvantages of penalty functions. But now I am having doubts. The distorted space affects termination behaviour. One termination criterion is the size of the simplex. By inflating the parameter space with atanh(x) we also inflate the simplex size.
Experiments with the algorithm also show that it does not work as intended. I do not yet understand how this happens, but I do get results that are out of bounds. I can say that almost half of the returned local minima are out of bounds.
As an example, take a look at nmkb() optimising the rosenbrook function when we gradually change the width of the box constraint:
...ANSWER
Answered 2020-May-02 at 18:43(This question has nothing to do with optimx, which is just a wrapper for R packages providing unconstrained optimization.)
The function in question is nmkb() in the dfoptim package for gradient-free optimization routines. The approach to transform bounded regions into unbounded spaces is a common one and can be applied with many different transformation functions, sometimes depending on the kind of the boundary and/or the type of the objective function. It may also be applied, e.g., to transform unbounded integration domains into bounded ones.
The approach is problematic if the optimum lies at the boundary, because the optimal point will be sent to (nearly) infinity and cannot ultimately be reached. The routine will not converge or the solution be quite inaccurate.
If you think the algorithm is not working correctly, you should write to the authors of that package and -- that is important -- add one or two examples for what you think are bugs or incorrect solutions. Without explicit code examples no one here is able to help you.
(1) Those transformations define bijective maps between bounded and unbounded regions and the theory behind this approach is obvious. You may read about possible transformations in books on multivariate calculus.
(2) The approach with penalties outside the bounds has its own drawbacks, for instance the target function will not be smooth at the boundaries, and the BFGS method may not be appropriate anymore.
(3) You could try the Hooke-Jeeves algorithm through function hjkb() in the same dfoptim package. It will be slower, but uses a different approach for treating the boundaries, no transformations involved.
EDIT (after discussion with Erwin Kalvelagen above)
There appear to be local minima (with some coordinates negative).
If you set the lower bounds to 0, nmkb() will find the global minimum (1,1,1,1,1,1) in any case.
Watch out: starting values have to be feasible, that is all their coordinates greater 0.
QUESTION
I'm learning Python (for fun) through the book ThinkPython. So far I'm really enjoying this new hobby. One of the recent exercise was crafting a Archimedes spiral.In order to test my skills I've been working on making a hyperbolic spiral but have been stumped!
according to the equation r=a+b*theta^(1/c)
- when c=1 we see a Archimedes spiral
- when c=-1 we should see a hyperbolic spiral
I'm using the following code and would appreciate any help towards the right direction.
- Extra points to whoever can help me in the right direction without directly giving away the answer. Unless it's simply a formatting issue.
- No points if the answer suggested involves using (x,y) coordinates to draw.
ANSWER
Answered 2020-May-01 at 05:51Simply adjusting the constants passed to draw_spiral() as arguments:
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