People doing Machine Learning [ML] experiments on their own Linux PCs or laptops know that the numerical training runs put a heavy load on the graphics cards and consume a lot of energy as a direct consequence. Especially in a hot summer like we have it in Germany right now, cooling of your systems may become a problem. And as energy has a high price tag here, any method to reduce the load and/or power consumption is welcome.

But I think that caring about energy consumption is a topic which we as a Linux and ML enthusiasts should keep in mind in general. Some big tech companies will probably not do it – as long as their money machinery works and as some heads follow fantasies about building small nuclear power plants for their big AI data centers. But we Opensource people would like to see more AI- and ML-services independent of the monopolists and their infrastructure, anyway. Not only for reasons of data and privacy protection.

As soon as we, however, proclaim and work for a development that favors local and resource optimized installations of AI and ML tools both for private people and companies, we have to care about side effects: We have to bring the energy consumption down for these many local installations substantially in parallel. Otherwise, centralized solutions may have a better energy efficiency than decentralized solutions.

For me as a retired person in Germany the general financial pressure is high enough to enforce a careful use of my private resources. With this post I want to draw your attention to two points which may help you, too, to save energy during your ML-experiments. (In addition to or aside of standard measures like saving certain model states during training runs to get better starting points for new runs.)

Working with Machine Learning and Deep Neural Networks not only requires GPU drivers, but in case of Nvidia GPUs also the installation of CUDA and cuDNN. This process is always a bit tricky as additional environment variables have to be set for IPython-based Jupyterlab or classic Jupyter Notebook. On an Opensuse system one must in addition take care of the right settings in /etc/alternatives.

I hope this helps people who want to use Leap 15.5 for Machine Learning with Nvidia GPUs, Keras/Tensorflow 2 and Jupyterlab.

Important addendum 01/27/2024:
Although the combination of CUDA 12.3, cuDNN 8.9.7, Tensorflow 2.15 and Nvidia drivers 545.29.06 works regarding AI-models, there is another major problem:
Nvidia’s driver 545.29.06 is buggy – at least for Leap 15.5, KDE/Plasma with multiple screens. The bug affects Suspend-to-RAM. Suspend-to-RAM seems to work in the suspend phase, and the system also comes up afterward in a seemingly proper state of your KDE/Plasma interface (on your screens).

However, the problems begin when you want to change to another virtual screen via Ctrl-Alt-Fx. You wait and wait and wait … The same for changing the run-level or systemd target state or when you want to shut the system down. This makes Suspend-to-RAM with driver 545.29.06 impossible to use.

Recommendation:
If you have a working older Nvidia driver (e.g. a stable 535 version) do not change to 545.29.06. Unfortunately, it is a mess on a multiscreen Leap 15.5 system to return to an older driver version. The Nvidia community repository does not offer you a choice. (Why by the way ????). Downloading an older proprietary driver from Nvidia and trying to install it afterward on a console terminal (after having stopped X11 or Wayland) did not work in my case – the screens displaying the terminal changed their resolution and froze afterward. So, you may have to completely uninstall the present driver 545 completely, go back to standard VGA and then try to install an older driver via Nvidias install mechanism. As I said: It is a mess …

we have clarified some basic properties of shear transformations [SHT]. We got interested in this topic, because Autoencoders can produce latent multivariate normal vector distributions, which in turn result from linear transformations of multivariate standard normal distributions. When we want to analyze such latent vector distributions we should be aware transformations of quadratic forms. An important linear transformation is a shear operation. It combines aspects of scaling with rotations.

The objects we applied SHTs to were so far only squares and cubes. Both (discrete) rotational and plane symmetries of the squares and cubes were broken by SHTs. We also saw that this symmetry breaking could not be explained by a pure scaling operation in another rotated Euclidean Coordinate System [ECS]. But cubes do not have a continuous rotational symmetry. The distances of surface points of a cube to its symmetry center show no isotropy.

However, already in the first post when we superficially worked with Blender we got the impression that the shearing of a sphere seemed to produce a figure with both plane and discrete rotational symmetries – namely ellipsoids, wich appeared to be rotated. We still have to prove this, mathematically. With this post we move a first step in this direction: We will apply a shear operation to a 2D-body with perfect continuous rotational symmetry in all directions, namely a circle. A circle is a special example of a quadratic form (with respect to the vector component values). We center our Euclidean Coordinate System [ECS] at the center of the circle. We know already that this point remains a fix-point of our transformations. As in the previous post I use Python and Matplotlib to produce visual results. But we support our impression also by some simple math.

We first check via plotting that the shear operations move an extremal point of the circle (with respect to the y-coordinate) along a line y_{max} = const. (Points of other layers for other values y_{l} = const also move along their level-lines.) We then have to find out whether the produced figure really is an ellipse. We do so by mathematically deriving its quadratic form with respect to the coordinates of the transformed points. Afterward, we derive the coordinate values of points with extremal y-values after the shear transformation.

In addition we calculate the position of the points with maximum and minimum distance from the center. I.e., we derive the coordinates of end-points of the main axes of the ellipse. This will enable us to calculate the angle, by which the ellipse is rotated against the x-axis.

The astonishing thing is that our ellipse actually can be created by a pure scaling operation in a rotated ECS. This seems to be in contrast to our insight in previous posts that a shear matrix cannot be diagonalized. But it isn’t … It is just the rotational symmetry of the circle that saves us.

Shearing a circle

We define a circle with radius r = a = 2.

I have indicated the limiting line at the extremal y-values. From the analysis in the last post we expect that a shear operation moves the extremal points along this line.

We now apply a shearing matrix with a x/y-shearing parameter λ = 2.0

Thus, we have indeed produced a rotated ellipse! We see this from the fact that the term mixing the x_{s} and the y_{l} coordinates does not vanish.

Position of maximum absolute y-values

We know already that the y-coordinates of the extremal points (in y-direction) are preserved. And we know that these points were located at x = 0, y = a. So, we can calculate the coordinates of the shifted point very easily:

In our case this gives us a position at (4, 2). But for getting some experience with the quadratic form let us determine it differently, namely by rewriting the above quadratic equation and by a subsequent differentiation. Quadratic supplementation gives us:

Let us also find the position of the end-points of the main axes of the ellipse. One method would be to express the ellipse in terms of the coordinates (x_{s}, y_{s}), calculate the squared radial distance r_{s} of a point from the center and set the derivative with respect to x_{s} to zero.

The “problem” with this approach is that we have to work with a lot of terms with square roots. Sometimes it is easier to just work in the original coordinates and express everything in terms of (x, y):

For our shearing in x-direction a_{s} gives us the longer axis. The rotational angle α between the longer main axis and the x-axis can be calculated via:

Plot of main axes, their end-points and of the points with maximum y-value

The coordinate data found above help us to plot the respective points and the axes of the produced ellipse. The diameters’ end-points are plotted in red, the points with extremal y-value in green:

It becomes very clear that the points with maximum y-values are not identical with the end-points of the ellipse’s main symmetry axes. We have to keep this in mind for a discussion of higher dimensional figures and vector distributions as multidimensional spheres, ellipsoids and multivariate normal distributions in later posts.

Rotated ECS to produce the ellipse?

The plot above makes it clear that we could have created the ellipse also by switching to an ECS rotated by the angle α. Followed by a simple scaling in x- and y-direction by the factors a_{s} and b_{s} in the rotated ECS. This seems to be a contradiction to a previous statement in this post series, which said that a shear matrix cannot be diagonalized. We saw that in general we cannot find a rotated ECS, in which the shear transformation reduces to pure scaling along the coordinate axes. We assumed from linear algebra that we in general need a first rotation plus a scaling and afterward a second different rotation.

But the reader has already guessed it: For a fully rotation-symmetric, i.e. isotropic body any first rotation does not change the figure’s symmetry with respect to the new coordinate axes. In contrast e.g. to squares or rectangles any rotated coordinate system is as good as any other with respect to the effect of scaling. So, it is just scaling and rotating or vice versa. No second rotation required. We shall in a later post see that this holds in general for isotropically shaped bodies.

Conclusion

Enough for today. We have shown that a shear transformation applied to a circle always produces an ellipse. We were able to derive the vectors to the points with maximum y-values from the parameters of the original circle and of the shear matrix. We saw that due to the circle’s isotropy we could reduce the effect of shearing to a scaling plus one rotation or vice versa. In contrast to what we saw for a cube in the previous post.