I am reading Cucker and Zhou’s “Learning Theory: An Approximation Theory Viewpoint”. I am trying to understand the accuracy versus precision separation in Digital Elevation Models (DEMs) by looking at how the bias/variance decomposition is done. Are these two different ways of decomposing error isomorphic? Is accuracy the same as bias, precision variance? I don’t think so. The accuracy error in a DEM could be defined by considering translations and other operations such as rotations and shearing. These operations being global could be used to define what one wants to call the accuracy of a model. What remains after these global operations would then be the precision error. This view of error decomposition means that it is not possible to look at something like the average error over a DEM and blindly state that it can also be decomposable into an accuracy and precision part. This is only possible for height estimates that have no uncertainty in their x,y locations. This is the case for the autonomous difference equations we described in the 8th Optical 3D conference. They are invariant under arbitrary rotations and rotations about axes that are parallel to the z-axis.

Getting back to Cucker and Zhou’s book. I found the following comment about “probably approximately correct” PAC learning interesting: “Extensions of PAC learning allowing for labeling mistakes with small probability exist.” Thus the “probably” part of PAC error estimates assumes that you have perfect training data. This is just another example of how little we know about error and how to properly take it into account in our estimation algorithms.

In the case of the DEM models, the autonomous difference equations can only detect an error that is invariant under the abelian operations of translations and rotations about a single axis. If I tried to decompose the height error taking into account x,y uncertainty — I would find that the average error is not decomposable into separable parts. Somehow, one must be able to express this non-separability as being technically due to the non-abelian character of 3D rotations.

This is my last semester teaching the Physics for Scientists and Engineers course at UMass. I’m currently covering Newton’s 3rd law in class. My training in Aikido and Iaido has given me plenty of examples of how physics applies to martial arts. One good example of this is sword cutting.

The most basic sword cut is called kiri-otoshi (“sword drop”) in Japanese. You bring the sword overhead and swing forward and down to perform a vertical cut. What may not be realized by those that perform the cut is that a lot of its effectiveness relies on the fact we live on a massive planet like Earth. This can be understood by thinking about Newton’s 3rd law: every force has an opposite and equal reaction.

This law is iron-clad. It always is true (or so we believe until proven otherwise by experiments) no matter how complicated the motion you are seeking to describe. If you punch someone, the force your strike exerts on them is exactly equal in magnitude to the force they exert on you during the strike. The ‘opposite’ in the law means that the forces are mirror images of each other. A force exerted forward on an object results in the object exerting a force backward.

A kiri-otoshi strike can be decomposed into two motions — an initial forward throw of the sword followed by a downward push on the sword. Therefore, by Newton’s 3rd law, the forward force you exert on the sword results in a backward force on you. This backward force does not push you back because of the frictional force the floor exerts forward on you (another application of the 3rd law).

A similar ‘chain’ of forces occurs during the downward phase of kiri-otoshi. The sword goes down because it is free-falling and it is also being pushed down. This downward force means, by the 3rd law, that the sword is exerting an upward force on you. Stated in mathematical notation:

$$F_{\text{you on sword}} = -F_{\text{sword on you}} $$

The outcome of all of this is that your apparent weight is thus reduced while you are cutting. The sword cut lifts you into the air but gravity keeps you planted. So the force of your strike depends on the strength of the planet’s gravity. Cutting in outer space would be less effective because there would be no external force to pull you down while the sword is pulling you up.