# Interpolating Classifications

In a previous article, I introduced a vectorized method for achieving interpolation, which works quite well, and is very fast. I abandoned it because I had other things getting my attention, including, maximizing the use of vectorization in my core algorithms, because it really changes their runtimes. I’ve also proven that you can’t beat the accuracy of my algorithms for a class of datasets that I’ve rigorously defined as, “locally consistent” datasets. The gist is, if classifications don’t change much over small distances, then the dataset is locally consistent. Because my core algorithms are also have linear runtimes, that’s pretty much the end of the story for locally consistent datasets, which probably covers a significant chunk of real world datasets.

So, I’ve started to think about datasets that are not locally consistent, and what’s interesting, is that the MNIST Fashion dataset seems to have some inconstant classes, because my algorithms have really high accuracy overall for MNIST Fashion, but have randomly horrible accuracy for certain rows (but not enough to push the overall accuracy down too much). So I’m now turning my attention to using interpolation, i.e., constructing a function that takes the features in a dataset, and calculates its class.

Before I decided to do this, I wanted to make sure that I wasn’t wasting my time, and creating something that is only superficially different to what I’ve already done, but mathematically equivalent. And it turns out, it is not the same as the distance-based clustering and prediction I’ve made use of so far. Specifically, it doesn’t require the same features to have similar values, which my current algorithms do –

It instead requires that the function returns the same value when evaluated for two elements of the same class. As a result, I’m going to spend a few days on this, because it is new, and perhaps it’s useful, and could provide a more efficient method of implementing what is a basically neural network.

# Efficiently Measuring Spatial Entropy

Spatial Entropy

I noted on Twitter that I came up with a really efficient way of calculating spatial entropy, and though I plan on writing something more formal on the topic, as it’s related to information theory generally, I wanted to at least write a brief note on the method, which is as follows:

Let $S$ be a set of $K$ points in Euclidean $N$-space. Sort them by row value, which will produce a sorted $K \times N$ matrix of vectors $M_S$. Now take the norm of the difference between all adjacent entries, $||M(row_{i}) -M(row_{i+1})||$, which will produce a $(1 \times K-1)$ vector $v$ of real numbers, that correspond to the norms of those differences. Let $T$ be the sum over the entries in the vector $v$, and take the entropy of the vector $\frac{v}{T}$, which will correspond to a distribution.

So for example, if you have $N$ points along a line, and all of them are equidistant from one another, then the entropy will be equal to the $\log(N-1)$, which corresponds to a uniform distribution, which is consistent with intuition, since you’d expect entropy to be maximized. In contrast, if all points are at the same location, then their distances would be zero, and the entropy would be zero.

You could do the same with a set of curves, but you’d need a comparison operator to first sort the curves and then compare their distances, and for this, you can use the operators I used in Section 1.3 of my A.I. summary (the “Expanding Gas Dataset”).

This allows for the measurement of the the entropy of an arbitrary dataset of vectors.

Information, Knowledge, and Uncertainty

The fundamental observation I’ve made regarding uncertainty, is that what you know about a system plus what you don’t, is all there is to know about the system. This sounds tautological, but so is the founding equation of accounting, which was also founded by an Italian mathematician, which it seems clear people are trying to forget.

Expressed symbolically, the equation states that:

$I = K + U$,

Where $I$ is the information content of the system, $K$ is your knowledge with respect to the system, and $U$ is your uncertainty with respect to the system. I go through some examples in this note on the matter, but the intuition should be straightforward.

Because we can now measure spatial entropy, we can measure the entropy of a dataset, which would also be the information content, $I$, of the dataset, but let’s table that for a moment.

My prediction algorithms return a set of vectors given an input vector, and it allows for blank entries in the input vector. So if you observe the first 50 out of 1,000 states of a system, you’ll get a set of vectors that reflect the information you have, and of course, that set winnows down in size as the number of observations increases (see Section 3.1 of A New Model of Artificial Intelligence).

Here’s a representative chart (from that paper) that shows how the set of possible paths of a random walk starts to tilt in favor of an upward path as data becomes available. In this case, 2,000 observations are available out of 10,000, and the input vector is in fact an upward tilting path.

We can also measure the entropy of this set of returned set of predictions, using the same method described above, which would in my opinion represent your uncertainty with respect to your prediction. This makes sense, because in the worst case, you get back the full training dataset, which would set your uncertainty equal to the information content of the dataset.  Expressed symbolically,

$I = U$.

This implies that your knowledge with respect to your prediction is zero, which is correct. That is, if your prediction algorithm says, my prediction is, the entire training dataset, then it doesn’t know anything at all. As observations become available, the set of returned predictions will shrink in size, which will reduce the entropy of the returned prediction set, implying non-zero knowledge. Eventually, you could be left with a single prediction, which would imply perfect knowledge, and no uncertainty, in the context of your training dataset, which is all you ever know about a given problem.

# Non-Locally Consistent Datasets

I’ve already written software that can solve basically all of the normal problems you have in machine learning, in low-degree polynomial time, for practical, real world datasets. The only requirement is that classifications are consistent near each point in the dataset, and then the accuracy of my algorithms cannot be beat (see my paper, Vectorized Deep Learning, for an explanation and examples). I’ve finalized the pseudo-code for a new set of algorithms that I’ve deliberately avoided publishing, because I’m worried they’re too powerful, because I know they can be used in thermodynamics, because that’s how I came up with them.

No one seems to care –

I’ve emailed the Congressional A.I. Caucus, the Bureau of Trade and Security, the International Trade Administration, and even a friend at the DOJ that I’m quite certain advises on defense issues, and no one cares.

# You and I, Third Movement

I’ve written a third movement to an experimental piece I’ve been working on, “You and I”, that combines electronic and classical techniques. However, the third movement is written entirely for acoustic instruments.

Enjoy!

# You and I (Music)

I’ve randomly decided to spend some time recording, and this is the result:

# A Note on Potential Energy

Introduction

In my main paper on physics, A Computational Model of Time-Dilation, I presented a theory of gravity that is consistent with the General Theory of Relativity (see Section 4), in that it implies the correct equations for time-dilation due to gravity, despite having no notion of space-time at all. I briefly discussed Potential Energy (see Section 4.2), hoping even then to address it formally in a follow up paper that would include other topics, but the economic pressure to focus on A.I. has led me to spend the vast majority of my time on A.I., not physics. As a result, I’ve decided to informally present my theory of potential energy, below, though I nonetheless plan to address the topic somewhat more formally in the work I’m now starting in earnest on thermodynamics, at the intersection of A.I. and physics.

Pressure

Imagine you have a wooden bookshelf. Even though the bookshelf is not visibly moving, in reality, there is some motion within the wood at very small scales, due to heat and intramolecular forces. Even if you cut into a plank, you still won’t see any motion inside of the wood at our scale of observation, because again, the motion is due to forces that are so small, you can’t see them with your eyes.

Now imagine you place a book on top of the shelf. Gravity will act on the book, and cause it to exert pressure on the surface of the bookshelf –

You know this must be the case, because it’s something you can perceive yourself, by simply picking up the book, and letting it lay in your hand.

Physicists discussing gravity typically ignore this aspect of pressure and focus instead on the notion of potential energy, saying that the book has some quantity of potential energy, because it is in the presence of a gravitational field, that would cause it to fall, but for the bookshelf.

We can instead make perfect physical sense of this set of facts without potential energy by assuming that the book is again actually moving at some very small scale, pushing the surface of the wood down, and in response, the intramolecular forces within the wood, which are now compressed past equilibrium, accelerate the surface of the wood back up, creating an oscillatory motion that is simply not perceptible at our scale of observation. In crude terms, the force of gravity pushes the book and the wood down, and the forces within the wood push the wood and the book back up, creating a faint oscillatory motion.

Note that this set of interactions doesn’t require potential energy at all, since if you push the book off of the shelf, it falls due to gravity, accumulating new kinetic energy. Therefore, the notion of potential energy is not necessary, at least for the purpose of explaining the accelerating effects of gravity.

The term I’ve used to describe this phenomenon is net effective momentum, which is simply the average total vector momentum of a system, over some period of time, or space. Returning to the example above, the net effective momentum of the book on the shelf would be roughly zero over any appreciable amount of time, since for every period of time during which the book moves up, there’s some period of time over which the book moves down, by a roughly equal distance, at roughly the same velocities, in each case, producing an average momentum that is very close to zero.

Now consider a baseball that’s just been hit by a bat –

Initially, the ball will likely be wobbly and distorted, if examined carefully over small periods of time, though its macroscopic velocity will probably stabilize quickly, and therefore, so will its net effective momentum, despite the internal tumult of the ball that certainly exists during its entire flight. The purpose of the net effective momentum is to capture the macroscopic motion of the ball, despite the complexities of the underlying motions.

As it turns out, my entire model of physics is statistical, and assumes that even elementary particles have complex motions, and the notion of net effective momentum is simply a macroscopic version of the definition of velocity that I present in Section 3.3 of my main paper.

# VeGa

I’ve completed the first part of my new book, VeGa.

Enjoy!

VeGa

# VeGa – Updated Draft

This is probably going to take me at least another year to finish, but here’s another draft of my book, VeGa.

# Analyzing Classification Consistency

Introduction

Real world data is often observably consistent over small distances. For example, the temperature on the surface of an object is probably going to be roughly the same over distances that are small relative to its total area. Similarly, the average color in an image of a real world object generally doesn’t change much over small distances, other than at its boundaries. Many commonly used datasets are also measurably consistent to some degree, in that classifications do not change over relatively small distances. In a previous article, I noted that if your dataset is what I call, “locally consistent”, which means, informally, that classifications don’t change over some fixed distance, then the accuracy of the nearest neighbor algorithm cannot be exceeded.

In this note, I’ll again present the idea of local consistency, though formally, through a series of lemmas. I’ve also attached code that makes it easy to measure the classification consistency of a dataset, which in turns implies that you can say, ex ante, whether nearest neighbor will in fact be the best the algorithm for the dataset, in terms of accuracy. Finally, because nearest neighbor can be implemented using almost entirely vectorized operations, it would have at worst a linear runtime as a function of the number of rows in the dataset, on a sufficiently parallel machine.

Local Consistency

A dataset $A$ is locally consistent if there is a value $\delta_x > 0$, for each $x\in A$, such that the set of all points for which $||x - y_i|| \leq \delta_x$, $B_x = \{y_1, \ldots, y_k\}$, has a single classifier, equal to the classifier of $x$, and $B_x$ is non-empty, for all $x$. That is, all of the points within $B_x$ are within $\delta_x$ of $x$, and are of a single class. We say that a dataset is locally consistent around a single point $x \in A$, if $B_x$ is non-empty.

For reference, the nearest neighbor algorithm simply returns the vector within a dataset that has the smallest Euclidean distance from the input vector. Expressed symbolically, if $f(x,A)$ implements the nearest neighbor algorithm, as applied to $x$, over $A$, and $y = f(x,A)$, for some $y \neq x$, then $||x - y|| \leq ||x - z||$, for all $z \in A$. Further, assume that if there is at least one $y_i \in A$, such that $||x - y|| = ||x - y_i||$, then the nearest neighbor algorithm will return a set of outputs, $\{y, y_1, \ldots, y_k\}$.

Lemma 1. If a dataset $A$ is locally consistent, then the nearest neighbor algorithm will never generate an error.

Proof. Assume that the nearest neighbor algorithm is applied to some dataset $A$, and that $A$ is locally consistent. Further, assume that for some input vector $x$, the output of the nearest neighbor algorithm, over $A$, contains some vector $y$, and that $y$ has a classifier that is not equal to the classifier of $x$, thereby generating an error. Since $y$ is contained in the output of the nearest neighbor algorithm, as applied to $x$, over $A$, it follows that $||x - y||$ is the minimum over the entire dataset. Because $A$ is locally consistent, there is some nonempty set $B_x$, that contains every vector $y_i$, for which, $||x - y_i|| \leq \delta_x$. Since $||x - y||$ is minimum, it must be the case that $||x - y|| \leq ||x - y_i|| \leq \delta_x$, for all $y_i$. But this implies that $y \in B_x$, which contradicts our assumption that the classifiers of $x$ and $y$ are not equal. Therefore, if we nonetheless assume their classifiers are not equal, then $A$ is not locally consistent, which leads to a contradiction, thereby completing the proof.

What Lemma 1 says is that the nearest neighbor algorithm will never generate an error when applied to a locally consistent dataset, since the nearest neighbor of an input vector $x$ is always an element of $B_x$, which guarantees accuracy. As a practical matter, if the nearest neighbor isn’t within $\delta_x$ of an input vector $x$, then we can flag the prediction as possibly erroneous. This could of course happen when your training dataset is locally consistent, and your testing dataset adds new data that does not fit into any $B_x$.

Corollary 2. A dataset is locally consistent if and only if the nearest neighbor algorithm does not generate any errors.

Proof. Lemma 1 shows that if a dataset $A$ is locally consistent, then the nearest neighbor algorithm will not generate any errors. So assume that the nearest neighbor algorithm generates no errors when applied to $A$, and for each $x\in A$, let $y_x$ denote a nearest neighbor of $x$, and let $\delta_x = ||x - y_x||$. Let $B_x = y_x$, and so each $B_x$ is non-empty. If more than one vector is returned by the nearest neighbor algorithm for a given $x$, then include all such vectors in $B_x$. By definition, $||x - y_x|| \leq \delta_x$. Therefore, there is a value $\delta_x > 0$, for each $x\in A$, such that the set of all points for which $||x - y_i|| \leq \delta_x$, $B_x$, has a single classifier, and $B_x$ is non-empty for all such $x$, which completes the proof.

Corollary 3. A dataset is locally consistent around $x \in A$, if and only if the nearest neighbor algorithm does not generate an error, when applied to $x$, over $A$.

Proof. Assume that the nearest neighbor algorithm does not generate an error when applied to $x \in A$. It follows that there is some $B = \{y_1, \ldots, y_k\} \subseteq A$, generated by the nearest neighbor algorithm, as applied to $x$, over $A$, for which $||x - y_i||$ is minimum, and the classifiers of $x$ and $y_i$ are all equal. Let $B_x = B$, and let $\delta_x = ||x - y_i||$. Since, the classifiers of $x$ and $y_i$ are all equal, this implies that $A$ is locally consistent around $x$.

Now assume that $A$ is locally consistent around $x$, and let $B = \{y_1, \ldots, y_k\} \subseteq A$ be the output of the nearest neighbor algorithm, as applied to $x$, over $A$. It follows that $||x - y_i||$ is minimum over $A$, and therefore, $B \subseteq B_x$. This implies that the classifier of $x$ equals the classifier of each $y_i$, which completes the proof.

Clustering

We can analogize the idea of local consistency to geometry, which will create intuitive clusters that occupy some portion of Euclidean space. We say that a dataset $A$ is clustered consistently, if for every $x \in A$, there exists some $\delta_x > 0$, such that all points contained within a sphere of radius $\delta_x$ (including its boundary), with an origin of $x$, denoted $S_x$, have the same classifier as $x$.

Lemma 4. A dataset is locally consistent if and only if it is clustered consistently.

Proof. Assume $A$ is locally consistent. It follows that for each $x \in A$, there exists a non-empty set of vectors $B_x$, each of which have the same classifier as $x$. Moreover, for each $y_i \in B_x$, it is the case that $||x - y_i|| \leq \delta_x$. This implies that all vectors within $B_x$ are contained within a sphere of radius $\delta_x$, with an origin of $x$. Therefore, $A$ is clustered consistently.

Now assume that $A$ is clustered consistently. It follows that all of the vectors $y_i$ within $S_x$ have a distance of at most $\delta_x$, from $x$. Let $B_x$ be the set of all vectors within $S_x$. This implies that for every $x \in A$, there exists a non-empty set of vectors $B_x$, all with the same classifier, such that for all $y_i \in B_x$, $||x - y_i|| \leq \delta_x > 0$. Therefore, $A$ is locally consistent, which completes the proof.

What Lemma 4 says, is that the notion of local consistency is equivalent to an intuitive geometric definition of a cluster that has a single classifier.

Lemma 5. If a dataset $A$ is locally consistent, and two vectors $x_1$ and $x_2$ have unequal classifiers, then there is no vector $y \in A$, such that $y \in C = (S_{x_1} \cap S_{x_2})$.

Proof. If the regions bounded by the spheres $S_{x_1}$ and $S_{x_2}$ do not overlap, then $C$ is empty, and therefore cannot contain any vectors at all. So assume instead that the regions bounded by the spheres $S_{x_1}$ and $S_{x_2}$ do overlap, and that therefore, $C$ is a non-empty set of points, and further, that there is some $y \in C$, such that $y \in A$.

There are three possibilities regarding the classifier of $y$:

(1) It is equal to the classifier of $x_1$;

(2) It is equal to the classifier of $x_2$;

(3) It is not equal to the classifiers of $x_1$ or $x_2$.

In all three cases, it follows that $y \in B_{x_1}$, and $y \in B_{x_2}$, which leads to a contradiction, since the Lemma assumes that the classifiers of $x_1$ and $x_2$ are not equal. Since these three cases cover all possible values for the classifier of $y$, it must be the case that $y$ does not exist, which completes the proof.

What Lemma 5 says, is that if two clusters overlap as spheres, then their intersecting region contains no vectors from the dataset.

Corollary 6. If $A$ is locally consistent, then any clustering algorithm that finds the correct value of $\delta_x$, for all $x \in A$, can generate clusters over $A$ with no errors.

Proof. It follows immediately from Lemma 5, since any such algorithm could simply generate the sphere $S_x$, for all $x \in A$.

You can easily approximate the value of $\delta$ for every row in a dataset using the code I’ve attached below, as applied to the MNIST Fashion Dataset, and it also generates the set $B_x$ for every row (it’s stored in the “final_rows” in the attached code). You can then count how many rows are locally consistent, which allows you to measure the classification consistency of the dataset.

Here’s a more formal PDF document that presents the same results: