Computational Processes that Appear to Model Human Memory

Abstract

This paper presents two computable functions, $\omega$ and $\varepsilon$, that map networks into networks. If all cognition occurs as an active neural network, then it is thought that $\omega$ models long-term memory consolidation and $\varepsilon$ models memory recall. A derived, intermediate network form, consisting of chordless cycles, could be the structural substrate of long-term memory; just as the double helix is the necessary substrate for genomic memory.

A Appendix

Too much formal mathematics makes a paper hard to read. Yet, it is important to be able to check some of the statements made in the body of the paper. In this appendix we provide a few propositions to formally prove some of our assertions.

The order in which nodes, or more accurately the singleton subsets, of $\mathcal{N}$ are encountered can alter which points are subsumed and subsequently deleted. Nevertheless, we show below that the reduced trace $\mathcal{T}= \mathcal{N}.\omega$ will be unique, up to isomorphism.

Proposition 1

Let $\mathcal{T}= \mathcal{N}.\omega$ and $\mathcal{T}' = \mathcal{N}.\omega '$ be irreducible subsets of a finite network $\mathcal{N}$, then $\mathcal{T}\cong \mathcal{T}'$.

Proof

Let $y_0 \in \mathcal{T}$, $y_0 \not \in \mathcal{T}'$. Then $y_0$ can be subsumed by some point $y_1$ in $\mathcal{T}'$ and $y_1 \not \in \mathcal{T}$ else because $y_0.\eta \subseteq y_1.\eta$ implies $y_0 \in \{y_1\}.\varphi$ and $\mathcal{T}$ would not be irreducible.

Similarly, since $y_1 \in \mathcal{T}'$ and $y_1 \not \in \mathcal{T}$, there exists $y_2 \in \mathcal{T}$ such that $y_1$ is subsumed by $y_2$. So, $y_1.\eta \subseteq y_2.\eta$.

Now we have two possible cases; either $y_2 = y_0$, or not.

Suppose $y_2 = y_0$ (which is often the case), then $y_0.\eta \subseteq y_1.\eta$ and $y_1.\eta \subseteq y_2.\eta$ or $y_0.\eta = y_1.\eta$. Hence $i(y_0) = y_1$ is part of the desired isometry, i.

Now suppose $y_2 \ne y_0$. There exists $y_3 \ne y_1 \in \mathcal{T}'$ such that $y_2.\eta \subseteq y_3.\eta$, and so forth. Since $\mathcal{T}$ is finite this construction must halt with some $y_n$. The points $\{ y_0, y_1, y_2, \dots y_n \}$ constitute a complete graph $Y_n$ with $\{y_i\}.\eta = Y_n.\eta$, for $i \in [0, n]$. In any reduction all $y_i \in Y_n$ reduce to a single point. All possibilities lead to mutually isomorphic maps.    $\square$

In addition to $\mathcal{N}.\omega$ being unique, we may observe that the transformation $\omega$ is functional because $\omega$ maps all subsets of N onto $N_{\omega }$. So we can have , thus “deleting” z. Similarly, $\varepsilon$ is functional because provides for the inclusion of new elements. Both $\omega$ and $\varepsilon$ are monotone, if we only modify its definition to be $X \subseteq Y$ implies $X.\varepsilon \subseteq Y.\varepsilon$, provided .

The following proposition characterizes the structure of irreducible traces.

Proposition 2

Let $\mathcal{N}$ be a finite symmetric network with $\mathcal{T}= \mathcal{N}.\omega$ being its irreducible trace. If $y \in \mathcal{T}$ is not an isolated point then either

• (1) there exists a chordless k-cycle C, $k \ge 4$ such that $y \in C$, or

• (2) there exist chordless k-cycles $C_1, C_2$ each of length $\ge 4$ with $x \in C_1$ $z \in C_2$ and y lies on a path from x to z.

Proof

 

• (1) Let $y \in N_{\mathcal{T}}$. Since y is not isolated, we let $y = y_0$ with $y_1 \in y_0.\eta$, so $(y_0, y_1) \in E$. Since $y_1$ is not subsumed by $y_0$, $\exists y_2 \in y_1.\eta , y_2 \not \in y_0.\eta$, and since $y_2$ is not subsumed by $y_1$, $\exists y_3 \in y_2.\eta$, $y_3 \not \in y_1.\eta$. Since $y_2 \not \in y_0.\eta$, $y_3 \ne y_0$. Suppose $y_3 \in y_0.\eta$, then $< y_0, y_1, y_2, y_3, y_0>$ constitutes a k-cycle $k \ge 4$, and we are done. Suppose $y_3 \not \in y_0.\eta$. We repeat the same path extension. $y_3.\eta \not \subseteq y_2.\eta$ implies $\exists y_4 \in y_3.\eta$, $y_4 \not \in y_2.\eta$. If $y_4 \in y_0.\eta$ or $y_4 \in y_1.\eta$, we have the desired cycle. If not $\exists \ y_5, \ldots$ and so forth. Because $\mathcal{N}$ is finite, this path extension must terminate with $y_k \in y_i.\eta$, where $0 \le i \le n-3$, $n = |N|$. Let $x = y_0, z = y_k$.

• (2) follows naturally.    $\square$

Finally, we show that $\omega$ preserves the shortest paths between all elements of the trace, $\mathcal{T}$.

Proposition 3

Let $\sigma (x, z)$ denote a shortest path between x and z in $\mathcal{N}$. Then for all $y \ne x, z, \in \sigma (x, z)$, if y can be subsumed by $y'$, then there exists a shortest path $\sigma '(x, z)$ through $y'$.

Proof

We may assume without loss of generality that y is adjacent to z in $\sigma (x, z)$.

Let $< x, \ldots , x_n, y, z>$ constitute $\sigma (x, z)$. If y is subsumed by $y'$, then $y.\eta = \{x_n, y, z\} \subseteq y'.\eta$. So we have $\sigma '(x, z) = < x. \ldots , x_n, y', z>$ of equal length. (Also proven in [23].)    $\square$

In other words, z can be removed from $\mathcal{N}$ with the certainty that if there was a path from some node x to z through y, there will still exist a path of equal length from x to z after y’s removal.

Fig. 9.

A network diamond

Figure 9 visually illustrates the situation described in Proposition 3, which we call a diamond. There may, or may not, be a connection between y and $y'$ as indicated by the dashed line. If there is, as assumed in Proposition 3, then either $y'$ subsumes y or vice versa, depending on the order in which y and $y'$ are encountered by $\omega$. This provides one example of the isomorphism described in Proposition 1. If there is no connection between y and $y'$ then we have two distinct paths between x and z of the same length.