This post describes a way of viewing deterministic automata from a combinatorial and topological lens, as covering spaces of certain graphs. Roughly, every word over a given set of symbols can be thought of as a path on a certain graph, and the execution of the automaton can be interpreted as path lifting. A quick search of the internet does not bring up anything related, but it is possible that this connection has already been made before....
Combinatory logic Combinatory logic is a simplified model of computation based on combinators - functions that take in other functions and combine them using only function application. It was introduced by Moses Schönfinkel and rediscovered by Haskell Curry. Interestingly, combinatory logic is free of variables - every combinator can be expressed in terms of just two special combinators known as $S$ and $K$. It is possible to encode natural numbers and intuitionistic propositional logic in the language of combinators....
A general equality problem for Abelian groups Consider the problem of deciding whether of deciding whether an equation, such as $$ (x + y) + z + -(x + z) = y$$ is true in all Abelian groups. This means that for any Abelian group $A$, and for any chosen values $x$, $y$ and $z$ in $A$, the equation should be true. This hardly poses a problem, for humans atleast - by "expanding", "cancelling", "rearranging" and "grouping", anyone well-versed in high-school algebra can solve an arbitrary instance of this problem without much thought....
The Unit Conjecture The Unit Conjecture is a fundamental and well-known question about group rings that goes back to Graham Higman and Irving Kaplansky in the mid-twentieth century. It is a part of a collection of several conjectures posed by Kaplansky - most notably the Zero-divisor conjecture and the Idempotent conjecture. These are collectively known as Kaplansky's conjectures. The statement Consider a field $K$ and a group $G$. In the group ring $K[G]$, the elements of the form $k \cdot g$, where $k \in K \backslash \{0\}$ and $g \in G$, are clearly invertible - with the inverse being $k^{-1} \cdot g^{-1}$....