# Bijections in de Bruijn Graphs

## TL;DR

This paper provides a new proof for the relationship between Eulerian cycles in T-nets and their doubling, constructs a bijection linking Eulerian cycles and binary sequences, and solves an open problem regarding de Bruijn sequences.

## Contribution

It introduces a novel proof of the Eulerian cycle count relation in T-nets and establishes a bijection connecting de Bruijn sequences with binary sequences, addressing an open question.

## Key findings

- Proved that the number of Eulerian cycles doubles with the doubling process.
- Constructed a bijection between Eulerian cycles and binary sequences.
- Solved an open problem relating de Bruijn sequences to binary sequences.

## Abstract

A T-net of order $m$ is a graph with $m$ nodes and $2m$ directed edges, where every node has indegree and outdegree equal to $2$. (A well known example of T-nets are de Bruijn graphs.) Given a T-net $N$ of order $m$, there is the so called "doubling" process that creates a T-net $N^*$ from $N$ with $2m$ nodes and $4m$ edges. Let $|X|$ denote the number of Eulerian cycles in a graph $X$. It is known that $| N^*|=2^{m-1}|N|$. In this paper we present a new proof of this identity. Moreover we prove that $|N|\leq 2^{m-1}$. Let $\Theta(X)$ denote the set of all Eulerian cycles in a graph $X$ and $S(n)$ the set of all binary sequences of length $n$. Exploiting the new proof we construct a bijection $\Theta(N)\times S(m-1)\rightarrow \Theta(N^*)$, which allows us to solve one of Stanley's open questions: we find a bijection between de Bruijn sequences of order $n$ and $S(2^{n-1})$.