Papers
Topics
Authors
Recent
Detailed Answer
Quick Answer
Concise responses based on abstracts only
Detailed Answer
Well-researched responses based on abstracts and relevant paper content.
Custom Instructions Pro
Preferences or requirements that you'd like Emergent Mind to consider when generating responses
Gemini 2.5 Flash
Gemini 2.5 Flash 74 tok/s
Gemini 2.5 Pro 39 tok/s Pro
GPT-5 Medium 16 tok/s Pro
GPT-5 High 13 tok/s Pro
GPT-4o 86 tok/s Pro
Kimi K2 186 tok/s Pro
GPT OSS 120B 446 tok/s Pro
Claude Sonnet 4 36 tok/s Pro
2000 character limit reached

Bijections in de Bruijn Graphs (1702.06906v1)

Published 22 Feb 2017 in math.CO

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 $S(2{n-1})$.

Summary

We haven't generated a summary for this paper yet.

List To Do Tasks Checklist Streamline Icon: https://streamlinehq.com

Collections

Sign up for free to add this paper to one or more collections.

Lightbulb On Streamline Icon: https://streamlinehq.com

Continue Learning

We haven't generated follow-up questions for this paper yet.

Authors (1)