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paths on a graph that visit each vertex exactly once
0
votes
1
answer
94
views
Hamiltonian path in the line graph of a connected graph
If $G = (V,E)$ is a finite, connected, simple, undirected graph, is there a Hamiltonian path in the line graph $L(G)$ of $G$?
1
vote
1
answer
55
views
Hamiltonian path in total graph
Let $G = (V, E)$ be a finite, simple, undirected graph with $V \cap E = \emptyset$. The total graph $T(G)$ is defined on the vertex set $V \cup E$ and its edge set is given by $$E(T(G)) = E \cup \big\ …
3
votes
1
answer
1k
views
Tournaments with exactly one directed Hamiltonian path
Every tournament contains a directed Hamiltonian path (a path visiting every vertex exactly once).
Suppose that $T$ is a tournament on $[n]:=\{1,\ldots,n\}$ for some integer $n\geq 2$ with exactly on …
4
votes
2
answers
203
views
Hamilton cycles in $\{0,1\}^n$ with fixed Hamming distance
Let $n>1$ be an integer. For $a,b\in \{0,1\}^n$ let $d_h(a, b)$ denote the Hamming distance of $a$ and $b$. For $k\in \{1,\ldots,n-1\}$ let $H(n,k)$ be the graph on $\{0,1\}^n$ given by the edge set $ …
1
vote
0
answers
186
views
Countable graph that is "as non-traceable as it gets"
If $\omega$ denotes the set of the natural numbers (= the first infinite ordinal), and if $E\subseteq\binom{\omega}{2}$ is any subset, we call a map $f\colon\omega\to\omega$ a walk in the graph $(\ome …
6
votes
1
answer
315
views
Are Gray codes in $\{0,1\}^n$ isomorphic?
Let $n\in\mathbb{N}$ be a positive integer. Two elements of $\{0,1\}^n$ form an edge if and only if their Hamming distance equals $1$. It is known that $\{0,1\}^n$ endowed with this graph structure po …
1
vote
2
answers
182
views
Hamiltonian cycle in $S_n$ with transpositions
For any set $X$, let $[X]^2=\{\{a,b\}:a\neq b \in X\}$. If $n\in\mathbb{N}$ is a positive integer, let $S_n$ denote the collection of bijections $\varphi:\{0,\ldots,n-1\}\to\{0,\ldots,n-1\}$. Let $E_n …
-1
votes
2
answers
192
views
Path of length $n$ but no Hamilton cycle [closed]
What is an example of a simple graph $G = (\{1,\ldots,n\}, E)$, where $n\in\mathbb{N}$ is a positive integer, with the following properties?
There is a path in $G$ of length $n$,
every vertex has at …
1
vote
2
answers
141
views
Does $\{0,1\}^{<\omega}$ have a Hamiltonian path?
Let $\{0,1\}^{<\omega}$ be the collection of $x \in \{0,1\}^\omega$ such that there is $N\in\omega$ with $x(k) = 0$ for all $k\geq N$. We say that $ x, y\in \{0,1\}^{<\omega}$ form an edge if they hav …
5
votes
1
answer
277
views
Integers with a Hamiltonian Square Path
Let $n>1$ be an integer and set $[n]=\{1,\ldots,n\}$. We say that $n$ has a "Hamiltonian Square Path" if there is a bijection $\varphi:[n]\to[n]$ such that for all $k\in [n-1]$ we have that $\varphi(k …
1
vote
3
answers
870
views
Hamiltonian paths in bipartite graphs with 2 sets of "almost" same cardinality
Suppose we have two finite disjoint sets $A, B \neq \emptyset$ such that $|A|$ and $|B|$ differ by at most $1$, and let $\Gamma = (A\cup B, E)$ where $E\subseteq \big\{\{a,b\}: a\in A, b\in B\big\}$ b …
1
vote
2
answers
316
views
Hamiltonicity and minimal degree in bipartite graphs
Given an integer $k>1$, is there a connected bipartite graph $\Gamma = (A, B, E)$ where $A\cap B = \emptyset$ and $E \subseteq \big\{\{a, b\}:a\in A, b\in B\big\}$ such that
$|A| = |B|$,
$\text{de …
2
votes
1
answer
117
views
$\omega$-Hamilton paths in $\mathbb{Z}^n$
For any set $X$ set $[X]^2 = \{\{x,y\}: x,y\in X, x\neq y\}$. If $n\geq 2$ is an integer, we endow $\mathbb{Z}$ with a graph structure in the following way. If $x,y\in \mathbb{Z}^n$ we say $x,y$ are d …
6
votes
2
answers
300
views
Hamiltonian path in bike-lock graph with $1$ known digit
Motivation. My youngest son has a bike lock with dials, and he forgot the unlocking combination completely, except that he remembered that digit $0$ appeared somewhere in the combination. So it was my …
7
votes
3
answers
2k
views
"Gray code" for building teams
Motivation. In a team of $n$ people, we had the task to build subteams of a fixed size $k<n$ such that every day, $1$ person of the subteam is replaced by another person in the team, but not in the su …