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2 votes
0 answers
116 views

Two more counterexamples to a conjecture from 1975 about hamiltonicity of digraphs

Question from 2013 gives one counterexample to Nash-Williams's conjecture 1975 about hamiltonicity of dense digraphs. In the linked answer, @LouisD "reverse engineered" the counterexample ...
16 votes
1 answer
696 views

A counterexample to a conjecture of Nash-Williams about hamiltonicity of digraphs?

Maybe I am missing something, but found potential counterexample to a conjecture of Nash-Williams. According to HAMILTONIAN DEGREE SEQUENCES IN DIGRAPHS The outdegree and indegree sequences of ...
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 ...
2 votes
1 answer
321 views

Extending perfect matchings into Hamiltonian cycles

Let $G$ be a simple cubic graph which has a Hamiltonian circuit $C$. In general, it is not possible to find a second Hamiltonian circuit which contains all the chords of $C$. For example, the Wagner ...
4 votes
0 answers
230 views

Is this case of Barnette's Conjecture known?

Context: Barnette's Conjecture is that every bipartite cubic polyhedral graph is Hamiltonian. I have been interested by this problem for a long time, and I recently came up with a result. From my ...
1 vote
1 answer
178 views

Is Hamiltonian cycle fixed parameter tractable with parameter clique cover?

Let $G$ be connected simple graph. Clique cover of graph $G$ is partition of the vertices of $G$ into $k$ disjoint cliques $D'_i$. Given $G$ and $k$-clique cover, can we solve Hamiltonian cycle in ...
4 votes
1 answer
109 views

Does 2-connectivity imply Hamiltoniancy for subgraphs of the rook graph

We say the rook graph, $R_n$, is the cartesian product of $K_n \times K_n$. Let $S$ be the set of graphs that are an induced subgraph of $R_n$ for some $n$. Does there exist some constant $c$ such ...
2 votes
1 answer
85 views

The number of Hamiltonian circuits on a convex polytope embedded in $\mathbb{R}^N$

Recently I wondered whether there might be a natural topological complexity measure for convex polytopes embedded in $\mathbb{R}^N$. After some reflection it occurred to me that the number of distinct ...
6 votes
0 answers
73 views

Normal colorings of bridgeless cubic graphs

Definition (informal) A normal edge-5-coloring of a bridgeless cubic graph $G$ is a proper 5 coloring of the edges of the graph, so that for each edge $e\in E(G)$, either $e$ and the four edges ...
10 votes
1 answer
513 views

What is the complexity of finding a third Hamilton Cycle in cubic graph?

According to Smith Theorem: if a cubic graph has a hamilton circuit then it must have a second one. SMITH : Given a Hamilton circuit in a 3-regular graph, find a second Hamilton circuit. It is known ...
14 votes
1 answer
783 views

What was Smith's proof of Smith's theorem on Hamilton cycles in cubic graphs?

In a short 1946 paper "On Hamiltonian Circuits", Tutte proved the famous result that an edge in a cubic graph lies in an even number of Hamilton circuits. He attributed the result to his friend CAB ...
3 votes
1 answer
138 views

Edge colorability and Hamiltonicity of certain classes of cubic graphs (MO graphs)

Let $G$ be a simple cubic graph (that is, 3-regular). A dominating circuit of $G$ is a circuit $C$ such that each edge of $G$ has an endvertex in $C$. The circuit $C$ is chordless if no edge which is ...
14 votes
1 answer
1k views

Are all cubic graphs almost Hamiltonian?

Call a graph $G$ $n$-almost-Hamiltonian if there is a closed walk in $G$ that visits every vertex of $G$ exactly $n$-times. So a Hamiltonian graph is $n$-almost-Hamiltonian for all $n$. Are all ...
6 votes
1 answer
335 views

What is the complexity of counting Hamiltonian cycles of a graph?

Since deciding whether a graph contains a Hamiltonian cycle is $NP$-complete, the counting problem which counts the number of such cycles of a graph is $NP$-hard. Is it also $PP$-hard in the sense ...
9 votes
1 answer
399 views

Are bipartite Moore graphs Hamiltonian?

This is motivated by a computer-generated conjecture that bipartite distance-regular graphs are hamiltonian. I decided to check the case of Moore graphs first. The cycles and complete bipartite graphs ...
7 votes
1 answer
210 views

Grinberg's uniquely hamiltonian 3-connected graphs (Russian paper)

Many years ago, Grinberg found some uniquely-hamiltonian $3$-connected graphs, and published his results in a paper that has been cited several times as follows. E. Grinberg, Three-connected graphs ...
4 votes
0 answers
143 views

Halin Graphs with Highest Number of Hamilton Cycles

Halin graphs contain a Hamilton cycle and have the interesting property, that, also in the case of arbitrary real edge weights, it is possible to report one of the shortest contained Hamilton cycles ...
3 votes
0 answers
107 views

Reference request: Bipartite symmetric graphs are hamiltonian

Does anyone know whether bipartite symmetric graphs are hamiltonian? I'm not sure whether anyone have proved it before, but a nonhamiltonian symmetric bipartite graph would lead to a counterexample to ...
6 votes
1 answer
299 views

Why is the number of Hamiltonian Cycles of n-octahedron equivalent to the number of Perfect Matching in specific family of Graphs?

In OEIS A003436, it is written that the number of inequivalent labeled Hamilton Cycles of an n-dimesnional Octahedron is the same as the number of Perfect Matchings in a the complement of the Cycle ...
3 votes
0 answers
145 views

Hamiltonian cycle polytope for the hypercube graph

Let $Q_n$ denote the $n$ dimensional hypercube graph (i.e., graph formed from the vertices and edges of an n-dimensional hypercube). Denote the set of edges and vertices of $Q_n$ by $E_n$ and $V_n$ ...
4 votes
0 answers
258 views

When is an induced subgraph of a Johnson graph hamilton-connected?

The Johnson graph $J(n,k)$ has as its vertices the $k$-subsets of $\{1, 2, \dots, n\}$ where two vertices are adjacent iff their intersection has size $k-1$. A graph is Hamilton-connected if every two ...
1 vote
0 answers
147 views

Properties of graphs with Hankel-like adjacency matrix

I am having undirected graphs with adjacency matrices which have a regular Hankel-like form, e.g., $$A=\begin{pmatrix}0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & (6\times 0 \text{ ...
5 votes
1 answer
1k views

How many hamiltonian cycles can be removed from a complete directed graph before it becomes disconnected?

The question started from a problem brought home by a friend's 5th grader: "How many ways can you seat 5 people around a round table so that the people sitting to the left of any person is different ...
1 vote
2 answers
321 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{...
1 vote
3 answers
883 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 $E\subseteq \big\{\{a,b\}: a\in A, b\in B\big\}$ ...
9 votes
2 answers
2k views

"Gray code" of all permutations

Informally asking, can we step through all permutations of the set $\{1,\ldots,n\}$ by just using transpositions? More formally: For any $n\in\mathbb{N}$ let $[n] = \{1,\ldots,n\}$ and let $S_n$ be ...
4 votes
3 answers
506 views

Does every finite bridgeless cubic planar simple undirected graph admit a 2-factorization with at most two components each of which has even order?

Consider simple bridgeless cubic planar graphs. Does each such graph admit a 2-factorization with $\leq 2$ components each of which has even order? If not, does anyone know of an counterexample? ...
3 votes
2 answers
123 views

Number Associated with Straight-line Drawings of Hamiltonian Graphs

Is there anything known about the maximum number of simple-polygonal Hamilton cycles that a straight-line drawing of a Hamiltonian graph can have? Put differently, if the vertices of a Hamilton ...
11 votes
1 answer
328 views

How many edges can be added to two circles before the graph becomes Hamiltonian?

Start with two $n$-circles $(v_1\cdots v_n)$ and $(w_1\cdots w_n)$ of vertice sets $V$ and $W$, where $n\ge 5$. Add a number of vertex-disjoint edges between $V$ and $W$ (thus no chords) in a way ...
9 votes
0 answers
245 views

Heuristic arguments regarding Sheehan's conjecture?

Sheehan conjectured that there are no 4-regular graphs that are uniquely hamiltonian (i.e. have exactly one hamilton cycle). Evidence that might be loosely seen to be in favour of this conjecture is: ...
8 votes
0 answers
123 views

Is there a non-bipartite hamiltonian cubic graph on $n$ vertices with no $(n-1)$-cycle?

Is there a cubic (3-regular) graph $G$ on $n$ vertices such that: $G$ is hamiltonian $G$ has no $(n-1)$-cycles $G$ is not bipartite My computer tells me that there are none on up to $24$ vertices.
6 votes
0 answers
129 views

Minimum number of hamilton cycles in a 4-connected planar triangulation?

I am currently interested in hamilton cycles (i.e. a cycle through every vertex) in planar triangulations (i.e. planar graphs with every face a triangle). There are non-hamiltonian planar ...
1 vote
1 answer
78 views

Graph gadget related to uniquely hamiltionian regular graphs (question #2)

Related to uniquely hamiltionian graphs. For natural numbers $a,b$ define $(a,b)$ gadget $G$: $G$ is finite simple graph. Two vertices $u,v$ are of degree $b$ and the rest of the vertices are of ...
5 votes
0 answers
99 views

Graph gadget related to uniquely hamiltionian regular graphs

A graph is uniquely hamiltonian if it has exactly one hamiltonian cycle. According to a conjecture there are no $r$-regular uniquely hamiltonian graphs for $r > 2$ and of special interest is the ...
6 votes
0 answers
76 views

Cage graphs and even cycles

Let $G$ be a $(\nu,g)$-cage graph of degree $\nu$ with girth $g$ and $n=n(\nu,g)$ vertices. Based on the known examples, I am wondering if the following can be proved/disproved: Is it true that ...
6 votes
0 answers
108 views

Localizing Bondy's metaconjecture on hamiltonicity

Definitions: Let $G$ be a graph on $n$ vertices. $G$ is Hamiltonian provided $G$ has a cycle of length $n$. $G$ is pancyclic provided $G$ has a cycle of length $\ell$ for every $3 \leq \ell \leq n$. ...
0 votes
1 answer
149 views

degG(x) + degG(y) >= n, show that the graph is hamiltonian [closed]

I'm trying to show that a connected graph which has order >=3, and having the following inequality is Hamiltonian: ...
3 votes
1 answer
266 views

What is the densest bipartite graph with unique Hamiltonian cycle?

In a prior post regarding perfect matching, it was stated that the densest graph with a unique perfect matching cannot have more than $n^2$ edges, if graph has $2n$ vertices. Analogously, what is the ...
0 votes
1 answer
1k views

Does this graph contain at least two Hamiltonian cycles?

Let $G$ be a simple graph which is a $2n$-cycle together with $n$ chords such that $G$ is $3$-regular. In other words, the set of $n$ chords is a perfect matching of $G$. I conjecture that for every ...
2 votes
1 answer
606 views

Simple decomposition of $K_{2n}-I$ into hamiltonian cycles

http://mathworld.wolfram.com/HamiltonDecomposition.html In the 1890s, Walecki showed that complete graphs K_n admit a Hamilton decomposition for odd n, and decompositions into Hamiltonian cycles ...
7 votes
1 answer
736 views

Refinement of Dirac's theorem on Hamiltonian graphs

Dirac's theorem states that if degree of each vertex of a graph $G=(V,E)$ is not less than $|V|/2$, then it has Hamiltonian cycle. It is less known, but still known and not so hard to prove (though I ...
4 votes
2 answers
349 views

Can we find 3 disjoint directed Hamiltonian cycles in the cube?

Let $D$ be the digraph on $2^d$ vertices with $d2^d$ edges that we obtain by directing each edge of the $d$-dimensional hypercube in both directions. Can we partition the edges of $D$ into $d$ ...
5 votes
0 answers
295 views

A digraph related to permutations

A finite sequence of distinct real numbers of length $n$ determines a linear order of $\{1,\ldots,n\}$, by mapping position to rank; call this the permutation of the sequence. Consider the following ...
3 votes
1 answer
117 views

Hamiltonicity of random graphs with high girth

We say that $G\sim G_{n,f}$ (for $f=f(n)$) if $G$ is chosen uniformly at random from all graphs on $n$ vertices with girth $g(G)\ge f(n)$. Is there any threshold function $F(n)$ such that when $f\ll F$...
8 votes
2 answers
2k views

How to efficiently find a Hamiltonian cycle in a graph whose closure is complete?

A graph whose closure is the complete graph is Hamiltonian by the Bondy-Chvátal theorem, but I haven't found a polynomial algorithm for finding a Hamiltonian cycle in such a graph. Is there one that ...
6 votes
1 answer
243 views

Hamiltonicity criteria for sparse graphs

Given a sparse graph, how can one go about proving that it is Hamiltonian? (Assuming it actually is, of course). There are three main classes of criteria for Hamiltonicity that I am aware of: Dirac-...
3 votes
1 answer
219 views

Regular graphs with $a$ and $b$ Hamiltonian edges

Special case of this question. Let $G$ be $r$-regular Hamiltonian graph. An $a$ edge is an edge which is on every Hamiltonian cycle. A $b$ edge is an edge which is on no Hamiltonian cycle. $a(G)$ ...
4 votes
1 answer
724 views

Minimum distance between Hamiltonian cycles in cubic Hamiltonian graph

It is $NP$-hard to find constant factor approximation of longest cycle in cubic Hamiltonian graphs. Therefore, finding a Hamiltonian cycle in a cubic Hamiltonian graph is NP-hard. By Smith's theorem, ...
5 votes
1 answer
224 views

Reconstructing the number of Hamiltonian cycles

As is common terminology in graph reconstruction, given a graph $G$, we call a vertex deleted subgraph of $G$, a card, and call the multiset of all cards, the deck of $G$. The graph reconstruction ...
2 votes
0 answers
184 views

Details about Kelman's equivalent form of Barnette's conjecture

Barnette's conjecture states that every cubic planar bipartite 3-connected graph admits Hamiltonian cycles. Kelman claims that this conjecture is equivalent to a stronger one, which imposes some ...