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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 ...
joro's user avatar
  • 25.4k
15 votes
2 answers
2k views

What is the smallest uniquely hamiltonian graph with minimum degree at least 3?

I would like to know more about uniquely hamiltonian graphs with minimum vertex degree at least 3, and in particular what is the smallest one. (Recall that a graph is hamiltonian if it has a cycle ...
Gordon Royle's user avatar
  • 12.7k
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 ...
user101010's user avatar
  • 5,349
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 ...
Gordon Royle's user avatar
  • 12.7k
13 votes
1 answer
1k views

Generalisation of this circular arrangement of numbers from $1$ to $32$ with two adjacent numbers being perfect squares

I posted this question on MSE, and failed to get the type of answer I wanted. That's why I would like to post it here and wait for the experts to reply. Here's the link to the MSE post, which I ...
Sayan Dutta's user avatar
12 votes
1 answer
424 views

Quantitatively characterizing the failure of the converse of Dirac's theorem

First, I am an undergraduate so I apologize if this is trivial and certainly understand if it is closed immediately. I am currently in a combinatorics and graph theory class and recently we have ...
1729's user avatar
  • 221
12 votes
1 answer
2k views

Hobbled rook tour – Hamiltonian cycle on square grid

Consider a square grid of even side length ($2n \times 2n$). It is easy to see that there must exist a Hamiltonian cycle on the corresponding grid graph. Such a cycle is called balanced if the number ...
John's user avatar
  • 121
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 ...
Wolfgang's user avatar
  • 13.4k
10 votes
2 answers
782 views

Graphs with many edges avoided by Hamiltonian cycles

Let $G$ be a $3$-connected Hamiltonian graph with at least one edge that belongs to each H-cycle of $G$. Some authors (e.g. in the link given here) call such an edge an a-edge and an edge that belongs ...
Wolfgang's user avatar
  • 13.4k
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 ...
user avatar
9 votes
2 answers
2k views

Is this graph Hamiltonian?

Let $G$ be a simple $2$-connected graph with $m+n$ vertices ($n>m \geq 3$) with degree sequence $(m-1)^m$, $(n-1)^n$; that is, $G$ is degree-equivalent to two disjoint cliques $K_m$, $K_n$ of ...
Valentin Brimkov's user avatar
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 ...
Dominic van der Zypen's user avatar
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 ...
LeechLattice's user avatar
  • 9,501
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: ...
Gordon Royle's user avatar
  • 12.7k
8 votes
2 answers
598 views

Orthogonal Hamiltonian cycles in (n x n x n) grids

Let $C_n$ be a cubical $n \times n \times n$ subset of the integer lattice, so consisting of $n^3$ vertices. I am interested in special Hamiltonian cycles in $C_n$, special in the sense that (a) each ...
Joseph O'Rourke's user avatar
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 ...
Alejandro Erickson's user avatar
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.
Gordon Royle's user avatar
  • 12.7k
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 ...
Dominic van der Zypen's user avatar
7 votes
7 answers
3k views

Efficient Hamiltonian cycle algorithms for graph classes

Generally speaking, finding a Hamiltonian cycle is NP-Hard and so tough. But if $G=L(H)$ is the line graph of $H$, then we can reduce the problem of finding a Hamiltonian cycle in $G$ to finding an ...
Felix Goldberg's user avatar
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 ...
Gordon Royle's user avatar
  • 12.7k
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 ...
Fedor Petrov's user avatar
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 ...
Mario Krenn's user avatar
6 votes
2 answers
304 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 ...
Dominic van der Zypen's user avatar
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-...
Felix Goldberg's user avatar
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 ...
T. D. Nguyen's user avatar
6 votes
1 answer
344 views

Hamiltonian cycles in power-graphs

I've stumbled across a short note from 1993 where a nice question was asked: Suppose you take a graph with vertices $\{1,2,\ldots,n\}$ and connect $i,j$ by an edge if and only if $i+j$ is a $k$th ...
Felix Goldberg's user avatar
6 votes
0 answers
164 views

Hamilton cycles in random graphs with just enough connectivity

What is the asymptotic probability that $G$ has a Hamilton cycle if $G$ is a random $n$ vertex $\frac{4}{3}n$ edge graph, with minimum degree 2 and without degree 2 vertices at distance 1 or 2 to each ...
Dmytro Taranovsky's user avatar
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 ...
EGME's user avatar
  • 1,018
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 ...
Gordon Royle's user avatar
  • 12.7k
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 ...
Wolfgang's user avatar
  • 13.4k
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 $3 \leq \ell \leq n$. ...
D. Ror.'s user avatar
  • 399
5 votes
2 answers
191 views

Number of Hamiltonian cycles on 24-cell graph

I asked Wolfram Alpha for the number of Hamiltonian cycles on the 24-cell graph. https://www.wolframalpha.com/input?i=number+of+hamiltonian+cycles+on+24-cell+graph It answers 114.9 billion but doesn't ...
Etienne's user avatar
  • 53
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 ...
Michael's user avatar
  • 2,205
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 ...
Gjergji Zaimi's user avatar
5 votes
1 answer
271 views

Approximation of Hamiltonian cycles

Let's define the $\texttt{MinHalfSimpCycle}$ search problem: Given $G=(V, E)$ a complete, undirected graph with weights on the edges. We want a simple cycle in $G$ (each vertex appears in it at most ...
Redbull's user avatar
  • 53
5 votes
1 answer
279 views

Infinitely many counterexamples to Nash-Williams's conjecture about hamiltonicity?

Question from 2013 gives one counterexample to Nash-Williams's conjecture about hamiltonicity of dense digraphs. Later, we found tens of counterexamples on more than 30 vertices and believe there are ...
joro's user avatar
  • 25.4k
5 votes
1 answer
119 views

Sufficient condition for a Hamilton cycle $C$ in a planar triangulation $G$ s.t. every triangle in $G$ has an edge in $C$

Let $G$ be a $k$-connected planar triangulation ($k\geq 4$) and let $C$ be a Hamilton cycle of $G$. Then: Which conditions would be sufficient to assure that every triangle of $G$ has at least one ...
Jose Antonio Martin H's user avatar
5 votes
0 answers
127 views

Do uniquely Hamiltonian graphs have cycles of a sufficiently long length?

Let $C$ be a Hamiltonian cycle of a graph $G$. Call an edge $e$ of $G$ a chord if $e\not\in C$. Let each edge of $C$ be weighted $1$ and each chord be weighted $2$. The weight of a path or cycle of ...
kabenyuk's user avatar
  • 673
5 votes
0 answers
154 views

How to construct 4-regular graphs with few Hamiltonian decompositions?

A Hamiltonian decomposition of a finite simple graph is a partition of its edge set so that each partition class forms a Hamiltonian cycle. This is only possible if the graph is $2k$-regular. ...
M. Winter's user avatar
  • 13.6k
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 ...
joro's user avatar
  • 25.4k
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 ...
David Feldman's user avatar
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 $d$ ...
domotorp's user avatar
  • 19.1k
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? ...
Jimmy Dillies's user avatar
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 ...
Zach Hunter's user avatar
  • 3,499
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, ...
Mohammad Al-Turkistany's user avatar
4 votes
1 answer
228 views

Probability problem in Sheehan's conjecture

As my first math project, I have been working on Sheehan's Conjecture and am stuck for weeks. I wonder if I am at a dead end. Sheehan's Conjecture states that every Hamiltonian 4-regular simple graph ...
Daniel Liu's user avatar
4 votes
0 answers
69 views

is a 4-connected planar graph still Hamiltonian after removing an edge?

We know that 4-connected planar graphs are Hamiltonian(by the known Tutte Theorem). Additionally, Thomas and Yu [1] proved that removing two vertices from a 4-connected planar graph still preserves ...
Licheng Zhang's user avatar
4 votes
0 answers
234 views

How many 20-vertex 2-connected 5-regular non-Hamiltonian graphs are there?

As for the question in title, I attempted to use nauty to obtain them, but it has been running on my computer for nearly three days without producing any results. <...
Licheng Zhang's user avatar
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 ...
Zach Hunter's user avatar
  • 3,499
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 ...
Manfred Weis's user avatar
  • 13.2k