7,989 research outputs found
Cubic graphs with large circumference deficit
The circumference of a graph is the length of a longest cycle. By
exploiting our recent results on resistance of snarks, we construct infinite
classes of cyclically -, - and -edge-connected cubic graphs with
circumference ratio bounded from above by , and
, respectively. In contrast, the dominating cycle conjecture implies
that the circumference ratio of a cyclically -edge-connected cubic graph is
at least .
In addition, we construct snarks with large girth and large circumference
deficit, solving Problem 1 proposed in [J. H\"agglund and K. Markstr\"om, On
stable cycles and cycle double covers of graphs with large circumference, Disc.
Math. 312 (2012), 2540--2544]
On almost hypohamiltonian graphs
A graph is almost hypohamiltonian (a.h.) if is non-hamiltonian, there
exists a vertex in such that is non-hamiltonian, and is
hamiltonian for every vertex in . The second author asked in [J.
Graph Theory 79 (2015) 63--81] for all orders for which a.h. graphs exist. Here
we solve this problem. To this end, we present a specialised algorithm which
generates complete sets of a.h. graphs for various orders. Furthermore, we show
that the smallest cubic a.h. graphs have order 26. We provide a lower bound for
the order of the smallest planar a.h. graph and improve the upper bound for the
order of the smallest planar a.h. graph containing a cubic vertex. We also
determine the smallest planar a.h. graphs of girth 5, both in the general and
cubic case. Finally, we extend a result of Steffen on snarks and improve two
bounds on longest paths and longest cycles in polyhedral graphs due to
Jooyandeh, McKay, {\"O}sterg{\aa}rd, Pettersson, and the second author.Comment: 18 pages. arXiv admin note: text overlap with arXiv:1602.0717
Thoughts on Barnette's Conjecture
We prove a new sufficient condition for a cubic 3-connected planar graph to
be Hamiltonian. This condition is most easily described as a property of the
dual graph. Let be a planar triangulation. Then the dual is a cubic
3-connected planar graph, and is bipartite if and only if is
Eulerian. We prove that if the vertices of are (improperly) coloured blue
and red, such that the blue vertices cover the faces of , there is no blue
cycle, and every red cycle contains a vertex of degree at most 4, then is
Hamiltonian.
This result implies the following special case of Barnette's Conjecture: if
is an Eulerian planar triangulation, whose vertices are properly coloured
blue, red and green, such that every red-green cycle contains a vertex of
degree 4, then is Hamiltonian. Our final result highlights the
limitations of using a proper colouring of as a starting point for proving
Barnette's Conjecture. We also explain related results on Barnette's Conjecture
that were obtained by Kelmans and for which detailed self-contained proofs have
not been published.Comment: 12 pages, 7 figure
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