Search Results

Tony's method works in the general case. Let $e_1, e_2, \ldots, e_N$ be the edges that are added to $G$, in the order they are added, to make the complete graph. Choose an arbitrary hamiltonian cycl …

Yes, every connected cubic graph is 3-almost-Hamiltonian. Replace each edge by two parallel edges then follow an Eulerian circuit. In the case of a bridgeless cubic graph, you can add a perfect match …

If a full set of $n$ edges is inserted between the cycles, what you have is a "cycle permutation graph". According to this paper, there are nonhamiltonian cycle permutation graphs for all odd $n\ge 9$ …

Bill Kocay found a more direct combinatorial method to reconstruct the number of hamiltonian cycles and some other spanning subgraphs. It is in his paper "Some new methods in reconstruction theory", …

Take any $k\ge 1$ and four disjoint sets of vertices $A,B,C,D$ with $|A|=|D|=k+2$, $|B|=|C|=k$. Completely join $A$ to $B$, $B$ to $C$ and $C$ to $D$. This gives a bipartite graph of minimum degree $ …

This is a NEW EDITION using the condition that two edges incident with a vertex not in the dominating circuit cannot be attached to consecutive vertices in the dominating circuit. I tried 200 million …

Carol Zamfirescu and Gunnar Brinkmann have informed me that this paper answers the question.

Gunnar Brinkmann informs me that this paper constructs planar triangulations where every hamiltonian cycle misses all the edges of many triangles. Some of the examples are even 5-connected.