In [this paper](http://www.sciencedirect.com/science/article/pii/S0020019013002214) (Construction 2.6 p860) the authors have built examples of 
connected $k$-regular graph without Hamiltonian path, but with a cut-vertex (i.e. it is not $2$-connected). 


*Question:* Is there a $2$-connected $k$-regular graph without Hamiltonian path?


*Remark*: every $2$-connected $k$-regular graph with at most $3k+3$ vertices admits a Hamiltonian cycle or is one of the two exceptions, which admit a Hamiltonian path. 



Is there a 2-connected k-regular graph without Hamiltonian path?
Take $k>3$ long paths between two vertices $a,b$. This graph is already 2-connected. Now add some edges to make it $k$-regular so that every edge joins only interior vertices of the same path (it is not hard). Then $G\setminus\{a,b\}$ has more connected components than any graph having Hamiltonian path with two vertices removed.


