NP-completeness for bridgeless undirected planar 3-regular bipartite graphs with face bounds

Question

the ... Hamiltonian cycle problems ... remain NP-complete even ... for bridgeless undirected planar 3-regular bipartite graphs ...

Fine. Are they still NP-complete when the minimum or/and maximum face degree $f_{\min / \max}$ is/are bound?

To give an example, let's say we exclude squares then, $f_\min=6$.

hmakholm left over Monica · Accepted Answer · 2013-02-07T23:58:18.760

Limiting $f_{\rm max}$ to about 20 without losing NP-hardness should be possible. If you have a face with high degree you can bisect it by replacing

    a   |                                       |    d
        |                                       |
--------O                c                      O----------
        |                                       |
    b   |                                       |    e

with

            |                                   |
         a  |           c1                      |
----O--N----O-----B---------B----A--A---.       |
    |  |           \         \   |  |    \      |
    N--O     x      A--A---.  \  A--A--A--A     |   d
    |  |            |  |    \  \  \    |  |     |
    O--O--A--A---.  A--A--A--A  \  `---A--A--O--O
    |     |  |    \  \    |  |   \           |  |   
b   |     A--A--A--A  `---A--A    \    z     O--N
    |      \    |  |          \    \         |  |
    |       `---A--A-----------B----B---O----N--O----
    |                 c2                |  e
    |                                   |

The three A complexes force the way a Hamiltonian cycle must take through the B nodes, ensuring that the horizontal BB edges can ever be used in a Hamiltonian cycle. The network of O and N nodes ensure that each of the three ways a Hamiltonian cycle to pass through the original node can be refined to one that passes through two the entire half of the new graph.

The construction increases the degree of each of face a, b, d, and e by 2, but if their degree get too high we can bisect them later by applying the construction recursively at one of the N nodes. This will increase the new faces x and z from degree 14 (or 10) to degree 16, but then they never need to be touched again.

(If the original nodes have different parity, turn the ON networks on one of the sides upside down).

On the other hand, $f_{\rm min}>5$ is impossible. It follows from the Euler characteristic of the plane that the average face degree of a 3-regular planar graph with $F$ faces is $6-12/F$, which means that every 3-regular planar graph has at least one face with degree $5$ or lower.

+1 for ASCII, but is the AAAA complex bipartite? But I think I got what you mean... — draks ..., Feb 07 '13 at 21:39
Whoops, completely forgot the "bipartite" requirement (in both parts of the answer). Let me see if I can save that. — hmakholm left over Monica, Feb 07 '13 at 21:52
it is saveable. I've seen such a subgraph somewhere. Looks like two connected squares $$ | \ / \backslash \ --- \ |;;\ |;;; | \ ---\ \backslash / \ | $$ But what about $f_\min$? — draks ..., Feb 07 '13 at 21:53
Much paper later, I have a reduction along these lines to graphs with $f_{\rm max}=14$. It seems almost possible to get down to $f_{\rm max}=12$, but the details are slippery. — hmakholm left over Monica, Feb 08 '13 at 04:17
Henning, does this mean that when we don't work in the plane, but on a surface with $\chi<0$ we could get graphs where $f_\min>6$? I posted this as a separate question here... — draks ..., Aug 04 '13 at 12:59

NP-completeness for bridgeless undirected planar 3-regular bipartite graphs with face bounds

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