Hamiltonian cycles in annular decomposable Barnette graphs

Let $G$ be an $n$-ADB graph. Then $H$ be a copy of $G$. Consider two consecutive edges $e_{G}$ and $e_{G}^{\prime}$ that lie in the interior of the outer annuals of $G$, that are adjacent to vertices $v_{G}^{1}$ and $v_{G}^{2}$ of $G$. Consider the two analogous edges $e_{H}$ and $e_{H}^{\prime}$ in $H$, that are adjacent to vertices $v_{H}^{1}$ and $v_{H}^{2}$ of $G$. Let $f_{G}$ and $f_{H}$ be the edges on the outer boundary of the outer annuli of $G$ and $H$, connecting $v_{G}^{1}$ with $v_{G}^{2}$ and $v_{H}^{1}$ with $v_{H}^{2}$ respectively. Note that each of the four vertices, $v_{G}^{1}$, $v_{G}^{2}$, $v_{H}^{1}$ and $v_{H}^{2}$, there is yet another edge adjacent to the vertices which we will denote as, $e_{v_{G}^{1}}$, $e_{v_{G}^{2}}$, $e_{v_{H}^{1}}$ and $e_{v_{H}^{2}}$ respectively. We now form a new graph $J$ by the following steps:

1. 1.

   Join $f_{G}$ and $f_{H}$ by drawing two edges across the existing edges

2. 2.

   Join $e_{v_{G}^{1}}$ and $e_{v_{H}^{1}}$ by drawing one edge across the existing edges

3. 3.

   Join $e_{v_{G}^{2}}$ and $e_{v_{H}^{2}}$ by drawing one edge across the existing edges

The resulting graph $K$ has $2x+1$ annuli and it is easy to see that $K$ is an ADB graph, where $x$ is the number of annuli in $G$. Note that $G$ can be viewed as a subgraph of $K$. Now we delete all the edges in the interior of the outer annulus of $G$, to obtain two graphs $K^{\prime}$ and $K^{\prime\prime}$ with $x+1$ and $x-1$ annuli respectively. Note that, $K^{\prime}$ is not a Barnette graph since it is not bipartite. (See Figure 2) ∎

Holton, Manvel and McKay proved that every Barnette graph can be constructed from the smallest Barnette graph $B_{0}$ (shown in Figure 1-(a)), by using the operations shown in Figure 3. Given that $B_{0}$ is $2$-edge connected, it is easy to see by induction that any Barnette graph $G$, would be $2$-edge connected. ∎

The proof of the Lemma is clear from Figure 4. Note that in Figure 4, the illustrated graph has an odd number of annuli. If we consider a graph with even number of annuli, the same strategy will still work. ∎

Let us create a graph $H$, by deleting all the edges in the interior of the outer annulus (along with their respective vertices). $H$ would be an ADB-AC graph, since $G$ is an ADB-AC graph. Thus, $F(O_{H})$ can be partitioned into subsets $F^{1}(O_{H})$ and $F^{2}(O_{H})$, such that no two faces in $F^{i}(O_{H})$, $i\in\{1,2\}$ are adjacent to each other.

We claim that, the edges in the interior of the outer annulus of $O_{G}$ are all attached to the outer boundary of faces in $O_{H}$, such that all such faces belong either in $F^{1}(O_{H})$ or in $F^{2}(O_{H})$, but not both. To prove this claim, let us assume otherwise. Then, there exits two edges $e_{1}$ and $e_{2}$ in the interior of $O_{G}$, such that $e_{1}$ is attached to the outer boundary of some face $f^{1}\in F^{1}(O_{H})$ and $e^{2}$ is attached to some face $f^{2}\in F^{2}(O_{H})$. Let, $v_{1}$ and $v_{2}$ be the two vertices by which $e_{1}$ and $e_{2}$ are attached to the outer boundary of $O_{H}$. Then there must be an odd number of vertices between $v_{1}$ and $v_{2}$, considering the fact that $f^{1}$ and $f^{2}$ are in different partitions. Thus, there is an odd cycle in $G$, which is impossible, since $G$ is bipartite.

Let us assume, without loss of generality, that all edges in interior of $O_{G}$ are attached to outer boundary of faces in $F^{1}(O_{H})$. Note that for an arbitrary face $f\in F^{1}(O_{H})$, to which some of such edges are attached, we can assert that there are only even number of edges attached to $f$, Otherwise, $f$ would transform into a odd cycle (face) $f^{\prime}$ in $G$. Thus, there would always be a sequence of odd number of $4$-cycles, $c_{1},\dots,c_{2k+1}$ (for some integer $k\geq 1$) formed in $O_{G}$ adjacent to $f^{\prime}$. Starting from $c_{1}$, if we consider all alternate faces in $O_{G}$, they will all be $4$-cycles and will form a face partition of $4$-cycles in $O_{G}$. Thus the statement holds. ∎

The proof of this statement also follows from Lemma 3.2. Note that if there exists such a graph $G$, then each of its annulus will have a $4$-cycle. ∎

Let $G$ have $x$ annuli for a particular planar annular embedding. For $x=1$, the result is obviously true.

Recall that, as per our convention, the $4$-face-only partition in the outer annulus of an ADB-AC graph $G^{\prime}$ is assumed to be $F^{1}(O_{G^{\prime}})$. Let $x>1$. By Lemma 3.3,say $F^{1}(O_{G})$ consists of only $4$-cycles. Let us select any two such $4$-cycles $m$ and $m^{\prime}$. All other edges in $O_{G}$ can be removed by performing a inverse of $alpha$-operations. Then, we perform an inverse $\alpha$-operation on the edges $e_{1}$ and $e_{2}$ in $m$ such that, $e_{1}$ and $e_{2}$ are not in the interior of $O_{G}$. Now, we perform another inverse of $\alpha$-operation by deleting edges $e^{\prime}_{1}$ and $e^{\prime}_{2}$ in $m^{\prime}$ such that, $e^{\prime}_{1}$ and $e^{\prime}_{2}$ were in the interior of $O_{G}$ in $G$. This gives us the $(x-1)$-annular restriction of $G$. By continuing this process we can ultimately reduce the graph to $1$-annular restriction of $G$. Thus it is clear that $G$ can be constructed from $B_{0}$ by using arbitrary number of $\alpha$-operations only. ∎

For the first statement, it is easy to construct the Hamiltonian cycle in $G$, in a recursive manner. We start by a planar annular projection of $G$. We include the interior face $f_{i}$ in the interior of a path connected region $R$. Recalling our convention that, the $4$-face-only partition in the outer annulus of an ADB-AC graph $G^{\prime}$ is assumed to be $F^{1}(O_{G^{\prime}})$, we can describe this strategy by extending $H$ through $k$-th annulus by coloring the faces in $F^{1}(O_{G_{k}})$ in grey, for every $k$-recursively. Thus we gradually extend $R$ following the strategy shown in Figure 5. It is easy to see that, when we recursively apply this strategy for all the annuli in $G$, the boundary of the final path connected region would manifest into a Hamiltonian cycle. We will call this strategy of constructing Hamiltonian cycles henceforth as the pyramid strategy.

Recall that, as per our convention, the $4$-face-only partition in the outer annulus of an ADB-AC graph $G^{\prime}$ is assumed to be $F^{1}(O_{G^{\prime}})$. Let $x>1$. For the second statement, note that for any $k$, $2\leq k\leq x$, if the edges interior to $k$-th annulus in $G$ are attached to faces that are not in $4$-cycle-only face partition in $O(H_{k-1})$, $F^{2}(O_{H})$, where $H$ is the $k-1$ annular restriction of $G$, then for the $k$-th annulus, there is a $4$-cycle. The reason is no edges in the $k+1$-th annulus are attached to the face partition with $4$-cycles of the $k$-th annulus. Thus by Lemma 3.2, $G$ has to be Hamiltonian. The Hamiltonian cycle can be constructed using the same strategy as depicted in Figure 3.2. We will call this strategy of constructing Hamiltonian cycles henceforth as the ring strategy. ∎

Let $G$ be an ADB-AC graph consisting only of ring annuli $A_{1},\dots,A_{x}$. Any arbitrary annulus $A_{i}$ must have $|F_{A_{i}}|$ $4$-cycles. One $4$-cycle from every annuli can be used to create a Hamiltonian cycle. Thus, it is clear that $G$ has at least $\frac{1}{2^{x}}\Pi_{i=1}^{x}(|F_{A_{i}}|)$ Hamiltonian cycles. Given that $|F_{A_{i}}|$ is at least $4$, we can derive that any such graph will have at least $2^{x}$ Hamiltonian cycles. ∎

Recall that, as per our convention, the $4$-face-only partition in the outer annulus of an ADB-AC graph $G^{\prime}$ is assumed to be $F^{1}(O_{G^{\prime}})$. Let $x>1$. Let us consider a planar projection of $G$ such that $f_{i}$ and $f_{o}$ are fixed and well defined. Note that such a graph $G$ must contain a specific substructure described as follows. This substructure contains one face from every annulus such that the faces are stacked on each other ($x$ faces stacked on one another). Let us denote these faces as $g_{1},\dots,g_{x}$ and observe that $g_{k}\in F^{1}(O_{G_{k}})$ for the $k$ annular restriction of $G$, $3\leq k\leq x$. Also the face $g_{x}$ in the $x$-th annulus is a $4$-cycle. Existence of such a structure is ensured due to the fact that $G$ has only block annuli. We call such a structure a pyramid. Now, note that, since $G$ is $2$-edge connected, there has to be at least two pyramids in $G$. We denote them as $B_{1}$ and $B_{2}$.

Note that, in the first annulus of $G$, there must exist a partition of faces with all $4$-cycles. We choose one from them and call it $m^{\prime}$. Our strategy to build a second Hamiltonian cycle in $G$ is based on these pyramids and the $4$-cycle $m^{\prime}$. The strategy is as follows:

1. 1.

   Build a Hamiltonian cycle $H_{1}$ in $G$ using the strategy mentioned in Theorem 3.7. For the sake of understanding let us assume that we color the interior of the Hamiltonian cycle in grey as shown in Figure 5.

2. 2.

   Note that all the faces in $B_{1}$ and $B_{2}$ are colored in grey, i.e. are interior to the cycle. Also, $m^{\prime}$ is colored in white, i.e. is exterior to the cycle.

3. 3.

   Remove $f_{i}$ from $H$, i.e color $f_{i}$ in white.

4. 4.

   Color all faces in annulus $1$ except for $m^{\prime}$. Call the new grey region $H_{2}$.

5. 5.

   Remove coloring from all faces in annulus $x-1$ except for the face in annulus $x-1$ that belongs in $B_{1}$.

6. 6.

   Color all faces in annulus $x$ except for the $4$-cycles attached to the face $h_{1}$, such that $h_{1}\in B_{1}\cap O(G_{x-1})$ and any one $4$-cycle $m$ attached to the face $h_{2}$, such that $h_{2}\in B_{2}\cap O(G_{x-1})$. Also, we choose $m$ such that the $m\in F^{1}(O_{G})$.

7. 7.

   Among the uncolored $4$-cycles mentioned in the step above, color the ones that are in $F^{2}(O_{G})$.

After Step $3$, the grey region gets path-disconnected but retains path-connectivity after Step $4$. Step $4$ also gives a Hamiltonian cycle $H_{2}$ in $G$. After Step $5$, note that only some vertices in the outer boundary of the $x-1$-th annulus are in the interior of the white region. After Step $6$ these vertices again lie in the boundary of the grey-white region. Note both the grey and white regions still remain path connected (the existence of $m$ ensures the path connectedness of the white region). After Step $7$ all the vertices lie in the boundary of the grey-white coloring and both the white and the grey region remain path connected. Thus the boundary of the grey-white path connected regions is a new Hamiltonian cycle $H_{3}$. We will call this strategy of constructing a Hamiltonian cycle in $G$ as pyramid-ring strategy or pr-strategy. The pr-strategy have been demonstrated in Figure 6. ∎

Recall that, as per our convention, the $4$-face-only partition in the outer annulus of an ADB-AC graph $G^{\prime}$ is termed as $F^{1}(O_{G^{\prime}})$. When $x=1,2$ we treat all the annuli as ring annuli. For $x\geq 3$ block annuli, we first employ the pr-strategy on $G_{x}$. Since the $x+1$-th annulus and all annuli afterwards are ring annuli, there must exist a $4$-cycle in the $x+1$-th annulus such that the edges of the $4$ face are attached to some face $f\in F^{2}(O_{G_{x}})$. Since all the faces in $F^{2}(O_{G_{x}})$ are marked in grey as per the pr-strategy, we can extend the grey color to $f$ and follow the ring strategy for the rest of the $y-1$ annuli after that to obtain a Hamiltonian cycle (whose interior is marked in grey). ∎

If $x=1,2$, then we can consider all the annuli as block annuli and use the pr-strategy for finding a Hamiltonian cycle. Recall that, as per our convention, the $4$-face-only partition in the outer annulus of an ADB-AC graph $G^{\prime}$ is termed as $F^{1}(O_{G^{\prime}})$. If $x>2$, first we can create a Hamiltonian cycle in $G_{x-1}$ using the ring strategy, and color it’s interior grey such that only one $4$-face in $O(G_{x-1})$ is colored in white. Until now, for $x$-th annulus in $G_{x-1}$, the $4$-face-only face partition has been conventionally $F^{1}(O_{G_{x-1}})$. We alter this convention from this annulus and call them $F^{2}(O_{G_{x-1}})$. Note that, this allows us to extend the grey region (Hamiltonian cycle) from $x-1$th annulus to the $x+y$th annulus, using the pr-strategy as described in Lemma 3.9. ∎