The girths of the cubic Pancake graphs

The Pancake graph $P_{n}=(\mathrm{Sym}_{n},PR),n\geqslant 2$, is the Cayley graph over the symmetric group $\mathrm{Sym}_{n}$ of permutations $\pi=[\pi_{1}\pi_{2}\ldots\pi_{n}]$ written as strings in one-line notation, where $\pi_{i}=\pi(i)$ for any $1\leqslant i\leqslant n$, with the generating set $PR=\{r_{i}\in\mathrm{Sym}_{n}:2\leqslant i\leqslant n\}$ of all prefix–reversals $r_{i}$ inversing the order of any substring $[1,i],2\leqslant i\leqslant n$, of a permutation $\pi$ when multiplied on the right, i.e. $[\pi_{1}\ldots\pi_{i}\pi_{i+1}\ldots\pi_{n}]r_{i}=[\pi_{i}\ldots\pi_{1}\pi_{i+1}\ldots\pi_{n}]$. This graph is well known because of the open combinatorial Pancake problem of finding its diameter [3].

The graph $P_{n}$ is a connected vertex–transitive $(n-1)$-regular graph of order $n!$ with no loops and multiple edges. It is almost pancyclic [4, 9] since it contains cycles of length $\ell$, $6\leqslant\ell\leqslant n!$, but doesn’t contain cycles of length $3,4$ or $5$. Since the length of the shortest cycle contained in the graph is six, hence we have $g(P_{n})=6$ for any $n\geqslant 3$, where $g(P_{n})$ is the girth of $P_{n}$. The girths of the burnt Pancake graphs over the hyperoctahedral group was considered in [2]. The (burnt) Pancake graphs are commonly used in computer science to represent interconnection networks [1, 10, 11].

Importance of fixed–degree Pancake graphs, in particular, cubic Pancake graphs as models of networks was shown in [1] by D. W. Bass and I. H. Sudborough. The authors have considered cubic Pancake graphs as induced subgraphs of the Pancake graph $P_{n}$. The necessary conditions for a set of three prefix–reversals to generate the symmetric group $\mathrm{Sym}_{n}$ were found. In particular, it was shown that the cubic Pancake graphs over the symmetric group $\mathrm{Sym}_{n}$, $n\geqslant 4$, are connected with the following generating sets:

The set $BS_{2}$ is known as ’big-3’ flips, and the corresponding cubic Pancake graph generated by this set is called as the Big-3 Pancake network [10]. J. Sawada and A. Williams have conjectured in [10] that this graph has cyclic Gray codes.

In this paper we study cubic Pancake graphs and their girths. Our main result is obtained for the cubic Pancake graphs $P_{n}^{i}=Cay(\mathrm{Sym}_{n},BS_{i})$ that are Cayley graphs over the symmetric group $\mathrm{Sym}_{n}$ generated by the sets $BS_{i},\ i=1,\ldots,5$.

For $5\leqslant n\leqslant 19$, the computational results on the girths of the cubic Pancake graph $P_{n}^{6}=Cay(\mathrm{Sym}_{n},BS_{6})$ are given as follows:

For any $19\leqslant n\leqslant 33$, the computational results show that $g(P_{n}^{6})=28$. One can conjecture that this is true for any $n\geqslant 19$.

The paper is organized as follows. The proof of Theorem 1 is based on the characterization of small cycles in the Pancake graphs. We present preliminary results with main definitions and notation in Section 2, where it is shown that for any $n\geqslant 7$ there are no $10$–cycles and $11$–cycles in $P_{n}^{5}$. Then we prove Theorem 1 in Section 3.

The first results on a characterization of small cycles in the Pancake graph were obtained in [5] where the following cycle representation via a product of generating elements was used. A sequence of prefix–reversals $C_{\ell}=r_{i_{0}}\ldots r_{i_{\ell-1}}$, where $2\leqslant i_{j}\leqslant n$, and $i_{j}\neq i_{j+1}$ for any $0\leqslant j\leqslant\ell-1$, such that $\pi r_{i_{0}}\ldots r_{i_{\ell-1}}=\pi$, where $\pi\in\mathrm{Sym}_{n},$ is called a form of a cycle $C_{\ell}$ of length $\ell$. A cycle $C_{\ell}$ of length $\ell$ is also called an $\ell$–cycle, and a vertex of $P_{n}$ is identified with the permutation which corresponds to this vertex. It is evident that any $\ell$–cycle can be presented by $2\,\ell$ forms (not necessarily different) with respect to a vertex and a direction. The canonical form $C_{\ell}$ of an $\ell$–cycle is called a form with a lexicographically maximal sequence of indices $i_{0}\ldots i_{\ell-1}$. We shortly write $C_{\ell}=(r_{a}r_{b})^{k}$ for a cycle form $C_{\ell}=r_{a}r_{b}\ldots r_{a}r_{b}$, where $\ell=2\,k,\ a\neq b$, $r_{a}r_{b}$ appears exactly $k$ times and $\pi\,r_{a}r_{b}\ldots r_{a}r_{b}=\pi$ for any $\pi\in\mathrm{Sym}_{n}$. The form $C_{\ell}=(r_{a}r_{b})^{k}$ is canonical if $a>b$. By using this description, the following results characterizing $6$– and $7$–cycles were obtained.

The complete characterization of $8$–cycles is given by the following theorem.

The complete characterization of $9$–cycles in the Pancake graphs were obtained in [6].

In general, the complete characterization of small cycles in the Pancake graphs presented in [5, 6] is based on the hierarchical structure of the Pancake graphs. The graph $P_{n},n\geqslant 4,$ is constructed from $n$ copies of $P_{n-1}(i),1\leqslant i\leqslant n,$ such that each $P_{n-1}(i)$ has the vertex set

where $\pi_{k}\in\{1,\dots,n\}\setminus\{i\}:1\leqslant k\leqslant n-1\},$ $|V_{i}|=(n-1)!,$ and the edge set

where $|E_{i}|=\frac{(n-1)!(n-2)}{2}.$ Any two copies $P_{n-1}(i)$ and $P_{n-1}(j),i\neq j,$ are connected by $(n-2)!$ edges presented as $\{[i\pi_{2}\dots\pi_{n-1}j],[j\pi_{(}n-1\dots\pi_{2}i]\},$ where $[i\pi_{2}\dots\pi_{n-1}j]r_{n}=[j\pi_{(}n-1\dots\pi_{2}i].$ Prefix-reversals $r_{j},$ $2\leqslant j\leqslant n-1,$ defines internal edges in $P_{n-1}(i),1\leqslant i\leqslant n$, and the prefix-reversal $r_{n}$ defines external edges between copies. Copies $P_{n-1}(i)$, or just $P_{n-1}$ when it is not important to specify the last element of permutations belonging to the copy, are also called $(n-1)-$copies.

The hierarchical structure of the Pancake graphs is used to prove the following two results.

Further, we consider all possible options for the distribution of vertices by copies. Without loss of generality we always put $\tau^{1}=I_{n}=[1\,2\,3\,\dots\,n-1\,n]$.

Case 1: 10-cycle within $P_{n}$ has vertices from two copies of $P_{n-1}$

Suppose that a sought $10$–cycle is formed on vertices from copies $P_{n-1}(i)$ and $P_{n-1}(j)$, where $1\leqslant i\neq j\leqslant n$. It was shown in [5, Lemma 2] that if two vertices $\pi$ and $\tau,$ belonging to the same $(n-1)-$copy, are at the distance at most two, then their external neighbours $\overline{\pi}$ and $\overline{\tau}$ should belong to distinct $(n-1)-$copies. Hence, a sought cycle cannot occur in situations when its two (three) vertices belong to one copy and eight (seven) vertices belong to another one. Therefore, such a cycle must have at least four vertices in each of the two copies. Hence, there are two following cases.

Case $(4+6).$ Suppose that four vertices $\pi^{1},\pi^{2},\pi^{3},\pi^{4}$ of a sought $10$–cycle belong to one copy, and other six vertices $\tau^{1},\tau^{2},\tau^{3},\tau^{4},\tau^{5},\tau^{6}$ belong to another copy. Let $\pi^{1}=\tau^{1}r_{n}$ and $\pi^{4}=\tau^{6}r_{n}$. Since $\tau^{1}=I_{n}$ then $\pi^{1}$ and $\pi^{4}$ should belong to $P_{n-1}(1)$. Herewith, the four vertices of $P_{n-1}(1)$ should form a path of length three whose endpoints should be adjacent to vertices from $P_{n-1}(n)$.

Consider all options for passing from $\tau^{1}$ to $\tau^{6}$ by internal edges in a copy $P_{n-1}(n)$. Since the generating set of $P_{n}^{5}$ consists of the elements $r_{n-1}$ and $r_{n-3}$ corresponding to internal edges then there are two ways to get paths of length five from $\tau^{1}$ to $\tau^{6}$ (see Figure 1).

The first path is presented as follows:

such that:

Since $\pi^{4}=\tau^{6}\,r_{n}$ then we have:

Note that $\pi^{1}=\tau^{1}\,r_{n}=[n\,n-1\,\dots\,2\,1]$ with $\pi_{n}^{1}=1$ for any $n\geqslant 5$. Hence, we immediately can conclude that $\pi^{4}$ given by (15) and $\pi^{1}$ belong to different copies of $P_{n}$ since $\pi_{n}^{4}\neq 1$ for any odd $n\geqslant 7$. For $n=5$ we get $\pi^{4}=[5\,3\,4\,2\,1]$ and there is no path of length three between $\pi^{4}$ and $\pi^{1}$, presented as $r_{n-3}\,r_{n-1}\,r_{n-3}$ or $r_{n-1}\,r_{n-3}\,r_{n-1}$, since $\pi^{4}\,r_{2}\,r_{4}\,r_{2}=[4\,2\,5\,3\,1]$ and $\pi^{4}\,r_{4}\,r_{2}\,r_{4}=[5\,3\,2\,4\,1]$. This gives a contradiction with an assumption that $\pi^{1}$ and $\pi^{4}$ belong to the same copy $P_{n-1}(1)$. Thus, a sought cycle cannot occur neither in $P_{n}$ nor in $P_{n}^{5}$.

The second path is presented as follows:

such that

and hence

which means that this permutation belongs to $P_{n-1}(3)$. However, $\pi^{1}$ belongs to $P_{n-1}(1)$ which gives a contradiction again. Thus, a sought $10$–cycle cannot occur in this case.

Case $(5+5).$ Suppose that five vertices $\pi^{1},\pi^{2},\pi^{3},\pi^{4},\pi^{5}$ of a sought $10$–cycle belong to a copy $P_{n-1}(1)$, and other five vertices $\tau^{1},\tau^{2},\tau^{3},\tau^{4},\tau^{5}$ belong to a copy $P_{n-1}(n)$, where $\tau^{1}=I_{n}$, $\pi^{1}=\tau^{1}r_{n}$, and $\pi^{5}=\tau^{5}r_{n}$. Then the five vertices of $P_{n-1}(1)$ should form a path of length four whose endpoints should be adjacent to the vertices from $P_{n-1}(n)$ (see Figure 1).

There are two ways to get paths of length four from $\tau^{1}$ to $\tau^{5}$.

The first way is given as follows:

more precisely, we have:

The second way is presented as follows:

and we have:

Since $\pi^{5}=\tau^{5}\,r_{n}$ then we have either:

or

which gives a contradiction with an assumption that $\pi^{1}$ and $\pi^{5}$ belong to the copy $P_{n-1}(1)$ for any odd $n\geqslant 5$. Thus, in this case a sought cycle cannot occur in $P_{n}$, and hence in $P_{n}^{5}$.

Case 2: 10-cycle within $P_{n}$ has vertices from three copies of $P_{n-1}$

There are four possible situations in this case.

Case $(2+2+6).$ Suppose that two vertices $\pi^{1},\pi^{2}$ of a sought $10$–cycle belong to one copy, other two vertices $\tau^{1},\tau^{2}$ belong to another copy, and remaining six vertices $\gamma^{1},\gamma^{2},\gamma^{3},\gamma^{4},\gamma^{5},\gamma^{6}$ belong to the third copy. Let $\pi^{1}=\tau^{2}r_{n}$, $\pi^{2}=\gamma^{6}r_{n}$, and $\tau^{1}=\gamma^{1}r_{n}=I_{n}$, then $\gamma^{1}$ and $\gamma^{6}$ should belong to $P_{n-1}(1)$ (see Figure 2).

It is evident that there are two ways to get paths of length one from $\tau^{1}$ to $\tau^{2}$:

or

Since $\pi^{1}=\tau^{2}\,r_{n}$, we have either

or

Similar, there are two ways to get a path of length one from $\pi^{1}$ to $\pi^{2}$ such that the first way is given by $\pi^{1}r_{n-3}=\pi^{2}$, where we have either

or

and the second way is given by $\pi^{1}r_{n-1}=\pi^{2}$, where we have either

or

and since $\gamma^{6}=\pi^{2}\,r_{n}$ we get:

To get a sought $10$-cycle there should be a path of length five between $\gamma^{6}$ and $\gamma^{1}$, where $\gamma^{1}=\tau^{1}\,r_{n}=[n\,n-1\,\dots\,2\,1]$. Let us check this. If $n=5$ the vertices $\gamma^{6}(B)$, $\gamma^{6}(C)$ and $\gamma^{1}=[5\,4\,3\,2\,1]$ belong to the copy $P_{n-1}(1)$, and there are two ways to get a path of length five from $\gamma^{6}$ and $\gamma^{1}$. Namely, applying $(r_{2}\,r_{4})^{2}r_{2}$ to $\gamma^{6}$ we have either:

or

and applying $(r_{4}r_{2})^{2}r_{4}$ to $\gamma^{6}$ we have either:

or

Hence, a path of length five does not occur between $\gamma^{6}$ and $\gamma^{1}$.

If $n=7$ the vertices $\gamma^{6}(A)$ and $\gamma^{1}=[7\,6\,5\,4\,3\,2\,1]$ belong to the copy $P_{n-1}(1)$, and the following two cases are possible:

or

Again, a path of length five does not occur between $\gamma^{6}$ and $\gamma^{1}$.

If $n\geqslant 9$, there is a contradiction with an assumption that $\gamma^{1}$ and $\gamma^{6}$ belong to the copy $P_{n-1}(1)$. Thus, a sought cycle cannot occur in this case.

Case $(2+3+5).$ Suppose that two vertices $\tau^{1},\tau^{2}$ of a sought $10$–cycle belong to one copy, other three vertices $\pi^{1},\pi^{2},\pi^{3}$ belong to another copy, and remaining five vertices $\gamma^{1},\gamma^{2},\gamma^{3},\gamma^{4},\gamma^{5}$ belong to the third copy. Let $\pi^{1}=\tau^{2}r_{n}$, $\pi^{3}=\gamma^{5}r_{n}$, and $\tau^{1}=\gamma^{1}r_{n}=I_{n}$, then $\gamma^{1}$ and $\gamma^{5}$ should belong to $P_{n-1}(1)$ (see Figure 2). There are two ways to get a path of length two from $\tau^{1}=I_{n}$ to $\pi^{1}$ such that either

or

Similar, there are two ways to get a path of length two from $\pi^{1}$ to $\pi^{3}$. The first one is presented as follows:

such that either

or

The second way is presented as follows:

such that either

or

Since $\pi^{3}=\gamma^{5}\,r_{n}$, then by (17) and (18) we obtain:

To get a sought $10$-cycle there should be a path of length four between $\gamma^{5}$ and $\gamma^{1}$, where $\gamma^{1}=\tau^{1}\,r_{n}=[n\,n-1\,\dots\,2\,1]$. Let us check this. If $n=5$ the vertices $\gamma^{5}(A)=I_{n}\,r_{n-3}\,r_{n}\,r_{n-3}\,r_{n-1}\,r_{n}=[2\,4\,5\,3\,1]$ and $\gamma^{1}=[5\,4\,3\,2\,1]$ belong to the copy $P_{n-1}(1)$, and the following cases are possible: