Shuffle-Exchange Conjecture (graph-theoretic form)

Importance: High ✭✭✭
Subject: Graph Theory
Keywords:
Recomm. for undergrads: no
Posted by: Vadim Lioubimov
on: October 30th, 2009

Given integers $k,n \ge 2$, the \emph{2-stage Shuffle-Exchange graph/network}, denoted $\text{SE}(k,n)$, is the simple $k$-regular bipartite graph with the ordered pair $(U,V)$ of linearly labeled parts $U:=\{u_0,\dots,u_{t-1}\}$ and $V:=\{v_0,\dots,v_{t-1}\}$, where $t:=k^{n-1}$, such that vertices $u_i$ and $v_j$ are adjacent if and only if $(j - ki) \text{ mod } t < k$ (see Fig.1).

Given integers $k,n,r \ge 2$, the \emph{$r$-stage Shuffle-Exchange graph/network}, denoted $(\text{SE}(k,n))^{r-1}$, is the proper (i.e., respecting all the orders) concatenation of $r-1$ identical copies of $\text{SE}(k,n)$ (see Fig.1).

Let $r(k,n)$ be the smallest integer $r\ge 2$ such that the graph $(\text{SE}(k,n))^{r-1}$ is rearrangeable.

\begin{problem} Find $r(k,n)$. \end{problem}

\begin{conjecture} $r(k,n)=2n-1$. \end{conjecture}

A \emph{mask} for the graph $G:=(\text{SE}(k,n))^{r-1}$ is a $k$-regular bipartite multigraph with the bipartition $\{U,V\}$. The graph $G$ is said to be \emph{rearrangeable} if for every its mask there exists a collection, called \emph{routing}, of corresponding mutually edge-disjoint paths in $G$ connecting its end parts. (For simplicity, we do not provide here a more general definition for rearrangeability of graphs.)

Note that $G$ is a simple $r$-partite graph with $r k^{n-1}$ vertices and $(r-1)k^{n}$ edges, and any route for it consists exactly of $k^{n}$ paths. Also, $r(k,n)\le r$ is equivalent to rearrangeability of $G$.

\Image{Garden-SE (red) 1.gif}

\textbf{Figure 1.} Examples of multistage Shuffle-Exchange graphs.

For example, according to the conjecture, the graph $(\text{SE}(2,3))^{4}$ (see Fig. 1) is rearrangeable, which is a well known result.

The problem and conjecture are equivalent "graph-theoretic" forms of remarkable \OPrefnum[Shuffle-Exchange (SE) problem and conjecture]{37167} due to the following identity (that is not hard to show by normal reasoning):

\begin{theorem} $r(k,n)=d(k,n)$. \end{theorem}

The definition of $d(k,n)$ and more on SE problem/conjecture including the other 2 main forms of them, combinatorial and group-theoretic, and a survey of results can be found \OPrefnum[here]{37167}.

Bibliography

*[S71] H.S. Stone, \emph {Parallel processing with the perfect shuffle}, IEEE Trans. on Computers \textbf {C-20} (1971), 153-161.

*[B75] V.E. Beneš, \emph {Proving the rearrangeability of connecting networks by group calculation}, Bell Syst. Tech. J. \textbf {54} (1975), 421-434.


* indicates original appearance(s) of problem.