Commit 78e91c37 authored by Eddie Schoute's avatar Eddie Schoute
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write problem statement

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\title{Sorting with Weighted Reversals}
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% pdfauthor = {Aniruddha Bapat, Eddie Schoute, Alexey V. Gorshkov and Andrew M. Childs},
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\author{Aniruddha Bapat, Andrew Childs, Alexey Gorshkov, Eddie Schoute}
\title{Routing with Weighted Reversals}
Hello World~\cite{Bapat2020}
Recently, we worked on a project for routing quantum information,
if you're interested you can read the paper at~\cite{Bapat2020}.
Here, we will try to abstract away the underlying quantum operations and operate at a higher level of routing.
We will consider the path graph $P_n=(V,E)$ in this project,
i.e., vertices $1,\dots,n$ connected as
\graph { 1 -- 2 -- "$\dots$" -- n };
\end{tikzpicture}, which helps simplify the problem significantly.
We assign each node $i\in V$ a \emph{token}, with a destination $\pi(i)$ given by a permutation $\pi \colon V \to V$.
To get the token at node $i$, we define $t(i)$ as the token at node $i$ and,
in an abuse of notation, let $\pi(t(i))$ be the destination of that token.
The goal is to route each token to its destination.
We can route tokens by performing \emph{reversals},
which exchange locations of tokens.
A reversal $\rev{i,j}$, for $i,j \in V$ and $i\leq j$, performs transpositions $\prod_{k=0}^{(j-i)/2}
\begin{pmatrix} i+k & j-k \end{pmatrix}$ of the tokens.
It is possible to implement any permutation using several reversals,
e.g., one could order tokens starting from the token at node 1 by performing $\rev{1,\pi(t(1))}$,
$\rev{2,\pi(t(2))}$, etc.
We are interested in doing this in a \emph{time} that is minimal
because, in fact, each reversal needs an amount of time to be implemented%
\footnote{The time is the number of swap gates that can be performed in the quantum model in the same amount of time.},
c(n) \coloneqq \sqrt{{(n+1)^2} - p(n)} / 3\,,
where $p(n) \coloneqq n \pmod{2}$ is the parity of $n$.
This is a bit of a mouthful to work with, so it may help to simply approximate this by $n/3$.
Additionally, we allow ourselves to perform reversals $\rev{i,j}$ and $\rev{k,l}$ with $i<j<k<l$ simultaneously,
thus only incurring the time cost once for two disjoint reversals.
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