A Simple and Efficient Algorithm for Sorting Signed Permutations by Reversals

TL;DR

The paper tackles the problem of sorting a signed permutation into the identity using a minimum-length sequence of reversals, a foundational task in computational genomics. It reinterprets the problem through the overlap graph and a recovery framework for unsafe reversals, linking reversals to local graph complementations and using a balanced BST (splay-tree) representation to efficiently manage good pairs. The main contribution is an $O(n \log n)$ worst-case algorithm that is lightweight to implement and relies on a recoverable sequence of good reversals, supported by a specialized data structure and a backtracking recovery mechanism. This work narrows the gap toward the conjectured lower bound and enhances practical feasibility for large-scale genome rearrangement studies.

Abstract

In 1937, biologists Sturtevant and Tan posed a computational question: transform a chromosome represented by a permutation of genes, into a second permutation, using a minimum-length sequence of reversals, each inverting the order of a contiguous subset of elements. Solutions to this problem, applied to Drosophila chromosomes, were computed by hand. The first algorithmic result was a heuristic that was published in 1982. In the 1990s a more biologically relevant version of the problem, where the elements have signs that are also inverted by a reversal, finally received serious attention by the computer science community. This effort eventually resulted in the first polynomial time algorithm for Signed Sorting by Reversals. Since then, a dozen more articles have been dedicated to simplifying the theory and developing algorithms with improved running times. The current best algorithm, which runs in $O(n \log^2 n / \log\log n)$ time, fails to meet what some consider to be the likely lower bound of $O(n \log n)$. In this article, we present the first algorithm that runs in $O(n \log n)$ time in the worst case. The algorithm is fairly simple to implement, and the running time hides very low constants.

Figures (6)

• Figure 1: A reversal sequence transforming $(0~\raisebox{.6mm}{\tiny\boldmath$-$}2~3~1~4)$ into the identity permutation. At the top, $(0~\raisebox{.6mm}{\tiny\boldmath$-$}2~3~1~4)$ is transformed into $(0~\raisebox{.6mm}{\tiny\boldmath$-$}2~\raisebox{.6mm}{\tiny\boldmath$-$}1~\raisebox{.6mm}{\tiny\boldmath$-$}3~4)$ by the first reversal $\rho(2,3) = \mu(\raisebox{.6mm}{\tiny\boldmath$-$}2,1)$. The pairs of points for the elements of a permutation appear in a line above the permutation, and an interval for each identity pair is indicated above those points by a black line; lines are solid for good pairs and dashed for bad pairs. The overlap graph for each permutation appears below it; good vertices are colored black and bad are colored white. The sequence of reversals on permutations can be represented by the sequence of vertex complementations indicated by arrows, yielding a graph with only isolated white vertices.
• Figure 2: The upper permutation $\pi = (0~\raisebox{.6mm}{\tiny\boldmath$-$}5~\raisebox{.6mm}{\tiny\boldmath$-$}2~\raisebox{.6mm}{\tiny\boldmath$-$}4~\raisebox{.6mm}{\tiny\boldmath$-$}7~\raisebox{.6mm}{\tiny\boldmath$-$}9~\raisebox{.6mm}{\tiny\boldmath$-$}10~\raisebox{.6mm}{\tiny\boldmath$-$}6~1~3~8~11)$ is transformed into $\pi' = \pi\rho(5, 10) = (0~\raisebox{.6mm}{\tiny\boldmath$-$}5~\raisebox{.6mm}{\tiny\boldmath$-$}2~\raisebox{.6mm}{\tiny\boldmath$-$}4~\raisebox{.6mm}{\tiny\boldmath$-$}3~\raisebox{.6mm}{\tiny\boldmath$-$}1~6~10~9~7~8~11)$ through the unsafe reversal $\rho(5, 10) = \mu(3, \raisebox{.6mm}{\tiny\boldmath$-$}4) = \mu(\raisebox{.6mm}{\tiny\boldmath$-$}7, 8)$. The overlap graph for $\pi$ is on the bottom-left, and the overlap graph for $\pi'$ is on the bottom-right. Vertices from the bad components of $OV(\pi')$ are labeled by $b$ and vertices from the good components are labeled by $g$. See the caption of Figure \ref{['fig:example']} for a more detailed description.
• Figure 3: On the left, a graph $H$ has a single connected component, before being split into graph $H/v$ having bad components $B_1, B_2, \ldots, B_{\ell_{1}}$ and good components $G_1, G_2, \ldots, G_{\ell_{2}}$, on the right. Good vertices are filled in black. Note that, due to the definition of vertex complementation, all vertices from $B = \cup_{i=1}^{\ell_{1}} B_i$ that are adjacent to $v$ must be good, while the others from $B$ must be bad. If, instead of $v$, a good vertex $u$ from $B$ is applied to $H$, then any good vertex in $H/u[B]$ will be adjacent to all the vertices of $X$.
• Figure 4: A splay tree a) before, and b) after the reversal $\mu(\raisebox{.6mm}{\tiny\boldmath$-$}7, 8)$ of Figure \ref{['fig:bigexample']}. Nodes are labeled by the values indicated in the bottom-left corner of the figure.
• Figure 5: A rotation promotes a node above its parent, while keeping the in-order traversal intact.

Theorems & Definitions (24)

• Theorem 1: hannenhalliTransformingCabbageTurnip1999
• Definition
• Theorem 2: tannierAdvancesSortingReversals2007
• Remark 1
• Remark 2
• Lemma 1
• proof
• Lemma 2
• proof
• Corollary 1