Sorting by Weighted Reversals and Transpositions

2.1. Cycle graph

Cycle graph is a tool to develop nontrivial bounds for sorting problems using reversals and transpositions (Bafna and Pevzner, 1998). Given a signed permutation π, we create its cycle graph \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$G ( \pi )$$ \end{document} as follows. For each element \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${ \pi _i}$$ \end{document} , with \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$1 \le i \le n$$ \end{document} , we add to the set \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$V ( G ( \pi ) )$$ \end{document} two vertices: \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$- { \pi _i}$$ \end{document} and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$+ { \pi _i}$$ \end{document} . Finally we add the vertices \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$+ { \pi _0}$$ \end{document} and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$- { \pi _{n + 1}}$$ \end{document} . The set of edges \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$E ( G ( \pi ) )$$ \end{document} is formed by two types of edges: black edges and gray edges. The set of black edges is \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\{ ( - { \pi _i} , + { \pi _{i - 1}} ) \vert 1 \le i \le n + 1 \} $$ \end{document} . The set of gray edges is \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\{ ( + ( i - 1 ) , - i ) \vert 1 \le i \le n + 1 \} $$ \end{document} . Since each vertex is incident to one gray edge and one black edge, there is a unique decomposition of edges in cycles. Note that \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$G ( \iota )$$ \end{document} has \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$( n + 1 )$$ \end{document} trivial cycles (i.e., cycles with one black edge and one gray edge).

The size of a cycle \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$c \in G ( \pi )$$ \end{document} is the number of black edges in c. A cycle c is odd if its size is odd, and it is even otherwise. A cycle is short if it contains three or less black edges, and it is long otherwise. We denote by \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$c ( G ( \pi ) )$$ \end{document} the number of cycles in \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$G ( \pi )$$ \end{document} and by \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${c_{odd}} ( G ( \pi ) )$$ \end{document} the number of odd cycles in \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$G ( \pi )$$ \end{document} .

We draw the cycle graph \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$G ( \pi )$$ \end{document} in a way that reveals important features (Fig. 1). For each element \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${ \pi _i} \in \pi$$ \end{document} , with \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$1 \le i \le n$$ \end{document} , we place the vertex \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$- { \pi _i}$$ \end{document} before the vertex \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${ \pi _i}$$ \end{document} . The vertex \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$+ { \pi _0}$$ \end{document} is located in the left of \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$- { \pi _1}$$ \end{document} and the vertex \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$- { \pi _{i + 1}}$$ \end{document} is located in the right of \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${ \pi _n}$$ \end{document} .

OPEN IN VIEWER

FIG. 1.

Cycle graph of permutation \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\pi = ( + 3 - 1 - 2 + 6 + 5 + 8 + 7 + 4 )$$ \end{document} .

Black edges of \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$G ( \pi )$$ \end{document} are labeled from 1 to \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$n + 1$$ \end{document} so that the black edge \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$( - { \pi _i} , { \pi _{i - 1}} )$$ \end{document} is labeled as i. We represent c as the sequence of labels of its black edges following the order that they appear in the path starting from the rightmost black edge being traversed from right to left. If a black edge i in this path is traversed from left to right we label it as \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$- i$$ \end{document} ; otherwise, we simply label it as i.

Two black edges in a cycle c are convergent if they have the same sign, and divergent otherwise. A cycle c is divergent if it has two black edges that are divergent, and it is convergent otherwise. Moreover, a convergent cycle \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$c = ( {i_1} , \ldots , {i_k} )$$ \end{document} , for all \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$k > 1$$ \end{document} , is nonoriented if \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${i_1} , \ldots , {i_k}$$ \end{document} is a decreasing sequence, and oriented otherwise. Two cycles \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${c_1} = ( {i_1} , \ldots , {i_k} )$$ \end{document} and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${c_2} = ( {j_1} , \ldots , {j_k} )$$ \end{document} with \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$k > 1$$ \end{document} are interleaving if either \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\vert {i_1} \vert > \vert {j_1} \vert > \vert {i_2} \vert > \vert {j_2} \vert > \ldots > \vert {i_k} \vert > \vert {j_k} \vert$$ \end{document} or \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\vert {j_1} \vert > \vert {i_1} \vert > \vert {j_2} \vert > \vert {i_2} \vert > \ldots > \vert {j_k} \vert > \vert {i_k} \vert$$ \end{document} .

Let g₁ and g₂ be two gray edges such that (1) g₁ is incident to black edges labeled as x₁ and y₁, (2) g₂ is incident to black edges labeled as x₂ and y₂, (3) \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\vert {x_1} \vert < \vert {y_1} \vert$$ \end{document} , and (4) \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\vert {x_2} \vert < \vert {y_2} \vert$$ \end{document} . We say that g₁ intersects g₂ if \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\vert {x_1} \vert < \vert {x_2} \vert < \vert {y_1} \vert < \vert {y_2} \vert$$ \end{document} or \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\vert {x_1} \vert < \vert {x_2} \vert = \vert {y_1} \vert < \vert {y_2} \vert$$ \end{document} . Two cycles c₁ and c₂ intersect if a gray edge \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${g_1} \in {c_1}$$ \end{document} intersects a gray edge \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${g_2} \in {c_2}$$ \end{document} .

We say that a gray edge g in a nontrivial cycle c is an open gate if it is not intersected by any other gray edge from c. An open gate g₁ from a cycle c₁ is closed if the gray edge g₁ is intersected by a gray edge g₂ from another cycle c₂. Given any permutation π, all open gates in \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$G ( \pi )$$ \end{document} are closed by cycles from \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$G ( \pi )$$ \end{document} .

Figure 1 shows the cycle graph for \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\pi = ( + 3 - 1 - 2 + 6 + 5 + 8 + 7 + 4 )$$ \end{document} . There are three cycles in \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$G ( \pi )$$ \end{document} : the oriented odd cycle \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${c_1} = ( + 9 , + 5 , + 7 )$$ \end{document} , the nonoriented even cycle \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${c_2} = ( + 8 , + 6 , + 4 , + 2 )$$ \end{document} , and the divergent even cycle \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${c_3} = ( + 3 , - 1 )$$ \end{document} (the black edge 1 has a minus sign because it is traversed from left to right). Moreover, gray edge \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${g_1} \in {c_3}$$ \end{document} , which links black edges 1 and 3, intersects with gray edge \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${g_2} \in {c_2}$$ \end{document} , which links black edges 2 and 4, so it follows that cycles c₂ and c₃ intersect.

A permutation π is simple if \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$G ( \pi )$$ \end{document} contains only short cycles. We can transform any permutation π into a simple permutation \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\hat \pi$$ \end{document} by adding new elements to break long cycles. This is a common approach for sorting permutations by reversals (Hannenhalli and Pevzner, 1999) and by transpositions (Elias and Hartman, 2006). Let c be a cycle of size \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$k \ge 4$$ \end{document} . We add a black edge (i.e., a new element in the permutation) to transform c into two cycles c₁ with three black edges and c₂ with \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$k - 2$$ \end{document} black edges, as shown in Figure 2.

OPEN IN VIEWER

FIG. 2.

Transformation of a long cycle of size \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$k = 5$$ \end{document} into two cycles of size 3. The transformation is made by removing edges g and b₃, adding two vertices w and b between w_b and v_b, adding two gray edges \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${g_1} = ( {w_g} , w )$$ \end{document} and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${g_2} = ( {v_g} , v )$$ \end{document} and two black edges \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$( {w_b} , w )$$ \end{document} and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$( v , {v_b} )$$ \end{document} .

3.1. Lower bound

Given a sequence \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${ \cal S}$$ \end{document} of operations applied on π, we denote by \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\Delta c ( \pi , { \cal S} ) = c ( G ( \pi \circ { \cal S} ) ) - c ( G ( \pi ) )$$ \end{document} and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\Delta {c_{odd}} ( \pi , { \cal S} ) = {c_{odd}} ( G ( \pi \circ { \cal S} ) ) - {c_{odd}} ( G ( \pi ) )$$ \end{document} the variation in the number of cycles and odd cycles, respectively, caused on \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$G ( \pi )$$ \end{document} by \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${ \cal S}$$ \end{document} .

Let w_c and w_o be non-negative numbers representing the weight of cycles and odd cycles in \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$G ( \pi )$$ \end{document} , respectively, we define an objective function \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${R_{ \cal S}}$$ \end{document} that relates the increase in the number of cycles and odd cycles (weighted by w_c and w_o) after applying a sequence \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${ \cal S}$$ \end{document} of operations, and the sum of the costs of operations in \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${ \cal S}$$ \end{document} given by \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_{ \cal S}} ( \pi , {w_ \rho } , {w_ \tau } )$$ \end{document} . \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} \begin{align*} { R_ { \cal S } } = { \frac { { w_c } \Delta c ( \pi , { \cal S } ) + { w_o } \Delta { c_ { odd } } ( \pi , { \cal S } ) } { { w_ { \cal S } } ( \pi , { w_ \rho } , { w_ \tau } ) } } . \end{align*} \end{document}

Note that \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\Delta c ( \pi , \rho ) \in \{ - 1 , 0 , 1 \} $$ \end{document} and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\Delta {c_{odd}} ( \pi , \rho ) \in \{ - 2 , 0 , 2 \} $$ \end{document} (Hannenhalli and Pevzner, 1999; Dias et al., 2014), so we define \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$ { R_ \rho } = \frac { { { w_c } + 2 { w_o } } } { 2 } $$ \end{document} as the maximum value that can be achieved by a reversal ρ. Similarly, \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\Delta c ( \pi , \tau ) \in \{ - 2 , - 1 , 0 , 1 , 2 \} $$ \end{document} and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\Delta {c_{odd}} ( \pi , \tau ) \in \{ - 2 , 0 , 2 \} $$ \end{document} (Bafna and Pevzner, 1998; Walter et al., 1998; Dias et al., 2014), so we define \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$ { R_ \tau } = \frac { { 2 { w_c } + 2 { w_o } } } { 3 } $$ \end{document} as the maximum value that can be achieved by a transposition.

The approximation factor of \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${ \cal S}$$ \end{document} can be computed by the formula \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$A { F_ { \cal S } } = { \frac { \max \ { { R_ \rho } , { R_ \tau } \ } } { { R_ { \cal S } } } } $$ \end{document} .

All these results lead to the lower bound described in Lemma 1.

Lemma 1. Given a permutation π we have that

Proof. Note that for any reversal \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\rho ( i , j )$$ \end{document} we have \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${R_{ \rho ( i , j ) }} \le {R_ \rho }$$ \end{document} , and for any transposition \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\tau ( i , j , k )$$ \end{document} we have \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${R_{ \tau ( i , j , k ) }} \le {R_ \tau }$$ \end{document} . Since \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$G ( \iota )$$ \end{document} has \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$n + 1$$ \end{document} odd cycles, a lower bound can be obtained by dividing the necessary gain regarding cycles and odd cycles to reach the identity permutation [in this case \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$( {w_c} + {w_o} ) ( n + 1 ) - ( {w_c} \times c ( \pi ) + {w_o} \times {c_{odd}} ( \pi ) )$$ \end{document} ] by the best ratio function between reversals and transpositions ( \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\max \{ {R_ \rho } , {R_ \tau } \} $$ \end{document} ).   ■

3.2. The 5/3-approximation algorithm

In 2002, Eriksen (2002) proposed a 1+ \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\varepsilon$$ \end{document} -approximation algorithm for sorting signed permutations by weighted reversals, transpositions, and transreversals. He considered that transpositions and transreversals have twice the weight of reversals. By arguing that two reversals can simulate a transreversal with the same cost, his algorithm never uses transreversals. We decreased the weight of transpositions to be a factor of 1.5 of the weight of a reversal instead of 2. Then, we analyzed the steps of his algorithm to compute its approximation factor.

Given a permutation π and its cycle graph \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$G ( \pi )$$ \end{document} , we call a component either a cycle that does not intersect with any other cycle in \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$G ( \pi )$$ \end{document} or a maximal group of cycles such that for each pair of cycles C_a and C_b in the component, either C_a intersects with C_b or there is a sequence \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${C_1} , {C_2} , \ldots , {C_m}$$ \end{document} of cycles such that C_i intersects with \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${C_{i + 1}}$$ \end{document} , \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$1 \le i < m$$ \end{document} , C_a intersects C₁, and C_b intersects C_m.

Let \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$( k , c )$$ \end{document} -component be a component with k black edges and c cycles. Eriksen's algorithm acts on specific components as follows: 1.

Apply a transposition on any (3,1)-component with an oriented convergent cycle. This transposition will create three odd cycles.

OPEN IN VIEWER

FIG. 3.

Two components in which Eriksen's algorithm reaches \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$A { F_ { \cal S } } = \frac { 5 } { 3 } $$ \end{document} . In (a) we have the \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$( 6 , 2 )$$ \end{document} -component with two nonoriented convergent cycles. In (b) we have a \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$( 5 , 1 )$$ \end{document} -component with the oriented cycle \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$c = ( 5 , 3 , 1 , 4 , 2 )$$ \end{document} .

To prove an approximation for the case where \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_ \rho } = 2$$ \end{document} and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_ \tau } = 3$$ \end{document} , we first performed a grid search to set appropriate values for w_c and w_o.

Lemma 2 shows the approximation factor for this algorithm using \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_o} = 1$$ \end{document} and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_c} = 2$$ \end{document} , values that minimizes the approximation factor.

Lemma 2. Eriksen's algorithm has an approximation factor of 5/3.

Proof. To prove the approximation, we set the parameters \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_o} = 1$$ \end{document} and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_c} = 2$$ \end{document} . Therefore, \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$ { R_ \rho } = \frac { { { w_c } + 2 { w_o } } } { 2 } = 2$$ \end{document} and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$ { R_ \tau } = \frac { { 2 { w_c } + 2 { w_o } } } { 3 } = 2$$ \end{document} . We now compute the approximation factor AF for each of the steps in Eriksen's algorithm.   ■

In the first step, the sequence of operations \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${ \cal S}$$ \end{document} is composed of a single transposition that increases by two the number of cycles and the number of odd cycles, which leads to \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$ { R_ { \cal S } } = { \frac { { w_c } \Delta c ( \pi , { \cal S } ) + { w_o } \Delta { c_ { odd } } ( \pi , { \cal S } ) } { { w_ { \cal S } } ( \pi , { w_ \rho } , { w_ \tau } ) } } = \frac { { 2 \times 2 + 1 \times 2 } } { 3 } = 2$$ \end{document} . Therefore, \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$A { F_ { \cal S } } = { \frac { \max \ { { R_ \rho } , { R_ \tau } \ } } { { R_ { \cal S } } } } = \frac { 2 } { 2 } = 1$$ \end{document} .

In the second step, two transpositions increase by four the number of cycles and the number of odd cycles, which leads to \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$ { R_ { \cal S } } = { \frac { { w_c } \Delta c ( \pi , { \cal S } ) + { w_o } \Delta { c_ { odd } } ( \pi , { \cal S } ) } { { w_ { \cal S } } ( \pi , { w_ \rho } , { w_ \tau } ) } } = { \frac { 2 \times 4 + 1 \times 4 } { 2 \times 3 } } = 2$$ \end{document} . Therefore, \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$A { F_ { \cal S } } = { \frac { \max \ { { R_ \rho } , { R_ \tau } \ } } { { R_ { \cal S } } } } = \frac { 2 } { 2 } = 1$$ \end{document} .

In the third step, for any \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$( k , c )$$ \end{document} -component, the algorithm will apply \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$k - c$$ \end{document} reversals if the component has at least one divergent cycle, and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$k - c$$ \end{document} reversals otherwise. We hereafter deal with the worst-case scenario that occurs when: 1.

All cycles are convergent, so the cost of the reversals in \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${ \cal S}$$ \end{document} is \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_{ \cal S}} = 2 ( k - c + 1 )$$ \end{document} .

That would lead to the highest approximation factor \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$A { F_ { \cal S } } = { \frac { \max \ { { R_ \rho } , { R_ \tau } \ } } { { R_ { \cal S } } } } = { \frac { 4 ( k - c + 1 ) } { 3 ( k - c ) } } $$ \end{document} . Note by the formula that the smaller the \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$k - c$$ \end{document} , the higher the AF. Since all cycles are odd, we know they have at least three black edges and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$k \ge 3c$$ \end{document} , which leads to \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$A { F_ { \cal S } } \le { \frac { 4 ( 3c - c + 1 ) } { 3 ( 3c - c ) } } = { \frac { 2 ( 2c + 1 ) } { 3c } } $$ \end{document} . Thus, the final piece of the puzzle is the lowest value for c. The cycles are odd and convergent, so one cycle cannot be a component because it would have open gates. Therefore, \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$c = 2$$ \end{document} that leads to the component in Figure 3a is the lowest value for c. In this case, \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$A { F_ { \cal S } } \le { \frac { 2 ( 2c + 1 ) } { 3c } } \le { \frac { 2 \times 5 } { 3 \times 2 } } = \frac { 5 } { 3 } $$ \end{document} . Figure 3b shows another component in which \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$A { F_ { \cal S } } = { \frac { 4 \times ( 5 - 1 + 1 ) } { 3 \times ( 5 - 1 ) } } = { \frac { 4 \times 5 } { 3 \times 4 } } = \frac { 5 } { 3 } $$ \end{document} .   ■

3.3. The 3/2-approximation algorithm

This algorithm starts by transforming π into a simple permutation \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\hat \pi$$ \end{document} , then a sorting sequence is computed for \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\hat \pi$$ \end{document} and later converted into a sorting sequence for π.

The procedure to transform π into a simple permutation \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\hat \pi$$ \end{document} (Section 2.1) does not guarantee that \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_{RT}} ( \hat \pi , {w_ \rho } , {w_ \tau } ) = {w_{RT}} ( \pi , {w_ \rho } , {w_ \tau } )$$ \end{document} , but it guarantees that π and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\hat \pi$$ \end{document} have the same lower bound given by Lemma 1. This is because each time a cycle is broken, it adds a new element to the identity permutation [so, instead of \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$( n + 1 )$$ \end{document} , we have \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$( n + 2 )$$ \end{document} in the first part of the dividend], and it adds a new odd cycle to the permutation [so both \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$c ( \pi )$$ \end{document} and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${c_{odd}} ( \pi )$$ \end{document} increase by one in the second part of the dividend]. Until the permutation is sorted, the algorithm searches for specific configurations and applies operations on them. Some steps of our algorithm consider that divergent cycles may be divided into two classes: (1) divergent cycles that have no open gate: \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$( + 3 , - 1 , + 2 )$$ \end{document} , \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$( + 3 , + 1 , - 2 )$$ \end{document} , and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$( + 3 , - 2 , - 1 )$$ \end{document} , as shown in Figure 4, and (2) divergent cycles that have open gates: \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$( + 3 , - 2 , + 1 )$$ \end{document} , \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$( + 3 , + 2 , - 1 )$$ \end{document} , and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$( + 3 , - 1 , - 2 )$$ \end{document} , as shown in Figure 5.

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FIG. 4.

Divergent cycles of size three with no open gate.

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FIG. 5.

Divergent cycles of size three with open gates.

The algorithm searches for configurations and applies operations using the following order of priority: i.

If there is an oriented cycle C, apply a transposition on C (Fig. 6i).

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FIG. 6.

Schema of operations applied and cycle representation in each of the 12 steps of the algorithm.

Lemma 3. If a permutation π is such that \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\pi \ne \iota$$ \end{document} then it is always possible to perform at least 1 of the 12 steps of the 3/2-approximation algorithm.

Proof. Since we are dealing with simple permutations, there are a limited number of classes that a given cycle can fit in. In some of the cases, classifying these cycles is enough to give a sequence of operations and that is done by some step in the 3/2-algorithm as listed hereunder. In some other cases, it is not possible to obtain enough information from a single cycle to determine which step of the algorithm will be performed.   ■

The classes listed as “not enough information” have open gates, which means they cannot appear in a graph unless intersected with another cycle. That is especially useful for the divergent cycle case, which is solved by simply intersecting with any other cycle marked as “not enough information.”

The unsolved classes are the convergent nonoriented cycle of size two, and the convergent nonoriented cycle of size three. The former has one open gate, and if this open gate is closed by another convergent nonoriented cycle of size two, then Step (iv) has a sequence of operations for it. Therefore, the unsolved combinations at this point have at least one convergent nonoriented cycle of size three. This cycle has more than one open gate that can be closed by another convergent nonoriented cycle of size three or by two other convergent nonoriented cycles with sizes two or three. These combinations are listed hereunder and the step of the algorithm is indicated.

Figure 6 illustrates each step and shows the cycles generated. For example, Step (v) starts with one cycle of size two, and one cycle of size three, which is converted into three trivial cycles (the cycle of size two is kept in the end). Therefore, \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\Delta c = 4 - 2 = 2$$ \end{document} , and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\Delta {c_{odd}} = 3 - 1 = 2$$ \end{document} . Since we used two reversals and each one costs 2, \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_{ \cal S}} = 2 \times 2 = 4$$ \end{document} . That leads to \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$ { R_ { \cal S } } = { \frac { { w_c } \Delta c ( \pi , { \cal S } ) + { w_o } \Delta { c_ { odd } } ( \pi , { \cal S } ) } { { w_ { \cal S } } ( \pi , { w_ \rho } , { w_ \tau } ) } } = \frac { { 2 { w_c } + 2 { w_o } } } { 4 } $$ \end{document} .

The best approximation factor is achieved when \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_c} = 2$$ \end{document} and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_o} = 1$$ \end{document} , so \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$ { R_ { \cal S } } = \frac { { 2 \times 2 + 2 \times 1 } } { 4 } = \frac { 6 } { 4 } = 1.5$$ \end{document} . The approximation factor is given by \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$A { F_ { \cal S } } = { \frac { \max \ { { R_ \rho } , { R_ \tau } \ } } { { R_ { \cal S } } } } $$ \end{document} , where \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$ { R_ \rho } = \frac { { { w_c } + 2 { w_o } } } { 2 } = 2$$ \end{document} and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$ { R_ \tau } = \frac { { 2 { w_c } + 2 { w_o } } } { 3 } = 2$$ \end{document} . Therefore, for this step \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$A { F_ { \cal S } } = \frac { 2 } { { 1.5 } } = \frac { 4 } { 3 } $$ \end{document} .

Similar computation can be made for each step and Table 1 summarizes the variation on cycles [ \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\Delta c ( \pi , { \cal S} )$$ \end{document} ], the variation on odd cycles [ \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\Delta {c_{odd}} ( \pi , { \cal S} )$$ \end{document} ], the sequence applied ( \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${ \cal S}$$ \end{document} ), the weight of the sequence applied [ \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_{ \cal S}} ( \pi , {w_ \rho } , {w_ \tau } )$$ \end{document} ], its objective function ( \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${R_{ \cal S}}$$ \end{document} ), and the approximation factor ( \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$A{F_{ \cal S}}$$ \end{document} ). The highest approximation factor in Table 1 is \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$ \frac { 3 } { 2 } $$ \end{document} . Therefore, this is the approximation factor of the algorithm.

We can find which step of the algorithm can be applied in quadratic time, and this process repeats at most n times. Thus, in the worst case, the algorithm runs in \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$O ( {n^3} )$$ \end{document} .

3.4. Experimental results

To compare the performance of the approximation algorithms presented in this article, we developed a database of permutations with different characteristics. The database is divided into two groups: GR1 and GR2. Each group has several sets of permutations of a specific size and every set has a total of 10,000 different permutations. The criteria used to obtain the groups are the following:

GR1: Permutation size ranged from 50 to 500 in intervals of 50. We generated the permutations by applying operations in the identity permutation. The number of reversals and transpositions were 8% and 12% of the permutation size, respectively.

Both databases represent scenarios where the occurrence of reversals is 50% greater than the use of transpositions.

Figure 7 shows the maximum and average approximation ratio provided by the algorithms. The approximation factor was computed using the lower bound presented in Section 3.1. We observe that the 3/2-approximation algorithm presented the best results. Considering both groups, the algorithm kept the average approximation factor ≤1.11. The 5/3-approximation algorithm presented an average approximation factor between 1.2 and 1.3 in GR1 and between 1.2 and 1.4 in GR2.

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FIG. 7.

Average and maximum approximation factor obtained by 5/3 and 3/2 approximation algorithms.

Figure 8 shows the average weighted distance estimated by both algorithms considering weights 3 and 2 for transpositions and reversals, respectively.

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FIG. 8.

Average weighted distance estimate provided by the 5/3 and 3/2 approximation algorithms.

3.5. Considering different weights for reversals and transpositions

We investigated the adoption of different values for \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_ \rho }$$ \end{document} and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_ \tau }$$ \end{document} and created a generic approximation algorithm based on the 3/2-approximation algorithm. We listed alternative sequences of operations that produce the same cycle graph as the original sequences for each step of the algorithm, except for steps (i), (ii), and (iii). Table 2 shows these alternative sequences.

Any transposition \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\tau ( i , j , k )$$ \end{document} can be mimicked by three reversals \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$\rho ( i , j - 1 ) \rho ( j , k - 1 ) \rho ( i , k - 1 )$$ \end{document} , so we stand as a limit that \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_ \tau } \le 3{w_ \rho }$$ \end{document} ; otherwise, the algorithm should apply only reversals. For each pair of \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_ \rho }$$ \end{document} and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_ \tau }$$ \end{document} , we compute the values of w_c and w_o that gives the best approximation factor. For each value of \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_ \tau } / {w_ \rho }$$ \end{document} , we kept the maximum value of objective function between the original and the alternative sequences. After that, the approximation factor AF was computed as usual. In our analysis, the approximation factor is always ≤2 for any \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_ \tau } / {w_ \rho }$$ \end{document} in the interval \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $$1 < {w_ \tau } / {w_ \rho } < 2$$ \end{document} .

As shown in Figure 9, the best approximation ratio reached was 1.5 when \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_ \tau } / {w_ \rho } = 1.5$$ \end{document} , but by allowing the alternative sequences from Table 2 we were able to provide approximations ≤ 2 for all different weight scenarios analyzed.

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FIG. 9.

Approximation factor for different values of \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_ \rho }$$ \end{document} and \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts}\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{pifont}\usepackage{stmaryrd}\usepackage{textcomp}\usepackage{portland, xspace}\usepackage{amsmath, amsxtra}\usepackage{upgreek}\pagestyle{empty}\DeclareMathSizes{10}{9}{7}{6}\begin{document} $${w_ \tau }$$ \end{document} .