We show that for any string $w$ of length $n$, $r_B = O(z\log^2 n)$, where $r_B$ and $z$ are respectively the number of character runs in the bijective Burrows-Wheeler transform (BBWT), and the number of Lempel-Ziv 77 factors of $w$. We can further induce a bidirectional macro scheme of size $O(r_B)$ from the BBWT. Finally, there exists a family of strings with $r_B = \Omega(\log n)$ but having only $r=2$ character runs in the standard Burrows--Wheeler transform (BWT). However, a lower bound for $r$ is the minimal run-length of the BBWTs applied to the cyclic shifts of $w$, whose time complexity might be $o(n^2)$ in the light that we show how to compute the Lyndon factorization of all cyclic rotations in $O(n)$ time. Considering also the rotation operation performing cyclic shifts, we conjecture that we can transform two strings having the same Parikh vector to each other by BBWT and rotation operations, and prove this conjecture for the case of binary alphabets and permutations.

翻译：我们证明对于任意长度为$n$的字符串$w$，有$r_B = O(z\log^2 n)$，其中$r_B$和$z$分别为双射Burrows-Wheeler变换（BBWT）中的字符游程数，以及$w$的Lempel-Ziv 77因子数。我们可进一步从BBWT推导出大小为$O(r_B)$的双向宏方案。最后，存在一类字符串满足$r_B = \Omega(\log n)$，但在标准Burrows-Wheeler变换（BWT）中仅具有$r=2$个字符游程。然而，$r$的下界是应用于$w$循环移位的所有BBWT的最小游程长度，其时间复杂度可能为$o(n^2)$——鉴于我们展示了如何在$O(n)$时间内计算所有循环旋转的Lyndon分解。考虑到执行循环移位的旋转操作，我们猜想可以通过BBWT和旋转操作将具有相同Parikh向量的两个字符串相互转换，并针对二进制字母表和排列情形证明了该猜想。