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arXiv 1907.01072math.CO
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$ω$ -林登的话

$ω$-Lyndon words

Mickaël Postic, Luca Q. Zamboni

摘要:

令$\A$是有限非空集,并且$\preceq $是$\A^\nats$上的全序,验证以下字典式条件:对于每个$n\in \nats$和$u, v\in \A^n,$,如果$u^ω\prec v^ω$则$ux\prec vy$对于所有$x, y \in \A^\nats.$一个字$x\in \A^\nats$称为$ω$ -林登,如果$x\prec y$对于$x.$的每个适当后缀$y$一个有限字$w\in \A^+$称为$ω$ -Lyndon如果对$w.$的每个真后缀$v$都有$w^ω\prec v^ω$本文证明了每个无限字可以唯一地写成$ω$-Lyndon字的非增积.

英文摘要:

Let $\A$ be a finite non-empty set and $\preceq $ a total order on $\A^\nats$ verifying the following lexicographic like condition: For each $n\in \nats$ and $u, v\in \A^n,$ if $u^ω\prec v^ω$ then $ux\prec vy$ for all $x, y \in \A^\nats.$ A word $x\in \A^\nats$ is called $ω$-Lyndon if $x\prec y$ for each proper suffix $y$ of $x.$ A finite word $w\in \A^+$ is called $ω$-Lyndon if $w^ω\prec v^ω$ for each proper suffix $v$ of $w.$ In this note we prove that every infinite word may be written uniquely as a non-increasing product of $ω$-Lyndon words.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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