FINITE COMPLETION OF COMMA-FREE CODES. Part II

FINITE COMPLETION OF COMMA-FREE CODES. Part II

Author(s) Lam, Nguyen Huong

Citation 数理解析研究所講究録 (2004), 1366: 129-140

Issue Date 2004-04

URL http://hdl.handle.net/2433/25370

Right

Type Departmental Bulletin Paper

Textversion publisher

Kyoto University

FINITE COMPLETION OF COMMA-FREE CODES. Part II

NGUYEN HUONG LAM*

Hanoi Institute of Mathematics

Keywords Comma-free Code, Completion, Finite Maximal Comma-ffee Code.

comma-free codes

in$\mathrm{g}\mathrm{e}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{l}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{o}\mathrm{r}\mathrm{y}\mathrm{o}\mathrm{f}^{\mathrm{c}\mathrm{o}\mathrm{m}}\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{b}\mathrm{l}\mathrm{e}\mathrm{m}\mathrm{o}\mathrm{c}\mathrm{o}\mathrm{d}\mathrm{e}_{\mathrm{s}[\mathrm{g}_{\mathrm{p}}\mathrm{j}\mathrm{a}_{\mathrm{h}\mathrm{a}}^{\mathrm{C}}\mathrm{o}\mathrm{d}_{\mathrm{h}\mathrm{a}}^{\mathrm{e}}\mathrm{o}\mathrm{f}}^{\mathrm{p}1\mathrm{e}}\mathrm{n}$

$\mathrm{S}\mathrm{o}_{\mathrm{o}\mathrm{m}\mathrm{e}}^{\mathrm{m}\mathrm{e}\mathrm{c}}1\mathrm{a}_{\mathrm{t}\mathrm{e}}^{\mathrm{s}}$$\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}_{0}^{\mathrm{i}\mathrm{n}}\mathrm{t}\mathrm{h}_{\mathrm{e}\mathrm{s}}^{\mathrm{i}\mathrm{s}}\mathrm{c}_{\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{h}\mathrm{e}}^{1\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{i}}\mathrm{s}_{\mathrm{S}}$

$\mathrm{m}^{0}\mathrm{n}\mathrm{g}\mathrm{p}\mathrm{T}_{\mathrm{y}}^{\mathrm{o}\mathrm{b}}1_{\mathrm{a}}^{\mathrm{e}}\mathrm{m}\mathrm{s}$

or Berstel and Perrin $[\mathrm{B}\mathrm{P}]).\mathrm{T}\mathrm{h}\mathrm{e}$ situationis samefor finite bifixcodes: there exist finite

finite outfix code is included in a finite maximal outfix code (a set $X$ is an outfix code

completion problem

con-tains a “short” $\mathrm{i}\mathrm{l}\mathrm{r}$-word with rich properties and starting from this word we construct

comma-free code (in \S 5).

emptyword 1 andasusual $\mathit{4}^{+}$

we use interchangeably the plus and minus signs to denote the union and difference of

$*$

word 1 as the unit For subsets $X$ and $X’$ of $A^{*}$ we denote

$XX’=\{xx’ : x\in X, x’\in X’\}$

$X^{0}=\{1\}$

$X^{i+1}=X^{i}X$, $i=0,1,2$,$\ldots$

$X^{*}=$ ,$J_{i\geq 0}Xi$.

A comma-ffee code is called$m$ aximal ifitis not a propersubset ofanyother

In view of Zorn’s lemma, every comma-free code always has completions

We shall usefrequentlythe following result (FineandWilf): If$u\{u, v\}$* and $\{u, v\}$ ’

have a common left factor of length at least $|u|+|\mathrm{t}$ $|$, in particular, if$uv=vu$, then $u$

for some non-empty words $s,t\in A^{+}$ and $w\in A^{+}$, or equivalently,

$us=tv$

$X^{*}= \bigcup_{i>0}X^{i}$.

Acomma-ffeecode is called maximal ifitis not a propersubset ofanyother

In view of Zorn’s lemma, every comma-ffee code always has completions

We shall useffequentlythe following result (FineandWilf): If$u\{u, v\}^{*}$ and $\{u, v\}^{*}$

have a common left factor of length at least $|u|+|v|$, in particular, if$uv=vu,$ then $u$

words $u$ and $v$, not necessarily distinct, overlap if

for some non-empty words $s$,$t\in A^{+}$ and $w\in A^{+}$, or equivalently,

$us=tv$

for some non-emptywords $s$,$t$ such that $|t$ and $|s|<|\mathrm{t}^{\mathrm{t}}|$. We call $w$ an overlap, $s$

paper [L] Let $N$ be apositive integer

paper [L] Let $N$ be apositive integer

;hat$\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{y}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{g}\mathrm{e}\mathrm{r}nn\leq|p|,|s|<n’+N\mathrm{n}0<n\mathrm{d}^{1:}|,s\in E(X),$ $\mathrm{o}\mathrm{r}\mathrm{j}\mathrm{u}\mathrm{s}\mathrm{t}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{i}\mathrm{s}\mathrm{a}\mathrm{l}\mathrm{e}\mathrm{f}\mathrm{t}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}$ ps$\mathrm{a}\mathrm{m}\mathrm{e}\mathrm{a}\mathrm{n},\mathrm{a}p$7$\mathrm{h}4*1^{\mathrm{C}}$(K$\mathrm{r}$

) and $s\not\in R(X)A^{*}$.

Ouraim nowis to prove that we can complete every finite $N$-canonical comma-free

1 $u\not\in A^{*}XA^{*}$.

2 $u\not\in F(X^{2})$.

3 $u\not\in A^{*}L(X)+R(X)A^{*}$.

4 $u\in$ I(X) $=\{u:u^{2}\not\in A^{*}XA^{*}\}$.

5 $A^{+}u\cap uP(X)=\emptyset$ and $uA^{+}$ $\mathrm{f}" 1$ $S(X)u=\emptyset$

(r) $u$ avoids $X$ (i.e $u$ has no factors in $X$), $u$ has no left factor in $R(X)$:

$u\in A^{+}-R(X)A^{*}-A^{*}XA^{*}$.

(1) $u$ avoids $X$, $u$ has no right factor in$L(X)$:

$u\in A^{+}-A^{*}L(X)-A^{*}XA^{*}$.

The good word $u$ is called $R$-good if$uv$ avoids $X$ for all $r$-words $v$. Similarly, $u$ is

We say that the word $u$ is an Lr-lr-word if it is an $1x$-word and for all $\mathrm{r}$ words $v$, $\mathrm{w}$

$v$. $uv$ avoids $X$.

factor of$u^{+}$ for some word $u$ oflength less or equal to $k$, $|u|\leq k,$ or equivalently, $\mathrm{m}$ is is

(ii) $u$ is a left factor of$x^{+}$,

(ii) $u$ is aright factor of$y^{+}$,

In the following simple statement we show that we can pick out of three special

integer

$eq^{2}u\mathrm{a}lt|v|\leq$No. $Supposethatu,uv_{1}en\mathrm{a}teastoneo$ $anduv_{1}v_{2}dothemisprim$intiovte.self-overlap with borders shorter or

LEMMA 3.3 Let $u$ and $v$ be words such that $|u|\geq 3N$,$0<|v|\leq N$, $u=\lambda^{m}$,$uv=\mu^{n}$

$with$ borders oflength shorter than or equal to$\overline{N}$

for some primitive words $\lambda$, $\mu$. Then $\mu=$ XX$n$

primitive word $\overline{\lambda}$ such that A is aleft factor of ) $+and$ $|$ ) $|< \frac{|v|}{2}$.

ilr-words,

of $w$, $w=uv$ and $v\neq 1$, which is a $k$-sesquipovver, that is, $u=$ $u\mathrm{j}\mathrm{T}\mathrm{J}2$ with $u_{2}$ a

properleft factor of$u_{1}$, $u_{1}$ primitive and $|\mathrm{t}\mathrm{t}_{1}$ $|\leq k.$ Then for every integer$t$ such that

$|u \mathrm{x}u_{2}|\geq\min$ $(| \mathrm{L}1\mathrm{S}t1u_{2}|, 2k)$ the $word$ us$t1u_{2}?7^{\cdot}\mathrm{s}$ not ak-sesquipower

$\mathrm{L}\mathrm{E}\mathrm{M}\mathrm{M}\mathrm{A}3.3.Letuandvwithprim\mathrm{i}tivewords\lambda,$

$\mu$ anbed$integersm\geq 2,nwordssuchthat|u>2.|\begin{array}{l}\geq 3N,0<I\mathrm{f}b\end{array}|$

not $othofu\mathrm{a}nduvseif-overlapv|\leq N,u=\lambda^{m}uv=\mu^{n}$

for some primitive words $\lambda$, $\mu$. Then $\mu=\lambda\overline{\lambda}^{n}$ for some positive integer $n$ and some

primitive word $\lambda-$ such that $\lambda$ is aleft factor of$\lambda-+$ and $| \overline{\lambda}|<\frac{|v|}{2}$.

ilr-words

properleft factor of$u_{1}$, $u_{1}$ primitive and $|u_{1}|\leq k.$ Then for every integer$t$ such that

$|u_{1}^{t}u_{2}| \geq\min(|u_{1}^{t}u_{2}|, 2k)$ the word $u_{1}^{t}u_{2}v$ is not ak-sesquipower

LEMMA 3.6. Let $p$ not be a factor of$q$ and $|q|\geq 2|p|$. Then $qp^{n}$ is primitive for all integers $n>0.$

$X\}\S 4 \mathrm{S}\mathrm{h}\mathrm{o}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{l}\mathrm{r}- \mathrm{w}\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{s}\mathrm{S}\mathrm{u}\mathrm{p}\mathrm{p}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}h\mathrm{i}\acute{\mathrm{s}}$ $\mathrm{a}\mathrm{p}_{\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}}$ $\mathrm{i}\mathrm{l}\mathrm{r}- \mathrm{w}\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{f}\mathrm{o}\mathrm{r}X\mathrm{o}\mathrm{f}\mathrm{l}\mathrm{e}\mathrm{f}\mathrm{f}\mathrm{i}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{e}N- \mathrm{c}\mathrm{a}\mathrm{n}\mathrm{o}\mathrm{n}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}$ wngorhd $\mathrm{w}\mathrm{i}\mathrm{g}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{h}_{\mathrm{m}}m\mathrm{m}:\{|x\mathrm{v}\mathrm{u}\mathrm{e}$

$x\mathrm{p}\mathrm{u}\mathrm{t}$

Proof. We first prove that $f$ has a factorization $f=f’f’$ such that either $f’$ is an

$\frac{|f|}{2}|-m<|f’|$, $|f’|<| \frac{|f|}{2}|+m.$

Let $f=f_{1}f_{2}$ be a factorization such that

$\lfloor\frac{|f|}{2}\rfloor+1\geq|f_{1}$ $|$, $|f_{2}|$ $\geq\lfloor\frac{|f|}{2}\rfloor$

$X$ or $E_{l}$ $(X)$ such that fiui contains afactor, not a right one, in $X$

$X$ or $E_{l}(X)$ such that fiui contains afactor, not a right one, in $X$

Since $f_{1}$ avoids $X$ and $u$ contains no proper factor in $X$, we see that $f_{1}$ has a right

$f_{1}=f_{1}’x_{1}$

Consider now the word $x_{1}f_{2}$ If there is some word $u_{2}$ in $X+E_{r}(X)$ such that

$u_{2}x_{1}f_{2}$ contains afactor, not a left one, $xEX$, that is

$u_{2}x_{1}f_{2}=$ llJxv

for some words $w\in A^{+}$ and $v\in A^{*}$. Since #1/2, being a factor of$f$, avoids $X$ and $u_{2}$

$x_{1}f_{2}=x_{2}f_{2}’$.

$f=f_{1}x_{2}f_{2}’$

with $|f_{1}x_{2}|< \lfloor\frac{|f|}{2}\rfloor+m$ and $|f_{1}x_{2}|=|f_{2}|$ $-|x2|>\lfloor_{2}^{\cup f}\rfloor-m$, because $0<|x_{2}|<m.$

Note that

$|x_{1}|<|x_{2}|$.

Consider now the word $x_{1}f_{2}$ If there is some word $u_{2}$ in $X+E_{r}(X)$ such that

$u_{2}x_{1}f_{2}$ contains afactor, not a left one, $x\in X,$ that is

$u_{2}x_{1}f_{2}=wxv$

for some words $w\in A^{+}$ and $v\in A^{*}$. Since $x_{1}f_{2}$, being a factor of$f$, avoids $X$ and $u_{2}$

$x_{1}f_{2}=x_{2}f_{2}’$.

$f=f_{1}x_{2}f_{2}’$

$f_{1}$ inthe former factorization for$f$ to obtain aleft factor $x_{3}$ of$x$ andsome factorization

$|x_{1}$ $|<|x_{2}$ $|<|x_{3}$ $|$

and so on. However we cannot iterate the argument infinitely, as the length of factors

$f=f’f’$

Suppose for definiteness that $f’$ is an Lr-lr-word Recall that $|f\prime\prime|>\lfloor \mathrm{j}^{\mathrm{L}}$ $\rfloor-m.$

$u$ $=$ $\mathrm{e}\mathrm{n}_{1}^{\mathrm{s}}u_{2}$

for $s\geq 0$ and $n_{2}$ is a proper left factor of$u_{1}$ Since $f$ is a power of a primitiveword, $h$,

$|u|<|$ $\mathrm{f}|+m$

otherwise $u_{1}\in h^{+}$, hence $|u1$ $|>m,$ a contradiction.

Put $u_{0}=u$ if $|u|<2m$ and $n_{0}$ $=u_{1}^{t}u_{2}$, where $t$ is the smallest integer such that

$|u\mathrm{i}u_{2}|\geq 2m,$ otherwise In any case, we have

$\min(|u|, 2m)\leq|u_{0}|<3m.$

for $s\geq 0$ and $u_{2}$ is aproper left factor of$u_{1}$ Since $f$ is apower ofaprimitive word, $h$,

$|u|<|f|+m$

otherwise $u_{1}\in h^{+}$, hence $|u_{1}|>m,$ acontradiction.

Put $u_{0}=u$ if $|u|<2m$ and $u_{0}=u_{1}^{t}u_{2}$, where $t$ is the smallest integer such that

$|u_{1}^{t}u_{2}|\geq 2m,$ otherwise In any case, we have

$\min(|u|, 2m)\leq|u_{0}|<3m.$

Note that $n_{0}$ is a right, and left, factor of$u$ Now let $u_{3}$ be the left factor of $f’f’f^{lJ}$ of

for some words $l\in A^{*}$, $r$,$v\in A^{+}$

where

$lu_{0}=u,$ $\mathrm{L}\mathrm{L}_{0?}=(L_{3}$.

If$u=u_{0}$, that is if $l=1,$ then

$\geq\frac{3}{2}|f|-4m$$\geq 9K+9N-4m>3N.$

If $|u|\geq 2K$ then $|u_{0}|\geq 2m$ and $|l|=|$ ?J $|-|u_{0}|<|7$ $|+$ $\mathrm{r}\mathrm{z}\mathrm{z}$ $-2m=|f|-m,$ hence

$>| \frac{|f|}{2}|-m\mathit{1}-$$m-3K$ $\geq\frac{|f|}{2}-3K$ $=3N.$

Nowwe usethe hypothesis Since$X$is $N$-canonical and$f^{2}\in$ L{X), for the factorization

$f^{2}=f’lu_{3}v$

with respect to the factor $v$ of length $|\mathrm{t}\mathrm{z}|\geq 3N,$ there exist three words $v_{1}$,VIV2 V3 such that $0<|1^{)}1$ $|$, $|\mathrm{t}^{)}21$ $|v_{3}|\leq N$ and $v_{1}$,$v_{1}v_{2}$,$\mathrm{V}1\mathrm{V}2\mathrm{V}3$ all are left factors of$v$ and

$f’lu_{3}v_{1}$, $f’lu_{3}v_{1}v_{2}$, $f’lu_{3}v_{1}v_{2}v_{3}$ $\not\in A^{*}L(X)$

$u_{3}v_{1}$, u$viV2, $\mathrm{U}3\mathrm{V}1\mathrm{V}2\mathrm{V}3\not\in A^{*}L(X)$.

$\mathrm{U}3\mathrm{V}1\mathrm{V}2\mathrm{V}3,\overline{\mathrm{h}\mathrm{e}}\mathrm{n}\mathrm{c}\mathrm{e}$all ofthemcannot be $m$-sesquipowers in view ofthe maximalityof $|u|$.

If $|u|\geq 2K$ then by Lemma 3.5 all ofthem cannot be $m$-sesquipowers either So in any

case

$u_{3}v_{1}$, u$viV2, $n_{3^{t)}1^{\mathit{1})}2^{\mathit{1})}3}$

$\mathrm{U}\mathrm{Z}\mathrm{V}1\mathrm{V}2\mathrm{V}3$, should be primitive

an $\mathrm{L}- \mathrm{g}\mathrm{o}\mathrm{o}\mathrm{d}\mathrm{o}\mathrm{n}\mathrm{e}.\mathrm{L}\mathrm{e}\mathrm{t}\mathrm{u}\mathrm{s}\mathrm{N}\mathrm{o}\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{s}\mathrm{r}\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{e}$t,ofovre!$\mathrm{t}\mathrm{a}\mathrm{b}\mathrm{c}\mathrm{e}\mathrm{a},g\mathrm{c}\mathrm{h}\mathrm{e}\mathrm{c}\mathrm{k}$$v\mathrm{t}\mathrm{L}^{2^{\mathrm{V}}}\mathrm{P}\mathrm{o}\mathrm{i}_{\mathrm{t}\mathrm{s}}^{\mathrm{S}\mathrm{a}}$ $\xi_{3),(4)\mathrm{a}\mathrm{n}\mathrm{d}(5)\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{d}\mathrm{e}\mathrm{f}\mathrm{i}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}^{\mathrm{o}\mathrm{o}\mathrm{d}\mathrm{w}\mathrm{o}\mathrm{r}\mathrm{d},\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{m}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}}$ ’

left factor of $f’$ if $n_{0}$ $=u$ and t&3 has $u_{0}$ as a left factor, which is a left factor of $u$ of

Certainly

$+|v3|$ $\leq 3K+3N$

what is desired to prove.

not exceeding $m$ Second, there is a primitive $\mathrm{i}\mathrm{l}\mathrm{r}$-word longer than $m$. This implies the existence ofan $\mathrm{L}$ or $\mathrm{R}$-good word, claimed in Theorem 4.1

answer is an easy consequence of the following results by Ito, Katsura, Shyr and Yu [IKSY] :

$o$(?l$R_{gth}\mathrm{C}onLX^{i}ns_{e}^{in}\mathit{2}\%_{e}el(,m\mathrm{a}nyp3n-$primitive words if and only ifit containsaprimitiveword

word

$3K<|g|\leq 3K+3N$

(a) If for almost all (i.e all but finitely many) primitive $\mathrm{i}\mathrm{l}\mathrm{r}$words $v$, $v$ contains a

factorin$X+g$, or,$vg$ contains a factor in$X$ or an occurencesof$g$dferent fromthelast

(b) We can effectively pick out aprimitive $\mathrm{i}\mathrm{l}\mathrm{r}$word $v$ such that

$|v|>2|g|$

state that $vg$ is both an L- and an $\mathrm{R}$-good word for $X$. Indeed,

onthe set ofilr-words

2 $vg$ is not in $F(X^{2})$, as $|vg|>|g|>3m>2m,$ too long to be afactor of$X^{2}$.

$xvg=vgy$

$\mathrm{W}\mathrm{d}\mathrm{o}^{\mathrm{e}\mathrm{r}\mathrm{e}x\in A^{+}}\mathrm{e}\mathrm{s}\mathrm{n}\mathrm{o}\mathrm{t}" \mathrm{o}\mathrm{n}\mathrm{t}\mathrm{a}\mathrm{i}\mathrm{n}’ \mathrm{a}!^{x|=|y|<}\mathrm{n}\mathrm{y}\mathrm{o}\mathrm{c}\mathrm{c}\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{c}$levsgol,fig$\mathrm{a}_{\mathrm{i}\mathrm{f}\mathrm{f}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{a}^{X}\{!\mathrm{o}\mathrm{m}\mathrm{t}\mathrm{h}\mathrm{e}1\mathrm{a}\mathrm{s}\mathrm{t}\mathrm{o}\mathrm{n}\mathrm{e}.\mathrm{T}\mathrm{h}\mathrm{u}\mathrm{s}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{b}\mathrm{o}\mathrm{r}_{\mathrm{d}\mathrm{e}\mathrm{r}\mathrm{s}\mathrm{a}\mathrm{r}}^{\mathrm{a}\mathrm{n}\mathrm{d}v}\mathrm{e}}\mathrm{o}\mathrm{W}\mathrm{S}|y|\mathrm{f}\mathrm{o}\mathrm{r}g\mathrm{o}\mathrm{e}\mathrm{s}\mathrm{n}\mathrm{c}\mathrm{o}\mathrm{a}\dot{\mathrm{m}}$,

(c) Put $p=vg.$ So$p$ is both an L- andan $\mathrm{R}$-good word and $|p|>3|g|>9K\geq 9m.$

Otherwise we can choose (again, effectively) an 1-word $q\in$ Ei(X) with $|$ ( $7|\geq 2|p|$ such

one.

$(n-1)|p|>|q|+6N.$

Let $G_{i}$, for every $i=0,1$,$\ldots$,$n-$ l, bethe set consisting of words of the form

$up^{i}qp^{n}$

(i) $|u|\geq|p|$

(i) $u$ is an 1-word and up avoids $X:u\in E_{l}(X)$, $up\not\in A^{*}XA^{*}$

(iiii) $up^{i}qp^{n}$ is primitive $up^{i}qp^{n}\in Q.$

REMARK 5.2 Since $|p|>9m>m$ and$p$is primitive, $|q|\geq 2|p|$ and$p$is not afactor of

$\mathrm{g}$, all words of $G_{i}$ are not m-sesquipowers

REMARK 5.3 All words of$G_{i}$ avoid $X$ and are not factors of$X^{2}$.

REMARK 5.4. All words of $G_{i}$ are $\mathrm{i}\mathrm{l}\mathrm{r}$words, $G_{i}\subseteq E(X)$, because $u$ is an 1-word and $p$

is an $\mathrm{R}$-good word

contain$p$, $n>2$ and$p$ is primitive

(gg) All $w$ ords of$G_{i}$ are not factors of$p^{n}qp^{n}$.

satisfying

(j) $|v|\geq|(\mathrm{j}|(\geq 2|p|>|p|)$

(jj) $v$ is 1-word and $vp$ avoids $X$, in other words, $vp$ is 1-word: $vp\in E_{l}(X)$.

(jjjj) $vp^{n}$ is primitive: $vp^{n}\in Q.$

5.6 are al $0$ valid for $H$ (insteadof$G_{i}$) Also, by the similar reasons, we have

(gg) All words of$G_{i}$ are not factors of$p^{n}qp^{n}$.

satisffing

(j) $|v|\geq|q|(\geq 2|p|>|p|)$

(jj) $v$ is 1-word and $vp$ avoids $X$, in other words, $vp$ is l-wOrd: $vp\in$ Ei(X).

(jjjj) $vp^{n}$ is primitive: $vp^{n}\in Q.$

5.6 are also valid for $H$ (insteadof$G_{i}$) Also, by the similar reasons, we have

REMARK 5.7. If $vp^{n}$ has another occurrence of$p^{n}$ different bom the last one, then it must be one in $vp$.

$\mathrm{b}^{1}\mathrm{e}\mathrm{t}$

$\overline{G}_{i}=G_{i}-A^{+}G_{i}$

$\overline{H}=H-A^{+}H$

“minimal” words are of bounded length, hence $\overline{G}_{i}$ and $\overline{H}$ are finite

PROPOSITION 5.8. (i) If$wp^{i}qp^{n}$ is a $lr$-word with $n>i\geq 0$, $|\mathrm{t}\mathrm{p}|\geq 6N+|p|$ and if$p$ is not a right factor of$w$ then $wp^{i}qp^{n}$ has aright factor in $G_{i}$, hence in $\overline{G}_{i}$.

(ii) If$wp^{n}$ is an $lr$-word with $|\mathrm{r}\mathrm{p}|\geq 6N+|q|$ and if both$p$, $q$ are not rightfactors of$w$

then $wp^{n}$ has aright factors in$H$, hence in $\overline{H}$.

Proof, (i) Since $|\mathrm{t}\mathrm{P}|\geq 6N+|p|$ and $X$ is $N$-canonical, we can write

$\mathit{4}l\mathit{1}$ $=w’w_{6}w_{5}w_{4}w_{3}w_{2}w_{1}w_{0}$

where $w’\in A^{*}$, $|w_{0}|=|p|$, $|w_{j}|$ $\leq N$ and

$UJ_{j}$ . $\mathrm{f}\mathrm{f}_{1^{\mathrm{j}\mathrm{j}7}}\mathrm{o}p^{i}qp^{n}$

is an 1-word (hence a $1\mathrm{r}$-word) for $j=1$, ’6. In view of Proposition 3.2, there exist

$1\leq s\leq 3<t\leq 6$

such that

$U\mathit{1}_{S}$ . 1111$u$ ) $\mathit{0}p^{i}qp^{n}$

and

$w_{t}$ .$w_{1}$ $u$ ) $0p^{i}qp^{n}$

where $w’\in A^{*}$, $|w_{0}|=|p|$, $|w_{j}|\leq N$ and

$w_{j}\ldots w_{1}w0p^{i}qp^{n}$

$1<s<3<t<6$

such that

$w_{s}\ldots w_{1}w_{0}p^{i}qp^{n}$

and

$w_{t}\ldots w_{1}w_{0}p^{i}qp^{n}$

bothareprimitive, for,first $|p^{i}qp^{n}|>3N,$ and, second all$w_{j}\ldots$$w_{1}w_{0}p^{\overline{t}}qp^{n}$ ,$j=1$,$\ldots$ ,6

$w_{s}$ .$w_{1}w_{0}p^{i}qp^{n}$

has no left factor$p$. Finally,

$w_{s}$ .$w_{1}n$ ) $0p^{i}qp^{n}$

as a factor ofan $1\mathrm{r}$-word, avoids $X$. All together, the facts above mean that

$\mathrm{U}^{\mathrm{j}}1_{\theta}$ . $\mathit{4}\mathit{1}J_{1}$ $UJ_{0p^{i}qp^{n}}$ $\in G_{i}$.

is no longer than $6N+$ $\mathrm{r}\mathrm{z}|p|+|(\mathrm{j}|$.