Context sensitive grammar for an odd number of copies of the same word

Question

Let $L = \{ w^m \mid m = 2k +1, k \ge 1 \}$. Please give some idea to write a Context sensitive grammar for $L$. Will it be like $L' = \{www \mid w \in \{a, b\}^*$? Then for each $w$ we have to introduce one nonterminal?

Answer 1

If $w$ in the question is a fixed string (or even from a finite set of finite strings) then you can write a context free grammar for it. Every context free grammar is also a context sensitive grammar. So following grammar suffices.

$S \rightarrow wwS\ |\ w$

If $w \in \{a,b\}^*$ then you will have to do something like (https://math.stackexchange.com/questions/163830/context-sensitive-grammar-for-the-copy-language) but with variable number of copies.

Also first try to solve for $L = \{w^m\ \ | \ \ m\geq 1, w \in \{a,b\}^*\}$.

It should be possible to write a context-sensitive-grammar because we can easily construct a linear bounded automata. Write $1^m$ non-deterministically in a separate track, copy it as $0^m$, $1^m$, $0^m$ through out the tape in that track, and then check it all contain the same copy of $w$ in the input track. (note you can also ensure odd number of copies here). Also note that even though we may not go right or left of input, we can still use different tracks by choosing a larger tape alphabet.

With this program if we apply the Myhill-Landweber-Kuroda Theorem (A language is accepted by a LBA iff it is context sensitive) we can easily get a context sensitive grammar.

Or taking ideas from the LBA, we can write the CSG in following way: first generate something like $BA_aA_bA_aA_a...A_aFDD...$ odd number $...DE$. Then whenever $A_a$ and $A_b$ transpose with $D$ they will leave behind $a$ and $b$. $D$ will start transposing only when it is touched by marker $F$ which will turn to $G$ to prevent further transposing. $D$'s will generate $a$' and $b$'s until it toucher $B$. $D$'s will create marker $H$ when they touch $B$, $H$ will move right and when it encounter $A_a$ or $A_b$ it will become $I$. $A_a$'s and $A_b$'s will have to move to the right of $a$'s and $b$'s to enable $I$ to get to $G$. This $I$ when it touches $G$, it will become $F$ which will start another round of substituting and generating $w$. $A_a$'s and $A_b$'s will vanish when they touch $E$. $B$ stands for beginning symbol, $D$ for delimiter symbol, and $E$ for end symbol. So combining all the ideas you get the following (We treat the case $w = \epsilon$ separately):

$S \rightarrow BTFUE \ | \ \epsilon$
$T \rightarrow TA_a\ |\ TA_b\ |\ A_a\ |\ A_b$
$U \rightarrow UDD\ |\ D$

Above is a straight forward CFG to generate the pattern that we want. Now are CSG rules. I will write the rules as non-contracting grammar.

$FD \rightarrow DG$
$A_aD \rightarrow DaA_a$
$A_bD \rightarrow DbA_b$
$aD \rightarrow Da$
$bD \rightarrow Db$
$A_aa \rightarrow aA_a$
$A_ab \rightarrow bA_a$
$A_bb \rightarrow aA_b$
$A_bb \rightarrow bA_b$
$BD \rightarrow BH$
$Ha \rightarrow aH$
$Hb\rightarrow bH$
$HA_b\rightarrow A_bI$
$HA_a\rightarrow A_aI$
$IA_a\rightarrow A_aI$
$IA_b\rightarrow A_bI$

Now a few more rules which do not follow the rules of non-contracting grammar.

$IG \rightarrow F$
$FE \rightarrow E$
$B \rightarrow \epsilon$
$A_aE \rightarrow E$
$A_bE \rightarrow E$
$E \rightarrow \epsilon$

If we convert the above grammar to a non-contracting grammer we can then convert it to a CSG for the language. This can be done if we do not use markers ($B$, $E$, $D$, $F$, $G$, $H$, and $I$, markers or their avatars need to be removed by replacing with $\epsilon$ at the end of production) but instead use real symbols for markers. The idea remains the same but the grammar becomes a little more complicated. $D$'s, $E$ and $F$ should stand for last letter of $w$, $B$ should stand for first letter of $w$. The last run should generate $w$ while vanishing $A_a$ and $A_b$.