The language is $L = \{a^{i} b^{j} c^{k} \;|\; k \neq 2j\}$. I'm trying to write a grammar for this language, what I have so far is:

$S \rightarrow AT_{1} \;|\; AT_{2} \;|\; AT_{3} \;|\; AB \;|\; AC$

$A \rightarrow aA \;|\; \varepsilon$

$B \rightarrow bB \;|\; \varepsilon$

$C \rightarrow cC \;|\; \varepsilon$

$T_{1} \rightarrow bbB'T_{1}c \;|\; \varepsilon $ (for $2j > k$)(1)

$B' \rightarrow bB' \;|\; b$

$T_{2} \rightarrow bT_{2}ccC'\;|\; \varepsilon$ (for $2j < k$)

$C' \rightarrow cC' \;|\; c$

$T_{3} \rightarrow bT_{3}c \;|\; \varepsilon$ (for $j = k$)

the problem that I am having is, the string $bbccc$ can't be generated although valid, in that case $j = 2$ and $k = 3$ so $2\times 2 > 3$ corresponding to production rule (1), how can I fix this?

  • $\begingroup$ Your grammar has several other problems. For example, $T_1$ doesn't produce any string of non-terminals, and $a^i$ is produced by your grammar. Also, what is case 3 for? $\endgroup$ Nov 1, 2012 at 4:50
  • $\begingroup$ @YuvalFilmus just fixed it, can you propose a solution ? case 3 is for when b's equal c's, which is accepted. $\endgroup$
    – Mike G
    Nov 1, 2012 at 4:57
  • 3
    $\begingroup$ Hint: can you think of two grammars? One for $L_> = \{a^{i} b^{j} c^{k} \;|\; k > 2j\}$, and the other for $L_< = \{a^{i} b^{j} c^{k} \;|\; k < 2j\}$? Divide and conquer. Think of each problem at a time and don't try to combine them at the beginning $\endgroup$
    – Ran G.
    Nov 1, 2012 at 5:55
  • 1
    $\begingroup$ Another hint: $a^i$ is clearly not the difficulty here. Leave it out for the time being. $\endgroup$
    – Raphael
    Nov 1, 2012 at 14:16

2 Answers 2


Production for $\{b^jc^k\; |\; j\neq 2k \}$ can be written as

$B\rightarrow bBcc\;|\;bB_1\;|\;cB_2$

$B_1\rightarrow bB_1\;|\;b\;|\;c\;|\;\epsilon$

$B_2\rightarrow cB_2\;|\;c\;|\;\epsilon$

You can see that it can't accept $bbcccc$. We can use $B\rightarrow bBcc$ twice but the final $B$ would have to be substituted with either b or c.

It accepts $bbccc$ as $B\rightarrow (bBcc) \rightarrow b(bB_1)cc \rightarrow bb(c)cc$

For even number of cs $B\rightarrow bBcc$ can be used. For odd number of cs, an extra $B_1\rightarrow c$ can be used.

So $\{a^ib^jc^k\; |\; j\neq 2k \}$ has the following grammar

$S\rightarrow AB$

$A\rightarrow aA \;|\; \epsilon$

$B\rightarrow bBcc\;|\;bB_1\;|\;cB_2$

$B_1\rightarrow bB_1\;|\;b\;|\;c\;|\;\epsilon$

$B_2\rightarrow cB_2\;|\;c\;|\;\epsilon$

Note: i can be 0 but j and k can not be 0 simultaneously.


Here is my version of a grammar for $\{a^ib^jc^k \mid k\neq 2j \}$. It is based on the very straight-forward grammar for $\{a^ib^jc^k \mid k= 2j \}$, but then it requires to have either an additional (non-empty) sequence of bs or an additional (non-empty) sequence of cs to show up in the proper place. There is on special rule that produces $a^*bc$.

$S\rightarrow AbBE \mid AECc \mid Abc$

$A\rightarrow aA \mid \varepsilon$

$B\rightarrow bB \mid \varepsilon$

$C\rightarrow cC \mid \varepsilon$

$E\rightarrow bEcc\mid \varepsilon$


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