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I have a homework assignment that requires me to create CFG $G$ for

$$L = \{a^i b^{i+j+k} c^j d^k\}$$

so that it can accept words like ab, aaabbbbd, abbbcd, but it should not accept abba, aabbbbbc, or abc since the number of b must be the number of a+c+d.

I need CFG of L(G)

  • I tried using chatgpt but the answers are not correct as I cross checked them with both gpt and myself.
  • I tried to cross check this pdf but they are not as complex as my question.
  • There is also a solution on chegg I can't see the whole solution as I don't have a subscription, but the final solution is also wrong as I checked it with both chatgpt and myself.
  • Also researched multiple sources but did not found the answer.

The best one i came up with this grammer but it doesn't feel right since it does not have corelation between b c and d.

1. attempt: 

S → AB 
A → aAb | ε
B -> bCEdD
C -> bB | ε
D -> dD | ε
E -> bEc | ε

2. attempt : 

S → AB 
A → aA | ε
B → bBC 
C → cC | Cd | ε 
D → dD | ε 

Is it possible to do it with cfg and how?

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  • $\begingroup$ cs.stackexchange.com/q/18524/755 $\endgroup$
    – D.W.
    Mar 15 at 17:09
  • $\begingroup$ I have already answered this question in the "Asnwers" section. I have problems understanding the accepted words, even after reading the proof of correctness @D.W. $\endgroup$ Mar 15 at 17:55

2 Answers 2

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Use the nesting structure of the language to obtain a context-free grammar.

$a^i\, b^{i+j+k}\, c^j d^k = a^i b^i\, b^k b^j c^j d^k $

We can see that the language is the concatenation of strings of the form $a^i\, b^i $ and strings of the form $b^{j+k}\, c^j d^k$ which each separately form context-free languages.

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  • $\begingroup$ The part I don't understand is when I use a cfg like : - S → ABcd - A → aAb | ε - B → bBc | ε - C → cCd | ε seems to be correct and also produces what I'm asking for, but it also accepts words like "abba, aabbbbc or abc" which I want to use specifically and only for the words I'm describing, so if the language doesn't have that, that's the part I don't understand $\endgroup$ Mar 15 at 16:38
  • $\begingroup$ Respect the nesting structure. Generate "bc"-pairs inside the "bd"-pairs. $\endgroup$ Mar 15 at 23:23
  • $\begingroup$ I also add this cfg to the main question but here's what i did : S → AB ---- A → aAb | ε ---- B -> bCEdD ---- C -> bB | ε ---- D -> dD | ε ---- E -> bEc | ε. I can't solve that point i don't know how to add bc in bd because there's also an bcd combination that i need to make recursive. $\endgroup$ Mar 16 at 9:00
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$S\rightarrow AD$
$A\rightarrow aAb,\varepsilon$
$D\rightarrow bDd, C$
$C\rightarrow bCc, \varepsilon$.

In this solution I divided the word into a prefix of $a^ib^i$ and a suffix of $b^{j+k}c^jd^k$. The latter part is generated by building $b^kCd^k$ and then replacing $C$ with $b^jc^j$.

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    $\begingroup$ Is the " , " the same as the " | " or does this symbol have a different purpose? $\endgroup$ Mar 16 at 16:33
  • $\begingroup$ yes it is the same, I just mean different choices of the right side. $\endgroup$ Mar 16 at 16:41

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