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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T12:28:49+00:00 2026-06-12T12:28:49+00:00

I need help with constructing a CFG for a given language. L = {

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I need help with constructing a CFG for a given language.

L = { x ∈ {a, b}* | x != x reversed }, in other words, L is the complement of all the palindromes in L.

I’m more interested in how to approach these kind of problems rather than the specific solution.

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  1. Editorial Team
    Editorial Team
    2026-06-12T12:28:50+00:00Added an answer on June 12, 2026 at 12:28 pm

    Well, I didn’t figure the common pattern for solving such problems yet, but I think I know how to solve this problem:

    First consider the CFG G(L) for palindrome language L (let’s consider binary alphabet):

    S -> ""
    S -> "0" S "0"
    S -> "1" S "1"
    S -> "1" // odd length case
    S -> "0" // odd length case
    

    The idea in constructing G(L) is to ensure the last symbol is equal to the first symbol in S, therefore, for every position i the ith symbol from the beginning is equal to the ith symbol from the end for the word produced by that grammar.

    For a word w which is not palindrome we want to ensure there is such a position i that the ith symbol from the beginning is not equal to the ith symbol from the end. So, let’s terminate word production only after we have put different letters in the beginning and the end of production:

    S -> "0" S "0"
    S -> "1" S "1"
    S -> "0" T "1"
    S -> "1" T "0"
    T -> ""
    T -> "0" T "0"
    T -> "1" T "1"
    T -> "0" T "1" // we are allowed to put different symbols more than once
    T -> "1" T "0" // we are allowed to put different symbols more than once
    T -> "0"
    T -> "1"
    

    You can give S the meaning of the state «we haven’t put different letters yet», and T the meaning of «we have put different letters». Note I’ve removed S -> "" rule in this CFG , so we will finish only from T, so we will definitely put different letters while producing word.

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