Prove that the grammar defines language












0












$begingroup$


Given grammar G, I have to prove it defines the language of all words which are not palindromes.



In other words, $win L(G) Leftrightarrow win L $ where $L =$ {w$in$ $sum^*$ | w is not palindrom}



G:



$R → URU | S$



$S → aT b | bT a$



$T → UT U|U |epsilon$



$U → a | b$



Tried to do so using induction for each side:



$win L(G) Leftarrow win L $ on the size of w ,|w|



$win L(G) Rightarrow win L $ on the number of steps in the grammar.



Had no luck so far, any help will do.



Thanks in advance.






computer-science formal-languages automata







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$endgroup$








  • 2




    $begingroup$
    Hint: $T$ matches $Sigma^*.$ (Assuming $Sigma={a,b}.$)
    $endgroup$
    – Thomas Andrews
    Dec 21 '18 at 17:47












  • $begingroup$
    Can you explain?
    $endgroup$
    – Avishai Yaniv
    Dec 21 '18 at 22:03
















0












$begingroup$


Given grammar G, I have to prove it defines the language of all words which are not palindromes.



In other words, $win L(G) Leftrightarrow win L $ where $L =$ {w$in$ $sum^*$ | w is not palindrom}



G:



$R → URU | S$



$S → aT b | bT a$



$T → UT U|U |epsilon$



$U → a | b$



Tried to do so using induction for each side:



$win L(G) Leftarrow win L $ on the size of w ,|w|



$win L(G) Rightarrow win L $ on the number of steps in the grammar.



Had no luck so far, any help will do.



Thanks in advance.






computer-science formal-languages automata







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Hint: $T$ matches $Sigma^*.$ (Assuming $Sigma={a,b}.$)
    $endgroup$
    – Thomas Andrews
    Dec 21 '18 at 17:47












  • $begingroup$
    Can you explain?
    $endgroup$
    – Avishai Yaniv
    Dec 21 '18 at 22:03














0












0








0





$begingroup$


Given grammar G, I have to prove it defines the language of all words which are not palindromes.



In other words, $win L(G) Leftrightarrow win L $ where $L =$ {w$in$ $sum^*$ | w is not palindrom}



G:



$R → URU | S$



$S → aT b | bT a$



$T → UT U|U |epsilon$



$U → a | b$



Tried to do so using induction for each side:



$win L(G) Leftarrow win L $ on the size of w ,|w|



$win L(G) Rightarrow win L $ on the number of steps in the grammar.



Had no luck so far, any help will do.



Thanks in advance.






computer-science formal-languages automata







share|cite|improve this question











$endgroup$




Given grammar G, I have to prove it defines the language of all words which are not palindromes.



In other words, $win L(G) Leftrightarrow win L $ where $L =$ {w$in$ $sum^*$ | w is not palindrom}



G:



$R → URU | S$



$S → aT b | bT a$



$T → UT U|U |epsilon$



$U → a | b$



Tried to do so using induction for each side:



$win L(G) Leftarrow win L $ on the size of w ,|w|



$win L(G) Rightarrow win L $ on the number of steps in the grammar.



Had no luck so far, any help will do.



Thanks in advance.






computer-science formal-languages automata




computer-science formal-languages automata






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 21 '18 at 17:50









Arthur

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123k7122211










asked Dec 21 '18 at 17:41









Avishai YanivAvishai Yaniv

173




173








  • 2




    $begingroup$
    Hint: $T$ matches $Sigma^*.$ (Assuming $Sigma={a,b}.$)
    $endgroup$
    – Thomas Andrews
    Dec 21 '18 at 17:47












  • $begingroup$
    Can you explain?
    $endgroup$
    – Avishai Yaniv
    Dec 21 '18 at 22:03














  • 2




    $begingroup$
    Hint: $T$ matches $Sigma^*.$ (Assuming $Sigma={a,b}.$)
    $endgroup$
    – Thomas Andrews
    Dec 21 '18 at 17:47












  • $begingroup$
    Can you explain?
    $endgroup$
    – Avishai Yaniv
    Dec 21 '18 at 22:03








2




2




$begingroup$
Hint: $T$ matches $Sigma^*.$ (Assuming $Sigma={a,b}.$)
$endgroup$
– Thomas Andrews
Dec 21 '18 at 17:47






$begingroup$
Hint: $T$ matches $Sigma^*.$ (Assuming $Sigma={a,b}.$)
$endgroup$
– Thomas Andrews
Dec 21 '18 at 17:47














$begingroup$
Can you explain?
$endgroup$
– Avishai Yaniv
Dec 21 '18 at 22:03




$begingroup$
Can you explain?
$endgroup$
– Avishai Yaniv
Dec 21 '18 at 22:03










1 Answer
1






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$begingroup$

First of all, let us note that
$$
L(T) = { xyz : |x| = |z|, |y| leq 1} = Sigma^*,
$$
since every natural number can be written as $2n$ or as $2n+1$.



Similarly,
$$
L(R) = { x sigma y tau z : |x| = |z|, sigma neq tau }.
$$
Since $(xsigma y tau z)^R = z^R tau y^R sigma x^R$ and $|x| = |z^R|$, each such word isn't a palindrome.



Conversely, suppose that $w$ isn't a palindrome. Then $w^R neq w$. Suppose that $w,w^R$ differ on the $i$th letter, that is, $w_i neq w^R_i = w_{n+1-i}$, where $n = |w|$. The two words also differ on the $(n+1-i)$th letter, so $i neq n+1-i$, and we can assume that $i < n+1-i$, implying $2i leq n$. We can therefore write $w = x sigma y tau z$, where $|x| = |z| = i-1$ and $|y| = n-2i geq 0$. Then $sigma = w_i neq w_{n+1-i} = tau$, showing that $w in L(R)$.






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    $begingroup$

    First of all, let us note that
    $$
    L(T) = { xyz : |x| = |z|, |y| leq 1} = Sigma^*,
    $$
    since every natural number can be written as $2n$ or as $2n+1$.



    Similarly,
    $$
    L(R) = { x sigma y tau z : |x| = |z|, sigma neq tau }.
    $$
    Since $(xsigma y tau z)^R = z^R tau y^R sigma x^R$ and $|x| = |z^R|$, each such word isn't a palindrome.



    Conversely, suppose that $w$ isn't a palindrome. Then $w^R neq w$. Suppose that $w,w^R$ differ on the $i$th letter, that is, $w_i neq w^R_i = w_{n+1-i}$, where $n = |w|$. The two words also differ on the $(n+1-i)$th letter, so $i neq n+1-i$, and we can assume that $i < n+1-i$, implying $2i leq n$. We can therefore write $w = x sigma y tau z$, where $|x| = |z| = i-1$ and $|y| = n-2i geq 0$. Then $sigma = w_i neq w_{n+1-i} = tau$, showing that $w in L(R)$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      First of all, let us note that
      $$
      L(T) = { xyz : |x| = |z|, |y| leq 1} = Sigma^*,
      $$
      since every natural number can be written as $2n$ or as $2n+1$.



      Similarly,
      $$
      L(R) = { x sigma y tau z : |x| = |z|, sigma neq tau }.
      $$
      Since $(xsigma y tau z)^R = z^R tau y^R sigma x^R$ and $|x| = |z^R|$, each such word isn't a palindrome.



      Conversely, suppose that $w$ isn't a palindrome. Then $w^R neq w$. Suppose that $w,w^R$ differ on the $i$th letter, that is, $w_i neq w^R_i = w_{n+1-i}$, where $n = |w|$. The two words also differ on the $(n+1-i)$th letter, so $i neq n+1-i$, and we can assume that $i < n+1-i$, implying $2i leq n$. We can therefore write $w = x sigma y tau z$, where $|x| = |z| = i-1$ and $|y| = n-2i geq 0$. Then $sigma = w_i neq w_{n+1-i} = tau$, showing that $w in L(R)$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        First of all, let us note that
        $$
        L(T) = { xyz : |x| = |z|, |y| leq 1} = Sigma^*,
        $$
        since every natural number can be written as $2n$ or as $2n+1$.



        Similarly,
        $$
        L(R) = { x sigma y tau z : |x| = |z|, sigma neq tau }.
        $$
        Since $(xsigma y tau z)^R = z^R tau y^R sigma x^R$ and $|x| = |z^R|$, each such word isn't a palindrome.



        Conversely, suppose that $w$ isn't a palindrome. Then $w^R neq w$. Suppose that $w,w^R$ differ on the $i$th letter, that is, $w_i neq w^R_i = w_{n+1-i}$, where $n = |w|$. The two words also differ on the $(n+1-i)$th letter, so $i neq n+1-i$, and we can assume that $i < n+1-i$, implying $2i leq n$. We can therefore write $w = x sigma y tau z$, where $|x| = |z| = i-1$ and $|y| = n-2i geq 0$. Then $sigma = w_i neq w_{n+1-i} = tau$, showing that $w in L(R)$.






        share|cite|improve this answer









        $endgroup$



        First of all, let us note that
        $$
        L(T) = { xyz : |x| = |z|, |y| leq 1} = Sigma^*,
        $$
        since every natural number can be written as $2n$ or as $2n+1$.



        Similarly,
        $$
        L(R) = { x sigma y tau z : |x| = |z|, sigma neq tau }.
        $$
        Since $(xsigma y tau z)^R = z^R tau y^R sigma x^R$ and $|x| = |z^R|$, each such word isn't a palindrome.



        Conversely, suppose that $w$ isn't a palindrome. Then $w^R neq w$. Suppose that $w,w^R$ differ on the $i$th letter, that is, $w_i neq w^R_i = w_{n+1-i}$, where $n = |w|$. The two words also differ on the $(n+1-i)$th letter, so $i neq n+1-i$, and we can assume that $i < n+1-i$, implying $2i leq n$. We can therefore write $w = x sigma y tau z$, where $|x| = |z| = i-1$ and $|y| = n-2i geq 0$. Then $sigma = w_i neq w_{n+1-i} = tau$, showing that $w in L(R)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 16 at 21:32









        Yuval FilmusYuval Filmus

        49k472148




        49k472148






























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