Prove that the grammar defines language
$begingroup$
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
edited Dec 21 '18 at 17:50
Arthur
123k7122211
asked Dec 21 '18 at 17:41
Avishai YanivAvishai Yaniv
173
$endgroup$
add a comment |
$begingroup$
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
edited Dec 21 '18 at 17:50
Arthur
123k7122211
asked Dec 21 '18 at 17:41
Avishai YanivAvishai Yaniv
173
$endgroup$
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
add a comment |
$begingroup$
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
edited Dec 21 '18 at 17:50
Arthur
123k7122211
asked Dec 21 '18 at 17:41
Avishai YanivAvishai Yaniv
173
$endgroup$
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
edited Dec 21 '18 at 17:50
Arthur
123k7122211
asked Dec 21 '18 at 17:41
Avishai YanivAvishai Yaniv
173
edited Dec 21 '18 at 17:50
Arthur
123k7122211
asked Dec 21 '18 at 17:41
Avishai YanivAvishai Yaniv
173
edited Dec 21 '18 at 17:50
Arthur
123k7122211
edited Dec 21 '18 at 17:50
Arthur
123k7122211
edited Dec 21 '18 at 17:50
Arthur
123k7122211
123k7122211
asked Dec 21 '18 at 17:41
Avishai YanivAvishai Yaniv
173
asked Dec 21 '18 at 17:41
Avishai YanivAvishai Yaniv
173
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $w in L(R)$.
answered Jan 16 at 21:32
Yuval FilmusYuval Filmus
49k472148
$endgroup$
add a comment |
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1 Answer
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1 Answer
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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 $w in L(R)$.
answered Jan 16 at 21:32
Yuval FilmusYuval Filmus
49k472148
$endgroup$
add a comment |
$begingroup$
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 $w in L(R)$.
answered Jan 16 at 21:32
Yuval FilmusYuval Filmus
49k472148
$endgroup$
add a comment |
$begingroup$
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 $w in L(R)$.
answered Jan 16 at 21:32
Yuval FilmusYuval Filmus
49k472148
$endgroup$
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 $w in L(R)$.
answered Jan 16 at 21:32
Yuval FilmusYuval Filmus
49k472148
answered Jan 16 at 21:32
Yuval FilmusYuval Filmus
49k472148
answered Jan 16 at 21:32
Yuval FilmusYuval Filmus
49k472148
answered Jan 16 at 21:32
Yuval FilmusYuval Filmus
49k472148
49k472148
add a comment |
add a comment |
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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