Proof verification of the language of all palindromes as being context-free
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Consider that the language L of all palindromes over $Sigma = {0,1}^*$ is not context-free. The following is my attempt at a proof by contradiction.
I am new to proof writing and I am wondering if the proof is correct, and if it proceeds in a connected logical sequence. I think I have all the cases covered, but I am not too sure.
proof-verification proof-writing context-free-grammar pumping-lemma palindrome
asked Dec 5 '18 at 23:57
SeesSoundSeesSound
859
$endgroup$
add a comment |
$begingroup$
Consider that the language L of all palindromes over $Sigma = {0,1}^*$ is not context-free. The following is my attempt at a proof by contradiction.
I am new to proof writing and I am wondering if the proof is correct, and if it proceeds in a connected logical sequence. I think I have all the cases covered, but I am not too sure.
proof-verification proof-writing context-free-grammar pumping-lemma palindrome
asked Dec 5 '18 at 23:57
SeesSoundSeesSound
859
$endgroup$
3
$begingroup$
Could you explain why $uwy = 0^k1^{2k}o^k$? If I set $w = 1, v = x = 1$, then neither $x$ nor $v$ equals $epsilon$, $|vwx| < p$ and $uv^iwx^iy in L$, but $uwy$ is not of the form you state. Apart from that, there's a very simple grammar for palindromes: $w := "" | "1" | "0" | "1" w "1" | "0" w "0"$ that looks pretty context free to me
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– Ronald
Dec 6 '18 at 0:33
$begingroup$
@Ronald could you elaborate? Are you saying that the language of palindromes is context free?
$endgroup$
– SeesSound
Dec 6 '18 at 1:02
add a comment |
$begingroup$
Consider that the language L of all palindromes over $Sigma = {0,1}^*$ is not context-free. The following is my attempt at a proof by contradiction.
I am new to proof writing and I am wondering if the proof is correct, and if it proceeds in a connected logical sequence. I think I have all the cases covered, but I am not too sure.
proof-verification proof-writing context-free-grammar pumping-lemma palindrome
asked Dec 5 '18 at 23:57
SeesSoundSeesSound
859
$endgroup$
Consider that the language L of all palindromes over $Sigma = {0,1}^*$ is not context-free. The following is my attempt at a proof by contradiction.
I am new to proof writing and I am wondering if the proof is correct, and if it proceeds in a connected logical sequence. I think I have all the cases covered, but I am not too sure.
proof-verification proof-writing context-free-grammar pumping-lemma palindrome
proof-verification proof-writing context-free-grammar pumping-lemma palindrome
asked Dec 5 '18 at 23:57
SeesSoundSeesSound
859
asked Dec 5 '18 at 23:57
SeesSoundSeesSound
859
asked Dec 5 '18 at 23:57
SeesSoundSeesSound
859
asked Dec 5 '18 at 23:57
SeesSoundSeesSound
859
asked Dec 5 '18 at 23:57
SeesSoundSeesSound
859
859
3
$begingroup$
Could you explain why $uwy = 0^k1^{2k}o^k$? If I set $w = 1, v = x = 1$, then neither $x$ nor $v$ equals $epsilon$, $|vwx| < p$ and $uv^iwx^iy in L$, but $uwy$ is not of the form you state. Apart from that, there's a very simple grammar for palindromes: $w := "" | "1" | "0" | "1" w "1" | "0" w "0"$ that looks pretty context free to me
$endgroup$
– Ronald
Dec 6 '18 at 0:33
$begingroup$
@Ronald could you elaborate? Are you saying that the language of palindromes is context free?
$endgroup$
– SeesSound
Dec 6 '18 at 1:02
add a comment |
3
$begingroup$
Could you explain why $uwy = 0^k1^{2k}o^k$? If I set $w = 1, v = x = 1$, then neither $x$ nor $v$ equals $epsilon$, $|vwx| < p$ and $uv^iwx^iy in L$, but $uwy$ is not of the form you state. Apart from that, there's a very simple grammar for palindromes: $w := "" | "1" | "0" | "1" w "1" | "0" w "0"$ that looks pretty context free to me
$endgroup$
– Ronald
Dec 6 '18 at 0:33
$begingroup$
@Ronald could you elaborate? Are you saying that the language of palindromes is context free?
$endgroup$
– SeesSound
Dec 6 '18 at 1:02
3
3
$begingroup$
Could you explain why $uwy = 0^k1^{2k}o^k$? If I set $w = 1, v = x = 1$, then neither $x$ nor $v$ equals $epsilon$, $|vwx| < p$ and $uv^iwx^iy in L$, but $uwy$ is not of the form you state. Apart from that, there's a very simple grammar for palindromes: $w := "" | "1" | "0" | "1" w "1" | "0" w "0"$ that looks pretty context free to me
$endgroup$
– Ronald
Dec 6 '18 at 0:33
$begingroup$
Could you explain why $uwy = 0^k1^{2k}o^k$? If I set $w = 1, v = x = 1$, then neither $x$ nor $v$ equals $epsilon$, $|vwx| < p$ and $uv^iwx^iy in L$, but $uwy$ is not of the form you state. Apart from that, there's a very simple grammar for palindromes: $w := "" | "1" | "0" | "1" w "1" | "0" w "0"$ that looks pretty context free to me
$endgroup$
– Ronald
Dec 6 '18 at 0:33
$begingroup$
@Ronald could you elaborate? Are you saying that the language of palindromes is context free?
$endgroup$
– SeesSound
Dec 6 '18 at 1:02
$begingroup$
@Ronald could you elaborate? Are you saying that the language of palindromes is context free?
$endgroup$
– SeesSound
Dec 6 '18 at 1:02
add a comment |
1 Answer
1
active
oldest
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$begingroup$
If a context free grammar exists that produces the language, then the language itself is context free.
A context free grammar is a 4-tuple $G = (V, Sigma, R, S)$ with the following properties:
$V$ is a finite set; $v in V$ is called a non-terminal symbol
$Sigma$ is a finite set of terminal symbols; $V cap Sigma =emptyset$
$R$ is a finite relation $V rightarrow (V cup Sigma)^*$
$S in V$ is the start symbol
With $V = {w}$, $Sigma = {0, 1}$, $S = w$ and the following map $R$:
$w rightarrow epsilon$
$w rightarrow 0$
$w rightarrow 1$
$w rightarrow 0w0$
$w rightarrow 1w1$
we've defined a context free grammar.
Note: if the empty string isn't considered to be a palindrome, the first production can be replaced by
$w rightarrow 00$
$w rightarrow 11$
Palindromes are, casually speaking, a special case of "well-formed parenthesis" (each opening parenthesis has a corresponding closing parenthesis) which is also known to be context free.
Context free languages can't "count" very well, but it isn't a problem to have an arbitrary number of matching pairs as this can be tested using a stack.
But the famous language $a^nb^nc^n$ isn't context free because there's no way to count to $n$.
Edit:
Strictly speaking I'll have to prove that the above grammar indeed produces the set of all palindromes.
First of all, if $w in L$, which means that $w$ is a palindrome, then also $0w0$ and $1w1$ are palindromes and therefore member of $L$.
Obviously $0, 1$ and $epsilon$ are palindromes.
In other words, the production rules only produce palindromes.
Now assume $p$ is a palindrome. Then $p = s_1s_2...s_nms_ns_{n-1}...s_1$, with $s_i in {0, 1}, i in {1, ..., n}$ and $m in {epsilon, 0, 1}$.
In order to produce this palindrome, we start with $w = m$ and then consecutively add the symbols $s_n, s_{n-1}, ..., s_1$ on either side of $w$, which is allowed by the last two production rules.
This proves that $p$ can be produced by the grammar above.
Together with the fact that $G$ produces palindromes only, this proves that $G$ is a valid grammar for $L$.
answered Dec 6 '18 at 9:08
RonaldRonald
673510
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
If a context free grammar exists that produces the language, then the language itself is context free.
A context free grammar is a 4-tuple $G = (V, Sigma, R, S)$ with the following properties:
$V$ is a finite set; $v in V$ is called a non-terminal symbol
$Sigma$ is a finite set of terminal symbols; $V cap Sigma =emptyset$
$R$ is a finite relation $V rightarrow (V cup Sigma)^*$
$S in V$ is the start symbol
With $V = {w}$, $Sigma = {0, 1}$, $S = w$ and the following map $R$:
$w rightarrow epsilon$
$w rightarrow 0$
$w rightarrow 1$
$w rightarrow 0w0$
$w rightarrow 1w1$
we've defined a context free grammar.
Note: if the empty string isn't considered to be a palindrome, the first production can be replaced by
$w rightarrow 00$
$w rightarrow 11$
Palindromes are, casually speaking, a special case of "well-formed parenthesis" (each opening parenthesis has a corresponding closing parenthesis) which is also known to be context free.
Context free languages can't "count" very well, but it isn't a problem to have an arbitrary number of matching pairs as this can be tested using a stack.
But the famous language $a^nb^nc^n$ isn't context free because there's no way to count to $n$.
Edit:
Strictly speaking I'll have to prove that the above grammar indeed produces the set of all palindromes.
First of all, if $w in L$, which means that $w$ is a palindrome, then also $0w0$ and $1w1$ are palindromes and therefore member of $L$.
Obviously $0, 1$ and $epsilon$ are palindromes.
In other words, the production rules only produce palindromes.
Now assume $p$ is a palindrome. Then $p = s_1s_2...s_nms_ns_{n-1}...s_1$, with $s_i in {0, 1}, i in {1, ..., n}$ and $m in {epsilon, 0, 1}$.
In order to produce this palindrome, we start with $w = m$ and then consecutively add the symbols $s_n, s_{n-1}, ..., s_1$ on either side of $w$, which is allowed by the last two production rules.
This proves that $p$ can be produced by the grammar above.
Together with the fact that $G$ produces palindromes only, this proves that $G$ is a valid grammar for $L$.
answered Dec 6 '18 at 9:08
RonaldRonald
673510
$endgroup$
add a comment |
$begingroup$
If a context free grammar exists that produces the language, then the language itself is context free.
A context free grammar is a 4-tuple $G = (V, Sigma, R, S)$ with the following properties:
$V$ is a finite set; $v in V$ is called a non-terminal symbol
$Sigma$ is a finite set of terminal symbols; $V cap Sigma =emptyset$
$R$ is a finite relation $V rightarrow (V cup Sigma)^*$
$S in V$ is the start symbol
With $V = {w}$, $Sigma = {0, 1}$, $S = w$ and the following map $R$:
$w rightarrow epsilon$
$w rightarrow 0$
$w rightarrow 1$
$w rightarrow 0w0$
$w rightarrow 1w1$
we've defined a context free grammar.
Note: if the empty string isn't considered to be a palindrome, the first production can be replaced by
$w rightarrow 00$
$w rightarrow 11$
Palindromes are, casually speaking, a special case of "well-formed parenthesis" (each opening parenthesis has a corresponding closing parenthesis) which is also known to be context free.
Context free languages can't "count" very well, but it isn't a problem to have an arbitrary number of matching pairs as this can be tested using a stack.
But the famous language $a^nb^nc^n$ isn't context free because there's no way to count to $n$.
Edit:
Strictly speaking I'll have to prove that the above grammar indeed produces the set of all palindromes.
First of all, if $w in L$, which means that $w$ is a palindrome, then also $0w0$ and $1w1$ are palindromes and therefore member of $L$.
Obviously $0, 1$ and $epsilon$ are palindromes.
In other words, the production rules only produce palindromes.
Now assume $p$ is a palindrome. Then $p = s_1s_2...s_nms_ns_{n-1}...s_1$, with $s_i in {0, 1}, i in {1, ..., n}$ and $m in {epsilon, 0, 1}$.
In order to produce this palindrome, we start with $w = m$ and then consecutively add the symbols $s_n, s_{n-1}, ..., s_1$ on either side of $w$, which is allowed by the last two production rules.
This proves that $p$ can be produced by the grammar above.
Together with the fact that $G$ produces palindromes only, this proves that $G$ is a valid grammar for $L$.
answered Dec 6 '18 at 9:08
RonaldRonald
673510
$endgroup$
add a comment |
$begingroup$
If a context free grammar exists that produces the language, then the language itself is context free.
A context free grammar is a 4-tuple $G = (V, Sigma, R, S)$ with the following properties:
$V$ is a finite set; $v in V$ is called a non-terminal symbol
$Sigma$ is a finite set of terminal symbols; $V cap Sigma =emptyset$
$R$ is a finite relation $V rightarrow (V cup Sigma)^*$
$S in V$ is the start symbol
With $V = {w}$, $Sigma = {0, 1}$, $S = w$ and the following map $R$:
$w rightarrow epsilon$
$w rightarrow 0$
$w rightarrow 1$
$w rightarrow 0w0$
$w rightarrow 1w1$
we've defined a context free grammar.
Note: if the empty string isn't considered to be a palindrome, the first production can be replaced by
$w rightarrow 00$
$w rightarrow 11$
Palindromes are, casually speaking, a special case of "well-formed parenthesis" (each opening parenthesis has a corresponding closing parenthesis) which is also known to be context free.
Context free languages can't "count" very well, but it isn't a problem to have an arbitrary number of matching pairs as this can be tested using a stack.
But the famous language $a^nb^nc^n$ isn't context free because there's no way to count to $n$.
Edit:
Strictly speaking I'll have to prove that the above grammar indeed produces the set of all palindromes.
First of all, if $w in L$, which means that $w$ is a palindrome, then also $0w0$ and $1w1$ are palindromes and therefore member of $L$.
Obviously $0, 1$ and $epsilon$ are palindromes.
In other words, the production rules only produce palindromes.
Now assume $p$ is a palindrome. Then $p = s_1s_2...s_nms_ns_{n-1}...s_1$, with $s_i in {0, 1}, i in {1, ..., n}$ and $m in {epsilon, 0, 1}$.
In order to produce this palindrome, we start with $w = m$ and then consecutively add the symbols $s_n, s_{n-1}, ..., s_1$ on either side of $w$, which is allowed by the last two production rules.
This proves that $p$ can be produced by the grammar above.
Together with the fact that $G$ produces palindromes only, this proves that $G$ is a valid grammar for $L$.
answered Dec 6 '18 at 9:08
RonaldRonald
673510
$endgroup$
If a context free grammar exists that produces the language, then the language itself is context free.
A context free grammar is a 4-tuple $G = (V, Sigma, R, S)$ with the following properties:
$V$ is a finite set; $v in V$ is called a non-terminal symbol
$Sigma$ is a finite set of terminal symbols; $V cap Sigma =emptyset$
$R$ is a finite relation $V rightarrow (V cup Sigma)^*$
$S in V$ is the start symbol
With $V = {w}$, $Sigma = {0, 1}$, $S = w$ and the following map $R$:
$w rightarrow epsilon$
$w rightarrow 0$
$w rightarrow 1$
$w rightarrow 0w0$
$w rightarrow 1w1$
we've defined a context free grammar.
Note: if the empty string isn't considered to be a palindrome, the first production can be replaced by
$w rightarrow 00$
$w rightarrow 11$
Palindromes are, casually speaking, a special case of "well-formed parenthesis" (each opening parenthesis has a corresponding closing parenthesis) which is also known to be context free.
Context free languages can't "count" very well, but it isn't a problem to have an arbitrary number of matching pairs as this can be tested using a stack.
But the famous language $a^nb^nc^n$ isn't context free because there's no way to count to $n$.
Edit:
Strictly speaking I'll have to prove that the above grammar indeed produces the set of all palindromes.
First of all, if $w in L$, which means that $w$ is a palindrome, then also $0w0$ and $1w1$ are palindromes and therefore member of $L$.
Obviously $0, 1$ and $epsilon$ are palindromes.
In other words, the production rules only produce palindromes.
Now assume $p$ is a palindrome. Then $p = s_1s_2...s_nms_ns_{n-1}...s_1$, with $s_i in {0, 1}, i in {1, ..., n}$ and $m in {epsilon, 0, 1}$.
In order to produce this palindrome, we start with $w = m$ and then consecutively add the symbols $s_n, s_{n-1}, ..., s_1$ on either side of $w$, which is allowed by the last two production rules.
This proves that $p$ can be produced by the grammar above.
Together with the fact that $G$ produces palindromes only, this proves that $G$ is a valid grammar for $L$.
answered Dec 6 '18 at 9:08
RonaldRonald
673510
edited Dec 6 '18 at 14:02
answered Dec 6 '18 at 9:08
RonaldRonald
673510
answered Dec 6 '18 at 9:08
RonaldRonald
673510
answered Dec 6 '18 at 9:08
RonaldRonald
673510
673510
add a comment |
add a comment |
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$begingroup$
Could you explain why $uwy = 0^k1^{2k}o^k$? If I set $w = 1, v = x = 1$, then neither $x$ nor $v$ equals $epsilon$, $|vwx| < p$ and $uv^iwx^iy in L$, but $uwy$ is not of the form you state. Apart from that, there's a very simple grammar for palindromes: $w := "" | "1" | "0" | "1" w "1" | "0" w "0"$ that looks pretty context free to me
$endgroup$
– Ronald
Dec 6 '18 at 0:33
$begingroup$
@Ronald could you elaborate? Are you saying that the language of palindromes is context free?
$endgroup$
– SeesSound
Dec 6 '18 at 1:02