Pumping Lemma - unregular expression
$begingroup$
How do prove that this expression is unregular, I know firstly you have to try prove that it is regular and work from there. I also know that $w=xuz$ and the three rules are needed
Let $M$ be the language over the alphabet ${a, b, c}$ given
by $M = {a^ib^jc^k mid i, j, k ≥ 0, j = i + k}$.
discrete-mathematics formal-languages context-free-grammar regular-expressions pumping-lemma
edited Jan 23 at 12:27
Wuestenfux
4,7941513
asked Jan 23 at 12:24
SueSue
125
$endgroup$
add a comment |
$begingroup$
How do prove that this expression is unregular, I know firstly you have to try prove that it is regular and work from there. I also know that $w=xuz$ and the three rules are needed
Let $M$ be the language over the alphabet ${a, b, c}$ given
by $M = {a^ib^jc^k mid i, j, k ≥ 0, j = i + k}$.
discrete-mathematics formal-languages context-free-grammar regular-expressions pumping-lemma
edited Jan 23 at 12:27
Wuestenfux
4,7941513
asked Jan 23 at 12:24
SueSue
125
$endgroup$
$begingroup$
What you provide is called a "language", not an "expression". Also "unregular" is a very uncommon term; usually one says "not regular" or "non-regular".
$endgroup$
– Peter Leupold
Jan 24 at 12:15
add a comment |
$begingroup$
How do prove that this expression is unregular, I know firstly you have to try prove that it is regular and work from there. I also know that $w=xuz$ and the three rules are needed
Let $M$ be the language over the alphabet ${a, b, c}$ given
by $M = {a^ib^jc^k mid i, j, k ≥ 0, j = i + k}$.
discrete-mathematics formal-languages context-free-grammar regular-expressions pumping-lemma
edited Jan 23 at 12:27
Wuestenfux
4,7941513
asked Jan 23 at 12:24
SueSue
125
$endgroup$
How do prove that this expression is unregular, I know firstly you have to try prove that it is regular and work from there. I also know that $w=xuz$ and the three rules are needed
Let $M$ be the language over the alphabet ${a, b, c}$ given
by $M = {a^ib^jc^k mid i, j, k ≥ 0, j = i + k}$.
discrete-mathematics formal-languages context-free-grammar regular-expressions pumping-lemma
discrete-mathematics formal-languages context-free-grammar regular-expressions pumping-lemma
edited Jan 23 at 12:27
Wuestenfux
4,7941513
asked Jan 23 at 12:24
SueSue
125
edited Jan 23 at 12:27
Wuestenfux
4,7941513
asked Jan 23 at 12:24
SueSue
125
edited Jan 23 at 12:27
Wuestenfux
4,7941513
edited Jan 23 at 12:27
Wuestenfux
4,7941513
edited Jan 23 at 12:27
Wuestenfux
4,7941513
4,7941513
asked Jan 23 at 12:24
SueSue
125
asked Jan 23 at 12:24
SueSue
125
asked Jan 23 at 12:24
SueSue
125
125
$begingroup$
What you provide is called a "language", not an "expression". Also "unregular" is a very uncommon term; usually one says "not regular" or "non-regular".
$endgroup$
– Peter Leupold
Jan 24 at 12:15
add a comment |
$begingroup$
What you provide is called a "language", not an "expression". Also "unregular" is a very uncommon term; usually one says "not regular" or "non-regular".
$endgroup$
– Peter Leupold
Jan 24 at 12:15
$begingroup$
What you provide is called a "language", not an "expression". Also "unregular" is a very uncommon term; usually one says "not regular" or "non-regular".
$endgroup$
– Peter Leupold
Jan 24 at 12:15
$begingroup$
What you provide is called a "language", not an "expression". Also "unregular" is a very uncommon term; usually one says "not regular" or "non-regular".
$endgroup$
– Peter Leupold
Jan 24 at 12:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
Suppose $M$ is regular. Then by the Pumping Lemma, there is some $p$ so that all words of length at least $p$ can be decomposed as $xyz$ in such a way that $xy^nzin M$ for all $n$.
Consider the word $ab^pc^{p-1}in M$, and say that its Pumping Lemma decomposition is $xyz$. What are $x$, $y$, and $z$?
Prove that if $xy^nzin M$ for all $ninmathbb{N}$, then necessarily a few properties must hold:
$x$ must contain $a$ and cannot contain any $c$
$z$ must contain all $p-1$ copies of $c$
This necessarily means that we have $x=ab^{m_x}$, $y=b^{m_y}$, and $z=b^{m_z}c^{p-1}$ for some $m_x,m_y,m_zgeq 0$ such that $m_x+m_y+m_z=p$ and $m_ygeq 1$. But then
$$
xy^nz=ab^{m_x+nm_y+m_z}c^{p-1}=ab^{p+(n-1)m_y}c^{p-1}.
$$
But, this word clearly cannot be in $M$, as $p+(n-1)m_y>1+(p-1)$ for all $ngeq 2$.
answered Jan 23 at 16:49
Nick PetersonNick Peterson
26.7k23962
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add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Hint:
Suppose $M$ is regular. Then by the Pumping Lemma, there is some $p$ so that all words of length at least $p$ can be decomposed as $xyz$ in such a way that $xy^nzin M$ for all $n$.
Consider the word $ab^pc^{p-1}in M$, and say that its Pumping Lemma decomposition is $xyz$. What are $x$, $y$, and $z$?
Prove that if $xy^nzin M$ for all $ninmathbb{N}$, then necessarily a few properties must hold:
$x$ must contain $a$ and cannot contain any $c$
$z$ must contain all $p-1$ copies of $c$
This necessarily means that we have $x=ab^{m_x}$, $y=b^{m_y}$, and $z=b^{m_z}c^{p-1}$ for some $m_x,m_y,m_zgeq 0$ such that $m_x+m_y+m_z=p$ and $m_ygeq 1$. But then
$$
xy^nz=ab^{m_x+nm_y+m_z}c^{p-1}=ab^{p+(n-1)m_y}c^{p-1}.
$$
But, this word clearly cannot be in $M$, as $p+(n-1)m_y>1+(p-1)$ for all $ngeq 2$.
answered Jan 23 at 16:49
Nick PetersonNick Peterson
26.7k23962
$endgroup$
add a comment |
$begingroup$
Hint:
Suppose $M$ is regular. Then by the Pumping Lemma, there is some $p$ so that all words of length at least $p$ can be decomposed as $xyz$ in such a way that $xy^nzin M$ for all $n$.
Consider the word $ab^pc^{p-1}in M$, and say that its Pumping Lemma decomposition is $xyz$. What are $x$, $y$, and $z$?
Prove that if $xy^nzin M$ for all $ninmathbb{N}$, then necessarily a few properties must hold:
$x$ must contain $a$ and cannot contain any $c$
$z$ must contain all $p-1$ copies of $c$
This necessarily means that we have $x=ab^{m_x}$, $y=b^{m_y}$, and $z=b^{m_z}c^{p-1}$ for some $m_x,m_y,m_zgeq 0$ such that $m_x+m_y+m_z=p$ and $m_ygeq 1$. But then
$$
xy^nz=ab^{m_x+nm_y+m_z}c^{p-1}=ab^{p+(n-1)m_y}c^{p-1}.
$$
But, this word clearly cannot be in $M$, as $p+(n-1)m_y>1+(p-1)$ for all $ngeq 2$.
answered Jan 23 at 16:49
Nick PetersonNick Peterson
26.7k23962
$endgroup$
add a comment |
$begingroup$
Hint:
Suppose $M$ is regular. Then by the Pumping Lemma, there is some $p$ so that all words of length at least $p$ can be decomposed as $xyz$ in such a way that $xy^nzin M$ for all $n$.
Consider the word $ab^pc^{p-1}in M$, and say that its Pumping Lemma decomposition is $xyz$. What are $x$, $y$, and $z$?
Prove that if $xy^nzin M$ for all $ninmathbb{N}$, then necessarily a few properties must hold:
$x$ must contain $a$ and cannot contain any $c$
$z$ must contain all $p-1$ copies of $c$
This necessarily means that we have $x=ab^{m_x}$, $y=b^{m_y}$, and $z=b^{m_z}c^{p-1}$ for some $m_x,m_y,m_zgeq 0$ such that $m_x+m_y+m_z=p$ and $m_ygeq 1$. But then
$$
xy^nz=ab^{m_x+nm_y+m_z}c^{p-1}=ab^{p+(n-1)m_y}c^{p-1}.
$$
But, this word clearly cannot be in $M$, as $p+(n-1)m_y>1+(p-1)$ for all $ngeq 2$.
answered Jan 23 at 16:49
Nick PetersonNick Peterson
26.7k23962
$endgroup$
Hint:
Suppose $M$ is regular. Then by the Pumping Lemma, there is some $p$ so that all words of length at least $p$ can be decomposed as $xyz$ in such a way that $xy^nzin M$ for all $n$.
Consider the word $ab^pc^{p-1}in M$, and say that its Pumping Lemma decomposition is $xyz$. What are $x$, $y$, and $z$?
Prove that if $xy^nzin M$ for all $ninmathbb{N}$, then necessarily a few properties must hold:
$x$ must contain $a$ and cannot contain any $c$
$z$ must contain all $p-1$ copies of $c$
This necessarily means that we have $x=ab^{m_x}$, $y=b^{m_y}$, and $z=b^{m_z}c^{p-1}$ for some $m_x,m_y,m_zgeq 0$ such that $m_x+m_y+m_z=p$ and $m_ygeq 1$. But then
$$
xy^nz=ab^{m_x+nm_y+m_z}c^{p-1}=ab^{p+(n-1)m_y}c^{p-1}.
$$
But, this word clearly cannot be in $M$, as $p+(n-1)m_y>1+(p-1)$ for all $ngeq 2$.
answered Jan 23 at 16:49
Nick PetersonNick Peterson
26.7k23962
answered Jan 23 at 16:49
Nick PetersonNick Peterson
26.7k23962
answered Jan 23 at 16:49
Nick PetersonNick Peterson
26.7k23962
answered Jan 23 at 16:49
Nick PetersonNick Peterson
26.7k23962
26.7k23962
add a comment |
add a comment |
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$begingroup$
What you provide is called a "language", not an "expression". Also "unregular" is a very uncommon term; usually one says "not regular" or "non-regular".
$endgroup$
– Peter Leupold
Jan 24 at 12:15