How can I prove given language is not regular?
$begingroup$
My first post here, so glad I found this great place. Hoping I could improve and learn a lot from you and contribute in the future if I can.
I have a problem with the following scenario:
Given $Sigma= {a,b,c,d}$, prove that $L$ is not regular, where $$L = {a^ib^jcd^k mid i geq 0; k > j > 0}$$
using: 1) pumping lemma 2) closure properties.
Regarding 1)
Assuming that $L$ is regular, using the pumping lemma there should exist some integer $n$ (pumping length). Choosing a word $w in L$ (not sure if I chose the right word). If $w=a^ib^jcd^k$ ($|w|>n$), but I cannot find a way to obtain $xy^i z notin L$, would very appreciate an explanation or example of how you found the term so it can be concluded that $L$ is not regular using the pumping lemma.
Regarding 2)
Assuming that $L$ is regular, it means that $a^ib^jcd^k$ is also regular. So it should be closed under intersection, however, the form of ${a^ib^jcd^k mid igeq 0;k>j>0}$ is not regular, and it can be achieved using completion, so contradiction and because of that $L$ is not regular.
Hoping to learn from my mistakes and improve.
Thank you very much for your aid.
computer-science automata regular-language
edited Dec 4 '18 at 6:20
J.-E. Pin
18.5k21754
asked Dec 2 '18 at 11:27
mathnoobiemathnoobie
93
$endgroup$
add a comment |
$begingroup$
My first post here, so glad I found this great place. Hoping I could improve and learn a lot from you and contribute in the future if I can.
I have a problem with the following scenario:
Given $Sigma= {a,b,c,d}$, prove that $L$ is not regular, where $$L = {a^ib^jcd^k mid i geq 0; k > j > 0}$$
using: 1) pumping lemma 2) closure properties.
Regarding 1)
Assuming that $L$ is regular, using the pumping lemma there should exist some integer $n$ (pumping length). Choosing a word $w in L$ (not sure if I chose the right word). If $w=a^ib^jcd^k$ ($|w|>n$), but I cannot find a way to obtain $xy^i z notin L$, would very appreciate an explanation or example of how you found the term so it can be concluded that $L$ is not regular using the pumping lemma.
Regarding 2)
Assuming that $L$ is regular, it means that $a^ib^jcd^k$ is also regular. So it should be closed under intersection, however, the form of ${a^ib^jcd^k mid igeq 0;k>j>0}$ is not regular, and it can be achieved using completion, so contradiction and because of that $L$ is not regular.
Hoping to learn from my mistakes and improve.
Thank you very much for your aid.
computer-science automata regular-language
edited Dec 4 '18 at 6:20
J.-E. Pin
18.5k21754
asked Dec 2 '18 at 11:27
mathnoobiemathnoobie
93
$endgroup$
add a comment |
$begingroup$
My first post here, so glad I found this great place. Hoping I could improve and learn a lot from you and contribute in the future if I can.
I have a problem with the following scenario:
Given $Sigma= {a,b,c,d}$, prove that $L$ is not regular, where $$L = {a^ib^jcd^k mid i geq 0; k > j > 0}$$
using: 1) pumping lemma 2) closure properties.
Regarding 1)
Assuming that $L$ is regular, using the pumping lemma there should exist some integer $n$ (pumping length). Choosing a word $w in L$ (not sure if I chose the right word). If $w=a^ib^jcd^k$ ($|w|>n$), but I cannot find a way to obtain $xy^i z notin L$, would very appreciate an explanation or example of how you found the term so it can be concluded that $L$ is not regular using the pumping lemma.
Regarding 2)
Assuming that $L$ is regular, it means that $a^ib^jcd^k$ is also regular. So it should be closed under intersection, however, the form of ${a^ib^jcd^k mid igeq 0;k>j>0}$ is not regular, and it can be achieved using completion, so contradiction and because of that $L$ is not regular.
Hoping to learn from my mistakes and improve.
Thank you very much for your aid.
computer-science automata regular-language
edited Dec 4 '18 at 6:20
J.-E. Pin
18.5k21754
asked Dec 2 '18 at 11:27
mathnoobiemathnoobie
93
$endgroup$
My first post here, so glad I found this great place. Hoping I could improve and learn a lot from you and contribute in the future if I can.
I have a problem with the following scenario:
Given $Sigma= {a,b,c,d}$, prove that $L$ is not regular, where $$L = {a^ib^jcd^k mid i geq 0; k > j > 0}$$
using: 1) pumping lemma 2) closure properties.
Regarding 1)
Assuming that $L$ is regular, using the pumping lemma there should exist some integer $n$ (pumping length). Choosing a word $w in L$ (not sure if I chose the right word). If $w=a^ib^jcd^k$ ($|w|>n$), but I cannot find a way to obtain $xy^i z notin L$, would very appreciate an explanation or example of how you found the term so it can be concluded that $L$ is not regular using the pumping lemma.
Regarding 2)
Assuming that $L$ is regular, it means that $a^ib^jcd^k$ is also regular. So it should be closed under intersection, however, the form of ${a^ib^jcd^k mid igeq 0;k>j>0}$ is not regular, and it can be achieved using completion, so contradiction and because of that $L$ is not regular.
Hoping to learn from my mistakes and improve.
Thank you very much for your aid.
computer-science automata regular-language
computer-science automata regular-language
edited Dec 4 '18 at 6:20
J.-E. Pin
18.5k21754
asked Dec 2 '18 at 11:27
mathnoobiemathnoobie
93
edited Dec 4 '18 at 6:20
J.-E. Pin
18.5k21754
asked Dec 2 '18 at 11:27
mathnoobiemathnoobie
93
edited Dec 4 '18 at 6:20
J.-E. Pin
18.5k21754
edited Dec 4 '18 at 6:20
J.-E. Pin
18.5k21754
edited Dec 4 '18 at 6:20
J.-E. Pin
18.5k21754
18.5k21754
asked Dec 2 '18 at 11:27
mathnoobiemathnoobie
93
asked Dec 2 '18 at 11:27
mathnoobiemathnoobie
93
asked Dec 2 '18 at 11:27
mathnoobiemathnoobie
93
93
add a comment |
add a comment |
1 Answer
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$begingroup$
For (1): You assume the language satisfies the pumping lemma for some $n$.
You want to choose a word that would have at least one of its letters related to $n$.
A good start would be taking one of your letters $lin w$ to be $l^n$, and using that to show that for some $i$, $xy^iznotin L$
For (2): You want to show that if $L$ is regular, then some other language $L_{not}$ which you know is not regular, is the intersection of another regular language $L_{reg}cap L=L_{not}$.
answered Dec 3 '18 at 4:34
NL1992NL1992
7311
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add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
For (1): You assume the language satisfies the pumping lemma for some $n$.
You want to choose a word that would have at least one of its letters related to $n$.
A good start would be taking one of your letters $lin w$ to be $l^n$, and using that to show that for some $i$, $xy^iznotin L$
For (2): You want to show that if $L$ is regular, then some other language $L_{not}$ which you know is not regular, is the intersection of another regular language $L_{reg}cap L=L_{not}$.
answered Dec 3 '18 at 4:34
NL1992NL1992
7311
$endgroup$
add a comment |
$begingroup$
For (1): You assume the language satisfies the pumping lemma for some $n$.
You want to choose a word that would have at least one of its letters related to $n$.
A good start would be taking one of your letters $lin w$ to be $l^n$, and using that to show that for some $i$, $xy^iznotin L$
For (2): You want to show that if $L$ is regular, then some other language $L_{not}$ which you know is not regular, is the intersection of another regular language $L_{reg}cap L=L_{not}$.
answered Dec 3 '18 at 4:34
NL1992NL1992
7311
$endgroup$
add a comment |
$begingroup$
For (1): You assume the language satisfies the pumping lemma for some $n$.
You want to choose a word that would have at least one of its letters related to $n$.
A good start would be taking one of your letters $lin w$ to be $l^n$, and using that to show that for some $i$, $xy^iznotin L$
For (2): You want to show that if $L$ is regular, then some other language $L_{not}$ which you know is not regular, is the intersection of another regular language $L_{reg}cap L=L_{not}$.
answered Dec 3 '18 at 4:34
NL1992NL1992
7311
$endgroup$
For (1): You assume the language satisfies the pumping lemma for some $n$.
You want to choose a word that would have at least one of its letters related to $n$.
A good start would be taking one of your letters $lin w$ to be $l^n$, and using that to show that for some $i$, $xy^iznotin L$
For (2): You want to show that if $L$ is regular, then some other language $L_{not}$ which you know is not regular, is the intersection of another regular language $L_{reg}cap L=L_{not}$.
answered Dec 3 '18 at 4:34
NL1992NL1992
7311
answered Dec 3 '18 at 4:34
NL1992NL1992
7311
answered Dec 3 '18 at 4:34
NL1992NL1992
7311
answered Dec 3 '18 at 4:34
NL1992NL1992
7311
7311
add a comment |
add a comment |
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