Is $L = {a^{i+j}b^{j+k}c^{i+k} | i,j,k > 0}$ context free?
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This is not an assignment question. Professor gave this for us to think about. I want to say its not context free since in the case $i=j=k$ then we have the language $a^nb^nc^n$ which we know is not a CFL.
formal-languages context-free-grammar
edited Dec 7 '18 at 21:03
Shaun
8,951113682
asked Dec 7 '18 at 19:42
sharprabbitzsharprabbitz
31
$endgroup$
add a comment |
$begingroup$
This is not an assignment question. Professor gave this for us to think about. I want to say its not context free since in the case $i=j=k$ then we have the language $a^nb^nc^n$ which we know is not a CFL.
formal-languages context-free-grammar
edited Dec 7 '18 at 21:03
Shaun
8,951113682
asked Dec 7 '18 at 19:42
sharprabbitzsharprabbitz
31
$endgroup$
$begingroup$
That doesn't make sense. you just showed that the given language contains another one which is not context free. where is the contradiction?
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– Federico
Dec 7 '18 at 19:51
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Your proof does not work
$endgroup$
– Federico
Dec 7 '18 at 20:40
$begingroup$
Noted. What if my proof is forgotten, is L context free? Im having a hard time thinking of a grammar (I know its not a valid reason).
$endgroup$
– sharprabbitz
Dec 7 '18 at 20:44
add a comment |
$begingroup$
This is not an assignment question. Professor gave this for us to think about. I want to say its not context free since in the case $i=j=k$ then we have the language $a^nb^nc^n$ which we know is not a CFL.
formal-languages context-free-grammar
edited Dec 7 '18 at 21:03
Shaun
8,951113682
asked Dec 7 '18 at 19:42
sharprabbitzsharprabbitz
31
$endgroup$
This is not an assignment question. Professor gave this for us to think about. I want to say its not context free since in the case $i=j=k$ then we have the language $a^nb^nc^n$ which we know is not a CFL.
formal-languages context-free-grammar
formal-languages context-free-grammar
edited Dec 7 '18 at 21:03
Shaun
8,951113682
asked Dec 7 '18 at 19:42
sharprabbitzsharprabbitz
31
edited Dec 7 '18 at 21:03
Shaun
8,951113682
asked Dec 7 '18 at 19:42
sharprabbitzsharprabbitz
31
edited Dec 7 '18 at 21:03
Shaun
8,951113682
edited Dec 7 '18 at 21:03
Shaun
8,951113682
edited Dec 7 '18 at 21:03
Shaun
8,951113682
8,951113682
asked Dec 7 '18 at 19:42
sharprabbitzsharprabbitz
31
asked Dec 7 '18 at 19:42
sharprabbitzsharprabbitz
31
asked Dec 7 '18 at 19:42
sharprabbitzsharprabbitz
31
31
$begingroup$
That doesn't make sense. you just showed that the given language contains another one which is not context free. where is the contradiction?
$endgroup$
– Federico
Dec 7 '18 at 19:51
$begingroup$
Your proof does not work
$endgroup$
– Federico
Dec 7 '18 at 20:40
$begingroup$
Noted. What if my proof is forgotten, is L context free? Im having a hard time thinking of a grammar (I know its not a valid reason).
$endgroup$
– sharprabbitz
Dec 7 '18 at 20:44
add a comment |
$begingroup$
That doesn't make sense. you just showed that the given language contains another one which is not context free. where is the contradiction?
$endgroup$
– Federico
Dec 7 '18 at 19:51
$begingroup$
Your proof does not work
$endgroup$
– Federico
Dec 7 '18 at 20:40
$begingroup$
Noted. What if my proof is forgotten, is L context free? Im having a hard time thinking of a grammar (I know its not a valid reason).
$endgroup$
– sharprabbitz
Dec 7 '18 at 20:44
$begingroup$
That doesn't make sense. you just showed that the given language contains another one which is not context free. where is the contradiction?
$endgroup$
– Federico
Dec 7 '18 at 19:51
$begingroup$
That doesn't make sense. you just showed that the given language contains another one which is not context free. where is the contradiction?
$endgroup$
– Federico
Dec 7 '18 at 19:51
$begingroup$
Your proof does not work
$endgroup$
– Federico
Dec 7 '18 at 20:40
$begingroup$
Your proof does not work
$endgroup$
– Federico
Dec 7 '18 at 20:40
$begingroup$
Noted. What if my proof is forgotten, is L context free? Im having a hard time thinking of a grammar (I know its not a valid reason).
$endgroup$
– sharprabbitz
Dec 7 '18 at 20:44
$begingroup$
Noted. What if my proof is forgotten, is L context free? Im having a hard time thinking of a grammar (I know its not a valid reason).
$endgroup$
– sharprabbitz
Dec 7 '18 at 20:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Look at it this way:
$$L = {a^ia^jb^jb^kc^kc^i mid i,j,k > 0}$$
Which is exactly the same (replacing the sum with concatenation is clearly not a change, and addition is commutative.)
answered Dec 7 '18 at 23:43
ricirici
37829
$endgroup$
$begingroup$
Ahh that makes it clear. Thanks!
$endgroup$
– sharprabbitz
Dec 8 '18 at 6:32
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Look at it this way:
$$L = {a^ia^jb^jb^kc^kc^i mid i,j,k > 0}$$
Which is exactly the same (replacing the sum with concatenation is clearly not a change, and addition is commutative.)
answered Dec 7 '18 at 23:43
ricirici
37829
$endgroup$
$begingroup$
Ahh that makes it clear. Thanks!
$endgroup$
– sharprabbitz
Dec 8 '18 at 6:32
add a comment |
$begingroup$
Look at it this way:
$$L = {a^ia^jb^jb^kc^kc^i mid i,j,k > 0}$$
Which is exactly the same (replacing the sum with concatenation is clearly not a change, and addition is commutative.)
answered Dec 7 '18 at 23:43
ricirici
37829
$endgroup$
$begingroup$
Ahh that makes it clear. Thanks!
$endgroup$
– sharprabbitz
Dec 8 '18 at 6:32
add a comment |
$begingroup$
Look at it this way:
$$L = {a^ia^jb^jb^kc^kc^i mid i,j,k > 0}$$
Which is exactly the same (replacing the sum with concatenation is clearly not a change, and addition is commutative.)
answered Dec 7 '18 at 23:43
ricirici
37829
$endgroup$
Look at it this way:
$$L = {a^ia^jb^jb^kc^kc^i mid i,j,k > 0}$$
Which is exactly the same (replacing the sum with concatenation is clearly not a change, and addition is commutative.)
answered Dec 7 '18 at 23:43
ricirici
37829
answered Dec 7 '18 at 23:43
ricirici
37829
answered Dec 7 '18 at 23:43
ricirici
37829
answered Dec 7 '18 at 23:43
ricirici
37829
37829
$begingroup$
Ahh that makes it clear. Thanks!
$endgroup$
– sharprabbitz
Dec 8 '18 at 6:32
add a comment |
$begingroup$
Ahh that makes it clear. Thanks!
$endgroup$
– sharprabbitz
Dec 8 '18 at 6:32
$begingroup$
Ahh that makes it clear. Thanks!
$endgroup$
– sharprabbitz
Dec 8 '18 at 6:32
$begingroup$
Ahh that makes it clear. Thanks!
$endgroup$
– sharprabbitz
Dec 8 '18 at 6:32
add a comment |
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$begingroup$
That doesn't make sense. you just showed that the given language contains another one which is not context free. where is the contradiction?
$endgroup$
– Federico
Dec 7 '18 at 19:51
$begingroup$
Your proof does not work
$endgroup$
– Federico
Dec 7 '18 at 20:40
$begingroup$
Noted. What if my proof is forgotten, is L context free? Im having a hard time thinking of a grammar (I know its not a valid reason).
$endgroup$
– sharprabbitz
Dec 7 '18 at 20:44