Timeline for Context Free Grammar for {a^ib^j | i,j ≥ 0; i ≠ 2j}
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 15, 2013 at 20:15 | comment | added | vonbrand | @user6885, better ask a new question with your new case, it is different enough from this one. | |
| Feb 15, 2013 at 20:13 | comment | added | user6885 | @Vonbrand the problem that i dont know i to do the extra bso that it will not be more then the a. | |
| Feb 15, 2013 at 18:11 | comment | added | Paresh | Aah ... I don't know how I missed that. Sorry! | |
| Feb 15, 2013 at 18:06 | comment | added | vonbrand | @Paresh, the part of extra $b$s is left to complete. And all $a$s is $S \Rightarrow A \Rightarrow^* a^k$. | |
| Feb 15, 2013 at 17:45 | comment | added | user6885 | If I change the question to i,j≥1 and i≠j and i<2j so how can i think of solution? | |
| Feb 15, 2013 at 13:49 | comment | added | Paresh | This will not generate strings of all $a$'s or all $b$'s, which are a part of $L$. | |
| Feb 15, 2013 at 13:37 | history | answered | vonbrand | CC BY-SA 3.0 |