Timeline for Context free grammar $\{0^n 1^m : n,m \geq 0\}$
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 4, 2023 at 13:46 | comment | added | SilvioM | I fixed your notation, however I suggest you to read this so that you can learn how to write mathematical notation on this website: math.stackexchange.com/help/notation | |
| S Dec 4, 2023 at 13:44 | history | suggested | SilvioM | CC BY-SA 4.0 |
Inserted math mode
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| Dec 4, 2023 at 13:44 | vote | accept | Jellyfish | ||
| Dec 4, 2023 at 13:43 | review | Suggested edits | |||
| S Dec 4, 2023 at 13:44 | |||||
| Dec 4, 2023 at 13:42 | answer | added | SilvioM | timeline score: 0 | |
| Dec 4, 2023 at 13:36 | comment | added | Jellyfish | @gnasher729 it's weird that in the youtube video from "Easy Theory" he uses a different approach which is overcomplicating the problem. He solves it with: S -> 0S | P P -> 1P | \epsilon. But thanks! | |
| Dec 4, 2023 at 13:34 | comment | added | gnasher729 | Looks like you asked ChatGPT. That step is absolute nonsense. There is no way to get to 0101. Yes, your answer was correct. | |
| Dec 4, 2023 at 13:33 | comment | added | Jellyfish | The longer I think about it, the more the derivation to 0101 seems incorrect. Could my answer be correct? | |
| Dec 4, 2023 at 13:26 | comment | added | Jellyfish | @gnasher729 That's the problem, I don't exactly get it but this is what I found on the internet. S -> 0S -> 0S1 -> 01S1 -> 010S1 -> 0101. What's bugging me is the step from 0S1 to 01S1, since there is no step from S to 1S | |
| Dec 4, 2023 at 13:23 | comment | added | gnasher729 | How exactly do you produce 0101? Show me. | |
| S Dec 4, 2023 at 13:09 | review | First questions | |||
| Dec 4, 2023 at 13:46 | |||||
| S Dec 4, 2023 at 13:09 | history | asked | Jellyfish | CC BY-SA 4.0 |