Timeline for Sandbox for Proposed Challenges
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 17, 2020 at 9:03 | history | edited | CommunityBot |
Commonmark migration
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| Sep 19, 2019 at 15:46 | history | edited | Night2 | CC BY-SA 4.0 |
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| Sep 19, 2019 at 15:39 | history | edited | Night2 | CC BY-SA 4.0 |
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| Sep 19, 2019 at 15:32 | history | edited | Night2 | CC BY-SA 4.0 |
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| Sep 19, 2019 at 15:25 | history | edited | Night2 | CC BY-SA 4.0 |
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| Sep 13, 2019 at 12:26 | comment | added | Night2 | @KevinCruijssen: You are absolutely correct, I have updated the question. Thank you for reading with such a high precision. | |
| Sep 13, 2019 at 12:22 | history | edited | Night2 | CC BY-SA 4.0 |
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| Sep 13, 2019 at 9:46 | comment | added | Kevin Cruijssen |
I think you meant to give the example "... so fixing "op op o o o open" would result in "op op open"." (so one less op) instead?
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| Sep 13, 2019 at 9:45 | comment | added | Kevin Cruijssen |
"You can assume that multiple valid sets of repeated stuttered words only happen from left to right, so fixing "op op op o o o open" would result in "op op op open"." Shouldn't this result in "op open" instead? In the first rule you mention "For example "ope" and "open" both can be a stuttered word for "open"." So since the entire word is valid as stutter, I would think "op op op o o o open" becomes "([op op] op) ([o o o] open)" (([this] ... ) being the stutter, and ([...] this) being the word), thus "op open".
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| Sep 11, 2019 at 15:00 | history | edited | Night2 | CC BY-SA 4.0 |
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| Sep 11, 2019 at 14:52 | history | answered | Night2 | CC BY-SA 4.0 |