Timeline for Is this Tic-Tac-Toe board valid?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Jun 17, 2020 at 9:04 | history | edited | CommunityBot |
Commonmark migration
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| Dec 13, 2016 at 14:59 | history | edited | zeppelin | CC BY-SA 3.0 |
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| Dec 13, 2016 at 9:32 | history | edited | zeppelin | CC BY-SA 3.0 |
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| Dec 13, 2016 at 9:27 | history | edited | zeppelin | CC BY-SA 3.0 |
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| Dec 13, 2016 at 8:55 | comment | added | Ismael Miguel | Oh, you are right. My testcase was the fail. Sorry! | |
| Dec 13, 2016 at 8:52 | comment | added | zeppelin | @Ismael Miguel - yep, it is invalid, as O and X both win | |
| Dec 12, 2016 at 23:42 | comment | added | Ismael Miguel |
Is OOO XXX OXO a fail?
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| Dec 12, 2016 at 23:06 | history | edited | zeppelin | CC BY-SA 3.0 |
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| Dec 12, 2016 at 23:03 | comment | added | zeppelin | @Neil, this will probably work, I'll give it a try tomorrow. Thx ! | |
| Dec 12, 2016 at 23:01 | history | edited | zeppelin | CC BY-SA 3.0 |
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| Dec 12, 2016 at 22:55 | history | edited | zeppelin | CC BY-SA 3.0 |
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| Dec 12, 2016 at 22:26 | history | edited | zeppelin | CC BY-SA 3.0 |
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| Dec 12, 2016 at 22:17 | history | edited | zeppelin | CC BY-SA 3.0 |
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| Dec 12, 2016 at 22:17 | comment | added | Neil |
(I started commenting before the latest edit) Can you a) write a+b+c+d+e+f+g+H+i instead of F.reduce((r,c)=>r+=c*1) (at which point you don't need F) b) write .includes(C) (and go on to inline C's value)?
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| Dec 12, 2016 at 22:12 | comment | added | zeppelin | Yep, I've lost the negation while golfing it. It is supposed to check that an opposite side is not the winner. Should be fixed now. Thx ! | |
| Dec 12, 2016 at 22:10 | history | edited | zeppelin | CC BY-SA 3.0 |
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| Dec 12, 2016 at 21:54 | comment | added | ETHproductions |
I may be wrong, but it looks like this only checks whether there is a winner. A valid board can have no winner; for example, [1,0,1,1,0,1,0,1,0] (XOX XOX OXO).
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| Dec 12, 2016 at 21:51 | history | answered | zeppelin | CC BY-SA 3.0 |