Timeline for Find the composite poison in the wine bottles
Current License: CC BY-SA 3.0
23 events
| when toggle format | what | by | license | comment | |
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| Apr 28, 2015 at 15:14 | comment | added | Timbo | @leoll2 I would prefer a generic solution. Something about how to double the tested bottles with 2 additional prisoner. 8 bottles with 6 prisoner is nearly the easy solution of 8 bottles with 7 prisoner. But 8 bottles with 6 prisoner is a non trivial first step | |
| Apr 28, 2015 at 14:04 | comment | added | leoll2 | @Timbo Would you accept as well 8 bottles with 6 prisoners? 16 bottles mean 120 cases to analyze, and it results a bit annoying. | |
| Apr 28, 2015 at 13:58 | comment | added | Timbo | Testing 4 bottles with 4 prisoner is not really difficult. I can't see a way to check 16 bottles with 8 prisoner if both poisons are in those 16 bottles. How do you argue that you need 2log(n) prisoner? I would say your way needs n prisoner. | |
| Apr 28, 2015 at 12:59 | comment | added | leoll2 | @JoeZ. Could you please check and evaluate my solution? I think that for some reasons it wasn't enough noticed. | |
| Apr 28, 2015 at 12:57 | comment | added | leoll2 | @user2357112 I've added explanation of the bitwise strategy, hope it's enough | |
| Apr 28, 2015 at 12:56 | comment | added | leoll2 | @goldPseudo user2357112 answered correctly to your question. | |
| Apr 28, 2015 at 12:55 | comment | added | leoll2 | @Falco explanation added, now it should be more clear to understand the strategy | |
| Apr 28, 2015 at 12:55 | history | edited | leoll2 | CC BY-SA 3.0 |
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| Apr 28, 2015 at 9:12 | comment | added | Falco | Please explain how you can find 2 Poisoned Bottles in 250 with 16 Prisoners ? You needed 1 Day and 4 Prisoners to get from 1000 bottles to 250, by induction another day and 4 prisoners will get you to 250/4 = 64... If you have a better approach for the second day, why don't you use it on the first ? | |
| Apr 28, 2015 at 6:00 | comment | added | user2357112 | @goldPseudo: You can; you know what poison each one has, and you can assign them to the groups that would have been guaranteed to take that poison anyway. | |
| Apr 28, 2015 at 1:26 | comment | added | goldPseudo | Can you recycle the first-day survivors in the "Two die" case? They both would've consumed one component of the poison in the first stage, and that component would remain in their systems to affect the second test. | |
| Apr 27, 2015 at 20:10 | comment | added | Ewan | why is it 4 and not 8? | |
| Apr 27, 2015 at 19:57 | comment | added | user2357112 | "Using a bitwise operation" isn't a very detailed explanation. How exactly do you use bitwise operations to locate two poisoned bottles? | |
| Apr 27, 2015 at 19:50 | history | edited | leoll2 | CC BY-SA 3.0 |
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| Apr 27, 2015 at 19:45 | comment | added | leoll2 | Correct, again. This was actually a remaining of the previous answer, which I forgot to fix. Thanks! | |
| Apr 27, 2015 at 19:42 | comment | added | JonTheMon | If all the poisoned bottles are in 1/4, doesn't that mean 3 die since they "drink all the bottles except the one with his name"? | |
| Apr 27, 2015 at 19:11 | comment | added | leoll2 | @JonTheMon I've edited and confirm that the best you can do is 18 prisoners. My strategy is basically the same you described... | |
| Apr 27, 2015 at 19:09 | history | undeleted | leoll2 | ||
| Apr 27, 2015 at 19:09 | history | edited | leoll2 | CC BY-SA 3.0 |
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| Apr 27, 2015 at 18:56 | history | deleted | leoll2 | via Vote | |
| Apr 27, 2015 at 18:54 | comment | added | JonTheMon | How can only 1 or 2 of the 8 die? Wouldn't it be 6 or 7 of them die? | |
| Apr 27, 2015 at 18:50 | history | edited | leoll2 | CC BY-SA 3.0 |
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| Apr 27, 2015 at 18:34 | history | answered | leoll2 | CC BY-SA 3.0 |