Timeline for 7 mathematicians around the clock in prison
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 22, 2023 at 22:08 | comment | added | theozh | +1, good answer, but a bit challenging to follow. Thanks for the Python code. | |
| Feb 18, 2023 at 21:14 | history | edited | loopy walt | CC BY-SA 4.0 |
added 206 characters in body
|
| Feb 18, 2023 at 3:06 | comment | added | loopy walt | @AndrewSavinykh Isn't it? "Then they must reencode using an order the prisoners have agreed on before because the order of assignment is not available to all of them." That said I agree that this post could be improved in many ways, | |
| Feb 18, 2023 at 2:33 | comment | added | Andrew Savinykh | I think that the last paragraph could be improved by explicitly specifying that we not just relabelling the rooms, but also reshuffling the mathematicians as per pre-agreed order. It took me a couple of minutes to figure this part out, as it is not called out explicitly. | |
| Feb 17, 2023 at 17:19 | history | edited | loopy walt | CC BY-SA 4.0 |
added 3114 characters in body
|
| Feb 17, 2023 at 16:02 | comment | added | loopy walt | Thanks @user16074 I suspected as much just wasn't sure about the details. | |
| Feb 17, 2023 at 15:57 | comment | added | loopy walt | @GeorgeMenoutis I'm trying to write a program that implements the procedure. | |
| Feb 17, 2023 at 15:55 | comment | added | loopy walt | @TheAverageHijano You need the order within the already assigned group, also, since that information is otherwise lost. Hence a factor of 3!. Easy to overlook in my messy write-up. | |
| Feb 17, 2023 at 15:20 | comment | added | TheAverageHijano | I am not sure who the 7 x 6-choose-3 / 2 x 3! = 420 calculation is performed. If you need to select 3 items from 6 posible items without specifying the order, there are $\frac{6!}{3!3!}=20$ posible combinations. You also mention that it is not necessary to specify which is the first and the last group, so this reduces the number by a factor of 2. In total there are $7\frac{6!}{3!3!2}=70$ posible states that need to be encoded. | |
| Feb 17, 2023 at 15:02 | comment | added | user16074 | Avpr! Guvf trarenyvmrf gb neovgenel $a$, ol rapbqvat gur beqre bs pryyf rkpyhqvat gur ynetrfg tnc. Lbh pna hfr guvf vqrn va rirel fgrc naq lbh qba'g rira arrq gur 'gevpx' sbe gur sbhegu cevfbare: Whfg yrg gurz rapbqr gur beqrevat bs gur svefg sbhe cevfbare'f pryyf naq gur svefg gjb tncf. | |
| Feb 17, 2023 at 14:37 | comment | added | George Menoutis | Seems interesting, can you add examples? | |
| Feb 17, 2023 at 12:16 | history | answered | loopy walt | CC BY-SA 4.0 |