Timeline for There are 2,001 doors leading to 2,001 hallways. How quickly can you find the way out?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| May 30, 2020 at 12:52 | comment | added | Florian F | @TannerSwett And what if you lost two nickel in the first hallway? | |
| May 29, 2020 at 22:24 | comment | added | aschepler | @BlackThorn After finding 1909 links to 1910, the next door to try must be 1925 or 1927, not 1915. | |
| May 29, 2020 at 17:04 | comment | added | BlackThorn | @FlorianF unless I'm mistaken, there are still cases where it takes 11 attempts to get out. E.g. if the escape is door 1939. The path is as follows, where you enter the first door of a pair: (1001,1002), (1502,1501), (1752,1751), (1877,1878), (1940,1941), (1909,1910), (1915,1916), (1928,1927), (1934,1933), (1937,1938), (1939). Please correct me if I did something wrong. | |
| S May 28, 2020 at 8:27 | history | suggested | Cireo | CC BY-SA 4.0 |
"the odd door" to "an odd door"
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| May 28, 2020 at 6:20 | review | Suggested edits | |||
| S May 28, 2020 at 8:27 | |||||
| May 28, 2020 at 3:23 | comment | added | Bit Chaser | rot13(Vs gur Rnfg-Jrfg unyyjnl vf irel ybat, vg znl or snfgre gb gel rneyvre qbbef.) Other than that possibility, you have the optimal algorithm. Fastest worst case, fastest average case if probabilities are equal. | |
| May 27, 2020 at 14:50 | comment | added | Tanner Swett | You should return via the other path, because what if you find a nickel on the floor? | |
| May 27, 2020 at 13:49 | comment | added | musefan | @athin: Doesn't matter?? My brain disagrees. I would probably lean towards the other path as I would feel bad the one path got visited twice and the other didn't get any visits. I wouldn't want the odd path to make fun of the even path for never getting used. Perhaps I am overthinking it... | |
| May 27, 2020 at 13:24 | comment | added | athin | @musefan doesn't matter as long as you remembered which hallway you are on right now. | |
| May 27, 2020 at 12:33 | comment | added | musefan | Yes, yes... now that's the easy part out of the way. The real conundrum is: when you hit a dead end, do you walk back down the path you just came from, or return via the other path? | |
| May 27, 2020 at 7:24 | comment | added | msh210 | @FlorianF you need to add the time spent walking east and west. | |
| May 27, 2020 at 7:22 | comment | added | Florian F | Actually, each failed attempt removes 2 doors and divides the rest in 2, rounded to the nearest odd number. The number of possible doors goes like this: 2001 -> 999 -> 499 -> 249 -> 123 -> 61 -> 29 -> 13 -> 5 -> 1. So, in the worst case, you have 9 failed attemps. The 10th attempt is guaranteed to lead you to freedom. You are free in 9.5 hours or less. PS: plus maybe half an hour to walk between the doors between attemps. | |
| May 27, 2020 at 2:30 | history | edited | athin | CC BY-SA 4.0 |
add visual help
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| May 27, 2020 at 2:28 | vote | accept | Tanner Swett | ||
| May 27, 2020 at 2:24 | history | answered | athin | CC BY-SA 4.0 |