Timeline for An averaging game on finite multisets of integers
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 4, 2020 at 22:38 | answer | added | Christopher Ryba | timeline score: 4 | |
| Oct 3, 2020 at 9:12 | comment | added | YCor | Challenge: from $[1,2,3,4,5,6,7,8,9]$ achieve stopping time 22 without computer aid. | |
| Oct 3, 2020 at 9:10 | comment | added | YCor | For $n=9$ by random walk on the graph I obtain the following stopping times (I don't claim these are the only possible ones). I checked once by choosing random $(a,b)$ among possibles, and once by choosing random $(a,b)$ among possible minimizing $b-a$. Did 30000 tests in each case. (Choosing random $(a,b)$ among possible maximizing $b-a$, in all 20000 tests, yields stopping time 4.) The obtained times: $[4,6,\dots,16,18,19,21,22,24,27,30]$. (With the minimizing $b-a$ option, I obtained stopping time 6 with probability about $1/12$, and the most likely is stopping time $24$: almost $1/2$.) | |
| Oct 3, 2020 at 7:37 | comment | added | YCor | For $n\le 8$ here's the set of possible stopping times starting from $\{1,\dots,n\}$. For $n=8$ it starts taking some time. $n\le 2$: $[0]$; $n=3$: $[1]$; $n=4$: $[2]$; $n=5$: $[2,5]$; $n=6$: $[2,5,8]$; $n=7$: $[3,5,6,8,11,14]$; $n=8$: $[4,6,8,9,11,12,14,17,20]$. | |
| Oct 3, 2020 at 1:25 | history | edited | Richard Stanley | CC BY-SA 4.0 |
added condition $a\neq b$
|
| Oct 3, 2020 at 1:25 | comment | added | Richard Stanley | @YCor: yes, I want $a\neq b$. This has been corrected. | |
| Oct 3, 2020 at 1:07 | comment | added | Sam Hopkins | If $n=2m+1$ is odd you can clearly terminate starting from $\{1,2,\ldots,n\}$ in $m$ steps by choosing the $a,b$ with $a+b=n+1$. | |
| Oct 3, 2020 at 0:56 | comment | added | WhatsUp | Have you gone through the experiment and pattern spotting procedures? | |
| Oct 3, 2020 at 0:14 | comment | added | YCor | You probably want to force $a\neq b$? or else one in most case, at some point, choose a pair of equal numbers to increase artificially the number of moves. So termination is when all numbers in $M$ of the same parity are equal. | |
| Oct 3, 2020 at 0:10 | history | edited | YCor |
edited tags
|
|
| Oct 2, 2020 at 23:55 | history | asked | Richard Stanley | CC BY-SA 4.0 |