Timeline for Make $n$ numbers equal using pairwise averages
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 7, 2022 at 9:56 | comment | added | Brendan McKay | @Ycor Sorry, I agree with you. I did one operation backwards! | |
| May 7, 2022 at 7:49 | comment | added | YCor | @BrendanMcKay no, [5,5,5,5,10] is not solvable because solvability is invariant under affine change and [0,0,0,0,1] is not solvable. | |
| May 7, 2022 at 7:47 | comment | added | YCor | More generally any non-2-power number $n$ has such an "irreversibility" phenomenon: if $n$ is odd start from $[2,4,\dots,2n]$ which obviously solves to $[n+1,\dots,n+1]$, but also gives rise to $[n,\dots,n,2n]$ which is not solvable. In general $n=2^km$ with $m$ odd and one runs the same argument with concatenating $2^k$ copies of $[2,4,\dots,2m]$. | |
| May 7, 2022 at 7:42 | history | edited | YCor | CC BY-SA 4.0 |
made title more precise, added top-level tag
|
| May 7, 2022 at 7:13 | comment | added | Brendan McKay | @jhw Clearly $n=5$ is solvable if one number equals the mean, but the converse is not true. For example 1,1,4,4,5 -> 3,1,4,4,3 and continue as before. So two numbers whose mean is the mean of all 5 is also ok. But that's not the end. I believe 5,5,5,5,10 (mean 6) is solvable. | |
| May 7, 2022 at 6:33 | comment | added | Brendan McKay | @YCor Yes, and from this one can conclude that sets of 3 numbers are solvable iff they are in arithmetic progression. Sets of 4 numbers are always solvable (as noted by jh w. What about 5 numbers? | |
| May 7, 2022 at 6:25 | comment | added | YCor | By a wrong step on can destroy the solvability. For instance $[2,4,6]$ is clearly solvable in one step. But if one averages 2-4 instead one gets $[3,3,6]$ which is not solvable. | |
| May 7, 2022 at 4:13 | answer | added | Brendan McKay | timeline score: 1 | |
| May 7, 2022 at 3:40 | comment | added | Brendan McKay | If the average is $x/2^k$ where $x$ is odd, I can prove that it is possible to reach a state where there are only two values, one of form $y/2^k$ for $y$ odd and one of form $z/2^k$ for $z$ even. | |
| May 6, 2022 at 20:04 | comment | added | Somos | A simple example to consider is $(0,0,1)$. | |
| May 6, 2022 at 12:05 | history | edited | Andrej Bauer |
edited tags
|
|
| May 3, 2022 at 19:04 | comment | added | Saúl RM | You can suppose the rationals are integers, in that case a necessary condition is that the denominator of the average is a power of $2$ | |
| May 3, 2022 at 14:57 | comment | added | jh w | if $n=2^k$, then we can easily construct an answer using divide-and-conquer. Let the average number of the $n$ numbers is $x$. A sufficient condition is if we can divide these $n$ numbers into several groups and for each group, its average number is equal to $x$ and its size is power of 2. | |
| S May 3, 2022 at 14:50 | review | First questions | |||
| May 3, 2022 at 15:03 | |||||
| S May 3, 2022 at 14:50 | history | asked | jh w | CC BY-SA 4.0 |