Timeline for Question for averaging the overall quantities by averaging a part
Current License: CC BY-SA 4.0
40 events
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| Jan 21, 2022 at 23:03 | answer | added | JetfiRex | timeline score: 0 | |
| Jan 21, 2022 at 22:43 | comment | added | JetfiRex | @PeterTaylor Thank you so much! I am going to write some of the proofs in my own answer, because it is too long to put everything in the comments/problems | |
| Jan 21, 2022 at 22:22 | comment | added | Peter Taylor | Graph search. I've played around with various functions to prioritise nodes to expand, but I'm coming to suspect that most of the small cases which don't resolve fairly quickly with breadth-first have a $p$-adic obstruction. (E.g. I suspect that $64 \stackrel?\succeq 26$ has a $13$-adic obstruction). Code and positive results | |
| Jan 21, 2022 at 19:17 | comment | added | JetfiRex | @PeterTaylor Also... can I have a look at how you find the method of averaging $54$ by $48$? Or how you searched for this? What algorithm have you used? Because currently all the positive results I found are with the same pattern... ( I think it is nearly impossible for me to figure out a solution by that...) Also, will it be efficient if a and b are different much, for example 50 and 64 case? | |
| Jan 21, 2022 at 18:37 | comment | added | JetfiRex | @PeterTaylor In my last thing in the problem, I have proved that, let $k=\lfloor a/b\rfloor$, if $(a\mod k)+k\times (b\mod k)\le b$, then $a,b$ can be done with $b$ equals and $a-b$ other equals. | |
| Jan 21, 2022 at 18:34 | history | edited | JetfiRex | CC BY-SA 4.0 |
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| Jan 21, 2022 at 9:34 | comment | added | Peter Taylor | $33$ can collapse $81$ into two numbers: average the first 33, the next 33, and the last 33. Then we have $x^{33}, y^{33}, z^{15}$. Perform two averages of $x^{16}, y^{16}, z$ to get $x, y, z^{13}, w^{66}$. One more average gives $w^{48} v^{33}$. Need to think how far this idea can be pushed... | |
| Jan 21, 2022 at 8:34 | answer | added | Peter Taylor | timeline score: 2 | |
| Jan 20, 2022 at 11:34 | comment | added | Peter Taylor | I found a bug in my search code, which explains my earlier incorrect negative claims. $(54,48)$ reduces to the case $\{0^{48}, 1^6\}$ and then $$\{0^{48}, 1^6\} \to \{0^4, 1^2, \tfrac1{12}^{48}\} \to \{0, 1, \tfrac1{12}^4, \tfrac7{72}^{48}\} \to \{0, \tfrac1{12}^3, \tfrac7{72}^2, \tfrac{25}{216}^{48}\} \to \{\tfrac1{12}^2, \tfrac7{72}^2, \tfrac{25}{216}^2, \tfrac{73}{648}^{48}\} \to \{\tfrac7{72}, \tfrac{25}{216}^2, \tfrac{73}{648}^3, \tfrac19^{48}\} \to \{\tfrac19^{54}\}$$ | |
| Jan 19, 2022 at 21:53 | history | edited | JetfiRex | CC BY-SA 4.0 |
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| Jan 19, 2022 at 21:08 | history | edited | JetfiRex | CC BY-SA 4.0 |
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| Jan 19, 2022 at 20:48 | history | edited | JetfiRex | CC BY-SA 4.0 |
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| Jan 19, 2022 at 18:50 | comment | added | JetfiRex | @PeterTaylor I just checked that $54$ actually can be averaged by $42$. | |
| Jan 19, 2022 at 18:21 | history | edited | Jon Bannon | CC BY-SA 4.0 |
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| Jan 19, 2022 at 18:13 | history | edited | JetfiRex | CC BY-SA 4.0 |
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| Jan 19, 2022 at 18:06 | history | edited | JetfiRex | CC BY-SA 4.0 |
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| Jan 19, 2022 at 18:05 | comment | added | JetfiRex | @PeterTaylor I have proved $8k$ can be averaged by $6k$ for all $k\ge 2$. So actually $56$ can be averaged by $42$, and $88$ can be averaged by $66$. Currently I can't say anything about $54$ and $42$, sorry (although I believe it can, but I don't have a proof. I very appreciate you if you can write a partial answer below or even for a general argument). Still, can I be informed how to average $54$ by $48$? Thanks! | |
| Jan 19, 2022 at 18:01 | comment | added | JetfiRex | @PeterTaylor Oh... The example $(63,21)$ means that if we have $(p^k,pr)$ where $\frac{p^{k-2}}{2}<r<p^{k-2}$, we can make $pr$ to $p^2r$ so that we can make into a configuration of two numbers. Similar example is $(729,237)$ we can make it into a configuration of two numbers (not sure whether it can work but making it into a configuration of two numbers is okay.) However, I don't really know whether $33$ can average $81$, and I don't even know whether we can make it into a configuration of two numbers ($33$ equal numbers + $48$ other equal numbers). Since it is small, but not that small. | |
| Jan 19, 2022 at 15:02 | comment | added | Peter Taylor | @JetfiRex, in that case you can arrive at a case which only contains two numbers. By your property 2, $63$ can be averaged by $21$. So average the first $63$ elements and then the last $21$ elements. | |
| Jan 17, 2022 at 3:05 | comment | added | JetfiRex | @IlyaBogdanov Not only if $b$ too close will cause problems, sometimes if $b$ is way smaller than $a$, for example, $(81,33)$ there might also be problems... If $b\le a\le 2b$ it may be better because we can easily make into the case with only two numbers, but if it small (but not that small, for example, in $(81,21)$ case we can multiple three two $21$) we can't arrive at a case that only contains two numbers. | |
| Jan 16, 2022 at 6:10 | history | edited | JetfiRex | CC BY-SA 4.0 |
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| Jan 16, 2022 at 5:57 | history | edited | JetfiRex | CC BY-SA 4.0 |
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| Jan 16, 2022 at 5:51 | history | edited | JetfiRex | CC BY-SA 4.0 |
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| Jan 16, 2022 at 5:39 | history | edited | JetfiRex | CC BY-SA 4.0 |
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| Jan 16, 2022 at 5:27 | comment | added | JetfiRex | @Vik78 I did that. | |
| Jan 16, 2022 at 5:27 | history | edited | JetfiRex | CC BY-SA 4.0 |
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| Jan 16, 2022 at 5:06 | comment | added | Vik78 | I would appreciate if you posed the problem in terms of typical mathematical notation, rather than talking about cups of water. | |
| Jan 16, 2022 at 4:45 | history | edited | JetfiRex | CC BY-SA 4.0 |
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| Jan 16, 2022 at 4:36 | history | edited | JetfiRex | CC BY-SA 4.0 |
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| Jan 14, 2022 at 17:41 | history | edited | JetfiRex | CC BY-SA 4.0 |
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| Jan 14, 2022 at 17:31 | history | edited | JetfiRex | CC BY-SA 4.0 |
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| Jan 13, 2022 at 9:21 | answer | added | Ilya Bogdanov | timeline score: 3 | |
| Jan 12, 2022 at 16:12 | comment | added | JetfiRex | @IlyaBogdanov Yes, you are right. In some sense, the "space to adjust" is limited if $a$ is not much larger than $b$. For another example, if $a=54$ and $b=48$, it is unknown (for me) whether this will work, but I guess it won't work... | |
| Jan 12, 2022 at 10:30 | comment | added | Ilya Bogdanov | Notice that, although $8\not\geq 6$, but $16\geq 12\geq 6$. So the trouble should arise when $a$ is not much larger than $b$ --- in some sense .. | |
| Jan 12, 2022 at 4:31 | history | edited | JetfiRex | CC BY-SA 4.0 |
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| Jan 12, 2022 at 4:28 | comment | added | JetfiRex | @user44191 It means if you factorize $x$ ad $y$ into primes (say $x=\pm 2^{p_2}3^{p_3}...$ and $y=\pm 2^{q_2}3^{q_3}...$) Then either $x=y$ or $p_3\ne q_3$. This can be proven by using induction. | |
| Jan 12, 2022 at 2:21 | comment | added | user44191 | Would you mind explaining "$x$ and $y$ has a distinct power exponential of $3$" a bit more? I'm guessing you mean that they will have different $3$-adic orders, but I don't see an immediate proof of that fact. | |
| Jan 11, 2022 at 23:21 | history | edited | JetfiRex | CC BY-SA 4.0 |
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| S Jan 11, 2022 at 23:16 | review | First questions | |||
| Jan 12, 2022 at 2:21 | |||||
| S Jan 11, 2022 at 23:16 | history | asked | JetfiRex | CC BY-SA 4.0 |