I recently asked a question On MSE, which was answered by one user, after I used bounty for first time. Later I re-bountied it for more clarification. Still I think it lacks more details, And can receive more attention from MathOverflow, What should I do? Can I ask the question again On MathOverflow, or can it be migrated?

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    $\begingroup$ Is this (math.stackexchange.com/questions/4774163/… ) the question you're talking about in your post? I cannot see any elements in it that would make it research level (which is what posts on MathOverflow are expected to be). $\endgroup$
    – postmortes
    Commented Nov 22, 2023 at 18:28
  • $\begingroup$ @postmortes, yes that was the question,Shall I consider Re-bounting it on MSE itself, As it seems, The question is has very less conversation activity on MSE, I think It can or will be well Answered On Overflow $\endgroup$ Commented Nov 22, 2023 at 18:36
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    $\begingroup$ Alex Ravksy's answer to your question seems comprehensive to me. If there's something in there that you don't understand I would suggest either asking him about it or posting another question on MSE that explicitly addresses your problem. I don't think that a MathOverflow answer, likely with less detail and using higher-level machinery, is going to help you. I also think you run the risk of having your question summarily sent to MSE by MO for not being of the required level. $\endgroup$
    – postmortes
    Commented Nov 22, 2023 at 18:40
  • $\begingroup$ @postmortes, I tried asking it To him in comments again, He replied first, I think is currently not visiting MSE very much $\endgroup$ Commented Nov 22, 2023 at 18:42
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    $\begingroup$ You should consider asking a new Question as an exercise in refining the problem. Perhaps there is a detail in Alex's Answer that you need help understanding, or perhaps there is another issue that you have encountered after reading and understanding what Alex wrote. In any case asking the same Question again is discouraged. If what you really want is "conversation activity", then Math.SE is not the right place for that. We are trying to collect and curate posts that ask and answer mathematically reasoned problems, so conversational ends are subordinated to arriving at definitive facts. $\endgroup$
    – hardmath
    Commented Nov 22, 2023 at 21:26
  • $\begingroup$ A related answer on MathOverflow Meta: Interesting (and not sufficiently answered) questions on math.SE. $\endgroup$ Commented Nov 23, 2023 at 6:14

1 Answer 1


The answer from Alex Ravsky completely answers your question, so there is no need to post on MathOverflow. Your question would not be appropriate for MO anyway, because your question is not about research-level mathematics.

You asked for further clarification in a comment, and then Alex edited his answer to add that clarification. You then posted another comment asking for more clarification, but Alex's edit already answered the question in your second comment. Specifically, this paragraph talks about why dividing into a different ratio, say $k$ to $n-k$, gives a sub-optimal answer.

On the other hand, suppose that at the first step we triggered exactly $k$ switches, say, from $1$ to $k$. If the bulb state was switched then one of these switches operates with the bulb, but if the bulb state was not switched then one of the remaining switches operates with the bulb. In the former case it remains to check switches from $1$ to $k$, whereas in the latter case it remains to check switches from $k$ to $n$. Thus after the first step in the worst case there can remain at least $\max\{k,n-k\}\ge \lceil n/2\rceil$ switches which can operate with the bulb so they have to be checked. Again by the induction hypothesis, they can be checked in at least $\lceil \log_2 \lceil n/2\rceil\rceil=\lceil \log_2 n\rceil-1$ steps.

This shows that, no matter what value of $k$ you choose, you do not get any additional benefit over a 50:50 split, so the 50:50 split is optimal.


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