@Jim I'm not sure whether you answered my questions "Why do you write stuff like ...?", "Is this supposed to be funny?", and "Cool, so my brute force search for SAT actually runs in constant time. Really?"

@Jim If this is a complete draft, and I were the reviewer, then I would simply reject it. No abstract, no introduction, no conclusion, no references, a wrong (or at least seriously misleading) "Proposition 3.2" whose numbering is unmotivated (to say the least), no real explanation what you want to do or why, ...

@ThomasKlimpel I tried , I will try them again, but I need some time.

@ThomasKlimpel " If this is a complete draft, and I were the reviewer, then I would simply reject it. " it is complete in technical sense, not for submission. ""Proposition 3.2" whose numbering is unmotivated (to say the least), no real explanation what you want to do or why"-- ... whose numbering is unmotivated ... would u please elaborate a little?

@ThomasKlimpel , I have asked a lot of question, I am reducing down to 2 question (1) I s the result (when x is a variable, note that in current draft x is a constant, i.e, x=3) is good enough to be on Undegrad Journal like SIURO) (2) Correctness of proposition 3.2 on page 5,6.

@ThomasKlimpel "no real explanation what you want to do or why, ..." I tried to give an explanation (before the matrix starts), but probably it is not enough. BTW: Do you want me to upload LATEX file so you can copy easily when interact?

@ThomasKlimpel , Correction: " (1) I s the result (when x is a variable, note that in current draft x is a constant, i.e, x=3) is good enough to be on Undegrad Journal like SIURO) " ----- I meant, initially I gave $\beta^x$ computational complexity, but now I "claim" $( x * \beta)^3$ complexity. If you consider $\beta^x$ computational complexity, is it good enough to be on Undegrad Journal like SIURO) ?

@Jim It would be your job as author to convince the reader that you have a nice result. My comment "... no real explanation what you want to do or why ..." indicates that you didn't manage to convince me. Your result doesn't need to be a big break through, but you must be able to explain to the reader why he should care about your result.

@ThomasKlimpel ok, which part is unclear in 3.2 ?

@Jim Why the number "3.2", for example?

@ThomasKlimpel I though this was the convention, if you write something under 3rd section, all of them(proposition, definition, lemma) will starts with 3.

@ThomasKlimpel Ok i think I know what you mean, I agree it is confusing, i will edit it soon.

@Jim OK, then you need to learn how to tell latex to write "1 Rearrangement of G" and "2 Construction of Search Tree for H" instead of "0.1 Rearrangement of G" and "0.2 Construction of Search Tree for H". Also you need to start section 3 with a proper heading then. And why is it proposition "3.2" and not proposition "3.1"?

I know how to do it, I am doing it now.

@Jim SIURO makes a lot more sense, at least with respect to the target audience.

Note however: "Papers written by undergraduate students (or teams of students) are now being accepted. Each paper must be submitted with a letter from a sponsor. The sponsor may be a faculty member at the student’s institution or at an institution the student is visiting, or someone associated with a non-academic organization or government lab who supervised the research."

@ThomasKlimpel I know, these conditions can be met.

@ThomasKlimpel , Please answer precisely, am I eligible even if the result is $\beta^x$ ?

@Jim No, if you are unable to explain why you have a nice result, then you are unlikely to get accepted. And for the original meaning of $\beta^x$, I can't see why it should be considered a nice result. Remember that you told me $x=O(n/\log n)$...

ok

@ThomasKlimpel , I have resolved the numbering issue, please see.

researchgate.net/publication/…

@Jim At the beginning of section 3, you indicate that $2^{\log n} \approx n$ and hence $\beta=O(1)$. However, we actually have $2^{\log_2 n} = n$, and hence can simplify $\beta$ to $\beta=n^{(\log_2 n-1)/2}$. So $\beta$ is still quasi-polynomial in $n$, and not $O(1)$.

@ThomasKlimpel if , $2^{\log{n}}=n \implies n^{log_2(n)}= (2^{\log_2{n}})^{log_2(n)}= (2^{(\log_2{n})^2})$ correct?

@Jim Yes, correct (mathematically).

@Jim No, the leading term (in the exponent) is $(\log_2 n)^2 / 2$, not $(\log_2 n)^2$.

@ThomasKlimpel your objection is a constant term 2, it does not mean much in Big Oh notation.

@Jim No, $2^{(\log_2 n)^2/2} \neq O(2^{(\log_2 n)^2})$. You are probably thinking of $2^{(\log_2 n)^2/2} = 2^{O((\log_2 n)^2)}$. Interestingly enough, it does matter...

@ThomasKlimpel what about $\beta$? is it $O(1)$ ?

@Jim No, $\beta$ is not $O(1)$.

@ThomasKlimpel how you asses $\beta = \frac{2^{(\log_2{n})^2}}{2^{O(\log_2{n})^2)}} $ ?

@Jim $\beta=n^{(\log_2 n-1)/2}$, so I don't need to assess such an imprecise expression...

@ThomasKlimpel ok, what about proposition 3.2 ?

@Jim I will take care of my laundry now, and then go to bed... The number "3.2" is OK now, but I still guess that the proposition itself is wrong...

@ThomasKlimpel ok, Goodnight.

@ThomasKlimpel "but I still guess that the proposition itself is wrong..." ---please, let me know if you find technical errors / inconsistencies.