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As CLRS book,page 260 stated,

Thus, the total time required for a successful search is $\Theta{\left(2+\alpha/2-\alpha/2n\right)}=\Theta{(1+\alpha)}$

I wouldn't have any problem if the author says the bound is eventually $\Theta{(2+\alpha/2-\alpha/2n)}$ or even $\Theta{(1+\alpha(1-\frac{1}{n}))}$. What kind of logics shall we apply to simplify the original result, i.e, cancelling the factor $1/n$ of $\alpha$. What i've missed? is anyone got the same confusion?

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It is easy to see that $\Theta(2 +\alpha/2 - \alpha/2n) \subseteq O(2+\alpha/2) \subseteq O(1+\alpha)$

For the other direction, note that $\alpha = \frac{n}{m}$, where $m$ is the size of the table and $m \geq 1$. Therefore, $\alpha/2n < 1/2$. It gives $\Theta(2 +\alpha/2 - \alpha/2n) \subseteq \Omega(3/2+\alpha/2) \subseteq \Omega(1+\alpha)$.

Therefore, $\Theta(2 +\alpha/2 - \alpha/2n) = \Theta(1+\alpha)$

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  • $\begingroup$ Let me note, that for subsets is used "$\subset$" symbol, not "$\in$", which is symbol for belonging. Especially, when you are meaning inequalities in your reasonings. $\endgroup$
    – zkutch
    Sep 14 at 7:58
  • $\begingroup$ @zkutch oops! I have made the edit. Also, please feel free to edit the answer. $\endgroup$ Sep 14 at 8:01

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