## Appendix B Sort II Error Analysis
\]\[=K\cdot\left[\frac{\varepsilon(1-\tau)}{\varepsilon-\tau}\cdot 2^{n\left[H( \tau)-T_{\varepsilon}(\tau)\right]}+\frac{\left(\delta_{\beta}+\frac{1}{n} \right)(1-\varepsilon)}{\delta_{\beta}-\varepsilon}\cdot 2^{-n\left[T_{\varepsilon} \left(\frac{\left\lfloor n\delta_{\beta}\right\rfloor+1}{n}\right)-H\left( \frac{\left\lfloor n\delta_{\beta}\right\rfloor+1}{n}\right)\right]}{\delta_{ \beta}-\varepsilon}\right] \tag{126}\] \[\stackrel{{(a)}}{{=}}2^{m\kappa}\cdot\left[\frac{ \varepsilon(1-\tau)}{\varepsilon-\tau}\cdot 2^{-n\left[T_{\varepsilon}(\tau)-H(\tau) \right]}+\frac{\left(\delta_{\beta}+\frac{1}{n}\right)(1-\varepsilon)}{\delta_ {\beta}-\varepsilon}\cdot 2^{-n\left[T_{\varepsilon}\left(\frac{\left\lfloor n \delta_{\beta}\right\rfloor+1}{n}\right)-H\left(\frac{\left\lfloor n\delta_{ \beta}\right\rfloor+1}{n}\right)\right]}{\delta_{\beta}-\varepsilon}\right]\] \[=\frac{\varepsilon(1-\tau)}{\varepsilon-\tau}\cdot 2^{-n\left[T_{ \varepsilon}(\tau)-H(\tau)-\kappa\right]}+\frac{\left(\delta_{\beta}+\frac{1} {n}\right)(1-\varepsilon)}{\delta_{\beta}-\varepsilon}\cdot 2^{-n\left[T_{ \varepsilon}\left(\frac{\left\lfloor n\delta_{\beta}\right\rfloor+1}{n} \right)-H\left(\frac{\left\lfloor n\delta_{\beta}\right\rfloor+1}{n}\right)- \kappa\right]}{\delta_{\beta}-\varepsilon},\]  
which implies that both the exponential factors given in (126) should yields strict positive exponents; that is, we obtain two separate upper bounds on the \(\kappa\) as follows:  
\[\kappa<T_{\varepsilon}(\tau)-H(\tau)\qquad\text{and}\qquad\kappa<T_{ \varepsilon}\left(\frac{\left\lfloor n\delta_{\beta}\right\rfloor+1}{n} \right)-H\left(\frac{\left\lfloor n\delta_{\beta}\right\rfloor+1}{n}\right), \tag{127}\]  
Therefore,  
\[\kappa<\min\Biggl{\{}T_{\varepsilon}(\tau)-H(\tau),\,T_{\varepsilon}\left( \frac{\left\lfloor n\delta_{\beta}\right\rfloor+1}{n}\right)-H\left(\frac{ \left\lfloor n\delta_{\beta}\right\rfloor+1}{n}\right)\Biggr{\}}. \tag{128}\]  
Now, we focus on the second argument in (128), and provide the following asymptotic behavior:  
\[\lim_{n\to\infty}T_{\varepsilon}\left(\frac{\left\lfloor n\delta_{\beta} \right\rfloor+1}{n}\right)-H\left(\frac{\left\lfloor n\delta_{\beta}\right \rfloor+1}{n}\right)=T_{\varepsilon}\left(\lim_{n\to\infty}\frac{\left\lfloor n \delta_{\beta}\right\rfloor+1}{n}\right)-H\left(\lim_{n\to\infty}\frac{\left \lfloor n\delta_{\beta}\right\rfloor+1}{n}\right), \tag{129}\]  
where the equality holds, since \(T_{\varepsilon}(\cdot)\) and \(H(\cdot)\) are continuous functions of \(\delta_{\beta}\). Now, observe that since \(\left\lfloor n\delta_{\beta}\right\rfloor-1<\left\lfloor n\delta_{\beta}\right\rfloor \leq n\delta_{\beta}\) for real \(n\delta_{\beta}\), we obtain  
\[\lim_{n\to\infty}\frac{n\delta_{\beta}-1+1}{n}\leq\lim_{n\to \infty}\frac{\left\lfloor n\delta_{\beta}\right\rfloor+1}{n}\leq\lim_{n\to \infty}\frac{n\delta_{\beta}+1}{n}\Rightarrow\\ \delta_{\beta}\leq\lim_{n\to\infty}\frac{\left\lfloor n\delta_{ \beta}\right\rfloor+1}{n}\leq\lim_{n\to\infty}\delta_{\beta}+\frac{1}{n} \stackrel{{(a)}}{{=}}\lim_{n\to\infty}\frac{\left\lfloor n\delta_{ \beta}\right\rfloor+1}{n}=\delta_{\beta}, \tag{130}\]  
where \((a)\) holds by the _squeeze theorem_.