Update on Overleaf.
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@@ -534,7 +534,7 @@ Use interactive version for now.
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% Theorem 2 Garay variable difficulty
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% Theorem 2 Garay variable difficulty
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For the common prefix to hold, we need $k \geq \frac{\theta \gamma m}{4 \tau}$. As $m = 2016$ and $\tau = 4$, we want to compute $\theta$ and $\gamma$ in order to get $k$:
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For the common prefix to hold, we need $k \geq \frac{\theta \gamma m}{4 \tau}$. As $m = 2016$ and $\tau = 4$, we want to compute $\theta$ and $\gamma$ in order to get $k$:
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% We can't right away compute \theta and \gamma with empirical values, as
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% We can't right away compute \theta and \gamma with empirical values, as \eta and \theta evaluation would match
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\begin{itemize}
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\begin{itemize}
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% Definition 5 Garay variable difficulty
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% Definition 5 Garay variable difficulty
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\item $\theta$: we have $f(T_r^{max}(E),n_r) \leq \theta f$ (with $f = 0.03$), so $\theta \geq \frac{f(T_r^{max}(E),n_r)}{f}$. As we can estimate $q$ as $\frac{diff(S)}{rounds(S)}$, with $S$ the number of blocks confirming the weather balloon, $diff(S)$ the expected number of nonces tested to mine the $S$ blocks and $rounds(S)$ the number of rounds occurring during the mining of the $S$ blocks. Then we have $\theta \geq \frac{1 - (1 - \frac{T}{2^\kappa})^{q n_r}}{f}$
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\item $\theta$: we have $f(T_r^{max}(E),n_r) \leq \theta f$ (with $f = 0.03$), so $\theta \geq \frac{f(T_r^{max}(E),n_r)}{f}$. As we can estimate $q$ as $\frac{diff(S)}{rounds(S)}$, with $S$ the number of blocks confirming the weather balloon, $diff(S)$ the expected number of nonces tested to mine the $S$ blocks and $rounds(S)$ the number of rounds occurring during the mining of the $S$ blocks. Then we have $\theta \geq \frac{1 - (1 - \frac{T}{2^\kappa})^{q n_r}}{f}$
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