## 6 Data Transmission Security and Reliability
** Since the shares, whose numbers belong to \(\widetilde{I}\), are known, then for all \(p_{i_{j}}\) such that \(i_{j}\in\widetilde{I}\), it is possible to recover the number \(\widetilde{s}\equiv s\bmod\widetilde{P}\) due to \(s=a\widetilde{P}+\widetilde{s}\). The only unknown secret parameter is \(a\in\mathbb{Z}\). Let us define the upper and lower bound of \(a\). \(s\) is defined with dynamic range \(P\). Consequently \(0\leq a\widetilde{P}+\widetilde{s}\leq P-1\), where  
\[-\frac{\widetilde{s}}{\widetilde{P}}\leq a\leq\frac{P-\widetilde{s}-1}{ \widetilde{P}}. \tag{12}\]  
Taking into account that \(s\) and \(a\) are non-negative and since \(\widetilde{s}<\widetilde{P}\) and \(\left\lfloor\frac{p}{P}\right\rfloor-1<\frac{P-\widetilde{s}-1}{\widetilde{P} }<\left\lfloor\frac{p}{P}\right\rfloor+1\), \(\left\lfloor\frac{p}{P}\right\rfloor-1\leq\left\lfloor\frac{p-\widetilde{s}-1 }{\widetilde{P}}\right\rfloor\leq\left\lfloor\frac{p}{P}\right\rfloor\). We have  
\[0=-\left\lceil\frac{\widetilde{s}}{\widetilde{P}}\right\rceil\leq a\leq \left\lfloor\frac{P-\widetilde{s}-1}{\widetilde{P}}\right\rfloor\text{.} \tag{13}\]  
That is, \(a\) lies in between \(\left[0,\left\lfloor\frac{P-\widetilde{s}-1}{\widetilde{P}}\right\rfloor\right]\), whose cardinality is \(f\left(\widetilde{I}\right)=\left\lfloor\frac{P-\widetilde{s}-1}{\widetilde{ P}}\right\rfloor+1\), \(\left\lfloor\frac{p}{P}\right\rfloor-1\leq\left\lfloor\frac{P-\widetilde{s}-1 }{\widetilde{P}}\right\rfloor\leq\left\lfloor\frac{p}{P}\right\rfloor\), and  
\[\left\lfloor\frac{P}{\widetilde{P}}\right\rfloor\leq f\left(\widetilde{I} \right)\leq\left\lfloor\frac{P}{\widetilde{P}}\right\rfloor+1\text{.} \tag{14}\]  
The statement is proven. \(\square\)  
Let us study how the cardinality of the enumeration sets of the Asmuth-Bloom scheme and DT-RRNS are related to RRNS parameters. The Asmuth-Bloom scheme is determined by the set of moduli \(p_{0}<p_{1}<p_{2}<\ldots<p_{k}<p_{k+1}<\ldots<p_{n}\). To ensure the asymptotic ideality of the Asmuth-Bloom SSS, we require that this sequence be compact with the initial value \(p_{0}\), or \(p_{n}<p_{0}+p_{0}^{\theta}\) for \(\theta\in(0,1)\). In this case, the sequence \(p_{1}<\ldots<p_{n}\) will be compact with the initial value \(p_{1}\). This RRNS will be used as the basis for the proposed DT-RRNS. The cardinality of the Asmuth-Bloom enumeration set is constant and equal to \(p_{0}\). \[\left\lfloor\frac{P}{\widetilde{P}_{max}}\right\rfloor\leq f\left(\widetilde{ I}_{max}\right)\leq\left\lfloor\frac{P}{\widetilde{P}_{max}}\right\rfloor+1<p_{0} \text{.} \tag{15}\]  
\(\widetilde{I}_{max}\) is a set of unauthorized subsets numbers with the largest range \(\widetilde{P}_{max}\), then,  
\[\widetilde{P}_{max}=\prod_{i=0}^{k-2}p_{n-i}\text{.} \tag{16}\]  
Establish a relationship between the value \(f\left(\widetilde{I}_{max}\right)\) and \(k\). Let us consider two SSSs with a threshold \(k\) and \(n-k+1\), assuming \(2\leq k\leq\frac{u}{2}\). Let us calculate \(\frac{P}{\widetilde{P}_{max}}\) for the second SSS:  
\[\frac{\prod_{i=1}^{n-k+1}p_{i}}{\prod_{i=0}^{n-k-1}p_{n-i}}=\frac{p_{1}\cdot p _{2}\cdot...\cdot p_{k}\cdot p_{k+1}\cdot...\cdot p_{n-k+1}}{\cdot p_{k+1} \cdot p_{k+2}\cdot...\cdot p_{n-k+1}\cdot p_{n-k+2}\cdot...\cdot p_{n}}= \frac{p_{1}\cdot p_{2}\cdot...\cdot p_{k}}{p_{n-k+2}\cdot...\cdot p_{n}}= \frac{\prod_{i=1}^{k}p_{i}}{\prod_{i=0}^{k-2}p_{n-i}}\text{.} \tag{17}\]The expression on the right is the value \(\frac{P}{P_{max}}\) for the first SSS. Therefore, the values \(\frac{P}{P_{max}}\) symmetrical in \(k\) regarding the meaning \(\lfloor\frac{n}{2}\rfloor\). Let now \(2\leq k<\lfloor\frac{n}{2}\rfloor\) and \(k_{1}=k+1\). Let now \(f_{k_{1}}=\frac{P}{P_{max}}\) for a threshold \(k_{1}\).