## 6 Data Transmission Security and Reliability
Let us estimate the value \(f_{k_{1}}\):  
\[f_{k_{1}}=\frac{\prod_{i=1}^{k_{1}}p_{i}}{\prod_{i=0}^{k_{1}-2}p_{n-i}}=\frac{ \prod_{i=1}^{k+1}p_{i}}{\prod_{i=0}^{k-1}p_{n-i}}=\frac{p_{n-k+2}}{p_{k+1}} \cdot\frac{\prod_{i=1}^{k}p_{i}}{\prod_{i=0}^{k-2}p_{n-i}}=\frac{p_{n-k+2}}{p_{ k+1}}\cdot f_{k}. \tag{18}\]  
where \(f_{k}\) represents the value \(\frac{P}{P_{max}}\) for SSS with threshold \(k\). Because \(2\leq k<\frac{n}{2}\), then, \(n-k+2>\frac{n}{2}\), therefore, \(n-k+2>k+1\). Considering the restrictions imposed on the RRNS moduli, we have \(p_{n-k+2}>p_{k+1}\), therefore, \(\frac{p_{n-k+2}}{p_{k+1}}>1\). Given the above considerations, we obtain that \(f_{k_{1}}<f_{k}\). In other words, the worst case in which \(\frac{P}{P_{max}}\) takes the smallest value is the case \(k=\lfloor\frac{n}{2}\rfloor\). Because of the symmetry \(\frac{P}{P_{max}}\) relative to this value, it is advisable to consider \(k\) within the borders \(\left[2,\frac{n}{2}\right]\), as for the interval \(\left[\frac{n}{2},n-1\right]\) reasoning proceeds in a similar way. A special case is SSS in which \(k=n\) and \(\frac{P}{P_{max}}=1\). Next, we prove several important statements that accurately estimate \(f\left(\widetilde{I}_{max}\right)\). **Statement 3**. : _For any sequence \(p_{1}<p_{2}<\ldots<p_{n}\) follows that  
Based on Statements 3 and 4, let us accurately determine the boundaries for the quantity \(\frac{P}{P_{max}}\), which directly depends on the value \(p_{1}\). Let us consider an example that shows how fast the value \(\frac{P}{P_{max}}\) converges to \(p_{1}\). Figure 4 shows \(R=\frac{f(\widetilde{I}_{max})}{p_{1}}\). \(R\) shows the relation of \(f\left(\widetilde{I}_{max}\right)\) and \(p_{1}\). This relation assesses how parameter \(a\) affects security by approaching the value of \(b\) by using compact sequences for various \(p_{1}\). We see that with increasing \(p_{1}\), \(R\) approaches 1, and, therefore, \(\frac{P}{P_{max}}\) approaches \(p_{1}\). In this case, Equation (22) estimates the lower bound of \(R\). Statement 4 estimates the proximity of the cardinality of possible secrets \(f\left(\widetilde{I}_{max}\right)\) to \(p_{1}\) depending on the \(p_{1}\), \(\theta\), \(k\), and \(n\) given before generating the sequence itself. Figure 4 shows that the higher the value of \(p_{1}\), the closer \(f\left(\widetilde{I}_{max}\right)\) to it. Thus, with a higher value of \(p_{1}\), the DT-RRNS has higher security. Statement 4 is important for estimating security. At fixed \(\theta\) and \(k\), magnitude \(f\left(\widetilde{I}_{max}\right)=\frac{P}{P_{max}}\) is within the following limits:  
\[\frac{P}{\widetilde{P}}>\frac{p_{1}^{k}}{p_{1}^{k-1}2^{k-1}}=\frac{p_{1}}{2^{k- 1}} \tag{10}\]0  
It is easy to show that for fixed \(0<\theta<1\) and \(2\leq k\leq n\)  
\[\frac{P}{\widetilde{P}}>\frac{p_{1}^{k}}{p_{1}^{k-1}2^{k-1}}=\frac{p_{1}}{2^{k- 1}} \tag{10}\]1  
Consequently,  
\[\frac{P}{\widetilde{P}}>\frac{p_{1}^{k}}{p_{1}^{k-1}2^{k-1}}=\frac{p_{1}}{2^{k- 1}} \tag{10}\]2  
where \(0<\varepsilon<1\). And the more \(p_{1}\), the closer \(\varepsilon\) to 0. Then from (24) it follows  
\[\frac{P}{\widetilde{P}}>\frac{p_{1}^{k}}{p_{1}^{k-1}2^{k-1}}=\frac{p_{1}}{2^{k- 1}} \tag{10}\]3  
Based on this expression, one can obtain the following estimate for \(f\left(\widetilde{I}_{max}\right)\):  
\[\frac{P}{\widetilde{P}}>\frac{p_{1}^{k}}{p_{1}^{k-1}2^{k-1}}=\frac{p_{1}}{2^{k- 1}} \tag{10}\]4  
The last inequality determines the degree of closeness of the quantity \(f\left(\widetilde{I}_{max}\right)\) to \(p_{1}\) without generating the sequence itself. Because \(p_{1}>p_{0}\) then due to restrictions imposedon \(\frac{p}{P_{max}}\), with an increasing number \(p_{0}\), the cardinality \(f\left(\widetilde{I}_{max}\right)\) of the enumeration set of DT-RRNS approaches \(p_{1}\). We can conclude that the cardinality of the brute force set for the DT-RRNS when choosing sufficiently large moduli is equivalent to the cardinality of the brute force set of the Asmuth-Bloom scheme, which is equal to \(p_{0}\).