# Continuous Variable Quantum Secret Sharing with Fairness
## 3 Security Analysis
### Internal Attack  
Generally, there are two kinds of attack considered in the QSS scheme. One is eavesdropping of [PERSON] from outside, the other one is the dishonest participant attack from inside. However the dishonest participant has more superiorities to steal the secret than [PERSON] from outside. Therefore, the scheme is secure with [PERSON]'s attack, so long as the protocol can resist inner attack. Consequently, the following security analysis primarily focuses on inner attack in noisy channel. For curiousness, [PERSON] and [PERSON] may guess the original key \(S\), which is hidden in \(X\). Although each participant has one share \(X^{A/B}\), they have no idea about the position of \(S\) in \(X\), which means they should guess the right \(P^{*}\) at first, then recover S by guess. According to the Equation (2) or Table 1, the authorized set \(X^{A}\cup X^{B}\) can recover the message \(X\) and from above caculations, \(H(X_{i})~{}=~{}H(X_{i}^{A})~{}+~{}H(X_{i}^{B})\), which means the unauthorized sets \(X_{i}^{A}\) and \(X_{i}^{B}\) are independent to each other. Apparently, \(H(X_{i})>H(X_{i}^{B})>H(X_{i}^{A})>0\), so, curious member cannot deduce \(X_{i}\) from unauthorized set \(X_{i}^{A}\) or \(X_{i}^{B}\), alone. Furthermore, from Table 1 and Equations (17) and (18), the conditional probabilities \(P(X_{i}\mid X_{i}^{A})\) and \(P(X_{i}\mid X_{i}^{B})\) can be concluded as Tables 2 and 3. Thus, according to [PERSON]'s share \(X_{i}^{A}\), the successful probabilities of guessing \(P^{*}\) and \(S\) are \((\frac{1}{3})^{L_{2}}\) and \((\frac{1}{3})^{L}\), which approaches to _zero_, when \(L,L_{2}>3\). Similarly, [PERSON] can successfully guess \(P^{*}\) and \(S\) with probabilities \((\frac{1}{2})^{L_{2}}\) and \((\frac{1}{2})^{L}\). So, it is hardly to accurately locate \(S\) in \(X\) and then recover it for [PERSON] and [PERSON]. Therefore, dishonest participant has to perform attacks to acquire more information for secret recovery. In this protocol, the usage of coherent states can resist intercepting and re-sending attack, so, dishonest participant named [PERSON] plays BS attack strategy [21; 26], which is shown in Figure 2. To avoid being detected, [PERSON] modulates the parameters of beam splitters to imitate the noisy quantum channels, i.e., the transmission coefficient of BS equals to the transmission efficiency of noisy quantum channel. In this way, the communicant may regard this attack as quantum channel lossy and noise. So, the mutual information between sender and receiver can be calculated as Equation (16). As for [PERSON], passing through the first BS, [PERSON] can get  
\[x_{E1}=\sqrt{\eta_{1}}x_{N1}-\sqrt{1-\eta_{1}}x_{2},~{}~{}p_{E1}=\sqrt{\eta_{ 1}}p_{N1}-\sqrt{1-\eta_{1}}p_{2}, \tag{19}\]  
where \(\eta_{1}\) is the transmission coefficient of BS, \(a_{N1}=(x_{N1},p_{N1})\) is a vacuum state and \(x_{N1},p_{N1}~{}\sim~{}N(0,\Sigma_{1}^{2})\). Similarly, using the second BS, [PERSON] can obtain  
\[x_{E2}=\sqrt{\eta_{2}}x_{N2}-\sqrt{1-\eta_{2}}x_{4},~{}~{}p_{E2}=\sqrt{\eta_{ 2}}p_{N2}-\sqrt{1-\eta_{2}}p_{4}, \tag{20}\]  
\begin{table}
\begin{tabular}{c c c c c c c} \hline \hline \(P(X_{i}\mid X_{i}^{A})\) & \(X_{i}=0\) & \(X_{i}=1\) & \(X_{i}=2\) & \(X_{i}=3\) & \(X_{i}=4\) & \(X_{i}=5\) \\ \hline \(X_{i}^{A}=0\) & \(1/3\) & \(0\) & \(1/3\) & \(0\) & \(1/3\) & \(0\) \\ \(X_{where \(\eta_{2}\) is the transmission coefficient of BS, \(a_{N2}=(x_{N2},p_{N2})\) is a vacuum state and \(x_{N2},p_{N2}\ \sim\ N(0,\Sigma_{2}^{2})\). Then, [PERSON] performs a gain amplification on them to obtain  
\[a_{E1}^{\prime}=g_{E1}a_{E1},\ \ a_{E2}^{\prime}=g_{E2}a_{E2}, \tag{21}\]  
where \(g_{E1}=\frac{1}{\sqrt{1-\eta_{1}}}\) and \(g_{E2}=\frac{1}{\eta_{1}(\sqrt{1-\eta_{2}})}\).