## 2 Results
### The Theoretical Description of the Protocol
In the unveil phase, since Equation (3) is satisfied, according to the HJW theorem, there exist two measurements \(M_{0}\) and \(M_{1}\) on \(\alpha\) such that, if [PERSON] applies \(M_{0}\) (\(M_{1}\)) on \(\alpha\), then \(\beta\) will collapse to a state belonging to the set described by \(\rho_{0}^{\delta}\) (\(\rho_{1}^{\delta}\)). Therefore, [PERSON] can unveil \(b\) as whatever value she likes in the unveil phase by choosing between the two measurements \(M_{0}\) and \(M_{1}\). However, in our protocol, the two sets of states take the forms in Equations (1) and (2). Suppose that dishonest [PERSON] prepares a bipartite system \(\alpha\otimes\beta\) in the state:  
\[|\Omega\rangle=\frac{1}{\sqrt{n-1}}\sum_{i=1}^{n-1}|\alpha_{i+}\rangle|\phi_{i+}\rangle \tag{4}\]  
so that she can cheat using the strategy in the no-go proofs. Here, \(\{|\alpha_{i+}\rangle,i=0,\ldots,n-1\}\) is an orthonormal basis of the \(n\)-dimensional system \(\alpha\). Let \(\rho_{0}^{\delta}\) and \(\rho_{1}^{\delta}\) be the density matrices corresponding to the sets \(\{\psi_{i}^{0}=|\phi_{i+}\rangle\}\) and \(\{\psi_{i}^{1}=|\phi_{i-}\rangle\}\), respectively. As shown in ref. [55], when \(n\rightarrow\infty\), Equation (3) is satisfied, so that it seems to meet the requirement of the HJW theorem. Now, let us see what happens if [PERSON] wants to cheat. Surely, if she wants to unveil \(b=0\), all she needs is simply to use \(\{|\alpha_{1+}\rangle,|\alpha_{2+}\rangle,\ldots,\)\(\left|\alpha_{(n-1)+}\right\rangle\}\) as the basis of the measurement \(M_{0}\) and apply it on her system \(\alpha\), which will make [PERSON]'s system \(\beta\) collapse into one of the state in \(\{|\phi_{i+}\rangle\}\) so that she can complete the protocol without being caught. Now, the question is, can she unveil \(b=1\) successfully? According to the HJW theorem, there should exist another measurement \(M_{1}\) with the basis \(\{|\alpha_{1-}\rangle,|\alpha_{2-}\rangle,\ldots,\left|\alpha_{(n-1)-}\right\rangle\}\), such that Equation (4) can be expressed as follows:  
\[|\Omega\rangle\simeq\frac{1}{\sqrt{n-1}}\sum_{i=1}^{n-1}|\alpha_{i-}\rangle| \phi_{i-}\rangle \tag{5}\]  
so that [PERSON]'s measuring \(\alpha\), in this basis, will make \(\beta\) collapse into a state in \(\{|\phi_{i-}\rangle\}\). To find the form of \(\{|\alpha_{i-}\rangle\}\), following ref. [56], we expand each \(|\phi_{i+}\rangle\) in Equation (4) using \(\{|\phi_{i-}\rangle\}\), and the result is below:  
\[|\Omega\rangle = \frac{1}{\sqrt{n-1}}\sum_{i=1}^{n-1}\sqrt{1-\frac{4}{n^{2}}}| \tilde{\alpha}_{i-}\rangle|\phi_{i-}\rangle \tag{6}\] \[+\sqrt{\frac{2}{n}}|\tilde{\alpha}_{n-}\rangle|\phi_{n-}\rangle,\]  
where  
\[|\phi_{n-}\rangle\equiv\frac{1}{\sqrt{n}}\Big{(}|0\rangle+\sum_{i=1}^{n-1}|i \rangle\Big{)}, \tag{7}\]  
\[|\tilde{\alpha}_{n-}\rangle\equiv\frac{1}{\sqrt{n-1}}\sum_{i=1}^{n-1}|\alpha _{i+}\rangle, \tag{8}\]\[|\tilde{\alpha}_{i-}\rangle\equiv\frac{1}{\sqrt{1-\frac{4}{n^{2}}}}\left(\frac{2-n} {n}|\alpha_{i+}\rangle+\frac{2}{n}\sum_{i^{\prime}=1,i^{\prime}\
eq i}^{n-1}| \alpha_{i^{\prime}+}\rangle\right), \tag{9}\]  
for \(i=1,\ldots,n-1\).