## 5 The GHZ\({}_{3}\)MQPEC Protocol in Action
* [PERSON]'s input register _[PERSON]'s_2_IR_ containing two qubits. * [PERSON]'s single-qubit output register _[PERSON]'s_2_OR_ initialized at \(|-\rangle\). * [PERSON]'s input register _[PERSON]'s_2_IR_ containing two qubits. * _trent's_0_IR, _[PERSON]'s_2_IR_, and _[PERSON]'s_2_IR_ are initialized in the entangled state \(|GHZ_{3}\rangle^{\otimes 2}\). * The three classical registers (each containing two bits) used by [PERSON]\({}_{2}\), [PERSON], and [PERSON] to store their measurements are collectively referred to as the _classicalRegister2_ with a total capacity of six bits. It is straightforward to verify that all outcomes satisfy the Hadamard Entanglement Property, as quantified by (19). To explain this point beyond any doubt, let us focus on the histogram of Figure 7 and, specifically, on the value of the label of the first bar, which is 000011 010101 110010. The label consists of three parts: 0000011, corresponding to the contents of _classicalRegister2_; 010101, corresponding to the contents of the _classicalRegister1_; and 110010, corresponding to the contents of the _classicalRegister0_. This ordering is a direct consequence of the way the quantum circuit in Figure 6 was constructed, i.e., the bits of _classicalRegister2_ are more significant than those of _classicalRegister1_, which in turn are more significant than those of _classicalRegister0_. * _classicalRegister0_ is the collective name for the three classical registers (each containing two bits) used by [PERSON], [PERSON], and [PERSON]\({}_{0}\) to store their measurements. Therefore, the contents of 110010 are analyzed as shown below: This means that \(\mathbf{y}_{2}=11\), \(\mathbf{y}_{1}=00\), and \(\mathbf{y}_{0}=10\). [PERSON] and [PERSON]\({}_{0}\) send to [PERSON] \(\mathbf{y}_{1}\) and \(\mathbf{y}_{0}\) via authenticated classical channels. So, [PERSON] has the complete information to compute the modulo 2 sum of \(\mathbf{y}_{2}\), \(\mathbf{y}_{1}\), and \(\mathbf{y}_{0}\) and conclude that \[\mathbf{s}\oplus\mathbf{f}_{0}=01\;,\] (22) according to (19). * Reasoning in exactly the same way for _classicalRegister1_ and _classicalRegister2_, it is straightforward to see that [PERSON] arrives at the additional conclusions \[\mathbf{s}\oplus\mathbf{f}_{1}=01\;,\;\text{and}\] (23) \[\mathbf{s}\oplus\mathbf{f}_{2}=11\;.\] (24)  
Ergo, [PERSON] ascertains that [PERSON]\({}_{0}\) and [PERSON]\({}_{1}\)'s fortunes are equal but different from [PERSON]\({}_{2}\)'s fortune. We emphasize that [PERSON], as stipulated in (A\({}_{3}\)), never reveals her secret number \(\mathbf{s}\), so [PERSON] cannot infer \(\mathbf{f}_{i}\) from \(\mathbf{s}\oplus\mathbf{f}_{i}\). Likewise, [PERSON] never divulges \(\mathbf{y}_{2}\) or the modulo 2 sum \(\mathbf{y}_{2}\oplus\mathbf{y}_{1}\oplus\mathbf{y}_{0}\) to any other in-game player or outside eavesdropper, as ordained by (A\({}_{4}\)) and (D\({}_{3}\)). Thus, no entity can gain any information about the millionaires' fortunes. Hopefully, our in-depth analysis demonstrates that it is very easy to extend the present example either by increasing the capacity \(m\) of the registers or by increasing the number of millionaires \(n\).