# Mdpli Mitigation of Routing Congestion on Data Networks: A Quantum Game Theory Approach
## 4 Results
### Pareto Optimality and Nash Equilibrium
#### 4.1.3 Mixed Quantum Strategies  
It is interesting to know if \(S_{1}=(\frac{\pi}{2},\frac{\pi}{4},0)\) is also a Nash equilibrium. For an \(S_{x}=(\varphi_{X},\varphi_{Y},\varphi_{Z})\) strategy to be a Nash equilibrium, no player will have any incentive to individually modify their strategy. It turns out that there exists a strategy \(S_{2}=(\frac{\pi}{2},\frac{\pi}{4},\frac{\pi}{2})\) that is beneficial to an individual player assuming that the rest apply \(S_{1}\), therefore \(S_{1}\) is not a Nash equilibrium. It also turns out that \(S_{2}\) is Pareto optimal and is not a Nash equilibrium, since there exists a \(S_{3}=(\frac{\pi}{2},\frac{\pi}{4},\pi)\) that is beneficial for an individual player assuming that the rest apply \(S_{2}\). Again, \(S_{3}\) is Pareto optimal and not a Nash equilibrium, since there exists another \(S_{4}=(\frac{\pi}{2},\frac{\pi}{4},\frac{3\pi}{2})\) that is beneficial for an individual player assuming that the rest apply \(S_{3}\). Furthermore, finally, \(S_{4}\) is Pareto optimal and not a Nash equilibrium since an individual player can apply \(S_{1}\) to benefit himself!  
In short, \(S_{1}\), \(S_{2}\), \(S_{3}\) and \(S_{4}\) are all pure strategies. All of them are Pareto optimal in our problem, since no player can improve their performance without worsening someone else's and if all the players apply the same strategy maximum performance can be obtained and the network congestion be avoided. However, none of them is a Nash equilibrium since there is always a dominant strategy that would increase the benefit of an individual player thus encouraging players to individually modify their strategy.  
Finally, we can build a mixed strategy \(S_{5}\) that applies the four strategies \(S_{1}\), \(S_{2}\), \(S_{3}\) and \(S_{4}\) with equal probability \(w=0.25\). \(S_{5}\) turns out to be Pareto optimal since no one can improve their own performance without worsening someone else's and it also appears to be Nash equilibrium since we could not find any \(S_{X}\) strategy that a player can apply individually to increase their own performance assuming that the other players are applying \(S_{5}\).  
\[S_{5}=\begin{cases}S_{1}=(\frac{\pi}{2},\frac{\pi}{4},0)&\text{with }w=0.25\\ S_{2}=(\frac{\pi}{2},\frac{\pi}{4},\frac{\pi}{2})&\text{with }w=0.25\\ S_{3}=(\frac{\pi}{2},\frac{\pi}{4},\pi)&\text{with }w=0.25\\ S_{4}=(\frac{\pi}{2},\frac{\pi}{4},\frac{3\pi}{2})&\text{with }w=0.25\end{cases} \tag{5}\]  
Figure 9: Trade-off barrier broken by quantum protocol. In red different pure quantum strategies, in blue different mixed classical strategies with \(p\) values between 0 and 0.99.  
The rigorous proof that \(S_{5}\) is a Nash equilibrium will be left for future work and its difficulty relies on the fact that the strategy space is infinite. Another interesting work that remains for the future is the study of the equilibrium strategies of the system formulated as a generic infinitely repeated quantum game [23]. \(\varphi_{X}\), \(\varphi_{Y}\) and \(\varphi_{Z}\) are already continuous variables, and for proving that \(S_{5}\) is a Nash equilibrium it is necessary to demonstrate that there is no mixed strategy in the entire space (three probability density function, over \(\varphi_{X}\), \(\varphi_{Y}\) and \(\varphi_{Z}\)) that has the incentive to deviate.  
Observing the outcome \(|\psi_{out}\rangle\) of the system when all players apply the mixed strategy \(S_{5}\), we see in Equation (6) how we avoid congestion. States \(|00\rangle\) and \(|11\rangle\) are absent, therefore, the congestion is avoided.  
\[|\psi_{out}\rangle=\begin{cases}\frac{|01\rangle+|10\rangle}{\sqrt{2}}&\text {with 50\%}\\ |10\rangle&\text{with 25\%}\\ |01\rangle&\text{with 25\%}\end{cases} \tag{6}\]