can prove that as long as $\lambda_{\text{path}} >\lambda_{\text{faults}}$, the optimal solution corresponding to the minimal number of faults will always have an $E_{\text{problem}}$ lower than any assignment of $\{x_i,y_i\}$ that is not in agreement with the observation. This is a consequence of the tree structure of the problem and the fact that there can only be penalties due to inconsistencies in the observations if the $i$-th ammeter is healthy ($y_i =1$), since whenever $y_i =0$, the contribution to $H_{\text{consist}}$ would be zero. It can be proved by considering that the energy of any inconsistent configuration incurring a penalty via $H_{\text{consist}}$ can always be lowered by flipping the sensor in its path; this will make the assignment a consistent one. Flipping this sensor to a faulty state will raise the energy by $\lambda_{\text{fault}}$, but at the same time the energy is lowered by a greater amount $\lambda_{\text{path}}$, as long as $\lambda_{\text{path}} >\lambda_{\text{faults}}$. In the example presented here, we set $\lambda_{\text{path}} = 3$ and $\lambda_{\text{faults}} =1$.

For the 21 CB, 6 fault problem, shown in Fig. \[fig:diagnosis\]a where $3$ of the $4$ outermost sets of $4$ sensors contain two $l_i=0$, the solution must contain a composite of $6$ components set to $0$. 