 Solution

The steady-state output is $q=\bar{a}\bar{c}+bc+\bar{c}d$. Its Karnaugh map, with rows $ab$ and columns $cd$ in Gray code order, isThe existing prime implicants cover adjacent ones except for some transitions in $c$. The consensus theorem supplies the two missing bridging implicants $\bar{a}b$ and $bd$, givingBoth are needed in a hazard-free sum of products retaining the original terms. At $a=0,b=1,d=0$, the two values of $c$ require a common implicant, necessarily covered by $\bar{a}b$; at $a=1,b=1,d=1$, the corresponding transition requires $bd$. Neither bridge covers the other's critical edge. All other adjacent-one pairs already share a term. Thus two is the least number of additional prime implicants. This addresses single-input static hazards; simultaneous input changes require separate analysis.