For his logical combination problem, I think the solution is to use enums: State nextState(bool a, bool b, bool c, bool d) { enum A { A_NO = 0, A_Y

For his logical combination problem, I think the solution is to use enums: State nextState(bool a, bool b, bool c, bool d) { enum A { A_NO = 0, A_YES = 1 } As[] = { A_NO, A_YES }; enum B { B_NO = 0, B_YES = 2 } Bs[] = { B_NO, B_YES }; enum C { C_NO = 0, C_YES = 4 } Cs[] = { C_NO, C_YES }; enum D { D_NO = 0, D_YES = 8 } Ds[] = { D_NO, D_YES }; switch (As[a]) + Bs[b] + Cs[c] + Ds[d]) { case A_NO + B_NO + C_NO + D_NO : return X; // 0 case A_NO + B_NO + C_NO + D_YES: return X; // 1 case A_NO + B_NO + C_YES + D_NO : return X; // 2 case A_NO + B_NO + C_YES + D_YES: return X; // 3 case A_NO + B_YES + C_NO + D_NO : return X; // 4 case A_NO + B_YES + C_NO + D_YES: return X; // 5 case A_NO + B_YES + C_YES + D_NO : return X; // 6 case A_NO + B_YES + C_YES + D_YES: return X; // 7 case A_YES + B_NO + C_NO + D_NO : return X; // 8 case A_YES + B_NO + C_NO + D_YES: return X; // 9 case A_YES + B_NO + C_YES + D_NO : return X; // 10 case A_YES + B_NO + C_YES + D_YES: return X; // 11 case A_YES + B_YES + C_NO + D_NO : return X; // 12 case A_YES + B_YES + C_NO + D_YES: return X; // 13 case A_YES + B_YES + C_YES + D_NO : return X; // 14 case A_YES + B_YES + C_YES + D_YES: return X; // 15 } } Let the compiler reorgonise the switch for efficiency. Do not: skip combination. Do not: reorder. Do not: remove the number that help see we got all cases. The compiler will error out if a case is duplicated. I wish we could make it detect missing cases too!