a → b → c → d, with a ← u1 → c, and a ← u2 → d, where we do not observe u1,u2, and u1,u2 are very complicated, then we can figure out the true graph exactly by independence type techniques without having to observe u1 and u2. Note: the marginal distribution p(a,b,c,d) that came from this graph has no conditional independences at all (checkable by d-separation on a,b,c,d), so typical techniques fail.
Irrelevant question: Isn’t (b || d) | a, c?
No, because b → c <-> a <-> d is an open path if you condition on c and a.
Ah, right.