Chapter recap
A ≤ₘ B transfers decidability from B to A and hardness from A to B; prove P undecidable by A_TM ≤ P. E_TM is co-RE not RE; NE_TM is RE; EQ_TM and TOTAL are neither. Rice: non-trivial properties of L(M) are undecidable; code properties and bounded runs are decidable. PCP is undecidable and gives CFG undecidability results. Master table: membership fails only at RE, emptiness/finiteness at CSL, equivalence/universality at CFL, containment/disjointness at DCFL.
Practice questions
- MCQ
A ≤ₘ B and A is not RE. B is:
- not RE
- RE
- decidable
- regular
- MSQ
Rice’s theorem shows undecidable:
- L(M) is infinite
- L(M) = ∅
- M has 4 states
- L(M) = {0, 1}
- MCQ
Which is decidable?
- CFG equivalence
- CSL emptiness
- DCFL universality
- TM halting
- MCQ
NE_TM = {⟨M⟩ : L(M) ≠ ∅} is:
- decidable
- RE, undecidable
- co-RE, not RE
- neither
- MSQ
Which are undecidable for CFGs?
- ambiguity
- membership
- L(G₁) ∩ L(G₂) = ∅
- L(G) = Σ*
- MCQ
To show P undecidable, a correct plan is:
- reduce P to HALT_TM
- reduce HALT_TM to P
- show P is RE
- show P̅ is RE
- NAT
In the master table (9 questions × 6 classes), how many entries are D for the CFL column?
Write your answer before opening the key.
- MCQ
PCP over a one-letter alphabet is:
- undecidable
- decidable
- not RE
- co-RE only
Answer key
Reveal answers when you’re ready
- A — Non-RE-ness flows forward.
- B: Contradiction.
- C: No.
- D: No.
- ABD — Language properties.
- C: Code property.
- C — Complement of a DCFL is a DCFL; test emptiness.
- A: U.
- B: U.
- D: U.
- B — Search for an accepted string.
- A: Rice.
- C: It is RE.
- D: It is RE.
- ACD — PCP-based.
- B: CYK.
- B — Known hard → new.
- A: Wrong direction.
- C: Not enough.
- D: Not enough.
- 3 — Membership, emptiness, finiteness.
- B — Compare lengths.
- A: Only for ≥ 2 letters.
- C: No.
- D: No.