Planning Questions and Answers

Answer: C

Achieving one conjunct may destroy the conditions needed for another, so a planner must be prepared to interleave the work rather than completing each in turn. That is exactly the positive interaction the additive heuristic ignores. Stated order, difficulty ordering and universal decomposability are all refuted by the example. An assumption that holds in easy cases and fails in a small counterexample is precisely the kind that needs an explicit test.

Answer: D

Interactions may be positive, where one action serves several goals, or negative, where achieving one undoes another, and both defeat a planner that treats goals independently. Mutual exclusion is the extreme negative case. Precedence and listing order are different notions. Recognising that both kinds of interaction exist explains why the additive heuristic overestimates and why the Sussman anomaly is hard, in one idea rather than two. A limit that survives making every external component ideal is an intrinsic one, and worth knowing for that reason.

Answer: A

If no goal's achievement undoes another, the problem decomposes and a planner may work through them in turn, which is why such domains are much easier. The Sussman anomaly is the standard example of a non-serialisable set. Fixed sequences, decomposition into separate problems and importance ordering are different notions. Identifying when decomposition is sound, rather than assuming it, is what separates a correct planner from a plausible one. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: B

Planners schedule spacecraft observations, control autonomous vehicles, route deliveries and sequence manufacturing operations, all domains where the actions and their effects are known and a sequence must be chosen. Question answering, learning and verification are served by other techniques. Identifying the shape of problem a technology suits, rather than its list of successes, is what allows it to be reached for at the right moment. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

Answer: C

Planning applies where the dynamics are known and written down, and the question is which sequence of known actions reaches a stated goal. Where the dynamics must be learned, reinforcement learning applies instead, and the other options describe verification and classification. Asking whether the model exists, before choosing between planning and learning, is the question that settles which family of techniques is appropriate.

Answer: D

Both seek a course of action maximising some objective, and the difference is whether the transition model is given or must be learned from experience. Model based reinforcement learning sits between them by learning a model and then planning with it. Neither contains the other, both handle discrete and continuous cases. Placing two apparently distinct fields on a single axis, according to what is known in advance, makes the relationship between them immediately clear.

Answer: A

Writing an axiom for every combination of action and fluent that is not affected would require a quadratic number of statements, most of them trivial, and the planning representation simply stipulates that unmentioned fluents persist. That stipulation is what makes the formalism concise. Reference frames, fluent selection and domain boundaries are different modelling questions. Adopting a convention that removes a quadratic bookkeeping burden is the single most important design decision in the representation.

Answer: B

The convention requires the effects to be complete, so an action with unknown or context dependent consequences cannot be described without extending the formalism. Conditional effects are a standard such extension. Negation is expressible, states list only true fluents, and conditional effects are supported by most modern languages. Knowing the assumption behind a convenience, and which extensions relax it, is what allows the formalism to be stretched safely.

Answer: C

Writing that moving a container also moves its contents, when it has any, requires an effect guarded by a condition evaluated in the state where the action occurs. It is how the ramification problem is handled within the formalism. Success conditions, numeric effects and delays are different extensions. Adding guarded effects, rather than writing a separate action for every case, is what keeps a domain description manageable.

Answer: D

Working backwards must consider both the case where the condition held and the case where it did not, which branches the search. Forward search simply evaluates the condition. Reversibility, action counts and unconditional requirements are different issues. A feature that is cheap in one search direction and expensive in the other is worth noting when choosing between them, since the choice is otherwise made on other grounds entirely. Checking the units of each quantity catches a wrong formula before the arithmetic begins.