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COMP90054 Cheatsheet · AI Planning for Autonomy

University of Melbourne · Cheatsheet for the final exam. Semester 2 examination period, 2–20 November 2026. Full COMP90054 study guide →

A4 landscape · closed-book memory sheet · 6 columns per side2 sides · COMP90054 · Semester 2, 2026
COMP90054
AI Planning for Autonomy
The University of Melbourne · Closed-book memory sheet
COMP90054 cheatsheet QR code – open the full COMP90054 study guide
Full guide + ask Sia
about COMP90054
SIDE 1 · RECALLPriority: definitions → equations → choice conditions → limitsNothing printed enters the exam

01 · Exam map COURSE · BRIEF

IPA 25Q 10IMA 15FINAL 50
AssessmentWtFormat
Individual Programming Assignment25%Week 6; ≈30 h
Online Quiz10%30 min; Week 7
Individual Modelling Assignment15%Weeks 11–12; ≈15 h
Closed-book Final Examination50%2 h; must be passed

“One closed book end-of-semester examination.” COMP90054 2026 Handbook assessment table (FETCHED_UNIMELB_2026_COMP90054_ASSESSMENT_DOM.txt)

02 · Model before solving C1

Autonomous system: active system that perceives, reasons and acts towards goals or rewards. A state contains information treated as relevant to future action choice. A planning model specifies actions, preconditions, effects, initial conditions and goals.

Deterministic transitions′ = γ(s,a)
Valid current state + action → one successor state.

Translation checklist

  1. Declare state variables and observability.
  2. List each action’s preconditions and effects.
  3. Write the initial state and goal.
  4. Represent safety constraints explicitly.
  5. Trace ordinary, boundary and invalid inputs.

Exam cueA planning bug may be a missing state variable. Omitted variables, wrong observability and abstraction choices can make a formally solved model unsafe.

State/action sanity audit

QuestionEvidence in the answer
What can change?State variables have domains; every action effect changes named variables
When is a move legal?Preconditions exclude invalid state–action pairs
What counts as success?Goal condition is testable against a state
What can the system know?Observability is stated rather than assumed
What must never occur?Safety is a constraint or safely handled transition, not an afterthought

Minimal complete explanation: “From state s, action a is applicable because its preconditions hold; γ produces s′ by its effects; s′ is then tested against the goal and safety conditions.” If the real task contains uncertainty, separate the known nominal plan from uncertain execution and say whether replanning or an uncertainty-aware policy is needed.

Model quality comes before algorithm choice. A faster solver does not recover variables, consequences or failure states omitted from the representation.

statemodelactionseffectsγ(s,a)goal?
Source-derived C1 route: representation controls every later guarantee.

03 · Uninformed search C2

A search node stores a state plus its path, cost and frontier information. The frontier is generated nodes awaiting expansion. Match its ordering to the objective.

Path costg(n) = Σᵢ₌₁ᵏ c(aᵢ)
MethodFrontier ruleUse / qualification
BFSShallowest firstMinimum depth; minimum cost only when step costs equal
DFSDeepest firstDepth order does not optimise total cost
UCSLowest g(n)Minimum total cost under non-negative costs; keep best-known repeated-state cost

Guarantee audit

  • Name the objective: depth or cost.
  • State frontier priority and tie handling.
  • For graph search, say how duplicates are handled.
  • Qualify completeness/optimality by branching and cost assumptions.

If you seevarying edge costs + minimum total cost → UCS, not BFS.

what is minimised?depthBFS / DFS orderpath costUCS: min g
C2 decision rule: frontier priority must express the promised objective.

05 · A-star C3

Evaluationf(n) = g(n) + h(n)
g = incurred path cost; h = estimated remaining cost.

Admissible: h(n) never overestimates true optimal remaining cost. Consistent: heuristic change across an edge does not exceed that edge cost.

Consistency testh(n) ≤ c(n,a,n′) + h(n′)
ClaimWhat to write
Next expansionCompute every frontier f; choose minimum f, not minimum h
OptimalityName heuristic property and graph-search handling conditions
g(n)cost so far+h(n)estimate=f(n)rank
C3 score anatomy; audit h before attaching an optimality claim.

07 · Delete relaxation C4

Delete relaxation ignores action delete effects: achieved facts persist. A relaxed plan solves the easier relaxed problem; a helpful action comes from an early relaxed-plan layer as promising guidance.

FF-style estimatehFF(s) = |RP(s,G)|

Relaxed planning graph

  1. Start from facts true in state s.
  2. Add applicable actions at the next action layer.
  3. Add their positive effects; ignored deletes do not remove facts.
  4. Continue until goals appear or no new facts arise.
  5. Extract a relaxed plan backwards; count its actions.
Useful becauseBlind because
Easier reachability gives search guidanceResource conflict and negative interaction can disappear
Extracted actions suggest promising successorsLow hFF is not an executable real plan

TrapRestore deletes before claiming feasibility, ordering or resource reuse.

factsF₀actionsfactspersistgoals?
C4 flow: ignored deletes make fact layers monotone, not the real world.

08 · Planning answer boundary C1 · C4

Always separate model validity, heuristic estimate and plan executability. An estimate can rank states well while its extracted relaxed sequence fails after real delete effects return.

Prompt verbRequired output
ConstructShow alternating fact/action layers and applicability
ExtractSelect actions supporting goals backwards through layers
EstimateCount extracted relaxed actions: hFF
GuideUse relaxed information to order real search
ValidateRestore deletes and check ordering/resources
HelpfulUse early relaxed-plan actions as guidance, not proof
QualifyIgnored deletes can hide resource conflict and subgoal interaction; a low estimate is not executable

09 · MDP + Bellman C5

An MDP has states, actions, transition probabilities, rewards and discounting. A policy maps states or observations to actions. Bellman reasoning combines immediate reward with discounted continuation value.

Bellman optimalityV*(s) = maxₐ Σs′ P(s′|s,a)[R(s,a,s′) + γV*(s′)]

Compute in this order

  1. For each action, enumerate successors s′.
  2. Inside each successor term add reward + γ continuation.
  3. Weight by P(s′|s,a) and sum.
  4. Only then maximise over actions.
SymbolRoleCommon error
PTransition uncertaintyTaking max over successors instead of probability-weighted sum
RImmediate encoded objectiveAssuming it contains unmodelled safety costs
γDiscountApplying it to the immediate reward
maxₐOptimal action choiceMaximising before the expectation

LimitThe Markov property, reward design, γ and transition model are assumptions. Correct arithmetic cannot repair a bad reward.

state saction a₁action a₂Σ P·(R+γV)
C5 Bellman order: expectation within each action, then maximum.

10 · Value vocabulary C5

Return is realised discounted reward; value is an expectation under a model or policy; optimal value uses the best available continuation. Keep sample, estimate and expectation distinct.

QuantityQuestion answered
Vπ(s)Expected return from s while following π
Qπ(s,a)Expected return after taking a in s, then following π
V*(s)Best expected return achievable from s
π(s)Action selected at state or observation s

Bellman answer audit

11 · MCTS / UCT C6

MCTS repeats selection → expansion → simulation → backup. Each cycle converts a rollout into updated visit counts and return estimates.

UCT selectionUCTᵢ = X̄ᵢ + c√(ln N / nᵢ)
X̄ᵢ = exploitation; bonus = exploration.
Term changesEffect
Higher X̄ᵢRaises evidence-based preference
Lower child visits nᵢRaises exploration bonus
Higher parent visits NRaises pressure to revisit underexplored children
Higher cWeights exploration more strongly

Four-stage answer

  1. Select recursively by UCT.
  2. Expand an available child.
  3. Simulate with the stated rollout policy.
  4. Back up sampled return through visited nodes.

QualificationFinite rollouts, weak simulation policies and variance can make the current leader unreliable.

selectexpandsimulatebackupvisits
C6 cycle: the answer must connect selection score to backed-up evidence.

12 · PUCT guidance C10

PUCT-style scoreScore(s,a) = Q(s,a) + cP(s,a)√N(s)/(1+N(s,a))

A strong prior can lift a little-visited action; additional child visits shrink its bonus. Learned policy and value guidance save search only if model errors remain discoverable under the compute budget.

Evidence sourceRoleFailure check
Q(s,a)Accumulated search evidenceFew or noisy rollouts
P(s,a)Learned policy priorDistribution shift or blind spot
Value networkLeaf outcome estimateCalibration and objective mismatch
Visit countsAllocate future searchCompute concentration
Self-playGenerate learning experienceNarrow coverage or changing opponents

13 · MC, TD and control C7 · C8

Monte Carlo returnGₜ = Σₖ₌₀ᵀ⁻ᵗ⁻¹ γᵏRₜ₊ₖ₊₁

MC waits for an episodic sampled return. TD learning updates before episode end from reward plus a bootstrapped next estimate.

SARSA · on-policyQ(s,a) ← Q(s,a)+α[r+γQ(s′,a′)−Q(s,a)]
a′ is the action actually selected.
Q-learning · off-policyQ(s,a) ← Q(s,a)+α[r+γ maxₐ′Q(s′,a′)−Q(s,a)]
SeeDo
Full episode rewardsDiscount by powers of γ; distinguish one Gₜ sample from expected value
Behaviour next action givenSARSA target uses that a′
Greedy bootstrap requestedQ-learning target uses max next Q
Update arithmeticWrite target → error → α step → new Q
bootstrap target?actual Q(s′,a′)SARSA / on-policymax Q(s′,·)Q-learn / off-policy
C8 split: the next-state bootstrap, not the outer update, distinguishes them.

14 · Policy gradient + control C9 · C11

Policy-gradient identity∇θJ(θ)=E[∇θ log πθ(a|s) Aπ(s,a)]

Actor selects actions; critic estimates value or advantage. Positive advantage drives ascent to increase the sampled action’s log probability locally. Variance, critic bias and unsafe exploration remain.

Best-response conditionuᵢ(aᵢ*,a₋ᵢ*) ≥ uᵢ(aᵢ,a₋ᵢ*) for every aᵢ

Equilibrium or reward optimality does not imply fairness, safety, legality or social desirability. Add stakeholder costs, hard constraints, monitoring, appeal and human intervention.

Bandit → prediction → control

A multi-armed bandit balances uncertain rewards without state transitions. An ε-greedy policy chooses a best action most of the time and a random action with probability ε. Monte Carlo prediction uses complete sampled returns; TD control bootstraps the unknown remainder.

Choice boundaryQuestion to ask
Bandit vs sequentialDo actions change a state that affects future rewards?

Stability and ethics audit

asksia.ai/cheatsheet/unimelb-comp90054
Forward this to your tutorial group · Shared by ________
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Semester 2 examination period, 2–20 November 2026.
COMP90054
Self-test drills · answers must show the path
AI Planning for Autonomy · The University of Melbourne
COMP90054 cheatsheet QR code – open the full COMP90054 study guide
Full guide + ask Sia
about COMP90054
SIDE 2 · DRILLProtocol: cover the answer → compute aloud → reveal → explain the limit4 worked paths + judgement cards

15 · Worked: A-star order C3 WORKED

Q. Frontier nodes have (g,h): (4,5), (7,1), (3,7). Which expands first?

Nodeg+hf
14+59
27+18
33+710
Answer pathCompute all f → compare 9, 8, 10 → expand (7,1).
Then qualify: ordering is not by h alone; final optimality still depends on heuristic and graph-search conditions.

Self-checkCan you state admissibility and consistency without looking?

frontierg4+h5f=9g7+h1f=8g3+h7f=10
C3 worked figure: minimum f expands; yellow marks the chosen node.

16 · Card: search choice C2 WORKED

Q. A graph has unit-depth distractors, widely varying edge costs, and the objective is minimum total cost.

  1. Reject depth as the objective.
  2. Use a priority frontier ordered by g.
  3. For repeats, keep the best-known cost.
  4. State non-negative-cost and finite-branching conditions.

Answer: uniform-cost search. BFS agrees only when step costs are equal.

Changed-case probes

ChangeWhat changes in the answer?
All step costs equalBFS and UCS can return the same minimum-cost result; their frontier keys remain different
Objective becomes fewest actionsDepth, not total cost, controls the choice
Repeated state found more cheaplyUpdate best-known cost and frontier treatment
Goal unreachableTermination/completeness needs finite branching and duplicate handling

Oral trace: say which node leaves the frontier, which successors enter, their new path costs, how repeats are treated, and the resulting order. If you cannot name the next frontier, you have not yet demonstrated the algorithm.

Heuristic property self-test

17 · Worked: Bellman choice C5 WORKED

Q. At state s, A gives reward 2 then value 10; B gives reward 5 then value 5. γ=0.9; transitions are certain.

A2 + 0.9×10 = 11
B5 + 0.9×5 = 9.5

Choose A under this model. The conclusion is conditional on state, reward and transition representation; inspect whether rewards encode all safety costs.

ACTION A2 + .9×1011ACTION B5 + .9×59.5
C5 worked figure: compare complete action values, then audit the model.

18 · Worked: MC return C7 WORKED

Q. From t, rewards are 3, 2, 4 and γ=0.5.

Answer pathGₜ = 3 + 0.5×2 + 0.25×4
= 3 + 1 + 1 = 5

This is one observed episodic return under a policy, not the value itself. More samples are needed to estimate expected value; later rewards carry increasing powers of γ.

19 · Return vs value C7

ObjectMeaning
GₜOne realised discounted sum
Vπ(s)Expected return from s under π
Qπ(s,a)Expected return after a, then π
Sample meanEstimate; sensitive to policy, coverage, rare events and non-stationarity

Change one assumption

If γ changes, recompute each reward’s power; do not reuse the old return. If transitions become uncertain, replace the certain Bellman continuation with a probability-weighted sum. If a reward is terminal, follow the task’s terminal handling rather than bootstrapping blindly. If a safety cost is absent, the larger numeric value is only the winner of the incomplete model.

Policy boundary: every MC sample is generated under a policy. State whose behaviour produced the episode before generalising from its return. Adequate coverage matters because rare but costly outcomes can be missed.

Bellman scratchpad discipline

LineWrite before calculating
1One expression per action

20 · Worked: Q-learning C8 WORKED

Q. Q(s,a)=4, r=3, γ=0.9, max next Q=6, α=0.2.

StageCalculationResult
Target3+0.9×68.4
Error8.4−44.4
Step0.2×4.40.88
Update4+0.884.88

Recitetarget → error → step → new value. This arithmetic alone does not prove convergence.

target8.4error4.4α step0.88new Q4.88
C8 worked update; separating stages prevents sign and target errors.

21 · Card: UCT judgement C6 WORKED

Q. A has mean 0.7 from 40 visits; B has mean 0.5 from 5 visits, under the same parent. No c or parent visit count is supplied.

  1. Write exploitation + exploration terms.
  2. B has the larger exploration bonus because nB is smaller.
  3. Do not invent a numeric winner: it depends on c and N.
  4. Report sensitivity to visit counts and rollout variance.

UCT can select B despite its lower mean. Limited visits make its uncertainty bonus larger.

22 · Card: advantage sign C9 WORKED

Positive estimated advantage → weight the sampled log-policy gradient positively → gradient ascent increases that action’s probability locally. Retain learning-rate, variance and safety controls; critic evidence is not ethical approval.

Target substitution test

Take the Q-learning numbers above but suppose the behaviour policy actually selects a next action valued below the maximum. The Q-learning target remains greedy; the SARSA target substitutes the selected action’s Q. Do not compute a numeric SARSA update unless that selected value is supplied.

Prompt wordingTarget evidence
“next action taken”Use Q(s′,a′): SARSA
“best/max next action”Use max Q(s′,·): Q-learning
“positive advantage”Increase sampled log probability locally
“strong prior, few visits”PUCT exploration may select it

UCT / PUCT retrieval strip

23 · Worked: relaxed plan C4 WORKED

Q. A relaxed solution uses four actions, but two require a reusable tool that the real model consumes.

  1. Report relaxed-plan length 4 as guidance.
  2. Restore the tool’s delete effect.
  3. Recheck ordering and re-acquisition.
  4. Do not execute the relaxed sequence without validation.

Answer: hFF may be 4 while the real plan needs extra actions or is infeasible. The ignored delete creates resource interaction.

RELAXEDtool persists4 actionsguidance onlyREALtool consumedre-acquire?or infeasible
C4 counter-case: delete effects can reverse the feasibility conclusion.

24 · Card: parcel robot C1 WORKED

Scenario. Robot moves among three rooms, carries one parcel and must deliver it without entering a blocked room.

1 mark eachRequired move
RepresentationRobot location, parcel location/carry state, blocked status
ActionsMove, pick, drop with explicit preconditions/effects
BoundaryInitial state and goal
SafetyBlocked-room entry prohibited or safely handled

TrapDo not bolt safety on after finding the nominal route.

25 · Method chooser C1–C10

If the task gives…Start with…
Known deterministic actions/goalsClassical planning/search
Costs + heuristicA-star; audit h
Transition probabilities/rewardsMDP/Bellman
Simulator + planning budgetMCTS/UCT
Episodic samplesMC prediction
Bootstrapped control samplesSARSA or Q-learning
Differentiable parameterised policyPolicy gradient/actor-critic

Integrated delivery-system frame

Answer route: Translate the task into states, actions, effects and goals. Under uncertainty, calculate values and select a policy; with a simulator, apply UCT. Finish by checking model and stakeholder controls.

26 · Judgement cards C2–C10

If you see XDo Y
Minimum depthUse a depth-ordered frontier; do not claim minimum weighted cost
Minimum cost, unequal edgesUse UCS / lowest g
A-star frontierCompute every g+h; expose ties
Heuristic guaranteeTest admissibility/consistency, not usefulness by intuition
Relaxed planUse to guide; restore deletes before execution claims
Stochastic successorsProbability-weight then sum before max
Low child visit countExpect larger UCT/PUCT exploration term
Sampled episodeCompute G; do not call one sample the value
Actual next actionSARSA target
Greedy next estimateQ-learning target
Positive advantageIncrease sampled action’s probability locally
Stable strategic profileTest external stakeholder harm anyway
modelassumechoosemethodshowworkingqualifyclaimcounter-case → revise model or claim
Integrated answer route: model, method, working, boundary, then counter-case.

27 · Classic traps C1–C11

  • State ≠ node: a node also carries path/search information.
  • h ≠ f: A-star ranks g+h.
  • Relaxed ≠ executable: ignored deletes matter.
  • Sum before max: stochastic Bellman backup.
  • Return ≠ value: sample versus expectation.
  • SARSA ≠ Q-learning: actual action versus maximum.
  • Convergence ≠ safety: performance claims stay inside model assumptions.
  • Equilibrium ≠ acceptability: incentives do not certify objectives.

Five-sentence answer skeleton

  1. Model: name state, actions, objective and uncertainty.
  2. Method: select the rule matching that model.
  3. Working: show frontier order, expectation, return or update.
  4. Decision: compare candidates and answer the actual question.
  5. Boundary: state the assumption or failure mode that could change it.

28 · Fast-recall matrix ALL CHAPTERS

MethodCore objectMust qualify
Plannings′=γ(s,a)State completeness; observability
UCSmin gCosts; duplicates; branching
A-starmin g+hAdmissibility/consistency
hFF|relaxed plan|Deletes/resources/interactions
BellmanΣP(R+γV), max aMarkov/reward/model
UCTmean + visit bonusBudget/policy/variance
MCdiscounted episode returnCoverage/non-stationarity
SARSAr+γQ(s′,a′)On-policy behaviour
Q-learningr+γmax Q(s′,·)Exploration/stability
Policy gradientlog-policy gradient × AVariance/bias/locality
PUCTQ + prior-weighted bonusPrior error/compute
Best responseutility under others’ actionsStakeholders and constraints
definecomputechoosequalify+ audit
Recall ladder: a complete response ends with the claim boundary.

29 · Exam-morning plan COURSE STRATEGY

  1. Retrieve: write all core equations from memory.
  2. Discriminate: explain BFS/UCS/A-star; MC/TD; SARSA/Q-learning; UCT/PUCT.
  3. Calculate: redo the four numeric drills without prompts.
  4. Transfer: change one assumption in each and revise the conclusion.
  5. Audit: state one modelling, compute, safety or stakeholder limit per method.

Final oral testFor any scenario: formalise → select → compute → compare → qualify. If one verb is missing, the answer is incomplete.

30 · Ethical-control close C11 WORKED

Routing-game drill: define parties/actions/utilities → test unilateral deviations → add external stakeholder costs and hard constraints → assign monitoring, appeal and intervention responsibility. A stable fleet equilibrium may still externalise unacceptable harm.

Technical resultQuestion still open
Search found a planWere relevant hazards represented?
asksia.ai/cheatsheet/unimelb-comp90054
Forward this to your tutorial group · Shared by ________
AskSiav2 · 2026-10-24
Your exam is closed book: nothing printed goes in. Use this sheet to revise and self-test, not to carry.
Semester 2 examination period, 2–20 November 2026.

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