真题逐题分析 — Complete Exam Analysis

Course: COMPSCI 713: AI Fundamentals, University of Auckland Instructor: Xinyu Zhang (mid-semester) / Thomas (final exam, partial) Scope: ALL available exam papers — S1 2025 Sample, S1 2025 Actual, S1 2026 Sample, S1 2024 Final Purpose: Question-by-question breakdown for exam preparation

How to Use This Document（使用指南）

1. First pass: Skim the tables at the end of each exam section to see topic/mark distribution
2. Second pass: Read the Learning Points for your weakest topics
3. Third pass: Use the Common Mistakes as a self-check before the exam
4. Final review: Jump to Cross-Exam Patterns at the bottom

💡 核心发现: 每一份试卷都考了 Symbolic Logic, LNN, Knowledge Graphs, 和 Decision Trees/Ensembles。这四个是绝对必考项。

Exam Paper 1: S1 2025 Sample Test

Format: 15 marks, 6 questions, 60 minutes (5 reading + 55 answering) Allowed aids: One double-sided handwritten A4 page

Q1 — Symbolic Logic [3 marks]

Question Summary

(a) Propositional Logic — Modus Tollens [~1.5m]

Scenario: A secure facility grants entry only if a person has a valid ID ($I$) AND fingerprint matches ($F$). Rule: $(I \wedge F) \rightarrow E$. Observed: person was NOT granted entry ($\neg E$).

Task: Deduce what must be true about $I$ and $F$.

(b) First-Order Logic [~1.5m]

Task: Write “Not all birds can fly” in FOL using $\text{Fly}(x)$. Give a realistic example.

Expected Answer

(a):

• By Modus Tollens: $(I \wedge F) \rightarrow E$ and $\neg E$ implies $\neg(I \wedge F)$
• By De Morgan: $\neg I \vee \neg F$
• Conclusion: Either the person lacked a valid ID, or the fingerprint didn’t match (or both)

(b):

• FOL: $\neg \forall x , \text{Fly}(x)$, equivalently $\exists x , \neg \text{Fly}(x)$
• Example: “Penguins are birds but cannot fly”

Analysis

Learning Points（学习要点）

• Modus Tollens 是本课程考试的第一推理模式: $(P \rightarrow Q), \neg Q \vdash \neg P$。务必熟练到条件反射
• “Not all” = $\neg \forall x$: 注意不是 $\forall x , \neg$（后者意为“所有都不“，语义完全不同）
• De Morgan 定律: $\neg(A \wedge B) = \neg A \vee \neg B$，推导结论时经常用到

⚠️ Common Mistake: Writing $\forall x , \neg \text{Fly}(x)$ which means “NO bird can fly” — much stronger than “not all birds can fly.”

Q2 — Logic Neural Networks (LNN) [2 marks]

Question Summary

Scenario: Smart home LNN rule: HeatingOn $\leftarrow$ Cold $\otimes$ AtHome (differentiable AND).

(a) Interpret in natural language. How does it differ from Boolean? [1m]

(b) Compute with Cold = 0.9, AtHome = 0.4. Discuss whether heating activates. [1m]

Expected Answer

(a):

• Natural language: “If it is cold AND someone is at home, turn on the heating.”
• Difference: Boolean AND requires both inputs strictly TRUE (1). LNN’s $\otimes$ works with continuous truth values in $[0, 1]$, producing intermediate results that capture partial truth and enable gradient-based learning.

(b):

• Product t-norm: $0.9 \times 0.4 = 0.36$
• Lukasiewicz: $\max(0, 0.9 + 0.4 - 1) = 0.3$
• Whether heating activates depends on threshold: if $\alpha = 0.3$, yes; if $\alpha = 0.7$, no.

Analysis

Learning Points

• 必背三个 t-norm:
  • Product: $a \times b$
  • Lukasiewicz: $\max(0, a + b - 1)$
  • Godel/min: $\min(a, b)$
• Boolean vs LNN 的关键差异: Boolean 是离散的 {0,1}；LNN 是连续的 [0,1]，支持梯度下降

⚠️ Common Mistake: Forgetting to discuss the threshold. Computing 0.36 is not enough — you must state what activation decision follows.

Q3 — Knowledge Graph Embeddings [2 marks]

Question Summary

Explain the role of entity/relation embeddings in KG completion. Introduce a common KG inference task with an example.

Expected Answer

• Embeddings: Map entities and relations to dense vectors in continuous space, enabling mathematical operations for reasoning
• Inference task: Link prediction — given $(h, r, ?)$, predict tail entity
• Example: $(Einstein, bornIn, ?) \rightarrow Germany$

Analysis

Learning Points

• TransE 核心公式: $h + r \approx t$（头实体向量 + 关系向量 ≈ 尾实体向量）
• 三种推理任务: tail prediction $(h, r, ?)$, head prediction $(?, r, t)$, relation prediction $(h, ?, t)$

⚠️ Common Mistake: Confusing “embedding” with “one-hot encoding.” Embeddings are dense, low-dimensional, learned vectors — not sparse indicator vectors.

Exam tip（答题技巧）: 永远给具体例子。“(Einstein, bornIn, ?) → Germany” 远比 “it predicts missing links” 好。

Q4 — Embodied AI / Robot Soccer [2 marks]

Question Summary

Robot soccer league: overhead camera, 225 features per frame, team of 5 robots, no inter-robot communication. Describe strategies/collective behaviours.

Expected Answer

Any of (1 mark each, max 2):

• Collective behaviours: passing strategy, interception prediction, passing point value assessment
• Positioning strategies: formations for attack/defense
• Role-based strategies: dynamic role assignment based on game situation
• Centralized control: overhead camera acts as single controller for all 5 robots

Analysis

Learning Points

• 三大策略类别: (1) 集体行为 — passing, (2) 位置策略 — formation, (3) 角色分配 — dynamic assignment
• 关键细节: No communication → centralized control via overhead camera → single decision-maker

⚠️ Common Mistake: Being too vague. “They work together” = 0 marks. Name specific strategies.

Q5 — Random Forest / Bagging [3 marks]

Question Summary

Dataset with 225 features.

(a) How are features selected per tree? How many? [2m]

(b) Why is feature bagging a good idea? [1m]

Expected Answer

(a):

• Random subset of features sampled per tree (not all 225)
• Typical: $\sqrt{p} = \sqrt{225} = 15$ features per tree
• Different trees see different feature subsets

(b):

• Prevents trees from being highly correlated (e.g., same dominant feature always at root)
• Decorrelated trees → ensemble averaging reduces variance more effectively

Analysis

Learning Points

• $\sqrt{p}$ rule: For $p$ features, sample $\sqrt{p}$ per tree. For 225 features → 15.
• Bagging vs Boosting:
  • Bagging → parallel trees → reduces variance
  • Boosting → sequential trees → reduces bias
• 两层随机化: (1) Bootstrap sampling of data rows, (2) Random sampling of features. Both reduce correlation.

⚠️ Common Mistake: Confusing bootstrap sampling (random data points) with feature bagging (random features). Both happen in Random Forest; they serve different purposes.

Q6 — MYCIN / Expert Systems [3 marks]

Question Summary

Medical diagnosis scenario using backward chaining. Patient has a runny nose. Possible diagnoses: common cold, allergies, measles. Demonstrate backward chaining reasoning.

Expected Answer

• Backward chaining: Start from hypothesis, work backward to check conditions
  1. Hypothesis: Common Cold → needs runny nose ✓, fever ?, cough ?
  2. Hypothesis: Allergies → needs runny nose ✓, sneezing ?, itchy eyes ?
  3. Hypothesis: Measles → needs runny nose ✓, rash ?, high fever ?
• Ask additional questions to discriminate between hypotheses
• Contrast with forward chaining: start from facts, derive conclusions

Analysis

Learning Points

• Backward chaining 三步法: (1) Start with hypothesis, (2) Check conditions, (3) Ask for missing info
• Forward vs Backward: Forward = data-driven (fact → conclusion); Backward = goal-driven (hypothesis → verify)
• MYCIN 特色: Uses certainty factors (CF) instead of probabilities; backward chaining for diagnosis

⚠️ Common Mistake: Describing forward chaining when asked for backward chaining. Direction matters!

S1 2025 Sample Test — Summary Table

Exam Paper 2: S1 2025 Actual Test

Format: 15 marks, 6 questions, 60 minutes This is the REAL exam that was sat

Q1 — Symbolic Logic [2 marks]

Question Summary

(a) Given $(P \vee Q) \rightarrow R$ and $\neg R$. Apply Modus Tollens. [1m]

(b) Given $\forall x(\text{Cheat}(x) \rightarrow \text{Disqualified}(x))$ and Alice is not disqualified. Conclude about Alice. [1m]

Expected Answer

(a):

• Modus Tollens: $(P \vee Q) \rightarrow R$ and $\neg R$ implies $\neg(P \vee Q)$
• By De Morgan: $\neg P \wedge \neg Q$
• Both P and Q must be false

(b):

• Universal instantiation: $\text{Cheat}(\text{Alice}) \rightarrow \text{Disqualified}(\text{Alice})$
• Given $\neg \text{Disqualified}(\text{Alice})$, by Modus Tollens: $\neg \text{Cheat}(\text{Alice})$
• Conclusion: Alice did not cheat

Analysis

Learning Points

• 这道题和 Sample 的区别: Sample 用 $(I \wedge F) \rightarrow E$，Actual 用 $(P \vee Q) \rightarrow R$。结论不同！
  • $\neg(A \wedge B) = \neg A \vee \neg B$（至少一个为假）
  • $\neg(A \vee B) = \neg A \wedge \neg B$（两个都假）
• FOL + Modus Tollens 组合拳: Universal instantiation 先把 $\forall x$ 具体化为 Alice，再用 Modus Tollens

⚠️ Common Mistake: For $\neg(P \vee Q)$, some students write “$P$ or $Q$ is false” — this is WRONG. BOTH must be false. De Morgan on disjunction gives conjunction of negations.

🔑 关键对比: AND 的否定 → 至少一个假 (disjunction); OR 的否定 → 全部假 (conjunction). 这是必须刻在脑子里的。

Q2 — LNN with Truth Bounds [3 marks]

Question Summary

Scenario: Autonomous vehicle collision alert system. Two sensors:

• Pedestrian detector: $P$ with bounds $[L_P, U_P] = [0.8, 0.9]$
• Obstacle detector: $Q$ with bounds $[L_Q, U_Q] = [0.3, 0.6]$

Rule: Alert $\leftarrow P \vee Q$ (disjunction, not conjunction!)

Alert threshold: $\alpha = 0.7$

(a) Determine alert status [2m]

(b) Why are bounds (instead of point estimates) useful in safety-critical applications? [1m]

Expected Answer

(a):

• Co-norm for OR (using Lukasiewicz):

  • Lower bound: $\min(1, L_P + L_Q) = \min(1, 0.8 + 0.3) = 1.0$
  • Upper bound: $\min(1, U_P + U_Q) = \min(1, 0.9 + 0.6) = 1.0$

• OR result bounds: $[1.0, 1.0]$

• Since lower bound $1.0 \geq \alpha = 0.7$: Alert ACTIVATES

  Alternative (product-based co-norm):

  • $P \vee Q = P + Q - P \cdot Q$
  • Lower: $0.8 + 0.3 - 0.8 \times 0.3 = 0.86$
  • Upper: $0.9 + 0.6 - 0.9 \times 0.6 = 0.96$
  • Bounds: $[0.86, 0.96]$, both $\geq 0.7$: Alert ACTIVATES

(b):

• Bounds capture epistemic uncertainty — we know the truth value lies somewhere in the interval
• In safety-critical systems, we can make conservative decisions: if even the lower bound exceeds threshold, we act
• Point estimates hide uncertainty; bounds let us reason about worst-case scenarios

Analysis

Learning Points

• 这是 LNN 的升级版考法: Sample 考 AND 的点值计算，Actual 考 OR 的区间计算
• AND vs OR t-norm/co-norm:
  • AND (t-norm): Product → $a \times b$; Lukasiewicz → $\max(0, a + b - 1)$
  • OR (co-norm): Product → $a + b - a \times b$; Lukasiewicz → $\min(1, a + b)$
• Safety-critical reasoning: 用 lower bound 做决策 = 最保守策略

⚠️ Common Mistake: Using AND formula when the question says OR! Read the operator carefully: $\otimes$ = AND, $\oplus$ = OR, $\vee$ = OR.

⚠️ 另一个常见错误: 忘记 bounds 是区间运算。不能只算一个值，要算 [lower, upper]。

Q3 — Knowledge Graphs / TransE [2 marks]

Question Summary

(a) Explain the TransE embedding model [1m]

(b) Write the TransE scoring function [1m]

Expected Answer

(a):

• TransE represents entities and relations as vectors in the same space
• Core idea: for a true triple $(h, r, t)$, the head plus relation should approximate the tail: $h + r \approx t$

(b):

• Scoring function: $f(h, r, t) = |h + r - t|$ (L1 or L2 norm)
• Lower score = more likely to be true
• Training: minimize score for true triples, maximize for false (negative sampling)

Analysis

Learning Points

• TransE 必背: $f(h,r,t) = |h + r - t|$，越小越可能是真三元组
• 与 Sample 的区别: Sample 考概念层面（什么是 embedding），Actual 考公式层面（TransE 具体怎么算）
• 局限性: TransE 无法建模 1-to-N 关系（如一个国家有多个城市）

⚠️ Common Mistake: Writing $h + r = t$ (equality) instead of $h + r \approx t$ (approximation). The model learns to minimize the distance, not enforce exact equality.

Q4 — Decision Trees / CART [2 marks]

Question Summary

What does “greedy” mean in the context of CART (Classification and Regression Trees)?

Expected Answer

• Greedy = at each node, CART picks the locally optimal split (maximum information gain or minimum Gini impurity) without considering future splits
• It does not evaluate all possible tree structures to find the global optimum
• This makes it computationally efficient but potentially suboptimal
• Why greedy? Finding the optimal tree is NP-hard

Analysis

Learning Points

• “Greedy“三要素: (1) 每步选当前最优 (2) 不回溯 (3) 不保证全局最优
• 为什么接受 greedy?: 找最优树是 NP-hard；greedy 在实践中效果够好
• Ensemble 弥补 greedy: Random Forest 通过多棵 greedy tree 的聚合来逼近更好的解

⚠️ Common Mistake: Saying greedy means “fast.” Greedy is about the optimization strategy (local vs global), not speed.

Q5 — Fuzzy Logic [3 marks]

Question Summary

Contrast traditional (Boolean) logic vs fuzzy logic for the rule: IF athlete is STRONG AND athlete is HEAVY THEN athlete is HAMMER_THROWER

Expected Answer

Traditional Logic:

• STRONG = {yes, no}, HEAVY = {yes, no} → HAMMER_THROWER = {yes, no}
• Sharp boundaries: an athlete is either strong or not
• AND = Boolean AND: both must be true for conclusion to hold

Fuzzy Logic:

• STRONG(x) ∈ [0, 1], HEAVY(x) ∈ [0, 1] → HAMMER_THROWER(x) ∈ [0, 1]
• Gradual membership: “somewhat strong” = 0.6, “very heavy” = 0.9
• AND = t-norm (e.g., min): HAMMER_THROWER ≥ min(0.6, 0.9) = 0.6
• Captures vagueness — no sharp cutoff between “strong” and “not strong”

Analysis

Learning Points

• 对比答题模板: 分三行写 — (1) Traditional: binary, (2) Fuzzy: continuous, (3) WHY fuzzy is better for this case
• Fuzzy logic 解决 vagueness: “Strong” 没有明确边界 → 需要 membership function
• 给具体数字: 说 “STRONG(athlete) = 0.6” 比抽象描述好得多

⚠️ Common Mistake: Confusing fuzzy logic with probability. Fuzzy = degree of membership (to what extent is this athlete “strong”?). Probability = likelihood of an event (what’s the chance this athlete wins?).

Q6 — GA / Embodied AI [3 marks]

Question Summary

Design a fitness function for a BigDog walking robot using Genetic Algorithm optimization.

Expected Answer

Fitness function components:

1. Distance traveled (primary): $f_1 = d / d_{max}$ — further is better
2. Stability (constraint): $f_2 = 1 - \text{angular_deviation} / \text{max_deviation}$ — less wobble is better
3. Energy efficiency (secondary): $f_3 = 1 - E_{used} / E_{max}$ — less energy is better
4. Penalty: $f_{penalty} = -C$ if robot falls

Combined: $F = w_1 f_1 + w_2 f_2 + w_3 f_3 + f_{penalty}$

Key design considerations:

• Must balance multiple objectives
• Weights reflect priority (distance > stability > efficiency typically)
• Penalties for catastrophic failure (falling) should be large

Analysis

Learning Points

• Fitness function 设计万能框架: (1) 定义主目标, (2) 加约束, (3) 加惩罚项, (4) 用加权求和合并
• 开放题没有唯一答案: 关键是逻辑自洽 + 覆盖关键方面
• 必须提到权衡: 速度 vs 稳定性 vs 能耗

⚠️ Common Mistake: Only considering one objective (e.g., just distance). Real fitness functions must balance multiple competing goals.

S1 2025 Actual Test — Summary Table

Exam Paper 3: S1 2026 Sample Test

Format: 20 marks, 6 questions, 60 minutes (5 reading + 55 answering) Note: Marks increased from 15 → 20. Same topics, more depth required.

Q1 — Symbolic Logic [5 marks]

Question Summary

(a) Propositional Logic — with Truth Table [3m]

Same scenario as S1 2025 Sample: $(I \wedge F) \rightarrow E$, given $\neg E$. But now explicitly requires a truth table for full marks.

(b) FOL — Birds [2m]

Same “not all birds can fly” question.

Expected Answer

(a):

Step 1: Truth table for $X \rightarrow E$ where $X = I \wedge F$:

When $E = 0$ and implication is TRUE: $X$ must be 0. [1 mark]

Step 2: Truth table for $I \wedge F$:

$I \wedge F = 0$ when at least one is 0. [1 mark]

Step 3: Conclusion — person either lacked valid ID, or fingerprint didn’t match, or both. [1 mark]

(b):

• $\neg \forall x , \text{Fly}(x)$ [1 mark]
• Example: penguins, ostriches, kiwi (kiwi 特别适合 UoA 的语境!) [1 mark]

Analysis

Learning Points

• 2026 版本更重视过程: 3 marks for truth table vs 2025’s ~1.5 marks. Show ALL steps.
• 真值表是得分保障: 即使你能直接用 Modus Tollens 推出结论，画真值表拿更稳的分

💡 策略提示: 5 marks = 25% of total. Spend proportional time (~14 minutes). Don’t rush the truth table.

Q2 — LNN [4 marks]

Question Summary

Same HeatingOn scenario as S1 2025 Sample, but 4 marks (was 2).

(a) Interpret rule + compare with Boolean [2m]

(b) Compute with Cold = 0.9, AtHome = 0.4 [2m]

Expected Answer

Same as S1 2025 Sample Q2 but more detail expected for the extra marks:

• (a): Need deeper comparison — mention gradient-based learning, continuous optimization, partial truth
• (b): Show at least two t-norms, discuss threshold selection, explain practical implications

Analysis

Learning Points

• More marks = more depth expected:
  • 2m version: basic computation + brief threshold mention
  • 4m version: multiple t-norms + threshold discussion + why LNN matters for AI
• 安全策略: 写出所有你知道的 t-norm 计算结果，对比它们

Q3 — Knowledge Graph Embeddings [2 marks]

Question Summary

Same as S1 2025 Sample Q3: explain entity/relation embeddings + KG inference task + example.

Expected Answer

Identical to S1 2025 Sample Q3. (See above.)

Learning Points

• 三年不变: 这道题完全一样。说明 KG embedding 是必考的固定题型。

Q4 — Robot Soccer [2 marks]

Question Summary

Same as S1 2025 Sample Q4: overhead camera, 225 features, describe strategies.

Expected Answer

Identical to S1 2025 Sample Q4. (See above.)

Learning Points

• 同样三年不变: Robot Soccer 策略也是固定考点。

Q5 — Random Forest [3 marks]

Question Summary

Same as S1 2025 Sample Q5: feature selection + why feature bagging.

Expected Answer

Identical to S1 2025 Sample Q5. (See above.)

Q6 — Vagueness vs Uncertainty [4 marks]

Question Summary

New question type (not in S1 2025 Sample):

Classify 4 scenarios:

1. Patient described as “high risk” → Vagueness
2. Security system estimates burglary → Uncertainty
3. Student rated “almost excellent” → Vagueness
4. Spam filter classifies email → Uncertainty

Expected Answer

Analysis

Learning Points

• 万能判断法则:
  • Vagueness → “To what degree?” → Fuzzy Logic (membership functions)
  • Uncertainty → “How likely?” → Bayesian Reasoning (probabilities)
• 快速测试: 概念边界模糊 → vagueness; 世界状态未知 → uncertainty
• 语言线索: “high”, “almost”, “kind of” → vagueness; “estimate”, “predict”, “classify” → uncertainty

⚠️ Common Mistake: Thinking fuzzy logic handles uncertainty. NO — fuzzy handles vagueness; Bayes handles uncertainty. This is THE most important distinction in W5.

S1 2026 Sample Test — Summary Table

Exam Paper 4: S1 2024 Final Exam (Thomas’s Section)

Note: This is the final exam (not mid-semester), with questions from a different instructor (Thomas). These topics may or may not appear in 2026’s mid-semester, but they are useful for final exam preparation and general knowledge.

Q1 — Continual Learning [4 marks]

Question Summary

Concept drift, replay methods, Gaussian Mixture Models in continual learning.

Expected Answer

• Concept drift: Data distribution changes over time; model must adapt
• Replay methods: Store subset of old data; replay during training on new data to prevent catastrophic forgetting
• GMM: Can be used to model data distributions; detect drift by comparing distributions
• Stability-plasticity tradeoff: Too much stability → can’t learn new; too much plasticity → forgets old

Analysis

Q2 — BFS vs UCS [3 marks]

Question Summary

Compare Breadth-First Search and Uniform-Cost Search.

Expected Answer

• BFS: Expands shallowest node first; optimal when all edge costs equal; uses FIFO queue
• UCS: Expands lowest-cost node first; optimal for any non-negative costs; uses priority queue
• Key difference: BFS = optimal for unweighted; UCS = optimal for weighted graphs

Analysis

Q3 — MCTS / UCB1 [3 marks]

Question Summary

Explain the components of the UCB1 formula used in Monte Carlo Tree Search.

Expected Answer

$$UCB1 = \bar{X}_j + C \sqrt{\frac{\ln N}{n_j}}$$

• $\bar{X}_j$: average reward of node $j$ (exploitation term)
• $N$: total visits to parent
• $n_j$: visits to node $j$
• $C$: exploration constant
• $\sqrt{\ln N / n_j}$: exploration term — favors less-visited nodes
• Balances exploration vs exploitation

Analysis

Q4 — RL for Pac-Man [1 mark]

Question Summary

Define state, action, policy, reward for RL applied to Pac-Man.

Expected Answer

• State: Current game board configuration (ghost positions, pellet locations, Pac-Man position)
• Action: Move direction (up, down, left, right)
• Policy: Mapping from state to action (which direction to move in each situation)
• Reward: +10 eating pellet, +200 eating ghost, -500 dying, -1 per time step

Analysis

Q5 — GNN [2 marks]

Question Summary

Explain permutation invariance and permutation equivariance in Graph Neural Networks.

Expected Answer

• Permutation invariance: Output doesn’t change when node ordering changes (graph-level prediction)
• Permutation equivariance: Output permutes consistently with input permutation (node-level embeddings)

Analysis

Q6 — Self-Supervised Learning [2 marks]

Question Summary

Distinguish pretext tasks from downstream tasks in self-supervised learning.

Expected Answer

• Pretext task: Artificial task designed to learn representations without labels (e.g., predict rotation, fill masked words)
• Downstream task: Actual target task the representations are used for (e.g., classification, NER)
• Relationship: Pretext → learn general features; fine-tune on downstream task with few labels

Analysis

Additional Topics from S1 2024 Final (Answer Key)

The following topics appeared in the S1 2024 final exam answer key:

Cross-Exam Patterns（跨卷规律总结）

Pattern 1: Repeated Questions（原题重复出现）

以下题目在多份试卷中几乎一模一样地出现：

💡 核心发现: Xinyu 喜欢在 Sample Test 和 Actual Test 之间做微调而非大改。Sample 就是 Actual 的预告片！

Pattern 2: Question Evolution（题目进化路径）

每个核心考点在不同年份有“升级版“:

Symbolic Logic 进化链:

S1 2025 Sample: (I∧F)→E, ¬E → 推理（无真值表要求）
S1 2025 Actual: (P∨Q)→R, ¬R → 推理 + FOL 组合
S1 2026 Sample: 同 2025 Sample 但要求画真值表，5 marks

→ 趋势: 从 “能推理” → “能推理 + 证明过程” → “能推理 + 证明 + 变体”

LNN 进化链:

S1 2025 Sample: 点值计算 (AND), 2 marks
S1 2025 Actual: 区间计算 (OR) + safety reasoning, 3 marks
S1 2026 Sample: 点值计算 (AND), 4 marks (deeper explanation)

→ 趋势: AND 和 OR 交替考，区间 vs 点值交替考

KG 进化链:

S1 2025 Sample: "Explain embeddings" (概念)
S1 2025 Actual: "Write TransE formula" (公式)
S1 2026 Sample: "Explain embeddings" (概念, 同2025 Sample)

→ 趋势: 概念和公式交替考。两手都要准备。

Pattern 3: Mark Allocation Trends

Key insight: Symbolic Logic + LNN consistently take 35-45% of total marks. These two topics alone are worth nearly half the exam.

Pattern 4: Cognitive Level Distribution

Topic Priority Matrix for 2026 Mid-Semester（2026 期中复习优先级）

Based on cross-exam analysis, here is the definitive priority ranking:

Exam Strategy Recommendations（应试策略建议）

Time Management（时间分配）

For a 20-mark, 55-minute exam:

• ~2.75 minutes per mark
• Q1 (5m): ~14 minutes
• Q2 (4m): ~11 minutes
• Q3 (2m): ~5.5 minutes
• Q4 (2m): ~5.5 minutes
• Q5 (3m): ~8 minutes
• Q6 (4m): ~11 minutes

Cheatsheet Priorities（A4 速查表优先写什么）

Your double-sided A4 page should include (in order of priority):

1. Truth table templates — implication, AND, OR truth tables pre-drawn
2. Modus Tollens + De Morgan — write the formulas
3. T-norm / Co-norm formulas — all 3 variants for AND and OR
4. LNN bounds computation — interval arithmetic rules
5. TransE formula — $f(h,r,t) = |h + r - t|$
6. $\sqrt{p}$ formula — for Random Forest feature bagging
7. Vagueness vs Uncertainty — decision table with examples
8. Backward vs Forward chaining — one-line definitions
9. Fitness function template — multi-objective weighted sum
10. Key FOL patterns — $\neg \forall x = \exists x \neg$

Answer Writing Tips（答题技巧）

1. Show your work: 2026 version gives more marks for process (truth tables, step-by-step computation)
2. Give concrete examples: “(Einstein, bornIn, ?) → Germany” > “it predicts missing links”
3. Use the scenario: Refer back to the specific context (smart home, autonomous vehicle, etc.)
4. Label your steps: “Step 1: … Step 2: … Therefore: …”
5. Quality over quantity: The exam explicitly states this. Be concise and precise.
6. When asked “why”: Give the mechanism, not just the outcome. “Feature bagging decorrelates trees, making ensemble averaging more effective at reducing variance.”

Appendix: Complete Question Index（完整题目索引）

For quick reference, every question across all papers: