COMP5318 Practice Exam 1 整卷逐题解析

Posted on 2026-11-03 In Course Notes , COMP5318 Word count in article: 7.6k Reading time ≈ 27 mins.

COMP5318 Semester 2 2026 官方 Practice Exam 1（Paper 1）整卷逐题解析：30 道选择题（kNN 预处理与加权投票、PRISM、MAE/MSE、L1 稀疏、最小二乘、 Laplace 平滑、调参与测试集、精确率召回率、信息增益、Bagging/Boosting、SVM 间隔与核、PCA、ReLU、感知机更新、CNN 参数、RNN、Dropout、 K-means、GMM、Viterbi、马尔可夫链、DQN 经验回放与目标网络、Q-learning）和 4 道计算题（信息增益、前向传播、K-means 一轮、HMM 前向算法）。答案以官方 Solutions 为准，附解题过程、错误选项分析与对应周次。

COMP5318 Practice Exam 1 整卷逐题解析

课程：COMP4318/COMP5318 — Machine Learning and Data Mining 材料：Exam Prep/COMP5318_Practice_Exam_1.pdf（题目，20 页）+ COMP5318_Practice_Exam_1_Solutions.pdf（官方答案与评分指引，10 页） 答案来源：所有答案都按官方 Solutions 给出。「为什么其他选项错」和知识点归属是我补充的。形式：2 小时 + 10 分钟阅读，闭卷，可用非编程计算器；Part A 30 道选择题 × 2 分 = 60 分，Part B 4 道计算题 × 10 分 = 40 分。

答案速查（官方）

Q	答	Q	答	Q	答	Q	答	Q	答
1	A	7	C	13	C	19	C	25	A
2	C	8	B	14	D	20	B	26	A
3	A	9	D	15	D	21	A	27	B
4	B	10	A	16	B	22	C	28	D
5	D	11	A	17	A	23	D	29	C
6	C	12	B	18	D	24	B	30	B

按周分布：W01 ×1、W02 ×2、W03 ×3、W04 ×3、W05 ×4（含规则学习 Q4）、W06 ×4、W07 ×3、W08 ×3、W09 ×2、W11 ×2、W12 ×3；Part B 覆盖 W05、W07、W09、W11。W10 Transformer 本卷没考。

Part A：选择题

Q1 学习任务类型 → A

Original question (English) · [2 marks] A property service uses past sales with known prices to predict the sale price of a new home. Which learning task is this?

A. Supervised regression.

B. Supervised classification.

C. Reinforcement learning.

D. Unsupervised clustering.

题干：用已知成交价的历史房屋数据预测新房价格。

答案 A 监督回归：有标签（价格已知）→ 监督；目标是连续数值 → 回归。
B 分类要求离散类别；C 强化学习没有「带标签样本」，靠奖励交互；D 聚类没有目标标签。
知识点：W01 机器学习任务分类

English answer (Official): A — Learning tasks. The target is a known numeric price. Learning from labelled examples to predict a continuous target is supervised regression.

Q2 kNN 流水线里标准化参数从哪来 → C

Original question (English) · [2 marks] A k-NN pipeline uses a standard scaler. When evaluating the pipeline on a held-out test set, how should the scaling parameters be obtained?

A. Estimate them from the combined training and test sets.

B. Standardise each test example separately using its own feature values.

C. Estimate means and standard deviations from the training set, then apply them to both sets.

D. Estimate them from the test set, then apply them to the training set.

答案 C：在训练集上估计均值和标准差，再用同一组参数变换训练集和测试集。
A、D 用到了测试集统计量 → 数据泄漏；B 每个测试样本用自己的值标准化，变换后特征没有意义（单个样本的「标准差」无从谈起），而且与训练集尺度不一致。
知识点：W02 标准化与 kNN、W04 评估流程

English answer (Official): C — k-NN preprocessing. Preprocessing must be fitted using training data only. Test-set statistics would leak information into model development.

Q3 不加权 vs 反距离平方加权投票 → A

Original question (English) · [2 marks] The three selected neighbours have (class, distance) pairs (Red, 2), (Blue, 3) and (Blue, 4). What do unweighted voting and inverse-square weighted voting predict, respectively?

A. Unweighted Blue; weighted Red.

B. Both Blue.

C. Unweighted Red; weighted Blue.

D. Both Red.

三个近邻 (Red, 2)、(Blue, 3)、(Blue, 4)。

不加权：Red 1 票，Blue 2 票 → Blue。
加权（权重）：Red ；Blue → Red。
所以「不加权 Blue、加权 Red」= A。离得最近的那个点在加权后权重翻盘，这是题目的设计点。
知识点：W02 Weighted Nearest Neighbour

图解

flowchart LR
  N(("新样本"))
  N -->|"d=2, w=1/4=0.25"| R1["Red"]
  N -->|"d=3, w=1/9≈0.111"| B1["Blue"]
  N -->|"d=4, w=1/16≈0.063"| B2["Blue"]
  R1 --> V1["不加权: Red 1 票, Blue 2 票 → Blue"]
  B1 --> V1
  B2 --> V1
  R1 --> V2["加权: Red 0.25 > Blue 0.174 → Red"]
  B1 --> V2
  B2 --> V2

English answer (Official): A — Weighted k-NN. Unweighted votes are Red = 1 and Blue = 2. Weighted votes are Red = 1/4 = 0.25 and Blue = 1/9 + 1/16 ≈ 0.174, so the weighted prediction is Red.

Q4 PRISM 为规则再加一个条件的目的 → B

Original question (English) · [2 marks] While growing a PRISM rule for class Yes, the current rule covers both Yes and No examples. What is the purpose of adding a further condition?

A. Make the rule cover every training example.

B. Restrict the covered set to improve its purity for class Yes.

C. Force the rule to include every available attribute.

D. Turn each Yes example into a separate output class.

答案 B：每加一个条件就是在合取式上多一个限制，缩小覆盖集，提高对 Yes 类的纯度（把混进来的 No 排除掉）。
A 正好相反（加条件只会让覆盖变少）；C PRISM 不要求用上所有属性，纯了就停；D 胡扯。
知识点：规则学习（1R / PRISM）。周总结里没有单独成节，最接近的是 W05 决策树（树的每条路径就是一条合取规则）。

English answer (Official): B — PRISM. A conjunction adds a restriction. PRISM seeks a rule whose covered examples are pure for the target class; adding conditions can remove covered negatives.

Q5 降低单个大残差影响 → D

Original question (English) · [2 marks] A regression model has one unusually large residual. Which change would reduce the relative influence of that residual on the reported loss?

A. Use MSE instead of MAE.

B. Square the residuals twice before averaging.

C. Report the largest squared residual as the loss.

D. Use MAE instead of MSE.

答案 D 用 MAE 代替 MSE：MAE 随残差线性增长，MSE 平方增长，离群点在 MSE 里被放大。
A 方向反了；B 四次方更放大；C 只看最大残差，离群点直接决定整个损失。
知识点：W03 三类误差

English answer (Official): D — Regression losses. MAE grows linearly with residual magnitude, whereas MSE grows quadratically. The outlier therefore has less relative influence under MAE.

Q6 L1 让多个系数恰好为 0 → C

Original question (English) · [2 marks] Two otherwise comparable linear models are fitted. One uses an L1 penalty and gives several coefficients exactly equal to zero. What property does this illustrate?

A. L1 regularisation requires all features to remain in the fitted model.

B. L1 regularisation always improves test accuracy.

C. L1 regularisation can produce sparse coefficient vectors.

D. L1 regularisation removes the need for choosing a penalty strength.

答案 C：L1（Lasso）能产生稀疏系数向量，相当于自动做特征选择。
A 反了；B 正则化不保证提高测试精度；D 惩罚强度 λ 仍需调（交叉验证）。
知识点：W03 正则化

English answer (Official): C — Regularisation. The L1 penalty can set coefficients exactly to zero. Its strength still needs to be selected, and better test performance is not guaranteed.

Q7 最小二乘截距与预测 → C

Original question (English) · [2 marks] For a simple least-squares fit, the mean of x is 3, the mean of y is 7, Sxy = 10 and Sxx = 5. What are the intercept and the prediction at x = 5?

A. Intercept 1; prediction 9.

B. Intercept 7; prediction 17.

C. Intercept 1; prediction 11.

D. Intercept 2; prediction 12.

答案 C（截距 1，预测 11）。A 的 9 是算成了之类；B、D 截距错。
知识点：W03 最小二乘

English answer (Official): C — Least squares. The slope is 10/5 = 2. The intercept is 7 - 2×3 = 1. At x = 5, the prediction is 1 + 2×5 = 11.

Q8 Laplace 平滑解决什么 → B

Original question (English) · [2 marks] A binary feature combination never occurs with class No in a small training sample. Without smoothing, one estimated likelihood is zero. What problem can Laplace smoothing address?

A. The classifier otherwise cannot accept categorical features.

B. A single unseen feature value otherwise makes the product likelihood zero for that class.

C. The class priors must otherwise sum to more than one.

D. Smoothing otherwise has to delete every correlated feature.

答案 B：朴素贝叶斯把各特征似然相乘，只要一个似然为 0，整类得分就是 0（零频问题）。平滑给未见取值一个非零估计。
A 朴素贝叶斯本来就能处理类别特征；C 先验之和永远是 1；D 平滑与删特征无关。
知识点：W04 零频问题

English answer (Official): B — Naive Bayes. Naive Bayes multiplies feature likelihoods. A zero factor makes the class score zero; smoothing supplies nonzero estimates for unseen categorical values.

Q9 调树深度同时保留独立的性能估计 → D

Original question (English) · [2 marks] A team wants to choose the maximum depth of a decision tree while retaining an independent estimate of final performance. Which procedure is appropriate?

A. Choose the depth that maximises test accuracy, and report that same accuracy.

B. Use the deepest tree because training accuracy is the final performance estimate.

C. Merge test labels into the training set before selecting depth.

D. Tune depth on validation data or training-set cross-validation, then evaluate once on the reserved test set.

答案 D：用验证集或训练集上的交叉验证选深度，最后在保留的测试集上只评估一次。
A 用测试集调参再报告同一个分数 → 乐观偏差；B 训练精度不是泛化估计；C 把测试标签并入训练，测试集就不独立了。
知识点：W04 评估流程

English answer (Official): D — Evaluation. Validation or training-set cross-validation supports model selection. An untouched test set is used once to estimate the selected model’s performance.

Q10 精确率与召回率 → A

Original question (English) · [2 marks] A binary classifier has TP = 12, FP = 4, FN = 8 and TN = 26. What are positive-class precision and recall?

A. Precision 0.75; recall 0.60.

B. Precision 0.24; recall 0.76.

C. Precision 0.60; recall 0.75.

D. Precision 0.75; recall 0.68.

答案 A。C 把两者颠倒了；TN 不进入这两个指标。
知识点：W04 性能度量

English answer (Official): A — Classification metrics. Precision = 12/(12+4) = 0.75. Recall = 12/(12+8) = 0.60. True negatives do not enter these two positive-class measures.

Q11 两个划分比较 → A

Original question (English) · [2 marks] Two candidate splits are applied to the same decision-tree parent node. Split X has lower size-weighted child entropy than split Y. What follows?

A. Split X has higher information gain.

B. The two splits must have equal information gain.

C. Split Y has higher information gain.

D. The parent entropy must be zero.

，父节点熵固定。加权子节点熵越小，减掉的越少，增益越大 → X 的 IG 更高，A。
B、C 与推理矛盾；D 父熵是否为 0 无法推出（为 0 时两个划分都没有增益）。
知识点：W05 熵与信息增益

English answer (Official): A — Decision trees. Information gain subtracts weighted child entropy from a fixed parent entropy. A smaller subtracted quantity gives a larger gain.

Q12 Bagging 的定义性操作 → B

Original question (English) · [2 marks] What is the defining training-data operation used in bagging?

A. Use the test set as an additional bootstrap training sample.

B. Fit base learners on bootstrap samples drawn with replacement.

C. Partition the features once and assign exactly one feature to each learner.

D. Fit each learner only on the examples misclassified by the previous learner.

答案 B：在有放回抽样的 bootstrap 样本上训练各个基学习器。
A 用测试集训练是泄漏；C 是一种（不常见的）随机子空间变体，不是 bagging 的定义；D 是 boosting 的思路。
知识点：W05 Bagging

English answer (Official): B — Bagging. Bagging trains base learners on bootstrap samples of the training observations. Sampling with replacement creates varied datasets.

Q13 平方损失下梯度提升树的训练目标 → C

Original question (English) · [2 marks] Under squared-error loss, a later regression tree is trained to correct what the current ensemble has not explained. What are the appropriate training targets for this later tree?

A. Current ensemble predictions minus the observed targets.

B. The current ensemble predictions themselves.

C. Observed targets minus the current ensemble predictions.

D. The original targets without accounting for current predictions.

答案 C 残差 ：平方损失的负梯度正好是「观测值 − 当前预测」，新树去拟合还没解释的部分。
A 符号反了（拟合它再相加会把误差放大）；B、D 没有利用当前误差。
知识点：W05 Boosting

English answer (Official): C — Gradient boosting. For squared-error gradient boosting, the residual target is observed value minus current prediction, so the learner aims to correct the remaining error.

Q14 两个支持平面之间的距离 → D

Original question (English) · [2 marks] A hard-margin SVM uses support planes wᵀx + b = +1 and wᵀx + b = -1, with w = (3, 4). What is the full distance between the planes?

A. 0.20.

B. 2.50.

C. 5.00.

D. 0.40.

答案 D。A 的 0.20 是决策面到一侧支持平面的距离，题目问的是两平面之间的全宽。B、C 是把放到分子上了。
知识点：W06 间隔与支持向量

图解

flowchart LR
  H1["支持平面 w·x+b=+1"] ---|"1/‖w‖ = 0.2"| H["决策面 w·x+b=0"]
  H ---|"1/‖w‖ = 0.2"| H2["支持平面 w·x+b=−1"]
  H1 -.->|"全宽 2/‖w‖ = 0.4（答案）"| H2

English answer (Official): D — SVM margin. The norm is √(3²+4²) = 5. The full margin width is 2/5 = 0.40; 1/5 is the distance from the decision plane to one support plane.

Q15 核 SVM 为什么不必显式存储映射后的向量 → D

Original question (English) · [2 marks] Why can a kernel SVM operate without explicitly storing each transformed feature vector?

A. Every nonlinear kernel creates a linear boundary in the original input space.

B. The kernel replaces the class labels with distances.

C. The kernel removes all dependence on training observations.

D. Its optimisation and predictions can use inner products supplied by the kernel.

答案 D：对偶形式的优化和预测只用到内积，核函数直接给出（核技巧）。
A 错：非线性核在原空间里给的是非线性边界；B 核不替换标签；C 预测仍依赖支持向量（训练样本）。
知识点：W06 核技巧

English answer (Official): D — Kernel methods. A kernel computes the inner product of implicit feature mappings. The SVM can use these values without forming the mapped vectors explicitly.

Q16 PCA 前缩放会改变第一主成分 → B

Original question (English) · [2 marks] PCA is applied to two measurements recorded in very different units. Why might scaling the measurements before PCA change the first component?

A. PCA uses class labels to choose which unit should dominate.

B. PCA is driven by variance, and changing units changes relative feature variances.

C. Scaling converts PCA from an unsupervised method into a supervised method.

D. PCA always gives the same direction under arbitrary feature rescaling.

答案 B：PCA 由方差驱动，换单位就改变各特征的相对方差，数值大的特征会主导第一主成分。
A、C PCA 是无监督的，不用标签；D 正好是错的。
知识点：W06 PCA

English answer (Official): B — PCA. Feature-wise rescaling changes the covariance structure. A large numerical scale can dominate a covariance-based PCA unless scaling is handled deliberately.

Q17 保留至少 85% 方差 → A

Original question (English) · [2 marks] There are exactly four principal components with variances 9, 5, 4 and 2, in descending order. What is the smallest number of leading components needed to retain at least 85% of total variance?

A. Three components, retaining 90%.

B. Four components, retaining 100%.

C. Three components, retaining 85%.

D. Two components, retaining 70%.

总方差。前两个，前三个 → 三个，保留 90%（A）。

C 的「85%」数字算错；B 不是最少；D 不达标。
知识点：W06 选几个主成分

English answer (Official): A — PCA retained variance. Total variance is 20. Two components retain 14/20 = 70%, whereas three retain 18/20 = 90%. Three is therefore the smallest qualifying number.

Q18 ReLU 负输入 → D 0

Original question (English) · [2 marks] A ReLU unit has a negative pre-activation. What is its output?

A. One.

B. The unchanged negative pre-activation.

C. A probability strictly between zero and one.

D. Zero.

。B 是恒等激活；C 是 sigmoid 的性质。

知识点：W07 激活函数

English answer (Official): D — Neural activations. ReLU(u) = max(0,u), so negative pre-activations produce zero. ReLU does not itself return a probability.

Q19 感知机更新 → C

Original question (English) · [2 marks] A perceptron has w = (0.4, -0.2), b = 0.1 and learning rate 0.2. For x = (2, 1), the supplied actual output is a = 1 and target is t = 0. What are the updated weights and bias?

A. w = (0.0, -0.4); b = 0.1.

B. w = (0.8, 0.0); b = 0.3.

C. w = (0.0, -0.4); b = -0.1.

D. w = (0.4, -0.2); b = -0.1.

：

答案 C。A 忘了更新偏置（偏置的「输入」恒为 1）；B 符号反了；D 只更新了偏置。
知识点：W07 感知机学习规则

English answer (Official): C — Perceptron update. The correction is η(t-a) = -0.2. Thus wnew = (0.4,-0.2) - 0.2(2,1) = (0,-0.4), and bnew = 0.1 - 0.2 = -0.1.

Q20 同一个边缘检测器用在整张图上 → B 权重共享

Original question (English) · [2 marks] A learned convolutional filter detects the same edge pattern at many image locations. Which design feature enables this with a small number of parameters?

A. Discarding the spatial arrangement before convolution.

B. Sharing the filter weights across spatial locations.

C. Using an independent parameter vector for every pixel location.

D. Fixing every convolution weight to the same constant.

A 卷积恰恰保留空间结构；C 每个位置独立参数就是全连接，参数量爆炸；D 权重是学出来的。
知识点：W08 卷积层

English answer (Official): B — CNNs. Weight sharing reuses one detector across the image. Each spatial position does not need its own independently learned filter.

Q21 RNN 中把信息带到下一步的是 → A 循环隐状态

Original question (English) · [2 marks] In a simple recurrent neural network, what carries information from an earlier time step into the next one?

A. The recurrent hidden state.

B. A distinct set of unshared recurrent weights for every sequence position.

C. A new independent label for each hidden unit.

D. A separate test set at every time step.

B 错：循环权重在各时间步共享；C、D 无关。
知识点：W08 简单 RNN

English answer (Official): A — RNNs. The hidden state is updated using the current input and previous hidden state. Recurrent weights are shared across time.

Q22 卷积层参数量 → C 76

Original question (English) · [2 marks] A convolutional layer has four 3 × 3 filters and receives two input channels. Each filter spans both input channels and has one bias. How many trainable parameters are there?

A. 72.

B. 22.

C. 76.

D. 40.

每个滤波器个权重 + 1 个偏置 = 19；4 个滤波器 → 。

A 72 忘了偏置；D 40 漏了输入通道维度，只算了；B 22 远小于正确值，同样是没把「每个滤波器跨越全部输入通道」算进去。
知识点：W08 多通道输入

图解

flowchart LR
  I["输入 2 通道"] --> F["每个滤波器 3×3×2 = 18 权重 + 1 偏置 = 19"]
  F --> T["4 个滤波器 × 19 = 76"]

English answer (Official): C — CNN parameters. Each filter has 3×3×2 = 18 weights and one bias, giving 19 parameters. Four filters give 4×19 = 76.

Q23 Dropout 的目的 → D

Original question (English) · [2 marks] During standard dropout training, why are different subsets of hidden activations temporarily disabled across updates?

A. To permanently delete every unit disabled at least once.

B. To make all hidden units produce exactly the same feature.

C. To replace supervised targets with random labels.

D. To reduce reliance on particular hidden-unit combinations and help limit overfitting.

答案 D：减少对特定隐藏单元组合的依赖（防止共适应），起正则化作用、抑制过拟合。
A 单元不会被永久删除；B 恰恰相反，要让单元各自有用；C 标签不变。
知识点：W07 现代深度网络的技巧

English answer (Official): D — Dropout. Dropout discourages co-adaptation by training with changing subsets of units. It is a regularisation technique; units are not permanently deleted.

Q24 平方欧氏距离下簇的最优代表点 → B 坐标均值

Original question (English) · [2 marks] A cluster contains several points and K-means uses squared Euclidean distance. Which representative minimises the sum of squared distances within that fixed cluster?

A. The coordinate-wise maximum.

B. The coordinate-wise mean of its points.

C. The most distant observed point.

D. Any randomly selected point, with the same objective value.

对固定簇，在处取最小（求导令其为 0 即得）。

知识点：W09 K-means

English answer (Official): B — K-means. For fixed assignments and squared Euclidean loss, the arithmetic mean minimises the within-cluster sum of squared distances.

Q25 GMM 后验隶属度 → A ≈ 0.462

Original question (English) · [2 marks] At observation x, a two-component GMM has π₁ = 0.3, π₂ = 0.7, f₁(x) = 0.4 and f₂(x) = 0.2. What is the posterior membership probability for component 1?

A. Approximately 0.462.

B. 0.120.

C. 0.300.

D. Approximately 0.538.

B 0.120 是没归一化的分子；C 0.300 是先验；D 0.538 是分量 2 的隶属度。
知识点：W09 GMM

English answer (Official): A — GMM membership. The component scores are 0.3×0.4 = 0.12 and 0.7×0.2 = 0.14. Normalising gives 0.12/(0.12+0.14) ≈ 0.462.

Q26 Viterbi 解决什么 → A 最可能的完整隐状态路径（解码问题）

Original question (English) · [2 marks] For a fixed HMM and an observed sequence, which task is solved by the Viterbi algorithm?

A. Find the most probable complete hidden-state path.

B. Sum probabilities over all possible observation alphabets.

C. Calculate a posterior mixture weight without using transitions.

D. Estimate the number of observed features by PCA.

B 无意义；C 不用转移就不是 HMM 了；D 与 HMM 无关。评估问题用 Forward，学习问题用 EM（Baum-Welch）。
知识点：W11 HMM 三个问题

English answer (Official): A — HMM tasks. Viterbi performs decoding: dynamic programming finds the single most probable hidden-state sequence for the observations and model.

Q27 两步后处于 B 的概率 → B 0.52

Original question (English) · [2 marks] A two-state Markov chain starts in A. Its transitions are P(A|A) = 0.6, P(B|A) = 0.4, P(A|B) = 0.3 and P(B|B) = 0.7. What is the probability of being in B after exactly two transitions?

A. 0.48.

B. 0.52.

C. 0.24.

D. 0.28.

两条路径：A→A→B ；A→B→B ；合计。

C、D 只算了其中一条路径；A 0.48 是两步后处于 A 的概率。
知识点：W11 马尔可夫链

图解

flowchart LR
  S(("t0: A")) -->|0.6| A1(("t1: A"))
  S -->|0.4| B1(("t1: B"))
  A1 -->|0.4| AB["A→A→B = 0.24"]
  B1 -->|0.7| BB["A→B→B = 0.28"]
  AB --> P["P(t2 = B) = 0.24 + 0.28 = 0.52"]
  BB --> P

English answer (Official): B — Markov chains. The two routes are A→A→B and A→B→B, with probabilities 0.6×0.4 and 0.4×0.7. Their sum is 0.24 + 0.28 = 0.52.

Q28 经验回放主要减少 → D 相邻训练样本的时间相关性

Original question (English) · [2 marks] A DQN trains on randomly sampled stored transitions rather than only the newest consecutive transitions. What does this sampling primarily help reduce?

A. All approximation error, regardless of network capacity.

B. The number of actions permitted by the environment.

C. The need for a reward signal from the environment.

D. Temporal correlation between training examples used in successive updates.

A 不可能消除全部近似误差；B 动作数由环境决定；C 仍需要奖励。
知识点：W12 DQN 经验回放

English answer (Official): D — Experience replay. Random minibatches from a replay buffer break up the strong correlations present in consecutive environment interactions and also reuse past experience.

Q29 目标网络参数固定若干步 → C 让自举目标更稳定

Original question (English) · [2 marks] A DQN holds the target-network parameters fixed for several online-network updates. What is the main intended benefit?

A. Force every action to be chosen with equal probability.

B. Remove rewards from the Q-learning target.

C. Make the bootstrap targets more stable during those updates.

D. Stop the online network from learning at all.

A 是探索策略的事；B 目标里仍有奖励；D 在线网络照常更新。
知识点：W12 目标网络

English answer (Official): C — Target networks. A fixed target network slows changes to the values used for bootstrapping. The online network still updates, and exploration is a separate issue.

Q30 Q-learning 更新 → B 3.50

Original question (English) · [2 marks] For a non-terminal transition, Q_old(s,a) = 1, α = 0.5, reward r = 2, γ = 0.8 and max Q(s′,a′) = 5. What is the updated value?

A. 2.50.

B. 3.50.

C. 6.00.

D. 4.00.

C 6.00 是目标值本身（相当于 α = 1）；A、D 算错了折扣或学习率。
知识点：W12 Q-learning 更新规则

English answer (Official): B — Q-learning. The target is 2 + 0.8×5 = 6. The update is 1 + 0.5(6-1) = 3.5.

Part B：计算题

Q31 熵与信息增益（10 分）

Original question (English) · Question 31. Entropy and information gain [10 marks]

A decision-tree node contains eight examples: four positive and four negative. A proposed split produces child L with three positives and one negative, and child R with one positive and three negatives. Use 0 log₂0 = 0.

[5 marks] Calculate the entropy of the parent and each child, in bits.

[5 marks] Calculate the size-weighted child entropy and the information gain. Round non-integer results to three decimal places.

父节点 4 正 4 负；划分后 L = 3 正 1 负，R = 1 正 3 负。

(a) 熵（5 分）

(b) 加权子熵与信息增益（5 分）

两个子节点各 4 个样本，权重都是：

评分：(a) 父熵 1 分 + 每个子节点 2 分；(b) 权重与加权熵 2 分、相减 2 分、最终取整 1 分。

知识点：W05 熵与信息增益

图解

flowchart TD
  P["父节点 4+ / 4−<br/>H = 1"] -->|"权重 4/8"| L["L: 3+ / 1−<br/>H = 0.811"]
  P -->|"权重 4/8"| R["R: 1+ / 3−<br/>H = 0.811"]
  L --> W["加权子熵 = 0.5×0.811 + 0.5×0.811 = 0.811"]
  R --> W
  W --> G["IG = 1 − 0.811 = 0.189"]

English answer (Official)

H(parent) = −0.5 log₂0.5 − 0.5 log₂0.5 = 1 bit. Each child has class proportions 0.75 and 0.25, so H(L) = H(R) = −0.75 log₂0.75 − 0.25 log₂0.25 = 0.75×0.4150 + 0.25×2 = 0.81125 ≈ 0.811 bits.
Each child holds 4 of the 8 examples, so both weights are 4/8 = 0.5. Weighted child entropy = 0.5×0.81125 + 0.5×0.81125 = 0.811 bits. IG = 1 − 0.81125 = 0.18875 ≈ 0.189 bits.

Q32 前向传播与换激活函数（10 分）

Original question (English) · Question 32. Forward pass and activation change [10 marks]

A network receives x₁ = 1 and x₂ = 3. Its hidden units use ReLU and its output uses a sigmoid. The equations (shown as an image in the paper) are u_A = x₁ + x₂ − 3, u_B = 2x₁ − x₂, h = ReLU(u), z = h_A + 2h_B, output = σ(z). Predict class 1 when the output is at least 0.5; otherwise predict class 0. Use e ≈ 2.718 if needed. The identity activation returns its input unchanged.

[5 marks] Calculate u_A, u_B, h_A and h_B.

[5 marks] Calculate z, the output probability and the predicted class. Then replace only unit B's ReLU by the identity activation h_B = u_B. Give the new probability and class. Round probabilities to three decimal places.

题目里的方程是图片，原式为：

输入，输出 ≥ 0.5 判为类 1。

(a)（5 分） ，；，。

(b)（5 分）

两个都是 ReLU：， → 类 1。
只把 B 换成恒等激活：，仍为 1，， → 类 0。分类结果翻转。

评分：(a) 预激活 2 分、激活 3 分；(b) 原输出与类别 2 分、修改后计算与类别 3 分。

要点：ReLU 把负的截成 0，等于把 B 这条通路「关掉」；换成恒等后负值被保留，乘以权重 2 把拉成负数。

知识点：W07 多层感知机与前向传播

图解

flowchart LR
  x1(("x₁ = 1")) -->|"+1"| uA["u_A = 1+3−3 = 1<br/>ReLU → h_A = 1"]
  x2(("x₂ = 3")) -->|"+1"| uA
  x1 -->|"+2"| uB["u_B = 2−3 = −1<br/>ReLU → h_B = 0"]
  x2 -->|"−1"| uB
  uA -->|"×1"| z["z = 1 + 2×0 = 1"]
  uB -->|"×2"| z
  z --> o["σ(1) ≈ 0.731 → 类 1"]
  uB -.->|"B 改恒等: h_B = −1"| z2["z = 1 − 2 = −1<br/>σ(−1) ≈ 0.269 → 类 0"]

English answer (Official)

u_A = 1 + 3 − 3 = 1; u_B = 2×1 − 3 = −1. Applying ReLU: h_A = 1, h_B = max(0, −1) = 0.
With both ReLUs: z = 1 + 2×0 = 1, output = 1/(1 + e⁻¹) ≈ 0.731 → class 1. With the identity activation at B only: h_B = −1 while h_A stays 1, so z = 1 + 2×(−1) = −1, output ≈ 0.269 → class 0. The predicted class changes, because ReLU had been clipping the negative contribution of unit B to zero.

Q33 K-means 一轮（10 分）

Original question (English) · Question 33. One K-means iteration [10 marks]

Run one complete K-means iteration with k = 2. Use initial centroids μ₁ = A and μ₂ = D. Assign every point using these initial centroids before updating either centroid. Points: A (0,1), B (2,1), C (6,3), D (8,3). Use the coordinate-wise mean.

[5 marks] Calculate the squared distance from each point to each initial centroid and give all cluster assignments.

[5 marks] Calculate the coordinates of both updated centroids.

点 A(0,1)、B(2,1)、C(6,3)、D(8,3)；k = 2，初始，。

(a) 平方距离与分配（5 分）

点	到 μ₁ = (0,1)	到 μ₂ = (8,3)	分配
A	0	64 + 4 = 68	1
B	4 + 0 = 4	36 + 4 = 40	1
C	36 + 4 = 40	4 + 0 = 4	2
D	64 + 4 = 68	0	2

簇 1 = {A, B}，簇 2 = {C, D}。

(b) 更新质心（5 分）

评分：(a) 距离对 4 分 + 分配 1 分；(b) 均值方法 2 分 + 两个质心 3 分。

易错：题目要求先用初始质心把所有点都分配完再更新（批量 K-means），不要分一个点就更新一次。

知识点：W09 K-means

图解

flowchart LR
  m1(("μ₁ = A(0,1)"))
  m2(("μ₂ = D(8,3)"))
  A["A(0,1)"] -->|"0"| m1
  B["B(2,1)"] -->|"4（vs 40）"| m1
  C["C(6,3)"] -->|"4（vs 40）"| m2
  D["D(8,3)"] -->|"0"| m2
  m1 --> n1["新 μ₁ = 均值(A,B) = (1,1)"]
  m2 --> n2["新 μ₂ = 均值(C,D) = (7,3)"]

English answer (Official)

Squared distances (to μ₁ = (0,1), to μ₂ = (8,3)): A (0, 68), B (4, 40), C (40, 4), D (68, 0). Assign each point to the nearer centroid: cluster 1 = {A, B}, cluster 2 = {C, D}.
μ₁,new = ((0+2)/2, (1+1)/2) = (1, 1); μ₂,new = ((6+8)/2, (3+3)/2) = (7, 3).

Q34 HMM 序列似然（10 分）

Original question (English) · Question 34. HMM sequence likelihood [10 marks]

An HMM has states A and B and observations x and y. The observed sequence is (x, y). Priors: π(A) = 0.7, π(B) = 0.3. Emissions: e_A(x) = 0.6, e_A(y) = 0.4, e_B(x) = 0.2, e_B(y) = 0.8. Transitions (rows = current, columns = next): A→A 0.8, A→B 0.2, B→A 0.3, B→B 0.7.

[5 marks] Calculate f₁(A) and f₁(B).

[5 marks] Use the Forward algorithm to calculate f₂(A), f₂(B) and P(x,y). Give final probabilities to four decimal places.

状态	π	e(x)	e(y)	→ A	→ B
A	0.7	0.6	0.4	0.8	0.2
B	0.3	0.2	0.8	0.3	0.7

观测序列 (x, y)。

(a)（5 分）初始化 ：

注意这是「第一个观测 + 处于该状态」的联合概率，不是归一化后的后验。

(b)（5 分）递推 ：

评分：(a) 「初始 × 发射」规则 1 分 + 每个值 2 分；(b) 每个前向概率 2 分 + 总和 1 分。

易错：Forward 是求和，Viterbi 是取最大；转移表行是当前状态、列是下一状态，别读反。

知识点：W11 Forward 算法

图解

flowchart LR
  s1A["t=1 A<br/>f₁(A) = 0.7×0.6 = 0.42"]
  s1B["t=1 B<br/>f₁(B) = 0.3×0.2 = 0.06"]
  s2A["t=2 A<br/>0.4×(0.336+0.018) = 0.1416"]
  s2B["t=2 B<br/>0.8×(0.084+0.042) = 0.1008"]
  s1A -->|"×0.8 = 0.336"| s2A
  s1B -->|"×0.3 = 0.018"| s2A
  s1A -->|"×0.2 = 0.084"| s2B
  s1B -->|"×0.7 = 0.042"| s2B
  s2A --> P["P(x,y) = 0.1416 + 0.1008 = 0.2424"]
  s2B --> P

English answer (Official)

f₁(A) = π(A)·e_A(x) = 0.7×0.6 = 0.42; f₁(B) = 0.3×0.2 = 0.06. (These are joint probabilities of the first observation and the state, not normalised posteriors.)
f₂(A) = e_A(y)·[f₁(A)a_AA + f₁(B)a_BA] = 0.4×(0.42×0.8 + 0.06×0.3) = 0.4×0.354 = 0.1416. f₂(B) = e_B(y)·[f₁(A)a_AB + f₁(B)a_BB] = 0.8×(0.42×0.2 + 0.06×0.7) = 0.8×0.126 = 0.1008. P(x,y) = f₂(A) + f₂(B) = 0.2424.

本卷的考法小结

计算型选择题占三分之一（Q3、7、10、14、17、19、22、25、27、30），全都是一两步就能算完的公式题，背熟公式就是送分。
干扰项的套路：只算一半（Q14 的 0.20、Q27 的 0.24/0.28）、漏掉偏置（Q19、Q22）、未归一化（Q25 的 0.12）、学习率当成 1（Q30 的 6.00）。
概念题围绕「为什么」：标准化在哪拟合、测试集只用一次、L1 稀疏、核技巧只用内积、经验回放去相关、目标网络稳目标。
Part B 四道都是固定套路：信息增益、前向传播、K-means 一轮、HMM Forward。另一套卷子见 Practice Exam 2 解析。

Q	答	Q	答	Q	答	Q	答	Q	答
1	A	7	C	13	C	19	C	25	A
2	C	8	B	14	D	20	B	26	A
3	A	9	D	15	D	21	A	27	B
4	B	10	A	16	B	22	C	28	D
5	D	11	A	17	A	23	D	29	C
6	C	12	B	18	D	24	B	30	B

Q	答	Q	答	Q	答	Q	答	Q	答
1	A	7	C	13	C	19	C	25	A
2	C	8	B	14	D	20	B	26	A
3	A	9	D	15	D	21	A	27	B
4	B	10	A	16	B	22	C	28	D
5	D	11	A	17	A	23	D	29	C
6	C	12	B	18	D	24	B	30	B

Q	答	Q	答	Q	答	Q	答	Q	答
1	A	7	C	13	C	19	C	25	A
2	C	8	B	14	D	20	B	26	A
3	A	9	D	15	D	21	A	27	B
4	B	10	A	16	B	22	C	28	D
5	D	11	A	17	A	23	D	29	C
6	C	12	B	18	D	24	B	30	B