[SOLVED] CS代考 COMP3308/3608 Introduction to Artificial Intelligenece (regular and advance

COMP3308/3608 Introduction to Artificial Intelligenece (regular and advanced)
semester 1, 2022 Information about the exam
• The exam will be online, via Canvas. It is set as a Quiz.
• The exam is proctored: Record+. Please see:

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o Preparing for online exams, select Record+: https://canvas.sydney.edu.au/courses/23380
o Help and FAQs about Record+ exams: https://canvas.sydney.edu.au/courses/23380/pages/record+-help-and-faqs
• The Canvas site for the exam is different that the Canvas site we use during the semester. There are 2 exam sites: one for COMP3308 called “Final Exam for: COMP3308” and another one for COMP3608 – “Final Exam for: COMP3608”. The Exams Office will give you access to your exam site.
• The duration of the exam is standard: 2 hours + 10 minutes reading time. In addition, there are 15 minutes for file upload. File upload is only relevant for the last question.
• You need to type your answers in the boxes in the Quiz or select the correct answer from a list for the multiple choice questions. An exception is the last question where you have 2 options: (1) type your answers and working in the Quiz (there is box) or (2) write them on paper, take a photo with your phone and upload the photo. If you choose the second option, the upload must be done during the upload time which is after the proctoring session has finished. The photo should be uploaded under “Assignments” in the Canvas exam site. You are not allowed to use your phone during the proctoring session.
• Permitted materials:
o 1 page of your own notes – double-sided A4 size, handwritten or typed o Calculator – handheld, non-programmable
o Blank scratch paper, multiple sheets
• No other materials or devices are allowed. No internet browsing is allowed. You can’t consult other people during the exam.
• The exam paper is confidential. You must not discuss the exam questions with other people, post or distribute the exam questions in any way, during or after the exam
• The exam is worth 60 marks ( =60% of your final mark). To pass the course you need at least 40% on the exam (i.e. 24 marks), regardless of what your mark during the semester is.
• All material is examinable except week 1, week 13a (recommender systems), historical context, Matlab and Weka.
• There are 3 types of questions: 1) questions requiring short answers, 2) problem- solving/calculation questions and 3) multiple-choice questions (small number).

Academic honesty (Very important!)
• All suspicious behavior during the proctoring session will be reviewed
• All file uploads will be compared for plagiarism
• Please do not cheat or copy! The consequences and penalties are very severe.
• If you copy, you will get caught. If you make your work available to another student to copy, this is also academic dishonesty and you will be investigated and penalized.
• The stress of going through the investigation is immense. The investigation takes many months and your mark will not be finalised until it is completed, your graduation may be delayed, you may have problems enrolling in other courses.
• It is not worth it. You will regret it all your life.

Sample exam questions
In addition to these questions please also see on Canvas:
Search: Weeks_2-3_Practice.pdf (prepared byBroom)
Bayesian networks: BN_practice_questions.pdf (prepared byBroom)
Question 1. (Type 2 – problem solving/calculation)
In the tree below the step costs are shown along the edges and the h values are shown next to each node. The goal nodes are double-circled: F and D.
Write down the order in which nodes are expanded using:
a) Breadth-first search
b) Depth-first search
c) Uniform cost search
d) Iterative deepening search
e) Greedy search
In case of ties, expand the nodes in alphabetical order.
a) Breadth-first search: ABCD
b) Depth-first search: ABEF
c) Uniform cost search: ABEF
d) Iterative deepening search:AABCD
e) Greedy search: AD
f) A*: ABEF
• BFS: Expands the shallowest unexpanded node • DFS: Expands the deepest unexpanded node

• UCS: Expands the node with the smallest path cost g(n)
• IDS: DFS at levels l = 0, 1, 2, etc. Expands the deepest unexpanded node within level l • Greedy: Expands the node with the smallest heuristic value h(n)
• A*: Expands the node with the smallest f(n)=g(n)+h(n)
Question 2. (Type 1 – short answers)
Answer briefly and concisely:
a) A* uses admissible heuristics. What happens if we use a non-admissible one? Is it still useful to use A* with a non-admissible heuristic?
b) What is the advantage of choosing a dominant heuristic in A* search?
c) What is the main advantage of hill climbing search over A* search?
a) Not optimal anymore. But it could still find a reasonably good solution in acceptable time, depending on how good the heuristic is.
b) Fewer nodes expanded. As a result, the optimal solution will be found quicker.
c) Space complexity – keeps only the current node in memory.
Question 3. (Type 2 – problem solving/calculation)
Consider the following game in which the evaluation function values for player MAX are shown at the leaf nodes. MAX is the maximizing player and MIN is the minimizing player. The first player is MAX.
a) What will be the backed-up value of the root node computed by the minimax algorithm?
b) Which move should MAX choose based on the minimax algorithm – to node B, C or D?

c) Assume that we now use the alpha-beta algorithm. List all branches that will be pruned, e.g. AB etc. Assume that the children are visited left-to-right (as usual).
a) The value of A is 4:
b) To node C
c) LD and MD will be pruned:
Question 4. (Type 1 – short answers)
Answer briefly and concisely:
a) The 1R algorithm generates a set of rules. What do these rules test?
b) Gain ratio is a modification of Gain used in decision trees. What is its advantage?
c) Propose two strategies for dealing with missing attribute values in learning algorithms.

d) Why do we need to normalize the attribute values in the k-nearest-neighbor algorithm?
e) What is the main limitation of the perceptrons?
f) Describe an early stopping method used in the backpropagation algorithm to prevent overfitting.
g) The problem of finding a decision boundary in support vector machine can be formulated as an optimisation problem using Lagrange multipliers. What are we maximizing?
h) In linear support vector machines, we use dot products both during training and during classification of a new example. What vectors are these products of?
During training:
During classification of a new example:
a) They test the values of a single attribute.
b) It penalizes highly-branching attributes by taking into account the number and the size of branches.
c) Strategy 1: Use the attribute mean to fill in the missing value.
Strategy 2: Use the attribute mean for all examples belonging to the same class.
d) As different attributes are measured on different scales, without normalization the effect of the attributes with smaller scale of measurement will be less significant than those with larger.
e) Can separate only linearly separable data.
f) Available data is divided into 3 subsets:
Training set – used for updating the weights.
Validation set – used for early stopping.
Training is stopped when the error on the validation set increases for a pre-specified number of iterations.
g) The margin of the hyperplane.
h) During training: Pairs of training examples.
During classification of a new example: The new example and the support vectors.

Question 5. (Type 2 – problem solving/calculation)
Consider the task of learning to classify mushrooms as safe or poisonous based on the following four features: stem = {short, long}, bell = {rounded, flat}, texture = {plain, spots, bumpy, ruffles} and number = {single, multiple}.
The training data consists of the following 10 examples:
Poisonous:
These examples are also shown in the table below:
rounded flat
flat rounded flat rounded flat rounded rounded rounded
spots ruffles ruffles plain plain plain plain bumpy spots bumpy
single single multiple single single single single single single single
safe poisonous poisonous poisonous poisonous poisonous
a) Use Naïve Bayes to predict the class of the following new example: stem=long, bell=flat, texture=spots, number=single. Show your calculations.
a) How would 3-Nearest Neighbor using the Hamming distance classify the same example as above? Explain your answer. (Hint: The Hamming distance is the number of different feature values).
b) Consider building a decision tree. Calculate the information gain for texture and number and briefly show your calculations Which one of these two features will be selected and why?

You may use this table:
-(x/y)*log2(x/y) x y -(x/y)*log2(x/y 0.50 4 5 0.26
0.53 1 6 0.43
0.39 5 6 0.22
0.5 1 7 0.40 0.31 2 7 0.52 0.46 3 7 0.52 0.53 4 7 0.46 0.44 5 7 0.35
y -(x/y)*log2(x/y 9 0.47
10 0.33 10 0.52 10 0.36 10 0.14
-(x/y)*log2(x/y x 0.19 5 0.38 7 0.53 8 0.42 1 0.17 3 0.35 7 0.48 9
d) Consider a single perceptron for this task. What is the number of inputs? What is the dimensionality of the weight space? Briefly explain your answer.
a) Naïve Bayes:
The new example is the evidence E.
E1=stem=long, E2=bell=flat, E3=texture=spots, E4=number=single
We need to compute P(safe|E) and P(poisonous|E) and compare them.
P(safe|E)= P(E1|safe)P(E2|safe)P(E3|safe)P(E4|safe)P(safe) P(E)
P(safe)=5/10=1/2 P(E1|safe)=P(stem=long|safe)=4/5 P(E2|safe)=P(bell=flat|safe)=3/5 P(E3|safe)=P(texture=spots|safe)=1/5 P(E4|safe)=P(number=single|safe)=4/5
P(poisonous)=5/10=1/2 P(E1|poisonous)=P(stem=long|poisonous)=2/5 P(E2|poisonous)=P(bell=flat|poisonous)=1/5 P(E3|poisonous)=P(texture=spots|poisonous)=1/5 P(E4|poisonous)=P(number=single|poisonous)=5/5
21151 5 P(poisonous|E)=55552= 625
43141 24 P(safe|E)=55552= 625
P(E) P(E) => Naïve Bayes predicts safe.
b) The three nearest neighbors have a distance of 1 and are examples 2 (safe), 5 (safe) and 9 (poisonous). The majority vote is safe, hence the new example will be classifed as safe.
c) Decision tree H(S)=I(5/10,5/10)= 1 bit
Split on texture: H(Sspots)=I(1/2,1/2)=1 bit H(Sruffles)=I(2/2,0/2)=0 bits H(Splain)=I(2/4,2/4)=1 bit

H(Sbumpy)=I(0/2,2/2)=0 bits
H(S|texture)=2/10*1 + 2/10*0 + 4/10*1 + 2/10*0=6/10=0.6 bits gain(texture)=1-0.6=0.4 bits
Split on number: H(Ssingle)=I(4/9,5/9)=-4/9log4/9-5/9log5/9=0.52+0.47=0.99 bits H(Smultiple)=I(1/1,0/1)=0 bits
H(S|number)=9/10*0.99 + 1/10*0=0.891 bits gain(number)=1-0.891=0.109 bits
gain(texture) > gain(number) => texture will be selected as it has a higher information gain
d) There should be 10 inputs, 1 for each attribute value. This means using binary encoding of the attributes and their values (e.g. 10 for stem=short and 01 for stem=long). Binary encoding is the most popular encoding for nominal attributes.
Hence, the weight space will have a dimensionality of 11 (10 weights +1 bias weight).
Question 6. Given the training data in the table below where credit history, debt, collateral and income are attributes and risk is the class, predict the class of the following new example using the 1R algorithm: credit history=unknown, debt=low, collateral=none, income=15-35K. Show your calculations.
credit debt collateral history
bad high none unknown high none unknown low none unknown low none unknown low none unknown low adequate bad low none
bad low adequate good low none good high adequate good high none good high none good high none bad high none
income risk
0-15k high 15-35k high 15-35k moderate 0-15k high over 35k low
over 35k low 0-15k high over 35k moderate over 35k low
over 35k low 0-15k high 15-35k moderate over 35k low 15-35k high
1. Attribute credit history
bad:0 low, 1 moderate, 3 high => risk=high, errors: 1/4 unknown: 2 low, 1 moderate, 2 high => risk=low, errors: 3/5 good: 3 low, 1 moderate, 1 high => risk=low, errors: 2/5 total errors: 6/14
2. Attribute debt

high: 2 low, 1 moderate, 4 high => risk=high, errors: 3/7 low: 3 low, 2 moderate, 2 high => risk=low, errors: 4/7 total errors: 7/14
3. Attribute collateral
none: 3 low, 2 moderate, 6 high => risk=high, errors: 5/11 adequate: 2 low, 1 moderate, 0 high => risk=low, errors: 1/3 total errors: 6/14
4. Attribute income
0-15K: 0 low, 0 moderate, 4 high => risk=high, errors: 0/6 15-35K: 0 low, 2 moderate, 2 high => risk=moderate, errors: 2/4 over 35K: 5 low, 1 moderate, 0 high => risk=low, errors: 1/6 total errors: 3/14
The rule based on the attribute income has the minimum number of errors =>1R produces the following rule:
if income=0-15K then risk=high
else if income=15-35K then risk=moderate else if income=over 35K then risk=low
The new example has income=15-35K and will be classified as risk=moderate. Question 7. (Type 2 – problem solving/calculation)
Use the k-means algorithm to cluster the following one dimensional examples into 2 clusters: 2, 5, 10, 12, 3, 20, 31, 11, 24. Suppose that the initial seeds are 2 and 5. The convergence criterion is met when either there is no change between the clusters in two successive epochs or when the number of epochs has reached 5.
Show the final clusters. How many epochs were needed for convergence? There is no need to show your calculations.
For simplicity let’s sort the data first: 2 3 5 10 11 12 20 24 31
K1={2}, K2={5}
end of epoch 1:
K1={2, 3}, mean_K1=2.5
K2={5, 10, 11, 12, 20, 24, 31}, mean_K2=16.1 Stopping criterion not met
End of epoch 2:
K1={2, 3, 5}, mean_K1=3.3
K2={10, 11, 12, 20, 24, 31}, mean_K2=18 Stopping criterion not met

End of epoch 3:
K1={2, 3, 5, 10}, mean_K1=5
K2={11, 12, 20, 24, 31}, mean_K2=19.6 Stopping criterion not met
End of epoch 4:
K1={2, 3, 5, 10, 11,12}, mean_K1=7.2 K2={20, 24, 31}, mean_K2=25 Stopping criterion not met
End of epoch 5:
K1={2, 3, 5, 10, 11,12}, mean_K1=7.2 K2={20, 24, 31}, mean_K2=25
Stopping criterion is met – no change in clusters
Note: The question asks you only to show the final clusters and number of epochs, so the answer is:
Clusters: K1={2, 3, 5, 10, 11,12}, K2={20, 24, 31} 5 epochs were needed.
Question 8. (Type 1 – short answers)
You task is to develop a computer program to rate chess board positions. You got an expert chess player to rate 100 different chessboards and then use this data to train a backpropagation neural network, using board features as the ones shown in the figure below.
Select the correct answer (“Yes” or “No”) in the questions below. Select “Yes” for all issues that could, in principle, limit your ability to develop the best possible chess program using this method. Select “No” for all issues that could not. Briefly explain your answer.
a) The backpropagation network may be susceptible to overfitting, since you tested its performance on the training data instead of using cross validation.
Yes No Explanation:
b) The backpropagation neural network can only distinguish between boards that are completely good or completely bad.

Yes No Explanation:
c) The backpropagation network will converge to the global minimum.
Yes No Explanation:
d) You should have used higher learning rate and momentum to guarantee convergence to the global minimum.
Yes No Explanation:
e) The topology of your neural net might not be adequate to capture the expertise of the human expert.
Yes No Explanation:
Note: This is not a multiple-choice question. You need to provide an explanation, otherwise you will receive 0 marks.
a) Yes. Testing the performance on the training data is an over-optimistic measure and it does not guard against overfitting. Cross-validation is a more reliable procedure.
b) No. Backpropagation neural networks can be applied for both classification and regression tasks.
c) No. The backprogarion algorithm implements gradient descent and is not guaranteed to converge to the global minimum – it finds the closest local minimum.
d) No. The use of momentum reduces the oscillations when using a higher learning rate but it doesn’t guarantee convergence to the global minimum.
e) Yes. Too few neurons – underfitting; too many – overfitting.

Question 9. (Type 2 – problem solving/calculation)
In the figure below, the circles are training examples and the squares are test examples, i.e. we are using the circles to predict the squares. Two algorithms are used: 1-Nearest Neighbour and 3-Nearest Neighbour.
We are given the following results about the squares:
Square Using 1-Nearest Neighbors Using 3-Nearest Neighbors 1-+
What will be the class of the following examples? Write +, – or U for cannot be determined.
1) Circle L:
2) Circle I:
3) Circle H:
4) Circle E:
5) Circle K:
6) Circle C:
7) Square 6 using 1-Nearest Neighbour:
8) Square 6 using 3-Nearest Neighbour:
9) Square 3 using 1-Nearest Neighbour?
10) Square 5 using 1-Nearest Neighbour?
Circle L: – Circle I: + Circle H: – Circle E: +

Circle K: U
Circle C: +
Square 6 using 1-Nearest Neighbour: U Square 6 using 3-Nearest Neighbour: – Square 3 using 1-Nearest Neighbour: + Square 5 using 1-Nearest Neighbour: U

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