2021 question paper

Data Mining

26 questions

Q1a. Discuss whether or not each of the following activities is a data mining task. Give briefly proper justification : (i) Dividing the customers of a company according to their profitability (ii) Monitoring and predicting failures in a hydropower plant20212m
Module 2: Data Mining and Association Rule Mining
Discuss whether or not each of the following activities is a data mining task. Give briefly proper justification :

(i) Dividing the customers of a company according to their profitability
(ii) Monitoring and predicting failures in a hydropower plant

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Worked Solution
Is it a Data Mining Task?

(i) Dividing customers according to profitability

Yes, this is a data mining task.
This is a clustering/segmentation problem — grouping customers into meaningful segments based on profitability-related attributes (purchase frequency, spend, margin) without predefined labels. Discovering these natural groupings from raw transaction/customer data is a core data mining functionality (unsupervised clustering).

(ii) Monitoring and predicting failures in a hydropower plant

Yes, this is a data mining task (specifically the "predicting" part).

Monitoring alone (just observing sensor readings in real time) is not mining — it's plain data collection/surveillance.

Predicting failures, however, requires learning patterns from historical sensor/operational data to forecast future failures — this is a classification/prediction task, a core data mining functionality.

So both qualify as data mining tasks, but for different reasons: (i) because segmentation without labels = clustering, and (ii) because predicting future failures from historical patterns = predictive modeling — not because "monitoring" itself is mining.
Q1b. Classify the following attributes as binary, discrete or continuous. Also classify them as qualitative (nominal or ordinal) or quantitative (interval or ratio). Some cases may have more than one interpretation, so briefly indicate your reasoning if you think there may be some ambiguity : (i) Distance from center of campus (ii) Ability to pass light in terms of opaque, translucent and transparent20212m
Module 2: Data Mining and Association Rule Mining
Classify the following attributes as binary, discrete or continuous. Also classify them as qualitative (nominal or ordinal) or quantitative (interval or ratio). Some cases may have more than one interpretation, so briefly indicate your reasoning if you think there may be some ambiguity :

(i) Distance from center of campus
(ii) Ability to pass light in terms of opaque, translucent and transparent

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Worked Solution
Attribute Classification

(i) Distance from center of campus

Type: Continuous (can take any real value, e.g., 0.5 km, 1.27 km).

Scale: Quantitative — Ratio (has a true zero — "0 distance" means literally at the campus center — and ratios are meaningful: 4 km is twice as far as 2 km).

(ii) Ability to pass light: opaque, translucent, transparent

Type: Discrete (a small, finite set of category values).

Scale: Qualitative — Ordinal. There is a natural order — opaque (passes no light) < translucent (passes some light) < transparent (passes most/all light) — but the "distance" between categories isn't numerically meaningful (we can't say translucent is exactly halfway between opaque and transparent in any measurable unit).

Ambiguity note

(ii) could arguably be seen as nominal if one ignores the light-passing order and treats the three labels as unordered categories, but since the attribute is explicitly defined by increasing "ability to pass light," the natural/expected interpretation is ordinal.

Attribute Discrete/Continuous Qualitative/Quantitative

Distance from campus center Continuous Quantitative — Ratio

Opaque/Translucent/Transparent Discrete Qualitative — Ordinal
Q1c. Differentiate between supervised and unsupervised classification.20212m
Module 3: Classification and Prediction
Differentiate between supervised and unsupervised classification.

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Q1d. In real-world data, tuples with missing values for some attributes are a common occurrence. List at least two methods for handling this problem.20212m
Module 2: Data Mining and Association Rule Mining
In real-world data, tuples with missing values for some attributes are a common occurrence. List at least two methods for handling this problem.

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Worked Solution
Handling Missing Values — At Least Two Methods

Ignore the tuple — Simply discard records with missing values. Only reasonable when the missing attribute is critical (e.g., the class label) and the dataset is large enough that dropping a few rows doesn't hurt.

Fill in the missing value manually — A domain expert examines and fills each gap. Accurate but extremely slow and impractical for large datasets.

Use a global constant — Replace all missing values with a fixed label like "Unknown" or −∞. Simple, but the mining algorithm may mistakenly treat "Unknown" as a meaningful, interesting category.

Use the attribute mean/median — Replace missing numeric values with the mean (or median for skewed data) of that attribute across all tuples.

Use the mean/median of the same class — Replace with the mean of tuples belonging to the same class as the tuple with the missing value (more accurate than a global mean).

Use the most probable value — Predict the missing value using inference methods like regression, decision-tree induction, or Bayesian formalism — the most sophisticated but most accurate approach.

Missing Value │ ├── Ignore tuple ─────────────▶ (drop row) ├── Manual fill ───────────────▶ (expert input) ├── Global constant ───────────▶ "Unknown" ├── Attribute mean/median ─────▶ statistical fill ├── Class-wise mean ───────────▶ context-aware fill └── Predicted value (regression/tree/Bayes) ─▶ model-based fill
Q1e. Describe briefly z-score normalization.20212m
Module 2: Data Mining and Association Rule Mining
Describe briefly z-score normalization.

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Worked Solution
Z-Score Normalization

Z-score (zero-mean) normalization rescales an attribute's values based on its mean (μ) and standard deviation (σ):

$v' = \frac{v - \mu}{\sigma}$

where:

$v$ = original value

$\mu$ = mean of the attribute

$\sigma$ = standard deviation of the attribute

$v'$ = normalized value

Effect

The transformed data has mean 0 and standard deviation 1.

Values are typically small, ranging roughly from −3 to +3 (assuming a roughly normal distribution), though technically unbounded.

Why use it

Useful when the min/max of an attribute are unknown or when there are outliers that would distort min-max normalization.

Puts differently-scaled attributes (e.g., income in lakhs vs age in years) on comparable footing before applying distance-based algorithms (k-NN, k-means, clustering).

Raw values: [200, 300, 400, 600, 1000] │ subtract mean, divide by std dev ▼ Z-scores: [negative, negative, ~0, positive, large positive]

A z-score of 0 means the value equals the attribute's mean; positive/negative z-scores indicate how many standard deviations above/below the mean the value lies.
Q1f. Describe briefly a priori principle.20212m
Module 2: Data Mining and Association Rule Mining
Describe briefly a priori principle.

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Worked Solution
The Apriori (A Priori) Principle

Statement: "If an itemset is frequent, then all of its subsets must also be frequent."

Equivalently, by contrapositive: "If an itemset is infrequent, then all of its supersets must also be infrequent" — this is the practical pruning rule actually used by the algorithm.

Why it holds

Support is anti-monotone: adding more items to an itemset can only keep or reduce the number of transactions containing it (support never increases as itemset size grows).

$X \subseteq Y \implies support(X) \geq support(Y)$

Diagram

{A,B,C,D} infrequent / | | \ {A,B,C} {A,B,D} {A,C,D} {B,C,D} ← if {A,B,C,D} infrequent, ALL its \ | | / supersets are pruned instantly (pruned without counting)

Why it matters

This principle is the backbone of the Apriori algorithm's efficiency: once an itemset is found infrequent, none of its supersets need to be generated or counted at all — dramatically shrinking the candidate search space compared to brute-force enumeration of all $2^d$ possible itemsets.
Q1g. List at least four general characteristics of hierarchical clustering methods.20212m
Module 4: Cluster Analysis
List at least four general characteristics of hierarchical clustering methods.

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Q1h. Compare and contrast k-medoids with k-means.20212m
Module 4: Cluster Analysis
Compare and contrast k-medoids with k-means.

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Attribute	Discrete/Continuous	Qualitative/Quantitative
Distance from campus center	Continuous	Quantitative — Ratio
Opaque/Translucent/Transparent	Discrete	Qualitative — Ordinal

Q1i. Describe the OLAP operation slice and dice.20212m

Module 1: Data Warehousing and Business Analysis

Describe the OLAP operation slice and dice.

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Worked Solution

Slice and Dice (OLAP Operations)

Think of a data cube as a 3D box of numbers indexed along multiple dimensions (e.g., Time, Product, Region).

Slice

Selects one specific value on one dimension, "cutting" a single 2D layer out of the cube.

        Product
       /|  /|  /|
      / | / | / |
     /__|/__|/__|
    |   |   |   |     <- Slice at Time = "2024"
Time|___|___|___|
    |   |   |   |
    |___|___|___|
        Region

Example: "Sales for Time = 2024" → gives a 2D Product × Region table.

Dice

Selects a sub-cube by picking a range/subset of values on two or more dimensions simultaneously.

Full cube ──dice──▶  smaller cube
(Time × Product × Region)   (Time∈{2023,2024} × Product∈{A,B} × Region=West)

Example: "Sales where Time ∈ {2023, 2024} AND Product ∈ {A, B} AND Region = West."

Key Difference

Operation	Dimensions restricted	Result
Slice	1 dimension fixed to a single value	One lower-dimension "layer" of the cube
Dice	2+ dimensions restricted to ranges/subsets	A smaller sub-cube (same dimensionality)

Both are used for interactive, ad-hoc drill-down analysis in OLAP tools without needing to re-query the source data.

Q1j. Why data mining is a misnomer?20212m
Module 2: Data Mining and Association Rule Mining
Why data mining is a misnomer?

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Worked Solution
Why "Data Mining" is a Misnomer

The term "data mining" literally suggests "mining data" — but strictly speaking, we don't mine data itself; we mine patterns/knowledge from data, much like gold mining extracts gold from rock, not "rock mining."

Reasoning

If taken literally, "mining coal from rocks" is called coal mining, not "rock mining" — the process is named after what is extracted, not the raw material it comes from.

Following that convention, the process should really be called "knowledge mining from data" or simply "knowledge mining" — since what we're actually after is the hidden knowledge/patterns, not the data itself (which we already have).

The formally correct term for the full pipeline is KDD — Knowledge Discovery in Databases — of which "data mining" is just one (pattern-extraction) step, yet the popular term "data mining" is loosely used for the entire process.

In short: "data mining" is a misnomer because it names the process after its raw input (data) rather than its actual output (knowledge/patterns) — the opposite convention of how "mining" terms normally work (e.g., gold mining, coal mining name the output, not the source rock).
Q2a. How does instance-based classifier work? List two instance-based classifiers.20216m
Module 3: Classification and Prediction
How does instance-based classifier work? List two instance-based classifiers.

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Q2b. What do you mean by under-fitting and over-fitting of a classification model?20214m
Module 3: Classification and Prediction
What do you mean by under-fitting and over-fitting of a classification model?

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Q2c. For the following vectors, x and y, calculate the indicated similarity or distance measures : \mathbf{x} = \{0, 1, 0, 1\}, \quad \mathbf{y} = \{1, 0, 1, 0\} Calculate: 1. Cosine similarity 2. Correlation 3. Euclidean distance 4. Jaccard similarity20214m

Module 2: Data Mining and Association Rule Mining

For the following vectors, x and y, calculate the indicated similarity or distance measures :

$\mathbf{x} = \{0, 1, 0, 1\}, \quad \mathbf{y} = \{1, 0, 1, 0\}$

Calculate:

Cosine similarity
Correlation
Euclidean distance
Jaccard similarity

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Worked Solution

Similarity/Distance Measures for x = {0,1,0,1}, y = {1,0,1,0}

Setup

$\mathbf{x} = (0, 1, 0, 1), \quad \mathbf{y} = (1, 0, 1, 0)$

1. Cosine Similarity

$\cos(\mathbf{x}, \mathbf{y}) = \frac{\mathbf{x} \cdot \mathbf{y}}{\|\mathbf{x}\|\|\mathbf{y}\|}$

Dot product: $(0)(1)+(1)(0)+(0)(1)+(1)(0) = 0$
$\|\mathbf{x}\| = \sqrt{0^2+1^2+0^2+1^2} = \sqrt{2}$
$\|\mathbf{y}\| = \sqrt{1^2+0^2+1^2+0^2} = \sqrt{2}$

$\cos(\mathbf{x},\mathbf{y}) = \frac{0}{\sqrt{2}\cdot\sqrt{2}} = \boxed{0}$

2. Correlation

Mean: $\bar{x} = \bar{y} = 0.5$

i	$x_i-\bar{x}$	$y_i-\bar{y}$	product	$(x_i-\bar{x})^2$	$(y_i-\bar{y})^2$
1	-0.5	0.5	-0.25	0.25	0.25
2	0.5	-0.5	-0.25	0.25	0.25
3	-0.5	0.5	-0.25	0.25	0.25
4	0.5	-0.5	-0.25	0.25	0.25
Σ			-1.0	1.0	1.0

$r = \frac{-1.0}{\sqrt{1.0 \times 1.0}} = \boxed{-1}$
(makes sense: y is exactly the complement of x, so they're perfectly negatively correlated)

3. Euclidean Distance

$d(\mathbf{x},\mathbf{y}) = \sqrt{(0-1)^2+(1-0)^2+(0-1)^2+(1-0)^2} = \sqrt{1+1+1+1} = \sqrt{4} = \boxed{2}$

4. Jaccard Similarity

Binary co-occurrence counts across the 4 positions:

Position	x	y	Match type
1	0	1	f01
2	1	0	f10
3	0	1	f01
4	1	0	f10

$f_{11}=0,\ f_{01}=2,\ f_{10}=2,\ f_{00}=0$

$J(\mathbf{x},\mathbf{y}) = \frac{f_{11}}{f_{01}+f_{10}+f_{11}} = \frac{0}{2+2+0} = \boxed{0}$

Summary

Measure	Value	Interpretation
Cosine similarity	0	Vectors are orthogonal
Correlation	-1	Perfectly negatively correlated
Euclidean distance	2	Maximum possible distance for these binary vectors
Jaccard similarity	0	No 1-1 overlap at all — completely dissimilar on "presence"

All four measures agree that x and y are maximally dissimilar — they're exact bitwise complements of each other.

Q3a. Build a decision tree using the training data in the table given below. Divide the height attribute into ranges as follows : \{0, 1.6\], (1.6, 1.7\], (1.7, 1.8\], (1.8, 1.9\], (1.9, 2.0\], (2.0, 5.0\] | Gender | Height (m) | Class | |:---:|:---:|:---:| | F | 1.6 | Short | | M | 2 | Tall | | F | 1.9 | Medium | | F | 1.88 | Medium | | F | 1.7 | Short | | M | 1.85 | Medium | | F | 1.6 | Short | | M | 1.7 | Short | | M | 2.2 | Tall | | M | 2.1 | Tall | | F | 1.8 | Medium | | M | 1.95 | Medium | | F | 1.9 | Medium | | F | 1.8 | Medium | | F | 1.75 | Medium |20218m

Module 3: Classification and Prediction

Build a decision tree using the training data in the table given below. Divide the height attribute into ranges as follows :
$\{0, 1.6\], (1.6, 1.7\], (1.7, 1.8\], (1.8, 1.9\], (1.9, 2.0\], (2.0, 5.0\]$

Gender	Height (m)	Class
F	1.6	Short
M	2	Tall
F	1.9	Medium
F	1.88	Medium
F	1.7	Short
M	1.85	Medium
F	1.6	Short
M	1.7	Short
M	2.2	Tall
M	2.1	Tall
F	1.8	Medium
M	1.95	Medium
F	1.9	Medium
F	1.8	Medium
F	1.75	Medium

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Q3b. What is over-fitting and under-fitting of a model? When do they happen?20216m
Module 3: Classification and Prediction
What is over-fitting and under-fitting of a model? When do they happen?

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Q4a. A university plans to build a data warehouse that would help them in analyzing the performance of the students in various courses in different academic sessions. They want to analyze if there is any relation between the average grade of a course and the number of students attending it. They would also like to know if there were some courses offered but did not have any students registered for them. Relative performance among boys and girls and average grades of students from various States and cities of the country for each course must be analyzed and also overall CGPA. Design a star schema for such a data warehouse clearly identifying the fact table(s) and dimension table(s), their primary key(s) and foreign key(s). Your schema should at least be able to satisfy the above-mentioned analysis requirements. You may consider other suitable attributes for the dimension table(s).202110m

Module 1: Data Warehousing and Business Analysis

A university plans to build a data warehouse that would help them in analyzing the performance of the students in various courses in different academic sessions. They want to analyze if there is any relation between the average grade of a course and the number of students attending it. They would also like to know if there were some courses offered but did not have any students registered for them. Relative performance among boys and girls and average grades of students from various States and cities of the country for each course must be analyzed and also overall CGPA.

Design a star schema for such a data warehouse clearly identifying the fact table(s) and dimension table(s), their primary key(s) and foreign key(s). Your schema should at least be able to satisfy the above-mentioned analysis requirements. You may consider other suitable attributes for the dimension table(s).

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Worked Solution

Star Schema Design — University Performance Data Warehouse

Requirements Analysis

We need to analyze: grade-vs-enrollment relation, courses with zero registrations, gender-wise performance, state/city-wise average grade per course, and overall CGPA. This means our grain (finest fact-table row) should be one student's performance in one course, in one academic session.

Fact Table

FACT_PERFORMANCE

Column	Type	Key
student_key	FK → DIM_STUDENT	PK (composite)
course_key	FK → DIM_COURSE	PK (composite)
session_key	FK → DIM_SESSION	PK (composite)
grade_point	Measure
sgpa	Measure
is_registered	Measure (flag, 1/0)

Using a is_registered flag (or simply having a row exist) lets us detect courses offered but with zero enrolled students via a LEFT JOIN from DIM_COURSE × DIM_SESSION against the fact table.

Dimension Tables

DIM_SESSION
| session_key (PK) | academic_year | semester_type (Odd/Even) |

Schema Diagram

                  DIM_STUDENT
                 (student_key PK)
                  gender, state, city, cgpa
                        │
                        │FK
                        ▼
DIM_COURSE ──FK──▶ FACT_PERFORMANCE ◀──FK── DIM_SESSION
(course_key PK)     (student_key,        (session_key PK)
course_name,         course_key,          academic_year,
department           session_key,         semester_type
                      grade_point,
                      sgpa,
                      is_registered)

How it satisfies the requirements

Grade vs enrollment relation → aggregate AVG(grade_point) and COUNT(student_key) grouped by course_key.
Courses with no students → DIM_COURSE LEFT JOIN FACT_PERFORMANCE, filter WHERE fact IS NULL.
Gender-wise performance → join DIM_STUDENT.gender, group by course & gender.
State/city-wise average grade → join DIM_STUDENT.state/city, group by course.
Overall CGPA → stored directly in DIM_STUDENT.cgpa (or derived as AVG(sgpa) across sessions per student).

This is a classic star schema: one fact table with numeric measures (grade_point, sgpa) surrounded by denormalized dimension tables, which keeps analytical queries fast (few joins, simple filters).

Q4b. Write an SQL query that runs on the schema designed in Q4a and returns the average SGPA of boys from the State of Karnataka for each Autumn (odd) semester during the years 2005–2009.20214m
Module 1: Data Warehousing and Business Analysis
Write an SQL query that runs on the schema designed in Q4a and returns the average SGPA of boys from the State of Karnataka for each Autumn (odd) semester during the years 2005–2009.

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Worked Solution
SQL Query — Avg SGPA of Boys from Karnataka, Odd Semesters (2005–2009)

Using the star schema from Q4a (FACT_PERFORMANCE, DIM_STUDENT, DIM_SESSION):

SELECT ds.academic_year, AVG(fp.sgpa) AS avg_sgpa FROM FACT_PERFORMANCE fp JOIN DIM_STUDENT st ON fp.student_key = st.student_key JOIN DIM_SESSION ds ON fp.session_key = ds.session_key WHERE st.gender = 'M' AND st.state = 'Karnataka' AND ds.semester_type = 'Odd' AND ds.academic_year BETWEEN 2005 AND 2009 GROUP BY ds.academic_year ORDER BY ds.academic_year;

How it maps to the requirement

st.gender = 'M' → restricts to boys.

st.state = 'Karnataka' → restricts to the given state.

ds.semester_type = 'Odd' → restricts to Autumn semesters only.

ds.academic_year BETWEEN 2005 AND 2009 → restricts the year range.

GROUP BY ds.academic_year → produces one average SGPA row per year, as required ("for each Autumn semester during 2005–2009").

The joins connect the fact table's measures (sgpa) to the descriptive attributes (gender, state, academic_year) stored in the dimension tables — this is exactly why a star schema keeps such analytical queries simple (only two joins, no deep nesting).

Q5a. The results of an exam are recorded along with some data about the students. The results can be found in the table below : | ID | Cell No. | Language | Passed all Assignments | GPA | Passed Exam | | :---: | :---: | :---: | :---: | :---: | :---: | | 1 | 93333-11101 | Java | No | 3.1 | Yes | | 2 | 93333-11112 | Java | No | 2.0 | No | | 3 | 93333-11102 | C++ | Yes | 3.5 | Yes | | 4 | 93333-11113 | Python | Yes | 2.5 | Yes | | 5 | 93333-11103 | Java | Yes | 3.9 | No | | 6 | 93333-11114 | C++ | No | 2.9 | No | | 7 | 93333-11104 | Java | No | 1.9 | No | | 8 | 93333-11115 | Python | Yes | 3.2 | Yes | In no more than one page of text, describe the design of a K-Nearest Neighbour classifier to predict if a student will fail or pass the exam.202110m

Module 3: Classification and Prediction

The results of an exam are recorded along with some data about the students. The results can be found in the table below :

ID	Cell No.	Language	Passed all Assignments	GPA	Passed Exam
1	93333-11101	Java	No	3.1	Yes
2	93333-11112	Java	No	2.0	No
3	93333-11102	C++	Yes	3.5	Yes
4	93333-11113	Python	Yes	2.5	Yes
5	93333-11103	Java	Yes	3.9	No
6	93333-11114	C++	No	2.9	No
7	93333-11104	Java	No	1.9	No
8	93333-11115	Python	Yes	3.2	Yes

In no more than one page of text, describe the design of a K-Nearest Neighbour classifier to predict if a student will fail or pass the exam.

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Q5b. Using the dataset from Q5a, use your K-NN classifier to predict whether the following student (who overslept and missed the original exam) will pass the re-exam : | ID | Cell No. | Language | Passed all Assignments | GPA | Passed Exam | | :---: | :---: | :---: | :---: | :---: | :---: | | 9 | 93333-11109 | C++ | Yes | 3.0 | ?? |20214m
Module 3: Classification and Prediction
Using the dataset from Q5a, use your K-NN classifier to predict whether the following student (who overslept and missed the original exam) will pass the re-exam :

ID Cell No. Language Passed all Assignments GPA Passed Exam

9 93333-11109 C++ Yes 3.0 ??

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Q6. Define any four of the following briefly : (a) Occam's Razor (b) ROC Curve (c) Vector Space Model (d) Data Marts (e) Multilevel Association Rules (f) OLAM202114m
Module 2: Data Mining and Association Rule Mining
Define any four of the following briefly :

(a) Occam's Razor
(b) ROC Curve
(c) Vector Space Model
(d) Data Marts
(e) Multilevel Association Rules
(f) OLAM

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Worked Solution
Six Key Terms Defined (pick any four for the exam)

(a) Occam's Razor

A principle stating that among competing hypotheses that fit the data equally well, the simplest one should be preferred. In data mining/ML, this motivates choosing simpler models (fewer parameters, shallower decision trees) over needlessly complex ones, since simpler models generalize better and are less prone to overfitting.

(b) ROC Curve

Receiver Operating Characteristic curve — plots the True Positive Rate (Sensitivity) against the False Positive Rate (1−Specificity) at various classification thresholds.

TPR 1 │ ●───● │ ● │ ● │ ● │● 0 └──────────────── FPR 0 1

The Area Under the Curve (AUC) summarizes overall classifier performance — AUC=1 is a perfect classifier, AUC=0.5 is random guessing (the diagonal line).

(c) Vector Space Model (VSM)

An algebraic model used mainly in text/document mining and information retrieval, where each document (or object) is represented as a vector of feature weights (e.g., TF-IDF term weights) in a high-dimensional space. Similarity between documents is then computed geometrically, typically using cosine similarity between their vectors.

(d) Data Marts

A subset of a data warehouse, scoped to a specific business line, department, or subject area (e.g., a Sales data mart, an HR data mart). Data marts are smaller, faster to query, and easier to maintain than the full enterprise warehouse, letting individual teams get focused analytical access without querying the entire warehouse.

(e) Multilevel Association Rules

Association rules mined at different levels of a concept hierarchy rather than only at the raw item level. For example, instead of just "{Bread} ⇒ {Butter}", multilevel mining can surface rules at higher abstraction levels like "{Bakery Items} ⇒ {Dairy Items}", or at lower levels like "{Whole Wheat Bread} ⇒ {Salted Butter}". This reveals patterns that might be too sparse (low support) at the finest level but become significant when items are generalized.

(f) OLAM (On-Line Analytical Mining)

Integrates OLAP with data mining, allowing mining algorithms to be applied directly on multidimensional data cubes rather than flat/raw data. This lets analysts interactively drill/roll/slice/dice to narrow down interesting regions of the cube, then apply mining functions (classification, clustering, association) at whatever granularity they've navigated to — combining OLAP's exploratory interactivity with mining's automatic pattern discovery.
Q7a. A database has five transactions. Let min_sup = 60% and min_conf = 80%. | TID | Items_bought | |---|---| | T100 | {M, O, N, K, E, Y} | | T200 | {D, O, N, K, E, Y} | | T300 | {M, A, K, E} | | T400 | {M, U, C, K, Y} | | T500 | {C, O, O, K, I, E} | (i) Find all frequent item sets using a priori algorithm. (ii) List all the strong association rules (with support s and confidence c) matching the following metarule, where X is a variable representing customers: \forall x \in \text{transactions}, \text{buys}(X, \text{item}_1) \wedge \text{buys}(X, \text{item}_2) \Rightarrow \text{buys}(X, \text{item}_3)\ [s, c]20218m
Module 2: Data Mining and Association Rule Mining
A database has five transactions. Let min_sup = 60% and min_conf = 80%.

TID Items_bought

T100 {M, O, N, K, E, Y}

T200 {D, O, N, K, E, Y}

T300 {M, A, K, E}

T400 {M, U, C, K, Y}

T500 {C, O, O, K, I, E}

(i) Find all frequent item sets using a priori algorithm.

(ii) List all the strong association rules (with support $s$ and confidence $c$ ) matching the following metarule, where $X$ is a variable representing customers:
$\forall x \in \text{transactions}, \text{buys}(X, \text{item}_1) \wedge \text{buys}(X, \text{item}_2) \Rightarrow \text{buys}(X, \text{item}_3)\ [s, c]$

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Worked Solution
(i) Frequent Itemsets via Apriori — MONKEY Dataset

Same dataset as before, min_sup = 60% (count ≥ 3), min_conf = 80%.

TID Items

T100 M, O, N, K, E, Y

T200 D, O, N, K, E, Y

T300 M, A, K, E

T400 M, U, C, K, Y

T500 C, O, K, I, E

L1 (count ≥3)

{M}:3, {O}:3, {K}:5, {E}:4, {Y}:3

L2 (count ≥3, from valid C2 candidates)

{K,M}:3, {K,O}:3, {O,E}:3, {K,E}:4, {K,Y}:3

L3

Only candidate surviving the apriori-prune test: {K,O,E} — check: {K,O}✅ {K,E}✅ {O,E}✅ all in L2 → count in DB (T100, T200, T500) = 3 → frequent.

All frequent itemsets:

{M} {O} {K} {E} {Y} {K,M} {K,O} {O,E} {K,E} {K,Y} {K,O,E}

(11 itemsets total — see full step-by-step derivation in Q65's solution for the same dataset.)

(ii) Strong Rules Matching the Metarule

Metarule: $\forall x, \text{buys}(X,\text{item}_1) \wedge \text{buys}(X,\text{item}_2) \Rightarrow \text{buys}(X,\text{item}_3)\ [s,c]$

This asks specifically for rules with exactly 2 items in the antecedent and 1 item in the consequent. The only frequent itemset large enough to produce such a rule is the 3-itemset {K, O, E} (support count = 3, support = 3/5 = 60%).

We test all three possible 2→1 splits of {K, O, E}:

Rule Confidence = support({K,O,E}) / support(antecedent) Strong? (≥80%)

{K,O} ⇒ E 3/3 = 100% ✅

{K,E} ⇒ O 3/4 = 75% ❌

{O,E} ⇒ K 3/3 = 100% ✅

Strong Association Rules (matching the metarule)

$\text{buys}(X, K) \wedge \text{buys}(X, O) \Rightarrow \text{buys}(X, E) \quad [s = 60\%,\ c = 100\%]$

$\text{buys}(X, O) \wedge \text{buys}(X, E) \Rightarrow \text{buys}(X, K) \quad [s = 60\%,\ c = 100\%]$

The rule {K,E} ⇒ O fails the min_conf = 80% threshold (only 75%), so it is not a strong rule and is excluded — even though {K,O,E} itself is a frequent (frequent) itemset, not every rule derived from it clears the confidence bar.
Q7b. Explain DBSCAN briefly.20216m
Module 4: Cluster Analysis
Explain DBSCAN briefly.

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Q8a. Why do we use ensemble methods? Describe an ensemble method.20217m
Module 3: Classification and Prediction
Why do we use ensemble methods? Describe an ensemble method.

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ID	Cell No.	Language	Passed all Assignments	GPA	Passed Exam
9	93333-11109	C++	Yes	3.0	??

TID	Items_bought
T100	{M, O, N, K, E, Y}
T200	{D, O, N, K, E, Y}
T300	{M, A, K, E}
T400	{M, U, C, K, Y}
T500	{C, O, O, K, I, E}

TID	Items
T100	M, O, N, K, E, Y
T200	D, O, N, K, E, Y
T300	M, A, K, E
T400	M, U, C, K, Y
T500	C, O, K, I, E

Rule	Confidence = support({K,O,E}) / support(antecedent)	Strong? (≥80%)
{K,O} ⇒ E	3/3 = 100%	✅
{K,E} ⇒ O	3/4 = 75%	❌
{O,E} ⇒ K	3/3 = 100%	✅

Q8b. Differentiate among OLAP, MOLAP and HOLAP.20217m

Module 1: Data Warehousing and Business Analysis

Differentiate among OLAP, MOLAP and HOLAP.

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Worked Solution

OLAP vs MOLAP vs HOLAP

These are three server architectures for implementing OLAP, differing in where and how the multidimensional data is physically stored.

ROLAP (Relational OLAP)

Stores data in standard relational tables (star/snowflake schema).
Multidimensional operations are translated into SQL at query time.
Pros: scales to very large data volumes, reuses existing RDBMS technology.
Cons: slower query performance (SQL translation overhead), limited by SQL's multidimensional expressiveness.

MOLAP (Multidimensional OLAP)

Stores data in a proprietary multidimensional array/cube structure (pre-computed).
Pros: very fast query response (data pre-aggregated), rich analytical functions.
Cons: cube-build time can be long, storage can explode for high-cardinality/sparse dimensions ("data explosion"), limited scalability for very large datasets.

HOLAP (Hybrid OLAP)

Combines both: detailed data stays in relational tables (ROLAP), while summary/aggregated data is stored in MOLAP cubes.
Pros: balances MOLAP's speed for aggregates with ROLAP's scalability for detailed data.
Cons: more complex architecture to build and maintain.

Comparison Table

Aspect	ROLAP	MOLAP	HOLAP
Storage	Relational tables	Multidimensional array/cube	Mixed (relational + cube)
Query speed	Slower (SQL translation)	Fastest (pre-computed)	Balanced
Scalability	Very high (large data)	Limited (cube size)	High
Storage overhead	Low	High (sparse data explosion)	Moderate
Best for	Very large, detailed datasets	Fast, smaller/aggregated cubes	Mix of both needs

ROLAP:  [Relational DB] ──SQL──▶ query engine ──▶ result
MOLAP:  [Precomputed Cube] ──direct lookup──▶ result   (fast!)
HOLAP:  [Detail: Relational] + [Summary: Cube] ──▶ result (best of both)

In short: ROLAP trades speed for scalability, MOLAP trades scalability for speed, and HOLAP tries to get both by splitting detail vs summary data across the two storage models.

Q9a. Describe classification accuracy. How do we measure it? Differentiate classification accuracy with precision.20217m
Module 3: Classification and Prediction
Describe classification accuracy. How do we measure it? Differentiate classification accuracy with precision.

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Q9b. Distinguish between noise and outliers. (i) Is noise ever interesting or desirable? Outliers? (ii) Can noise objects be outliers? (iii) Can noise make a typical value into an unusual one or vice versa?20217m
Module 4: Cluster Analysis
Distinguish between noise and outliers.

(i) Is noise ever interesting or desirable? Outliers?
(ii) Can noise objects be outliers?
(iii) Can noise make a typical value into an unusual one or vice versa?

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Data Mining

Is it a Data Mining Task?

(i) Dividing customers according to profitability

(ii) Monitoring and predicting failures in a hydropower plant

Attribute Classification

(i) Distance from center of campus

(ii) Ability to pass light: opaque, translucent, transparent

Ambiguity note

Handling Missing Values — At Least Two Methods

Z-Score Normalization

Effect

Why use it

The Apriori (A Priori) Principle

Why it holds

Diagram

Why it matters

Slice and Dice (OLAP Operations)

Slice

Dice

Key Difference

Why "Data Mining" is a Misnomer

Reasoning

Similarity/Distance Measures for x = {0,1,0,1}, y = {1,0,1,0}

Setup

1. Cosine Similarity

2. Correlation

3. Euclidean Distance

4. Jaccard Similarity

Summary

Star Schema Design — University Performance Data Warehouse

Requirements Analysis

Fact Table

Dimension Tables

Schema Diagram

How it satisfies the requirements

SQL Query — Avg SGPA of Boys from Karnataka, Odd Semesters (2005–2009)

How it maps to the requirement

Six Key Terms Defined (pick any four for the exam)

(a) Occam's Razor

(b) ROC Curve

(c) Vector Space Model (VSM)

(d) Data Marts

(e) Multilevel Association Rules

(f) OLAM (On-Line Analytical Mining)

(i) Frequent Itemsets via Apriori — MONKEY Dataset

L1 (count ≥3)

L2 (count ≥3, from valid C2 candidates)

L3

(ii) Strong Rules Matching the Metarule

Strong Association Rules (matching the metarule)

OLAP vs MOLAP vs HOLAP

ROLAP (Relational OLAP)

MOLAP (Multidimensional OLAP)

HOLAP (Hybrid OLAP)

Comparison Table