CS 7641 UL Unit Quiz| Questions and Answers Latest
Updated 2026/2027 | Georgia Institute of Technology
UL Unit Quiz
CS 7641: Machine Learning
Solutions
1 Question 1 - Clustering
Part 1. Clustering Scenario (MCMA)
You are given a 2D dataset with the following characteristics:
• There are two elongated, curved clusters shaped like crescents (i.e., non-spherical).
• The clusters are close together at one end but clearly separated in density.
You are tasked with selecting a clustering method to analyze this data.
Which of the following statements are true in this scenario? Select all that apply.
Correct Answers: A, B, C, F
A. K-Means is likely to struggle with this dataset because it assumes clusters are
spherical and uses Euclidean distance.
True. K-Means assumes isotropic (spherical) clusters and uses Euclidean distance for assign-
ment. It struggles with non-spherical clusters like crescents, often misclassifying points along
the curved boundaries.
B. Single linkage is likely to merge the two clusters prematurely due to chaining,
especially at the close end.
True. Single linkage uses the minimum pairwise distance between clusters and is sensitive to
chaining effects. It may incorrectly merge nearby points at the close ends of the crescents
into one cluster.
C. EM with Gaussian Mixture Models may succeed if enough components are used,
as it allows overlapping densities and soft assignments.
True. Although GMMs assume ellipsoidal distributions, using multiple components can ap-
proximate non-spherical clusters. Soft assignments also allow modeling uncertainty in bound-
ary regions.
D. K-Means will perform well as long as the number of clusters is set to 2.
False. Even with the correct number of clusters, K-Means performs poorly on non-convex
1
, shapes like crescents due to its assumption of spherical clusters and Euclidean distance min-
imization.
E. Single linkage will always produce better clusters than K-Means in this scenario.
False. Single linkage may perform better in some cases, but it is sensitive to noise and
chaining, which can lead to incorrect early merges. There is no guarantee it will outperform
K-Means in all scenarios.
F. Soft clustering methods like EM can assign a point near the overlapping region to
both clusters with non-zero responsibility.
True. EM assigns a probability distribution over clusters to each point. This is particularly
useful near overlapping or ambiguous regions, where hard clustering would force a less accurate
binary decision.
Part 2. Choosing k (MCMA)
You are clustering a dataset of 500 points that visually appear to form 5 compact, roughly
spherical clusters, with some noise around the edges.
You are experimenting with different values of k in K-Means and EM-based Gaussian
Mixture Models (GMMs).
Which of the following statements are true as you vary k? Select all that apply.
Correct Answers: A, B, C, E
A. If k is set too low (e.g., k = 2), both K-Means and EM are likely to merge multiple
true clusters into one.
True. Setting k too low forces the algorithm to combine multiple actual clusters, resulting in
loss of structure and underfitting.
B. If k is set too high (e.g., k = 20), K-Means may split a true cluster into multiple
artificial parts.
True. Overestimating k leads to fragmentation of true clusters into unnecessary sub-clusters.
C. For very high k, the log-likelihood in EM for GMMs will generally increase, even
if the clustering becomes meaningless.
True. EM optimizes log-likelihood, which generally increases as k increases, but this can lead
to overfitting and poor interpretability.
D. Choosing a high k improves interpretability, as each cluster becomes more distinct
and informative.
False. Excessive k values can reduce interpretability by fragmenting meaningful clusters and
producing noisy results.
E. If k is too low, EM may still assign soft probabilities that capture structure better
than K-Means.
True. EM provides soft clustering, which can represent uncertainty and overlapping distri-
butions even when k is low.
F. Both K-Means and EM are guaranteed to produce globally optimal clusterings for
any fixed k.
2
Updated 2026/2027 | Georgia Institute of Technology
UL Unit Quiz
CS 7641: Machine Learning
Solutions
1 Question 1 - Clustering
Part 1. Clustering Scenario (MCMA)
You are given a 2D dataset with the following characteristics:
• There are two elongated, curved clusters shaped like crescents (i.e., non-spherical).
• The clusters are close together at one end but clearly separated in density.
You are tasked with selecting a clustering method to analyze this data.
Which of the following statements are true in this scenario? Select all that apply.
Correct Answers: A, B, C, F
A. K-Means is likely to struggle with this dataset because it assumes clusters are
spherical and uses Euclidean distance.
True. K-Means assumes isotropic (spherical) clusters and uses Euclidean distance for assign-
ment. It struggles with non-spherical clusters like crescents, often misclassifying points along
the curved boundaries.
B. Single linkage is likely to merge the two clusters prematurely due to chaining,
especially at the close end.
True. Single linkage uses the minimum pairwise distance between clusters and is sensitive to
chaining effects. It may incorrectly merge nearby points at the close ends of the crescents
into one cluster.
C. EM with Gaussian Mixture Models may succeed if enough components are used,
as it allows overlapping densities and soft assignments.
True. Although GMMs assume ellipsoidal distributions, using multiple components can ap-
proximate non-spherical clusters. Soft assignments also allow modeling uncertainty in bound-
ary regions.
D. K-Means will perform well as long as the number of clusters is set to 2.
False. Even with the correct number of clusters, K-Means performs poorly on non-convex
1
, shapes like crescents due to its assumption of spherical clusters and Euclidean distance min-
imization.
E. Single linkage will always produce better clusters than K-Means in this scenario.
False. Single linkage may perform better in some cases, but it is sensitive to noise and
chaining, which can lead to incorrect early merges. There is no guarantee it will outperform
K-Means in all scenarios.
F. Soft clustering methods like EM can assign a point near the overlapping region to
both clusters with non-zero responsibility.
True. EM assigns a probability distribution over clusters to each point. This is particularly
useful near overlapping or ambiguous regions, where hard clustering would force a less accurate
binary decision.
Part 2. Choosing k (MCMA)
You are clustering a dataset of 500 points that visually appear to form 5 compact, roughly
spherical clusters, with some noise around the edges.
You are experimenting with different values of k in K-Means and EM-based Gaussian
Mixture Models (GMMs).
Which of the following statements are true as you vary k? Select all that apply.
Correct Answers: A, B, C, E
A. If k is set too low (e.g., k = 2), both K-Means and EM are likely to merge multiple
true clusters into one.
True. Setting k too low forces the algorithm to combine multiple actual clusters, resulting in
loss of structure and underfitting.
B. If k is set too high (e.g., k = 20), K-Means may split a true cluster into multiple
artificial parts.
True. Overestimating k leads to fragmentation of true clusters into unnecessary sub-clusters.
C. For very high k, the log-likelihood in EM for GMMs will generally increase, even
if the clustering becomes meaningless.
True. EM optimizes log-likelihood, which generally increases as k increases, but this can lead
to overfitting and poor interpretability.
D. Choosing a high k improves interpretability, as each cluster becomes more distinct
and informative.
False. Excessive k values can reduce interpretability by fragmenting meaningful clusters and
producing noisy results.
E. If k is too low, EM may still assign soft probabilities that capture structure better
than K-Means.
True. EM provides soft clustering, which can represent uncertainty and overlapping distri-
butions even when k is low.
F. Both K-Means and EM are guaranteed to produce globally optimal clusterings for
any fixed k.
2
