CS 7643 DEEP LEARNING QUIZ
QUESTIONS AND ANSWERS
ALREADY GRADED A+. 100%
LATEST MOCK PRACTICE SET
248 Questions with Answers and Detailed Rationales
100 PERCENT GUARANTEED PASS
INSTANT DOWNLOAD ANSWERS INCLUDED
IMPORTANCE OF THIS DOCUMENT
This comprehensive examination preparation guide has been meticulously developed to help you succeed in the
CS 7643 DEEP LEARNING QUIZ QUESTIONS AND ANSWERS ALREADY GRADED A+. 100% VERIFIED
SOLUTIONS | UPDATED PER LATEST GUIDELINES | GRADED A+. It contains 248 carefully selected questions
that reflect the most current exam content and testing strategies. Each question is accompanied by a correct
answer and a detailed rationale that explains the underlying pathophysiology, pharmacology, or clinical reasoning.
Self-Assessment – Test your knowledge and Exam Preparation – Familiarize yourself with the
identify areas requiring further question format and content
study areas
Concept Reinforcement – Deepen your Confidence Building – Develop test-taking
understanding through strategies and reduce
evidence-based exam anxiety
rationales
Time Management – Practice answering
questions under simulated
exam conditions
Review Summary 248 Questions
Foundations - Application - CS 7643 DEEP Learning AND Already A 100 Solutions Updated PER
Guidelines A DEEP Learning CS 7643 Graduate
All answers with rationales
,Table of Contents
Section A - Neural Networks Basics Section B - Convolutional Neural
Questions 1 to 62 Networks CNNS
Questions 63 to 124
Section C - Recurrent Neural Section D - Optimization AND
Networks RNNS AND Lstms Regularization
Questions 125 to 186 Questions 187 to 248
,Section A - Neural Networks Basics
Q1.
In a deep residual network with bottleneck blocks, the residual function F(x) = W_2 (W_1 x)
where is ReLU. If the input x has 256 channels, the bottleneck reduces to 64 channels in
the first convolution, and the output of the block has 256 channels, what is the total
number of parameters in the block (excluding biases)?
A. 256*64*1*1 + 64*64*3*3 + 64*256*1*1 = B. 256*64*1*1 + 64*64*3*3 + 64*256*1*1 =
16,384 + 36,864 + 16,384 = 69,632 16,384 + 36,864 + 16,384 = 69,632, plus
shortcut if dimensions change
C. 256*64*1*1 + 64*64*3*3 + 64*256*1*1 = D. 256*64*1*1 + 64*64*3*3 + 64*256*1*1 =
16,384 + 36,864 + 16,384 = 69,632, plus 16,384 + 36,864 + 16,384 = 69,632, and no
shortcut projection if needed shortcut parameters since dimensions
match
Correct: D - 256*64*1*1 + 64*64*3*3 + 64*256*1*1 = 16,384 + 36,864 + 16,384 = 69,632, and
no shortcut parameters since dimensions match
Rationale:The bottleneck block has three convolutions: 1x1 reducing 256 to 64, 3x3 with 64
input/output, and 1x1 expanding back to 256. Total parameters = 256*64 + 64*64*9 + 64*256
= 16,384 + 36,864 + 16,384 = 69,632. Since input and output channels match (256), no
shortcut projection is needed, so no additional parameters.
Q2.
Consider training a deep neural network with Batch Normalization (BN) after each
convolutional layer. During inference, BN uses running averages of mean and variance
computed during training. If the training data has a covariate shift such that the test batch
has mean _test and variance ²_test significantly different from the running averages,
which of the following is the most likely effect on the model's test accuracy?
A. The model will adapt quickly due to BN's B. The model's accuracy will degrade
ability to handle covariate shift because BN uses population statistics that
may not match the test distribution
C. The model's accuracy will remain D. The model will fail entirely because BN
unchanged because BN layers learn affine cannot be used with covariate shift
parameters that compensate for any shift
Correct: B - The model's accuracy will degrade because BN uses population statistics that
may not match the test distribution
Page 3
, Section A - Neural Networks Basics
Rationale: During inference, BN uses running averages of mean and variance from training. If
test data has a different distribution, the normalized activations will be miscalibrated, leading
to degraded accuracy. While affine parameters can scale and shift, they are learned on
training distribution and may not compensate for large shifts. Option A is false because BN
does not adapt at test time; option C is false because affine parameters are fixed; option D is
too extreme.
Q3.
In a variational autoencoder (VAE) with a Gaussian prior and a Gaussian encoder q(z|x) =
N((x), ²(x)I), the reparameterization trick samples z = + * , where ~ N(0, I). Which of the
following correctly describes the gradient of the ELBO with respect to the encoder
parameters ?
A. _ ELBO = E_{~N(0,I)} [ _ (log p(x|z) + log B. _ ELBO = E_{~N(0,I)} [ _ (log p(x|z) + log
p(z) - log q(z|x)) ] where z is a function of p(z) - log q(z|x)) ] but the gradient does not
through and flow through the sampling because of
reparameterization
C. _ ELBO = E_{~N(0,I)} [ _ (log p(x|z) + log D. _ ELBO = E_{~N(0,I)} [ _ (log p(x|z) + log
p(z) - log q(z|x)) ] and the gradient flows p(z) - log q(z|x)) ] but the gradient through
through both the encoder parameters and the sampling is zero because is
the sampling operation directly independent of
Correct: A - _ ELBO = E_{~N(0,I)} [ _ (log p(x|z) + log p(z) - log q(z|x)) ] where z is a
function of through and
Rationale:The reparameterization trick expresses z as a deterministic function of Æ (via ¼ and
) and , allowing gradients to flow through the sampling operation. Thus, the gradient of the
ELBO with respect to includes terms from log p(x|z), log p(z), and log q(z|x) where z depends
on . Option B is incorrect because reparameterization enables gradient flow; option C is
incorrect because sampling is not directly differentiable; option D is incorrect because the
gradient is not zero.
Q4.
In a transformer model with multi-head attention, the output of the attention layer is
computed as: MultiHead(Q, K, V) = Concat(head_1, ..., head_h) W^O. Each head_i =
Attention(QW_i^Q, KW_i^K, VW_i^V). If the model dimension d_model = 512 and the
number of heads h = 8, what is the dimension of each head's output (d_v) and the total
number of parameters in the projection matrices W_i^Q, W_i^K, W_i^V, and W^O?
A. d_v = 64; total parameters = 8 * (512*64 B. d_v = 64; total parameters = 8 * (512*64
+ 512*64 + 512*64) + 512*512 = 8*98,304 + + 512*64 + 512*64) + 512*512 = 8*98,304 +
262,144 = 1,048,576 262,144 = 1,048,576, but note that W_i^K
and W_i^V are shared across heads
Page 4
QUESTIONS AND ANSWERS
ALREADY GRADED A+. 100%
LATEST MOCK PRACTICE SET
248 Questions with Answers and Detailed Rationales
100 PERCENT GUARANTEED PASS
INSTANT DOWNLOAD ANSWERS INCLUDED
IMPORTANCE OF THIS DOCUMENT
This comprehensive examination preparation guide has been meticulously developed to help you succeed in the
CS 7643 DEEP LEARNING QUIZ QUESTIONS AND ANSWERS ALREADY GRADED A+. 100% VERIFIED
SOLUTIONS | UPDATED PER LATEST GUIDELINES | GRADED A+. It contains 248 carefully selected questions
that reflect the most current exam content and testing strategies. Each question is accompanied by a correct
answer and a detailed rationale that explains the underlying pathophysiology, pharmacology, or clinical reasoning.
Self-Assessment – Test your knowledge and Exam Preparation – Familiarize yourself with the
identify areas requiring further question format and content
study areas
Concept Reinforcement – Deepen your Confidence Building – Develop test-taking
understanding through strategies and reduce
evidence-based exam anxiety
rationales
Time Management – Practice answering
questions under simulated
exam conditions
Review Summary 248 Questions
Foundations - Application - CS 7643 DEEP Learning AND Already A 100 Solutions Updated PER
Guidelines A DEEP Learning CS 7643 Graduate
All answers with rationales
,Table of Contents
Section A - Neural Networks Basics Section B - Convolutional Neural
Questions 1 to 62 Networks CNNS
Questions 63 to 124
Section C - Recurrent Neural Section D - Optimization AND
Networks RNNS AND Lstms Regularization
Questions 125 to 186 Questions 187 to 248
,Section A - Neural Networks Basics
Q1.
In a deep residual network with bottleneck blocks, the residual function F(x) = W_2 (W_1 x)
where is ReLU. If the input x has 256 channels, the bottleneck reduces to 64 channels in
the first convolution, and the output of the block has 256 channels, what is the total
number of parameters in the block (excluding biases)?
A. 256*64*1*1 + 64*64*3*3 + 64*256*1*1 = B. 256*64*1*1 + 64*64*3*3 + 64*256*1*1 =
16,384 + 36,864 + 16,384 = 69,632 16,384 + 36,864 + 16,384 = 69,632, plus
shortcut if dimensions change
C. 256*64*1*1 + 64*64*3*3 + 64*256*1*1 = D. 256*64*1*1 + 64*64*3*3 + 64*256*1*1 =
16,384 + 36,864 + 16,384 = 69,632, plus 16,384 + 36,864 + 16,384 = 69,632, and no
shortcut projection if needed shortcut parameters since dimensions
match
Correct: D - 256*64*1*1 + 64*64*3*3 + 64*256*1*1 = 16,384 + 36,864 + 16,384 = 69,632, and
no shortcut parameters since dimensions match
Rationale:The bottleneck block has three convolutions: 1x1 reducing 256 to 64, 3x3 with 64
input/output, and 1x1 expanding back to 256. Total parameters = 256*64 + 64*64*9 + 64*256
= 16,384 + 36,864 + 16,384 = 69,632. Since input and output channels match (256), no
shortcut projection is needed, so no additional parameters.
Q2.
Consider training a deep neural network with Batch Normalization (BN) after each
convolutional layer. During inference, BN uses running averages of mean and variance
computed during training. If the training data has a covariate shift such that the test batch
has mean _test and variance ²_test significantly different from the running averages,
which of the following is the most likely effect on the model's test accuracy?
A. The model will adapt quickly due to BN's B. The model's accuracy will degrade
ability to handle covariate shift because BN uses population statistics that
may not match the test distribution
C. The model's accuracy will remain D. The model will fail entirely because BN
unchanged because BN layers learn affine cannot be used with covariate shift
parameters that compensate for any shift
Correct: B - The model's accuracy will degrade because BN uses population statistics that
may not match the test distribution
Page 3
, Section A - Neural Networks Basics
Rationale: During inference, BN uses running averages of mean and variance from training. If
test data has a different distribution, the normalized activations will be miscalibrated, leading
to degraded accuracy. While affine parameters can scale and shift, they are learned on
training distribution and may not compensate for large shifts. Option A is false because BN
does not adapt at test time; option C is false because affine parameters are fixed; option D is
too extreme.
Q3.
In a variational autoencoder (VAE) with a Gaussian prior and a Gaussian encoder q(z|x) =
N((x), ²(x)I), the reparameterization trick samples z = + * , where ~ N(0, I). Which of the
following correctly describes the gradient of the ELBO with respect to the encoder
parameters ?
A. _ ELBO = E_{~N(0,I)} [ _ (log p(x|z) + log B. _ ELBO = E_{~N(0,I)} [ _ (log p(x|z) + log
p(z) - log q(z|x)) ] where z is a function of p(z) - log q(z|x)) ] but the gradient does not
through and flow through the sampling because of
reparameterization
C. _ ELBO = E_{~N(0,I)} [ _ (log p(x|z) + log D. _ ELBO = E_{~N(0,I)} [ _ (log p(x|z) + log
p(z) - log q(z|x)) ] and the gradient flows p(z) - log q(z|x)) ] but the gradient through
through both the encoder parameters and the sampling is zero because is
the sampling operation directly independent of
Correct: A - _ ELBO = E_{~N(0,I)} [ _ (log p(x|z) + log p(z) - log q(z|x)) ] where z is a
function of through and
Rationale:The reparameterization trick expresses z as a deterministic function of Æ (via ¼ and
) and , allowing gradients to flow through the sampling operation. Thus, the gradient of the
ELBO with respect to includes terms from log p(x|z), log p(z), and log q(z|x) where z depends
on . Option B is incorrect because reparameterization enables gradient flow; option C is
incorrect because sampling is not directly differentiable; option D is incorrect because the
gradient is not zero.
Q4.
In a transformer model with multi-head attention, the output of the attention layer is
computed as: MultiHead(Q, K, V) = Concat(head_1, ..., head_h) W^O. Each head_i =
Attention(QW_i^Q, KW_i^K, VW_i^V). If the model dimension d_model = 512 and the
number of heads h = 8, what is the dimension of each head's output (d_v) and the total
number of parameters in the projection matrices W_i^Q, W_i^K, W_i^V, and W^O?
A. d_v = 64; total parameters = 8 * (512*64 B. d_v = 64; total parameters = 8 * (512*64
+ 512*64 + 512*64) + 512*512 = 8*98,304 + + 512*64 + 512*64) + 512*512 = 8*98,304 +
262,144 = 1,048,576 262,144 = 1,048,576, but note that W_i^K
and W_i^V are shared across heads
Page 4
