Endsem 2025 solved

This page solves the IIIT Kottayam endsem paper of November 2025 for IoE-414. It shows every number in full detail for backprop, Markov chains, RNN, GAN, and RBM. After this page you can repeat each method on any similar exam question.

IIIT Kottayam held this paper in November 2025. The course is IoE-414 Deep Learning. Maximum marks are 100. Batch 2022 attempted it. The paper has five questions. Each question carries 20 marks.

Q1 Fix training symptoms Search on YouTube

Q1 tests diagnosis. Each part gives one symptom. You must name the fix and give the reason.

Q1(a) Oscillation and covariate shift Search on YouTube

Large oscillation across minibatches shows unstable layer inputs. The distribution of inputs shifts during training. batch norm fixes this shift.

\hat{x} = \frac{x - \mu}{\sqrt{\sigma^2 + \epsilon}}, \quad y = \gamma \hat{x} + \beta

$\mu$ is minibatch mean. $\sigma^2$ is minibatch variance. $\epsilon$ is a small constant for stability. $\gamma$ and $\beta$ are learnable scale and shift.

Normalization keeps each layer input stable. Gradients stay steadier. Training oscillates less.

Q1(b) Overfitting gap Search on YouTube

Train accuracy 99 percent and validation accuracy 70 percent show overfitting. The gap widens with more epochs. The network memorises training labels. dropout is the primary fix.

y = \frac{m \odot a}{1 - p}, \quad m \sim \text{Bernoulli}(1-p)

$a$ is layer activation. $m$ is a random binary mask. $p$ is drop probability.

Dropout drops random units during training. It breaks co-adaptation between units. It acts like an ensemble average at test time. Early stopping and L2 regularization also help. Early stopping halts before memorisation grows. L2 regularization penalises large weights. Pick dropout as the main answer.

Q1(c) Slow tiny updates Search on YouTube

Tiny updates show slow progress in flat regions. The optimizer needs momentum. optimizers solve this problem.

v = \beta v + g, \quad \theta = \theta - \eta v

$g$ is current gradient. $v$ is velocity from past gradients. $\beta$ is momentum factor. $\eta$ is learning rate. $\theta$ is parameter vector.

Momentum accumulates past gradients. Nesterov momentum looks ahead before the step. Adam adapts the rate per weight. A higher learning rate also helps.

Q1(d) Smooth but low accuracy Search on YouTube

Smooth loss with low train and validation accuracy shows underfitting. Capacity is too small. The optimizer or setting may also limit learning.

Tune learning rate, minibatch size, and epoch count. Add layers or neurons for more capacity. Change optimizer or initialization.

Write symptom, fix, and one-line reason for full marks. Name one primary fix per part. Mention one alternative only when time permits.

Q2 Two backprop iterations Search on YouTube

Q2 asks for two full updates on a 2-2-1 sigmoid network. Learning rate $\eta$ is 0.5. Input $X$ is $[1, 0]$ and label $y$ is 1. Initial weights are $W_1$ , $W_2$ , and $W_3$ . All biases start at 0. The backprop rule updates each weight.

\sigma(z) = \frac{1}{1 + e^{-z}}, \quad \sigma^{\prime}(z) = \sigma(z)(1 - \sigma(z))

$\sigma$ is sigmoid activation function. $z$ is weighted input. $\sigma^{\prime}$ is its derivative.

L = (O - y)^2, \quad \delta_o = 2(O - y) O (1 - O)

$O$ is network output. $y$ is label. $L$ is per-sample loss. $\delta_o$ is output delta.

\theta_{new} = \theta_{old} - \eta g

$\theta$ is parameter vector. $g$ is gradient of loss. $\eta$ is 0.5 here.

Keep the sign of $\delta_o$ . The gradient is negative when $O$ is less than $y$ . The update subtracts the gradient, so weights grow. Dropping the minus sign reverses the update.

Do not confuse bias with decision threshold. Bias is a learnable weight. Decision threshold 0.5 decides the label. The value $w_0$ equals negative threshold in perceptron notation.

Q2 Iteration 1 forward pass Search on YouTube

Initial data are $X = [1, 0]$ , $y = 1$ , $W_1 = [[0.1, 0.2], [0.3, 0.4]]$ , $W_2 = [[0.5, 0.6], [0.7, 0.8]]$ , $W_3 = [0.9, 1.0]$ . All biases are 0.

Step 1. First hidden input. Multiply $W_1$ by $X$ .

\begin{aligned} z_1 &= W_1 X + b_1 \\ &= [0.1 \cdot 1 + 0.2 \cdot 0, \; 0.3 \cdot 1 + 0.4 \cdot 0] \\ &= \mathbf{[0.1, 0.3]} \end{aligned}

Step 2. First hidden activation. Apply sigmoid.

\begin{aligned} a_1 &= [\sigma(0.1), \; \sigma(0.3)] \\ &= [1/(1+e^{-0.1}), \; 1/(1+e^{-0.3})] \\ &= \mathbf{[0.5250, 0.5744]} \end{aligned}

Step 3. Second hidden input. Multiply $W_2$ by $a_1$ .

\begin{aligned} z_{2,1} &= 0.5 \cdot 0.5250 + 0.6 \cdot 0.5744 = 0.2625 + 0.3446 = \mathbf{0.6072} \\ z_{2,2} &= 0.7 \cdot 0.5250 + 0.8 \cdot 0.5744 = 0.3675 + 0.4595 = \mathbf{0.8270} \end{aligned}

Step 4. Second hidden activation. Apply sigmoid.

\begin{aligned} a_2 &= [\sigma(0.6072), \; \sigma(0.8270)] \\ &= \mathbf{[0.6473, 0.6957]} \end{aligned}

Step 5. Output and loss. Combine with $W_3$ .

\begin{aligned} z_3 &= 0.9 \cdot 0.6473 + 1.0 \cdot 0.6957 = 0.5826 + 0.6957 = \mathbf{1.2783} \\ O &= \sigma(1.2783) = \mathbf{0.7822} \\ L &= (0.7822 - 1)^2 = (-0.2178)^2 = \mathbf{0.04745} \end{aligned}

Q2 Iteration 1 backward pass and update Search on YouTube

Loss derivative uses factor 2 because $L = (O-y)^2$ . Sigmoid derivative is $O(1-O)$ at output.

Step 1. Output delta. Combine loss slope with sigmoid slope.

\begin{aligned} \delta_o &= 2(O-y) O (1-O) \\ &= 2 \cdot (-0.2178) \cdot 0.7822 \cdot 0.2178 \\ &= \mathbf{-0.07423} \end{aligned}

Step 2. Gradients for $W_3$ and $b_3$ . Multiply delta by $a_2$ .

\begin{aligned} g_{W3} &= \delta_o \cdot a_2 = -0.07423 \cdot [0.6473, 0.6957] = \mathbf{[-0.04805, -0.05165]} \\ g_{b3} &= \mathbf{-0.07423} \end{aligned}

Step 3. Second hidden deltas. Push delta back through $W_3$ .

\begin{aligned} a_2(1-a_2) &= [0.6473 \cdot 0.3527, \; 0.6957 \cdot 0.3043] = [0.2283, 0.2117] \\ d_2 &= \delta_o \cdot W_3 \odot a_2(1-a_2) \\ &= [-0.07423 \cdot 0.9 \cdot 0.2283, \; -0.07423 \cdot 1.0 \cdot 0.2117] \\ &= \mathbf{[-0.01525, -0.01571]} \end{aligned}

Step 4. Gradients for $W_2$ . Outer product of $d_2$ and $a_1$ .

\begin{aligned} g_{W2} &= \begin{bmatrix} -0.01525 \cdot 0.5250 & -0.01525 \cdot 0.5744 \\ -0.01571 \cdot 0.5250 & -0.01571 \cdot 0.5744 \end{bmatrix} \\ &= \mathbf{ \begin{bmatrix} -0.00801 & -0.00876 \\ -0.00825 & -0.00903 \end{bmatrix}} \end{aligned}

Step 5. First hidden deltas. Push $d_2$ back through $W_2$ .

\begin{aligned} W_2^T d_2 &= [0.5 \cdot (-0.01525) + 0.7 \cdot (-0.01571), \; 0.6 \cdot (-0.01525) + 0.8 \cdot (-0.01571)] \\ &= [-0.01862, -0.02172] \\ a_1(1-a_1) &= [0.2494, 0.2445] \\ d_1 &= [-0.01862 \cdot 0.2494, \; -0.02172 \cdot 0.2445] = \mathbf{[-0.00465, -0.00531]} \end{aligned}

Step 6. Gradients for $W_1$ . Outer product of $d_1$ and $X$ .

\begin{aligned} g_{W1} &= \begin{bmatrix} -0.00465 \cdot 1 & -0.00465 \cdot 0 \\ -0.00531 \cdot 1 & -0.00531 \cdot 0 \end{bmatrix} = \mathbf{ \begin{bmatrix} -0.00465 & 0 \\ -0.00531 & 0 \end{bmatrix}} \end{aligned}

Step 7. Update with $\eta = 0.5$ . Subtract half the gradient.

\begin{aligned} W_1 &= [[0.1, 0.2],[0.3, 0.4]] - 0.5 \cdot g_{W1} = \mathbf{[[0.10232, 0.2],[0.30266, 0.4]]} \\ b_1 &= [0,0] - 0.5 \cdot d_1 = \mathbf{[0.00232, 0.00266]} \\ W_2 &= [[0.5, 0.6],[0.7, 0.8]] - 0.5 \cdot g_{W2} = \mathbf{[[0.50400, 0.60438],[0.70412, 0.80451]]} \\ b_2 &= [0,0] - 0.5 \cdot d_2 = \mathbf{[0.00763, 0.00786]} \\ W_3 &= [0.9, 1.0] - 0.5 \cdot g_{W3} = \mathbf{[0.92403, 1.02582]} \\ b_3 &= 0 - 0.5 \cdot (-0.07423) = \mathbf{0.03712} \end{aligned}

Q2 Iteration 2 forward and backward pass Search on YouTube

The second iteration starts from updated weights. It repeats the same chain rule.

Step 1. Forward pass. Use updated $W_1$ and $b_1$ .

\begin{aligned} z_1 &= [0.10232 + 0.00232, \; 0.30266 + 0.00266] = \mathbf{[0.10464, 0.30532]} \\ a_1 &= [\sigma(0.10464), \; \sigma(0.30532)] = \mathbf{[0.5261, 0.5757]} \\ z_2 &= [0.6207, 0.8415] \\ a_2 &= [\sigma(0.6207), \; \sigma(0.8415)] = \mathbf{[0.6504, 0.6988]} \\ z_3 &= 0.92403 \cdot 0.6504 + 1.02582 \cdot 0.6988 + 0.03712 = \mathbf{1.3550} \\ O &= \sigma(1.3550) = \mathbf{0.7949} \\ L &= (0.7949-1)^2 = \mathbf{0.04205} \end{aligned}

Step 2. Backward pass. Recompute deltas at new point.

\begin{aligned} \delta_o &= 2 \cdot (0.7949-1) \cdot 0.7949 \cdot 0.2051 = \mathbf{-0.06688} \\ g_{W3} &= -0.06688 \cdot [0.6504, 0.6988] = \mathbf{[-0.04350, -0.04673]} \\ d_2 &= \mathbf{[-0.01405, -0.01444]} \\ g_{W2} &= \mathbf{ \begin{bmatrix} -0.00739 & -0.00809 \\ -0.00760 & -0.00831 \end{bmatrix}} \\ d_1 &= \mathbf{[-0.00430, -0.00491]} \\ g_{W1} &= \mathbf{ \begin{bmatrix} -0.00430 & 0 \\ -0.00491 & 0 \end{bmatrix}} \end{aligned}

Step 3. Update again. Subtract half the new gradient.

\begin{aligned} W_1 &= \mathbf{[[0.10447, 0.2],[0.30511, 0.4]]} \\ b_1 &= \mathbf{[0.00447, 0.00511]} \\ W_2 &= \mathbf{[[0.50770, 0.60842],[0.70792, 0.80867]]} \\ b_2 &= \mathbf{[0.01465, 0.01508]} \\ W_3 &= \mathbf{[0.94577, 1.04918]} \\ b_3 &= \mathbf{0.07055} \end{aligned}

Step 4. Final check. Run forward once more.

\begin{aligned} O &= \mathbf{0.8060} \\ L &= (0.8060-1)^2 = \mathbf{0.03762} \end{aligned}

Loss falls across iterations: 0.04745 to 0.04205 to 0.03762.

Write $z$ , $a$ , $O$ , $L$ , then $\delta_o$ , $d_2$ , $d_1$ in order. Keep the factor 2 from $(O-y)^2$ . Examiners give marks per line.

Q3 Markov chains Search on YouTube

Q3 tests two Markov chains. Part (a) evolves a distribution. Part (b) cubes a transition matrix.

Q3(a) Streetlight distribution Search on YouTube

States are ON and OFF in order $[ON, OFF]$ . Rows of $T$ are current state. Columns are next state. Start vector $\mu_0$ is $[0, 1]$ , so light is OFF.

\mu_{t+1} = \mu_t T

$\mu_t$ is row distribution at time $t$ . $T$ is transition matrix. Each row of $T$ sums to 1.

\pi = \pi T, \quad \pi_{ON} + \pi_{OFF} = 1

$\pi$ is stationary distribution. It stays unchanged after one step.

Step 1. One step from OFF. Pick OFF row of $T$ .

\begin{aligned} \mu_1 &= [0,1] \begin{bmatrix} 0.75 & 0.25 \\ 0.15 & 0.85 \end{bmatrix} \\ &= \mathbf{[0.15, 0.85]} \end{aligned}

Step 2. Second step. Multiply $\mu_1$ by $T$ .

\begin{aligned} \mu_{2,ON} &= 0.15 \cdot 0.75 + 0.85 \cdot 0.15 = 0.1125 + 0.1275 = \mathbf{0.24} \\ \mu_{2,OFF} &= 0.15 \cdot 0.25 + 0.85 \cdot 0.85 = 0.0375 + 0.7225 = \mathbf{0.76} \\ \mu_2 &= \mathbf{[0.24, 0.76]} \end{aligned}

Stationary solution. Solve balance plus sum 1.

\begin{aligned} \pi_{ON} &= 0.75 \pi_{ON} + 0.15 \pi_{OFF} \\ 0.25 \pi_{ON} &= 0.15 \pi_{OFF} \\ \pi_{ON} &= \frac{0.15}{0.15+0.25} = \frac{0.15}{0.40} = \mathbf{0.375} \\ \pi_{OFF} &= \mathbf{0.625} \end{aligned}

Stationary vector is $\mathbf{[0.375, 0.625]}$ .

For two states use $q/(p+q)$ . Here $p = 0.25$ is ON to OFF. Here $q = 0.15$ is OFF to ON. Then $\pi_{ON} = q/(p+q)$ .

Q3(b) Thermostat cube Search on YouTube

States are Cooling and Heating. Matrix $P$ holds one-hour moves. The query spans three hours. It needs P^3.

P^2 = P P, \quad P^3 = P^2 P

$P^2$ holds two-step probabilities. $P^3$ holds three-step probabilities.

Step 1. Square $P$ . Multiply rows by columns.

\begin{aligned} P^2 &= \begin{bmatrix} 0.65 & 0.35 \\ 0.45 & 0.55 \end{bmatrix} \begin{bmatrix} 0.65 & 0.35 \\ 0.45 & 0.55 \end{bmatrix} \\ &= \begin{bmatrix} 0.65\cdot0.65+0.35\cdot0.45 & 0.65\cdot0.35+0.35\cdot0.55 \\ 0.45\cdot0.65+0.55\cdot0.45 & 0.45\cdot0.35+0.55\cdot0.55 \end{bmatrix} \\ &= \mathbf{ \begin{bmatrix} 0.58 & 0.42 \\ 0.54 & 0.46 \end{bmatrix}} \end{aligned}

Step 1. Cube $P$ . Multiply $P^2$ by $P$ .

\begin{aligned} P^3 &= \begin{bmatrix} 0.58 & 0.42 \\ 0.54 & 0.46 \end{bmatrix} \begin{bmatrix} 0.65 & 0.35 \\ 0.45 & 0.55 \end{bmatrix} \\ &= \begin{bmatrix} 0.58\cdot0.65+0.42\cdot0.45 & 0.58\cdot0.35+0.42\cdot0.55 \\ 0.54\cdot0.65+0.46\cdot0.45 & 0.54\cdot0.35+0.46\cdot0.55 \end{bmatrix} \\ &= \mathbf{ \begin{bmatrix} 0.566 & 0.434 \\ 0.558 & 0.442 \end{bmatrix}} \end{aligned}

Step 2. Read the query. Cooling at 02PM to Cooling at 05PM is three steps.

\begin{aligned} P(\text{Cooling 05PM} \vert \text{Cooling 02PM}) = \mathbf{0.566} \end{aligned}

It is row 1, column 1 of $P^3$ .

Count steps first. From 02PM to 05PM is 3 moves. Square first, then cube. Read entry (start, end).

Q4 Sequence models Search on YouTube

Q4 tests recurrent models. Part (a) asks for RNN and BPTT. Part (b) asks for bidirectional models.

Q4(a) RNN and BPTT Search on YouTube

An RNN reads one token per time step. It keeps a hidden state. The same weights apply at every step.

h_t = \tanh(U x_t + W h_{t-1} + b), \quad o_t = V h_t + c

$h_t$ is hidden state at step $t$ . $x_t$ is input. $U$ , $W$ , $V$ are shared weights. $b$ and $c$ are biases.

Backprop through time unfolds the loop. It sums gradients over steps. Repeated multiplication shrinks or grows gradients. Shrinking causes vanishing gradients.

\frac{\partial L}{\partial W} = \sum_t \frac{\partial L_t}{\partial h_t} \frac{\partial h_t}{\partial W}

$L_t$ is loss at step $t$ . The sum shares one $W$ across time.

Q4(b) Bidirectional models Search on YouTube

A bidirectional model reads both ways. Forward pass reads past to future. Backward pass reads future to past. Translation needs both sides because later words fix earlier meaning.

h_t = [\overrightarrow{h}_t ; \overleftarrow{h}_t]

$\overrightarrow{h}_t$ is forward hidden state. $\overleftarrow{h}_t$ is backward hidden state. Brackets mean concat. Sum is an alternative merge.

Bi-LSTM uses two LSTM chains. One moves forward. One moves backward. The merge holds past plus future context.

Draw unfolded boxes with shared $U$ , $W$ , $V$ for RNN. Draw two opposite arrows for bidirectional. Label concat or sum.

Q5 Generative models Search on YouTube

Q5 tests generative ideas. Part (a) asks for GAN. Part (b) asks for RBM.

Q5(a) GAN game Search on YouTube

A GAN has a generator and a discriminator. The generator makes fake samples. The discriminator scores real against fake. Training alternates the two updates.

V(G,D) = E_x[\log D(x)] + E_z[\log(1 - D(G(z)))]

$D(x)$ is score of real sample. $G(z)$ is fake sample from noise $z$ . $V$ is minimax value.

\min_G \max_D V(G,D)

Discriminator maximises $V$ . Generator minimises $V$ .

Value trace from l14. Watch $V$ rise then fall.

\begin{aligned} V_0 &= \mathbf{-0.5798} \\ V_1 &= \mathbf{-0.3285} \quad \text{after discriminator update} \\ V_2 &= \mathbf{-1.0217} \quad \text{after generator update} \end{aligned}

Discriminator step raises $V$ . Generator step lowers $V$ . Alternation drives the game.

State roles in one line each. Write minimax $V$ exactly. Say discriminator maximises and generator minimises.

Q5(b) RBM structure Search on YouTube

An RBM is bipartite, symmetric, and generative. Visible units link only to hidden units. No links stay inside one layer. Weights are symmetric in both directions.

E(v,h) = -b^T v - c^T h - v^T W h

$E$ is energy. $v$ is visible vector. $h$ is hidden vector. $b$ and $c$ are biases. $W$ holds symmetric weights.

P(h_j = 1 \vert v) = \sigma(c_j + v^T W_{:,j}), \quad P(v_i = 1 \vert h) = \sigma(b_i + W_{i,:} h)

Each conditional uses sigmoid activation function. Columns of $W$ drive hidden units. Rows of $W$ drive visible units.

Contrastive divergence trains the RBM. It runs one Gibbs step from data. It lowers energy of data and raises energy of reconstruction. A feedforward network is discriminative and directed. It maps input to label. An RBM models the joint distribution instead.

Draw two layers with cross links only. Write energy first, then conditionals. Contrast with feedforward network in one line.

Recap Search on YouTube

Use batch norm for shifting inputs and dropout for overfitting gap.
Use momentum, Nesterov, or Adam for tiny slow updates.
Add layers, neurons, or better settings for underfitting.
Apply sigmoid chain rule with sign check for backprop.
Evolve with row rule and solve balance for stationary distribution.
Square then cube $P$ and read entries for multi-step queries.
Alternate GAN minimax updates and contrast RBM with feedforward maps.

Q1 Fix training symptomsSearch on YouTubeBookmarkSaved

Q1(a) Oscillation and covariate shiftSearch on YouTubeBookmarkSaved

Q1(b) Overfitting gapSearch on YouTubeBookmarkSaved

Q1(c) Slow tiny updatesSearch on YouTubeBookmarkSaved

Q1(d) Smooth but low accuracySearch on YouTubeBookmarkSaved

Q2 Two backprop iterationsSearch on YouTubeBookmarkSaved

Q2 Iteration 1 forward passSearch on YouTubeBookmarkSaved

Q2 Iteration 1 backward pass and updateSearch on YouTubeBookmarkSaved

Q2 Iteration 2 forward and backward passSearch on YouTubeBookmarkSaved

Q3 Markov chainsSearch on YouTubeBookmarkSaved

Q3(a) Streetlight distributionSearch on YouTubeBookmarkSaved

Q3(b) Thermostat cubeSearch on YouTubeBookmarkSaved

Q4 Sequence modelsSearch on YouTubeBookmarkSaved

Q4(a) RNN and BPTTSearch on YouTubeBookmarkSaved

Q4(b) Bidirectional modelsSearch on YouTubeBookmarkSaved

Q5 Generative modelsSearch on YouTubeBookmarkSaved

Q5(a) GAN gameSearch on YouTubeBookmarkSaved

Q5(b) RBM structureSearch on YouTubeBookmarkSaved

RecapSearch on YouTubeBookmarkSaved

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