# R23 Q5b Subject: Deep Learning Exam: TEE Reader: https://deep-learning-study-room.vercel.app/#tee/read/16/tee-c16-s12 Canonical: https://deep-learning-study-room.vercel.app/read/dl/tee/section/tee-c16-s12/ ## Question **Question (10 marks).** Justify multiple ANN layers. Design a fraudulent/non-fraudulent transaction classifier for 10,000 samples with 4 features; choose layers/units and count parameters. ## Answer — step by step ### Given 10,000 samples; 4 features per sample; two target classes: fraudulent/non-fraudulent. Hidden layer sizes are design choices, not fixed in the question. ### To find Explain multiple layers, propose an appropriate classifier and count parameters for that explicit design. ### Formula — which one and why **Dense-layer parameters.** Every output neuron receives every input, so multiply inputs by outputs for weights. Add one bias per receiving neuron. Sum separate layers. n\_in = incoming features; n\_out = receiving neurons. The input layer itself has no learned receiving weights or biases. $$ P=(n_{\mathrm{in}}+1)n_{\mathrm{out}} $$ **Yaad rakhne ka cue:** Har output ke liye saare inputs plus ek bias. ### Solution — step by step A single perceptron forms a weighted sum and applies a threshold. Its decision boundary is linear: w·x+b = 0. It can learn linearly separable patterns such as AND and OR under suitable learning conditions, but cannot represent XOR with one linear separator. ![Opposite corners for XOR cannot be split by one straight line](https://deep-learning-study-room.vercel.app/assets/figures/xor.svg) Diagram: Opposite corners for XOR cannot be split by one straight line. Labels: x1 | x2 | 0 | 1 | 1 | 0 | XOR labels occupy opposite corners | 0 | 1 | 0 | 1 | One straight decision boundary cannot separate the two classes. Opposite corners for XOR cannot be split by one straight line Tap the diagram to enlarge. For binary AND, w₁ = w₂ = 1 and b = −1.5 give preactivations −1.5, −0.5, −0.5 and 0.5 for 00, 10, 01 and 11. A step that returns 1 at nonnegative inputs produces 0,0,0,1. For XOR, use h₁ = step(x₁+x₂−0.5) for OR and h₂ = step(x₁+x₂−1.5) for AND. Then XOR = step(h₁−2h₂−0.5). For inputs 00,01,10,11, the hidden pairs are (0,0), (1,0), (1,0), (1,1), producing 0,1,1,0. XNOR = step(−h₁+2h₂+0.5), producing 1,0,0,1. This is a multilayer construction, not a claim that a single perceptron solves XOR. A multilayer feedforward network adds hidden nonlinear features before the final boundary. Without hidden nonlinearities, adding linear layers still produces one affine map. **Chosen design:** 4 → 8 → 4 → 1 with ReLU hidden layers, one sigmoid output and binary cross-entropy. Hidden widths are explicit design assumptions. Architecture: 4 → 8 → 4 → 1. Each non-input neuron has one bias. ### Layer 4 → 8 **Step 1: count connections.** $$ 4(8)=32 $$ **Step 2: count receiving-neuron biases.** There are 8 receiving neurons, so 8 biases. **Step 3: add weights and biases.** $$ 32+8=40 $$ ### Layer 8 → 4 **Step 1: count connections.** $$ 8(4)=32 $$ **Step 2: count receiving-neuron biases.** There are 4 receiving neurons, so 4 biases. **Step 3: add weights and biases.** $$ 32+4=36 $$ ### Layer 4 → 1 **Step 1: count connections.** $$ 4(1)=4 $$ **Step 2: count receiving-neuron biases.** There are 1 receiving neurons, so 1 biases. **Step 3: add weights and biases.** $$ 4+1=5 $$ **Total = 81**. Input values and activations are not trainable parameters. Use stratified train/validation/test partitions; handle class imbalance using appropriate training weights and evaluate precision/recall or a suitable decision threshold alongside accuracy. More layers are useful only if they learn needed nonlinear structure; the problem statement does not prescribe one uniquely correct architecture. ### Verify input (0,0) **Step 1: calculate hidden OR preactivation.** $$ z_1=0+0-0.5=-0.5 $$ **Step 2: apply its threshold.** $$ h_1=0 $$ **Step 3: calculate hidden AND preactivation.** $$ z_2=0+0-1.5=-1.5 $$ **Step 4: apply its threshold.** $$ h_2=0 $$ **Step 5: calculate XOR output preactivation.** $$ z_{XOR}=0-2(0)-0.5=-0.5 $$ **Step 6: apply the XOR threshold.** $$ y_{XOR}=0 $$ **Step 7: calculate XNOR output preactivation.** $$ z_{XNOR}=-0+2(0)+0.5=0.5 $$ **Step 8: apply the XNOR threshold.** $$ y_{XNOR}=1 $$ ### Verify input (0,1) **Step 1: calculate hidden OR preactivation.** $$ z_1=0+1-0.5=0.5 $$ **Step 2: apply its threshold.** $$ h_1=1 $$ **Step 3: calculate hidden AND preactivation.** $$ z_2=0+1-1.5=-0.5 $$ **Step 4: apply its threshold.** $$ h_2=0 $$ **Step 5: calculate XOR output preactivation.** $$ z_{XOR}=1-2(0)-0.5=0.5 $$ **Step 6: apply the XOR threshold.** $$ y_{XOR}=1 $$ **Step 7: calculate XNOR output preactivation.** $$ z_{XNOR}=-1+2(0)+0.5=-0.5 $$ **Step 8: apply the XNOR threshold.** $$ y_{XNOR}=0 $$ ### Verify input (1,0) **Step 1: calculate hidden OR preactivation.** $$ z_1=1+0-0.5=0.5 $$ **Step 2: apply its threshold.** $$ h_1=1 $$ **Step 3: calculate hidden AND preactivation.** $$ z_2=1+0-1.5=-0.5 $$ **Step 4: apply its threshold.** $$ h_2=0 $$ **Step 5: calculate XOR output preactivation.** $$ z_{XOR}=1-2(0)-0.5=0.5 $$ **Step 6: apply the XOR threshold.** $$ y_{XOR}=1 $$ **Step 7: calculate XNOR output preactivation.** $$ z_{XNOR}=-1+2(0)+0.5=-0.5 $$ **Step 8: apply the XNOR threshold.** $$ y_{XNOR}=0 $$ ### Verify input (1,1) **Step 1: calculate hidden OR preactivation.** $$ z_1=1+1-0.5=1.5 $$ **Step 2: apply its threshold.** $$ h_1=1 $$ **Step 3: calculate hidden AND preactivation.** $$ z_2=1+1-1.5=0.5 $$ **Step 4: apply its threshold.** $$ h_2=1 $$ **Step 5: calculate XOR output preactivation.** $$ z_{XOR}=1-2(1)-0.5=-1.5 $$ **Step 6: apply the XOR threshold.** $$ y_{XOR}=0 $$ **Step 7: calculate XNOR output preactivation.** $$ z_{XNOR}=-1+2(1)+0.5=1.5 $$ **Step 8: apply the XNOR threshold.** $$ y_{XNOR}=1 $$ **Final answer:** XOR = [0,1,1,0]; XNOR = [1,0,0,1] for 00,01,10,11. The hidden nonlinear units make this possible.