# Drill 35 Complete small CNN

Subject: Deep Learning

Exam: MT2

Reader: https://deep-learning-study-room.vercel.app/#mt2/read/17/mt2-c17-s35

Canonical: https://deep-learning-study-room.vercel.app/read/dl/mt2/section/mt2-c17-s35/

## Question

**Question.** Input 10 × 10 × 1 → Conv 3 × 3,4 filters, stride 1, valid → Pool 2 stride 2 → Flatten → Dense 3. Find all shapes and parameters.

## Answer — step by step

### Given

Input = 10 × 10 × 1.

Conv: 3 × 3, 4 filters, stride 1, valid padding.

Pool: 2 × 2, stride 2; then flatten and dense layer with 3 outputs.

Include one bias per filter/dense neuron.

### To find

Find all output shapes, layer parameter counts and their sum.

### Formula — which one and why

**Complete CNN calculation.** Carry each output shape into the next layer. Convolution changes spatial size and channel count; pooling changes spatial size; flattening converts all activations into dense-layer inputs.

H,W,C = feature-map height, width and channels; K = kernel size; S = stride; P\_pad = padding. One bias is assumed for each learned filter or dense neuron unless specified otherwise.



$$
N_{\mathrm{out}}=\left\lfloor\frac{N+2P_{\mathrm{pad}}-K}{S}\right\rfloor+1
$$





$$
P_{\mathrm{conv}}=(K^2C_{\mathrm{in}}+1)C_{\mathrm{out}}
$$





$$
n_{\mathrm{flat}}=HWC
$$





$$
P_{\mathrm{dense}}=(n_{\mathrm{in}}+1)n_{\mathrm{out}}
$$





$$
P_{\mathrm{pool}}=P_{\mathrm{flatten}}=0
$$



**Yaad rakhne ka cue:** Ek layer ka output, next layer ka input; shape aur parameters alag count.

### Solution — step by step

### Formulas and assumptions

“Valid” means zero padding. The convolution and dense layers each have one bias per receiving filter or neuron. Pooling and flattening have no learned parameters.



$$
N_{\mathrm{out}}=\left\lfloor\frac{N+2P-K}{S}\right\rfloor+1
$$





$$
P_{\mathrm{conv}}=(K^2C_{\mathrm{in}}+1)C_{\mathrm{out}}
$$





$$
n_{\mathrm{flat}}=HWC,\qquad P_{\mathrm{dense}}=(n_{\mathrm{in}}+1)n_{\mathrm{out}}
$$



### Step 1: convolution shape



$$
N_{\mathrm{conv}}=\frac{10+0-3}{1}+1=7+1=8
$$



Four filters produce four channels: **8 × 8 × 4**.

### Step 2: convolution parameters



$$
\text{weights}=3\times3\times1\times4=36
$$





$$
\text{biases}=4,\qquad P_{\mathrm{conv}}=36+4=40
$$



### Step 3: pooling shape

Pooling now receives size 8, not the original size 10.



$$
N_{\mathrm{pool}}=\left\lfloor\frac{8-2}{2}\right\rfloor+1=3+1=4
$$



Channels remain 4: **4 × 4 × 4**. Pooling parameters = 0.

### Step 4: flatten



$$
n_{\mathrm{flat}}=4\times4\times4=64
$$



Flatten creates a vector of 64 numbers. It rearranges values, so its parameter count is 0.

### Step 5: dense layer with 3 outputs



$$
\text{weights}=64\times3=192
$$





$$
\text{biases}=3,\qquad P_{\mathrm{dense}}=192+3=195
$$



### Step 6: add learned parameters



$$
P_{\mathrm{total}}=40+0+0+195=235
$$



**Final answer:** shapes 8 × 8 × 4 → 4 × 4 × 4 → 64 → 3; total 235 parameters.