# Drill 51 Inception concatenation and parameter count

Subject: Deep Learning

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/#tee/read/17/tee-c17-s51

Canonical: https://deep-learning-study-room.vercel.app/read/dl/tee/section/tee-c17-s51/

## Question

**Question.** Input 10 × 10 × 8. Parallel branches: A uses 1 × 1 conv 4 filters ; B uses 3 × 3 same Conv 6 filters; C max pool 3 same, stride 1 then 1 × 1 conv 2 filters. Find combined shape and trainable parameters.

## Answer — step by step

### Given

Input = 10 × 10 × 8.

Branch A: 1 × 1 convolution, 4 filters.

Branch B: 3 × 3 same convolution, 6 filters.

Branch C: 3 × 3 same max pool with stride 1, then 1 × 1 convolution, 2 filters.

### To find

Find each branch shape and parameters, then the concatenated shape and total parameters.

### Formula — which one and why

**Inception channel concatenation.** Parallel branches retain their separate features. Add their channel counts while keeping the common height and width. Add trainable parameters across branches.

C₁,C₂,… = branch output channels. Concatenation requires matching spatial dimensions.



$$
C_{\mathrm{out}}=\sum_k C_k
$$





$$
P_{\mathrm{total}}=\sum_k P_k
$$



**Yaad rakhne ka cue:** Channels jodo; height-width same rakho.

**Convolution parameters.** Every filter spans all input channels. Count its spatial coefficients across those channels, add one bias and multiply by the number of filters.

K\_h,K\_w = filter height/width; C\_in = input channels; C\_out = number of filters and output channels. Spatial output positions reuse these weights.



$$
N_w=K_hK_wC_{\mathrm{in}}C_{\mathrm{out}}
$$





$$
N_b=C_{\mathrm{out}}
$$





$$
P=N_w+N_b
$$



**Yaad rakhne ka cue:** Filter area × input channels × filters; phir ek bias per filter.

### Solution — step by step

### Meaning and formulas

Each parallel branch receives the same original 10 × 10 × 8 input. “Same” with stride 1 preserves height and width. Concatenation joins the channel lists without adding the feature values.



$$
C_{\mathrm{combined}}=C_A+C_B+C_C
$$





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



Assume one bias per filter. Pooling has no learned parameters.

**Step 1: Branch A.**

Its 4 filters produce shape 10 × 10 × 4.



$$
P_A=(1\times1\times8+1)\times4=9\times4=36
$$



**Step 2: Branch B.**

Its 6 filters produce shape 10 × 10 × 6.



$$
P_B=(3\times3\times8+1)\times6=73\times6=438
$$



**Step 3: Branch C.**

Pooling preserves 10 × 10 × 8 here because stride is 1 and padding is same. Its parameter count is zero. The following 2-filter projection creates 10 × 10 × 2.



$$
P_C=0+(1\times1\times8+1)\times2=18
$$



**Step 4: Join the shapes and add the parameters.**



$$
C_{\mathrm{combined}}=4+6+2=12
$$





$$
P_{\mathrm{total}}=36+438+18=492
$$



**Final answer:** combined shape 10 × 10 × 12; 492 parameters. Branch B does not take branch A's four channels as input: they are parallel branches.