Deep Learning · TEE · section

Faculty Assignment2 Q5 · 32×32 three-convolution CNN

Faculty Assignment 2 · worked practice

Question

Question. Input32×32×1; conv3,8same; pool2stride2; conv5,16valid; pool2stride2; conv3,32same; dense64; output5. Find shapes and parameters. (Faculty assignment: no marks printed)

Answer — step by step

Given

Input=[32, 32, 1]

Layer	Kernel	Stride	Padding	Filters / units
Conv	3	1	1	8
Pool	2	2	0	0
Conv	5	1	0	16
Pool	2	2	0	0
Conv	3	1	1	32
Dense	0	0	0	64
Dense	0	0	0	5

To find

Each output shape and all trainable weights plus biases.

Formula: which one and why

$O=\left\lfloor\frac{N+2P-F}{S}\right\rfloor+1$ LaTeX: O=\left\lfloor\frac{N+2P-F}{S}\right\rfloor+1

$P_{conv}=(F_hF_wC_{in}+1)C_{out},\quad P_{dense}=(n_{in}+1)n_{out}$ LaTeX: P_{conv}=(F_hF_wC_{in}+1)C_{out},\quad P_{dense}=(n_{in}+1)n_{out}

Each filter/unit has one bias. Pooling has zero learned parameters. Flatten preserves the number of values, then a dense layer connects every input value to every output unit.

Step 1. Work layer by layer

Layer	Input	Full substitution and reason	Output	Parameters
Conv	32×32×1	⌊(32+2×1−3)/1⌋+1=32; parameters=(3×3×1+1)×8=80	32×32×8	80
Pool	32×32×8	⌊(32−2)/2⌋+1=16; channels stay 8; no trainable weights	16×16×8	0
Conv	16×16×8	⌊(16+2×0−5)/1⌋+1=12; parameters=(5×5×8+1)×16=3216	12×12×16	3216
Pool	12×12×16	⌊(12−2)/2⌋+1=6; channels stay 16; no trainable weights	6×6×16	0
Conv	6×6×16	⌊(6+2×1−3)/1⌋+1=6; parameters=(3×3×16+1)×32=4640	6×6×32	4640
Dense	6×6×32	Flatten/input length=6×6×32=1152; parameters=(1152+1)×64=73792	64	73792
Dense	1×1×64	Flatten/input length=1×1×64=64; parameters=(64+1)×5=325	5	325

Step 2. Sum only trainable parameters

80 + 3216 + 4640 + 73792 + 325 = 82053

Final answer

Total parameters=82053. Shape and parameter count are separate results.