Convolution — feature-map size

Background

Sliding a K×K filter over an H×W image (with padding P and stride S) shrinks the spatial size in a predictable way. Per dimension:

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

For "valid" convolution ($P = 0$, $S = 1$) this is just $N - K + 1$.

Problem statement

Implement conv_output_size(N, K, stride, padding) returning the output size along one dimension.

Input

• N — int, input size (height or width).
• K — int, kernel size.
• stride — int $\ge 1$.
• padding — int $\ge 0$.

Output

Returns an int: $\lfloor (N + 2P - K)/S \rfloor + 1$.

Examples

Example 1 — valid conv

Input:  N = 5, K = 3, stride = 1, padding = 0
Output: 3

Explanation: $5 - 3 + 1 = 3$.

Example 2 — same padding keeps the size

Input:  N = 5, K = 3, stride = 1, padding = 1
Output: 5

Explanation: $(5 + 2 - 3)/1 + 1 = 5$.

Constraints

• Use floor division for the stride term.
• Return a plain int.

Notes

• Stride 2 roughly halves the size; "same" padding ($P=\lfloor K/2\rfloor$ at stride 1) keeps it unchanged — both are everyday CNN design choices.