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L2 Vector Norm in Triton

Linear AlgebraGPU Programming

Easy

Compute the L2 (Euclidean) norm of a 1D float tensor in Triton. For a tensor $x$ of length $N$ , write the scalar norm into $\texttt{out}[0]$ :

\texttt{out}[0] = \sqrt{\sum_{i=0}^{N-1} x[i]^2}

The reduction is a sum-of-squares; the square root is applied once on the host after the kernel returns. Each program reduces its tile in registers, combines partials via a single atomic per program on a scratch scalar, and the launcher takes the square root.

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Examples

Input: x = [3.0, 4.0]

Output: [5.0]

Input: x = [1.0, 2.0, 2.0]

Output: [3.0]

Hint 1

Square inside the tile, not afterwards. Fusing the multiply into the reduction with $\texttt{tl.sum(x * x)}$ halves the register pressure.

Hint 2

Mask with $\texttt{other=0.0}$ so the tail lanes contribute zero. The square of zero is still zero, so no special handling at the boundary.

Hint 3

The square root is a single host op after the kernel: $\texttt{out.copy\_(torch.sqrt(sumsq\_buf))}$ . Keep the per-lane work to the reduction only.

Requirements

Compute the sum of squares in a single pass, squaring lanes in registers before the reduction
Accumulate partials across programs with one atomic add per program on a scratch scalar
Apply the square root on the host after the kernel; do not square-root per lane in the kernel
Mask the tail so masked lanes contribute zero to the sum of squares
The $\texttt{solve}$ signature must not change

Constraints

$1 \le N \le 10^6$ .
Input tensor is torch.float32 on cuda and contiguous; output has shape (1,).
Inputs are bounded such that the sum of squares stays representable in float32 (no overflow).
Tolerance: combined atol=1e-2, rtol=1e-2 to absorb large- $N$ accumulation drift.

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L2 Vector Norm in Triton

Linear AlgebraGPU Programming

Easy

Compute the L2 (Euclidean) norm of a 1D float tensor in Triton. For a tensor $x$ of length $N$ , write the scalar norm into $\texttt{out}[0]$ :

\texttt{out}[0] = \sqrt{\sum_{i=0}^{N-1} x[i]^2}

Loading visualization...

Examples

Input: x = [3.0, 4.0]

Output: [5.0]

Input: x = [1.0, 2.0, 2.0]

Output: [3.0]

Hint 1

Square inside the tile, not afterwards. Fusing the multiply into the reduction with $\texttt{tl.sum(x * x)}$ halves the register pressure.

Hint 2

Mask with $\texttt{other=0.0}$ so the tail lanes contribute zero. The square of zero is still zero, so no special handling at the boundary.

Hint 3

The square root is a single host op after the kernel: $\texttt{out.copy\_(torch.sqrt(sumsq\_buf))}$ . Keep the per-lane work to the reduction only.

Requirements

Compute the sum of squares in a single pass, squaring lanes in registers before the reduction
Accumulate partials across programs with one atomic add per program on a scratch scalar
Apply the square root on the host after the kernel; do not square-root per lane in the kernel
Mask the tail so masked lanes contribute zero to the sum of squares
The $\texttt{solve}$ signature must not change

Constraints

$1 \le N \le 10^6$ .
Input tensor is torch.float32 on cuda and contiguous; output has shape (1,).
Inputs are bounded such that the sum of squares stays representable in float32 (no overflow).
Tolerance: combined atol=1e-2, rtol=1e-2 to absorb large- $N$ accumulation drift.

Try Similar Problems

x =

Accepts: array

Examples

Requirements

Constraints

Kernel performance

L2 Vector Norm in Triton

Examples

Requirements

Constraints

Kernel performance