# Introduction

This example shows fusing two back-to-back GEMMs/Convolutions into one kernel.

When running two unfused GEMM/Conv operations, each operation loads one input
activation matrix, one weight matrix (or filter matrix) from the memory and then
stores the result activation matrix back to the memory.

When the two GEMM/Conv operations are fused together, the mainloops of the two
GEMMs/Convs run back to back in a single kernel. The output accumulator of the
1st GEMM/Conv will be stored in the register file and reused as the activation
input of the 2nd GEMM/Conv. This saves a round trip to memory for the activation
matrix.

This example computes the following:
- 1st GEMM/Conv: D0 = relu(alpha0 .\* A0 \*\* B0)
- 2nd GEMM/Conv: D1 = relu(alpha1 .\* D0 \*\* B1 + beta1 .\* C1)

In the above equation, operator \*\* can be matrix multiplication or convolution operation.

# Implementation Details

In order to run two GEMM/Convs in a single kernel, the example requires the same number of
threadblocks are used across 2 GEMMs/Convs. This also ensures the same threadblock tile M across
2 GEMMs/Convs.

In order to reuse the output accumulator (stored in register-file) of the 1st GEMM as the
input activation, the example enforces the following two constraints:

- thread_block_tile_N = problem_N

This constraint ensures that each threadblock loads the entire weight/filter matrix in
addition to its own input activation tile. Therefore the input activation tile of the
2nd GEMM/Conv only depends on the output activation tile of the 1st GEMM/Conv, and the
operation can be fully block-resident.

- warp_tile_N = thread_block_tile_N

This constraint ensures that each warp loads the entire weight/filter kBlock in
addition to its own input activation tile. Therefore the input activation warp tile of the
2nd GEMM/Conv only depends on the output warp accumulator of the 1st GEMM/Conv in the
register file, and the operation can be fully register-file-resident.

On the other hand, this constraint can be relaxed if the output accumulator of the 1st GEMM/CONV
is staged in the shared memory and then used as input for the 2nd GEMM/CONV. In this case, the
input of each warp tile can be loaded from the shared memory so they do not need to be RF-resident,
therefore each warp does not need to store the entire input matrix of 2nd GEMM in its RF. This is
illustrated in the diagram below.

When applying the above constraint to convolutions, it is required that the 2nd Convolution
kernel doesn't have halos such that data used by each threadblock doesn't depend on any other
threadblock. Typically this requires the 2nd Convolution uses 1x1 filter without any paddings.

# Build and run

- Run cmake at top-level CUTLASS
- `make 13_two_tensor_op_fusion`
- Run individual benchmarks
  - `./examples/13_two_tensor_op_fusion/13_fused_two_convs_f16_sm75_rf`
  - `./examples/13_two_tensor_op_fusion/13_fused_two_convs_f16_sm75_shmem`
  - `./examples/13_two_tensor_op_fusion/13_fused_two_convs_f16_sm80_rf`
  - `./examples/13_two_tensor_op_fusion/13_fused_two_convs_f16_sm80_shmem`
  - `./examples/13_two_tensor_op_fusion/13_fused_two_convs_s8_sm75_rf`
  - `./examples/13_two_tensor_op_fusion/13_fused_two_convs_s8_sm75_shmem`
  - `./examples/13_two_tensor_op_fusion/13_fused_two_convs_s8_sm80_rf`
  - `./examples/13_two_tensor_op_fusion/13_fused_two_convs_s8_sm80_shmem`
  - `./examples/13_two_tensor_op_fusion/13_fused_two_gemms_f16_sm75_rf`
  - `./examples/13_two_tensor_op_fusion/13_fused_two_gemms_f16_sm75_shmem`
  - `./examples/13_two_tensor_op_fusion/13_fused_two_gemms_f16_sm80_rf`
  - `./examples/13_two_tensor_op_fusion/13_fused_two_gemms_f16_sm80_shmem`
  - `./examples/13_two_tensor_op_fusion/13_fused_two_gemms_s8_sm75_rf`
  - `./examples/13_two_tensor_op_fusion/13_fused_two_gemms_s8_sm75_shmem`
  - `./examples/13_two_tensor_op_fusion/13_fused_two_gemms_s8_sm80_rf`
  - `./examples/13_two_tensor_op_fusion/13_fused_two_gemms_s8_sm80_shmem`


# Copyright

Copyright (c) 2017 - 2025 NVIDIA CORPORATION & AFFILIATES. All rights reserved.
SPDX-License-Identifier: BSD-3-Clause

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