CUDA Standard Algorithms » Parallel Reduction

Include the Header

You need to include the header file, taskflow/cuda/algorithm/reduce.hpp, for using the parallel-reduction algorithm.

#include <taskflow/cuda/algorithm/reduce.hpp>

Reduce a Range of Items with an Initial Value

tf::cuda_reduce performs a parallel reduction over a range of elements specified by [first, last) using the binary operator bop and stores the reduced result in result. It represents the parallel execution of the following reduction loop on a GPU:

while (first != last) {
  *result = bop(*result, *first++);
}

The variable result participates in the reduction loop and must be initialized with an initial value. The following code performs a parallel reduction to sum all the numbers in the given range with an initial value 1000:

const size_t N = 1000000;
int* res = tf::cuda_malloc_shared<int>(1);  // result
int* vec = tf::cuda_malloc_shared<int>(N);  // vector

// initializes the data
*res = 1000;
for(size_t i=0; i<N; i++) 
  vec[i] = i;
} 

// create an execution policy
tf::cudaStream stream;
tf::cudaDefaultExecutionPolicy policy(stream);

// queries the required buffer size to reduce N elements using the given policy
auto bytes  = policy.reduce_bufsz<int>(N);
auto buffer = tf::cuda_malloc_device<std::byte>(bytes);

// *res = 1000 + (0 + 1 + 2 + 3 + 4 + ... + N-1)
tf::cuda_reduce(policy,
  vec, vec + N, res, [] __device__ (int a, int b) { return a + b; }, buffer
);

// synchronize the execution
stream.synchronize();

// delete the memory
cudaFree(buffer);
cudaFree(res);
cudaFree(vec);

The reduce algorithm runs asynchronously through the stream specified in the execution policy. You need to synchronize the stream to obtain correct results. Since the GPU reduction algorithm may require extra buffer to store the temporary results, you need to provide a buffer of size at least larger or equal to the value returned from tf::cudaDefaultExecutionPolicy::reduce_bufsz.

Reduce a Range of Items without an Initial Value

tf::cuda_uninitialized_reduce performs a parallel reduction over a range of items without an initial value. This method represents a parallel execution of the following reduction loop on a GPU:

*result = *first++;  // no initial values to participate in the reduction loop
while (first != last) {
  *result = bop(*result, *first++);
}

The variable result is directly assigned the reduced value without any initial value participating in the reduction loop. The following code performs a parallel reduction to sum all the numbers in the given range without any initial value:

const size_t N = 1000000;
int* res = tf::cuda_malloc_shared<int>(1);  // result
int* vec = tf::cuda_malloc_shared<int>(N);  // vector

// initializes the data
for(size_t i=0; i<N; i++) 
  vec[i] = i;
} 

// create an execution policy
tf::cudaStream stream;
tf::cudaDefaultExecutionPolicy policy(stream);

// queries the required buffer size to reduce N elements using the given policy
auto bytes  = policy.reduce_bufsz<int>(N);
auto buffer = tf::cuda_malloc_device<std::byte>(bytes);

// *res = 0 + 1 + 2 + 3 + 4 + ... + N-1
tf::cuda_uninitialized_reduce(policy,
  vec, vec + N, res, [] __device__ (int a, int b) { return a + b; }, buffer
);

// synchronize the execution
stream.synchronize();

// delete the buffer
cudaFree(res);
cudaFree(vec);
cudaFree(buffer);

Reduce a Range of Transformed Items with an Initial Value

tf::cuda_transform_reduce performs a parallel reduction over a range of transformed elements specified by [first, last) using a binary reduce operator bop and a unary transform operator uop. It represents the parallel execution of the following reduction loop on a GPU:

while (first != last) {
  *result = bop(*result, uop(*first++));
}

The variable result participates in the reduction loop and must be initialized with an initial value. The following code performs a parallel reduction to sum all the transformed numbers multiplied by 10 in the given range with an initial value 1000:

const size_t N = 1000000;
int* res = tf::cuda_malloc_shared<int>(1);  // result
int* vec = tf::cuda_malloc_shared<int>(N);  // vector

// initializes the data
*res = 1000;
for(size_t i=0; i<N; i++) {
  vec[i] = i;
} 

// create an execution policy
tf::cudaStream stream;
tf::cudaDefaultExecutionPolicy policy(stream);

// queries the required buffer size to reduce N elements using the given policy
auto bytes  = policy.reduce_bufsz<int>(N);
auto buffer = tf::cuda_malloc_device<std::byte>(bytes);

// *res = 1000 + (0*10 + 1*10 + 2*10 + 3*10 + 4*10 + ... + (N-1)*10)
tf::cuda_transform_reduce(policy,
  vec, vec + N, res, 
  [] __device__ (int a, int b) { return a + b; },
  [] __device__ (int a) { return a*10; },
  buffer
);

// synchronize the execution
stream.synchronize();

// delete the buffer
cudaFree(res);
cudaFree(vec);
cudaFree(buffer);

Reduce a Range of Transformed Items without an Initial Value

tf::cuda_transform_uninitialized_reduce performs a parallel reduction over a range of transformed items without an initial value. This method represents a parallel execution of the following reduction loop on a GPU:

*result = *first++;  // no initial values to participate in the reduction loop
while (first != last) {
  *result = bop(*result, uop(*first++));
}

The variable result is directly assigned the reduced value without any initial value participating in the reduction loop. The following code performs a parallel reduction to sum all the transformed numbers multiplied by 10 in the given range without any initial value:

const size_t N = 1000000;
int* res = tf::cuda_malloc_shared<int>(1);  // result
int* vec = tf::cuda_malloc_shared<int>(N);  // vector

// initializes the data
for(size_t i=0; i<N; i++) {
  vec[i] = i;
} 

// create an execution policy
tf::cudaStream stream;
tf::cudaDefaultExecutionPolicy policy(stream);

// queries the required buffer size to reduce N elements using the given policy
auto bytes  = policy.reduce_bufsz<int>(N);
auto buffer = tf::cuda_malloc_device<std::byte>(bytes);

// *res = 0*10 + 1*10 + 2*10 + 3*10 + 4*10 + ... + (N-1)*10
tf::cuda_uninitialized_reduce(policy,
  vec, vec + N, res, 
  [] __device__ (int a, int b) { return a + b; },
  [] __device__ (int a) { return a*10; },
  buffer
);

// synchronize the execution
stream.synchronize();

// delete the data
cudaFree(res);
cudaFree(vec);
cudaFree(buffer);