`std::reduce()`

can produce non-deterministic results when used with non-commutative or non-associative operators because it allows elements to be combined in any order, especially in parallelÂ execution.

To ensure deterministic results, consider theseÂ strategies:

`std::accumulate()`

If you need deterministic results, and your operation is non-commutative, consider using `std::accumulate()`

instead of `std::reduce()`

. `std::accumulate()`

processes elements sequentially from left to right, ensuring a consistentÂ order.

```
#include <iostream>
#include <numeric>
#include <vector>
int main() {
std::vector<int> numbers{1, 2, 3, 4, 5};
int result = std::accumulate(
numbers.begin(), numbers.end(), 0,
[](int a, int b) {
return a - b;
});
std::cout << "Result: " << result;
}
```

`Result: -15`

When using `std::reduce()`

, you can specify the `std::execution::seq`

execution policy to enforce serial execution. This ensures elements are combined in a specific order, similar to `std::accumulate()`

.

```
#include <execution>
#include <iostream>
#include <numeric>
#include <vector>
int main() {
std::vector<int> numbers{1, 2, 3, 4, 5};
int result = std::reduce(std::execution::seq,
numbers.begin(), numbers.end(), 0,
[](int a, int b) {
return a - b;
});
std::cout << "Result: " << result;
}
```

`Result: -15`

For more complex scenarios, you might need to implement a custom reduction logic that manually controls the order ofÂ operations.

This can be done by splitting the data into chunks, processing them individually, and then combining the results in a controlledÂ manner.

```
#include <iostream>
#include <numeric>
#include <vector>
#include <thread>
void accumulateRange(
const std::vector<int>& numbers,
int start, int end, int& result
) {
result = std::accumulate(
numbers.begin() + start,
numbers.begin() + end,
0,
[](int a, int b) {
return a - b;
}
);
}
int main() {
std::vector<int> numbers{1, 2, 3, 4, 5};
int result1 = 0, result2 = 0;
std::thread t1(
accumulateRange,
std::ref(numbers),
0,
numbers.size() / 2,
std::ref(result1)
);
std::thread t2(
accumulateRange,
std::ref(numbers),
numbers.size() / 2,
numbers.size(),
std::ref(result2)
);
t1.join();
t2.join();
std::cout << "Result 1: " << result1;
std::cout << "\nResult 2: " << result2;
int finalResult = result1 - result2;
std::cout << "\nFinal Result: " << finalResult;
}
```

```
Result 1: -3
Result 2: -12
Final Result: 9
```

**Use the**`std::accumulate()`

**Algorithm**: For non-commutative operations where deterministic results are required.**Serial Execution Policy**: Apply`std::execution::seq`

with`std::reduce()`

to enforce a specific order.**Custom Logic**: Implement manual reduction logic to control the order of operations.

These strategies help ensure deterministic results even with non-commutative operations in reductionÂ algorithms.

Answers to questions are automatically generated and may not have been reviewed.

This Question is from the Lesson:### The Reduce and Accumulate Algorithms

A detailed guide to generating a single object from collections using the `std::reduce()`

and `std::accumulate()`

algorithms