[BUG] Min and Max aggregations involving infinity produce incorrect results

[ttnghia]

Currently, min and max aggregations for numeric types firstly initialize the results by an identity value then call a device operator. The identity is:

• cuda::std::numeric_limits<T>::max(), if the aggregation is min, and
• cuda::std::numeric_limits<T>::lowest(), if the aggregation is max.

When the input is floating-point data that has infinity, the min and max are computed incorrectly. That's because min(max, infinity) == max and max(lowest, -infinity) == lowest. In other words, for floating-point numbers the lowest value is still bigger than -infinity and max is still smaller than infinity.

We should specialize the identity value for floating-point numbers.

[bdice]

cudf/cpp/include/cudf/detail/utilities/device_operators.cuh

Lines 168 to 177 in 8faf2b0

	template <typename T,
	std::enable_if_t<!std::is_same_v<T, cudf::string_view> && !cudf::is_dictionary<T>() &&
	!cudf::is_fixed_point<T>()>* = nullptr>
	static constexpr T identity()
	{
	// chrono types do not have std::numeric_limits specializations and should use T::min()
	// https://eel.is/c++draft/numeric.limits.general#6
	if constexpr (cudf::is_chrono<T>()) return T::min();
	return cuda::std::numeric_limits<T>::lowest();
	}

This should probably read something like:

  template <typename T,
            std::enable_if_t<!std::is_same_v<T, cudf::string_view> && !cudf::is_dictionary<T>() &&
                             !cudf::is_fixed_point<T>()>* = nullptr>
  static constexpr T identity()
  {
    // chrono types do not have std::numeric_limits specializations and should use T::min()
    // https://eel.is/c++draft/numeric.limits.general#6
    if constexpr (cudf::is_chrono<T>()) {
        return T::min();
    } else if constexpr (cuda::std::numeric_limits<T>::has_infinity) {
        return -cuda::std::numeric_limits<T>::infinity();
    }
    return cuda::std::numeric_limits<T>::lowest();
  }

and similarly for the min identity with +inf.