Unexpected result when summing sorted (and unsorted) positive floating point numbers

Question

I am exploring Higham's excellent Accuracy and Stability of Numerical Algorithms and chapter 4 is dedicated to summation.

So I decided to test the most basic thing. Summing positive random numbers uniformly distributed between $0$ and $1$ (exactly $2^{20}$ of them). I was expecting that first sorting (increasing order of course) them would give me a better relative error (I am also counting how many no-operations are there and with the sorted version I get 0, while thousands with the unsorted version). However this is not the case, many times, the unsorted version gives a smaller error by $1-2$ order of magnitudes (despite many no-operations). Is this normal? Is there any clear manufactured example where this is the case?

I am accumulating IEEE floats (c++) for the approximate computations while computing the exact solution in a 16 bytes long double (so casting each float to long double and accumulate on the long double result).

test_recursive_sum
Relative error:
1.1244650891750177666438375817657311598468661273386715038213878870e-06
Noops:
10974
test_sort_recursive_sum
Relative error:
1.3188994459459007621524197348059082024462895788019523024559020996e-04
Noops:
0
Sample:
1048576

Answer 1

Very interesting problem! I might have a partial answer.

To start, I replicated a simple C++ demo that can reproduce the effect

#include <iostream>
#include <vector>
#include <random>
#include <algorithm>
#include <numeric>
#include <iomanip>
#include <boost/multiprecision/cpp_dec_float.hpp>
using namespace boost::multiprecision;
template <typename T>
T KahanSum(const std::vector<T>& numbers) {
    T sum = 0.0f;
    T c = 0.0f;
for (const T&amp; num : numbers) {
    T y = num - c;
    T t = sum + y;
    c = (t - sum) - y;
    sum = t;
}

return sum;

}
int main() {
    // Define the number of random numbers
    const int N = 1048576;
// Create a Mersenne Twister pseudo-random generator
std::random_device rd;
std::mt19937 gen(rd());
std::uniform_real_distribution&lt;float&gt; dis(0.0f, 1.0f);

// Generate random numbers and store them in a vector
std::vector&lt;float&gt; numbers(N);
for (int i = 0; i &lt; N; ++i) {
    numbers[i] = dis(gen);
}

float sum_float = std::accumulate(numbers.begin(), numbers.end(), 0.0f);
double sum_double = std::accumulate(numbers.begin(), numbers.end(), 0.0);
auto sum_double_k = KahanSum(numbers);

// Calculate sum using higher precision (cpp_dec_float_100)
using high_precision_float = cpp_dec_float_100;
high_precision_float sum_high_precision = std::accumulate(numbers.begin(), numbers.end(), high_precision_float(0));

std::sort(numbers.begin(), numbers.end());

float sum_float_sorted = std::accumulate(numbers.begin(), numbers.end(), 0.0f);
double sum_double_sorted = std::accumulate(numbers.begin(), numbers.end(), 0.0);
auto sum_double_k_sorted = KahanSum(numbers);

std::cout &lt;&lt; &quot;Sum using float accumulator:                &quot; &lt;&lt; std::scientific &lt;&lt; std::setprecision(30) &lt;&lt; sum_float &lt;&lt; std::endl;
std::cout &lt;&lt; &quot;Sum using float accumulator after sorting:  &quot; &lt;&lt; std::scientific &lt;&lt; std::setprecision(30) &lt;&lt; sum_float_sorted &lt;&lt; std::endl;
std::cout &lt;&lt; &quot;Sum using double accumulator:               &quot; &lt;&lt; std::scientific &lt;&lt; std::setprecision(30) &lt;&lt; sum_double &lt;&lt; std::endl;
std::cout &lt;&lt; &quot;Sum using double accumulator after sorting: &quot; &lt;&lt; std::scientific &lt;&lt; std::setprecision(30) &lt;&lt; sum_double_sorted &lt;&lt; std::endl;
std::cout &lt;&lt; &quot;Sum using high precision:                   &quot; &lt;&lt; std::scientific &lt;&lt; std::setprecision(30) &lt;&lt; sum_high_precision &lt;&lt; std::endl;
std::cout &lt;&lt; &quot;Sum using KahanSum:                         &quot; &lt;&lt; std::scientific &lt;&lt; std::setprecision(30) &lt;&lt; sum_k &lt;&lt; std::endl;
std::cout &lt;&lt; &quot;Sum using KahanSum sorted:                         &quot; &lt;&lt; std::scientific &lt;&lt; std::setprecision(30) &lt;&lt; sum_k_sorted &lt;&lt; std::endl;

float float_diff = std::abs(static_cast&lt;double&gt;(sum_float) - static_cast&lt;double&gt;(sum_high_precision));
float float_sorted_diff = std::abs(static_cast&lt;double&gt;(sum_float_sorted) - static_cast&lt;double&gt;(sum_high_precision));
std::cout &lt;&lt; &quot;Error of float accumulator:                &quot; &lt;&lt; float_diff/sum_high_precision &lt;&lt; std::endl;
std::cout &lt;&lt; &quot;Error of float accumulator after sorting:  &quot; &lt;&lt; float_sorted_diff/sum_high_precision &lt;&lt; std::endl;
std::cout &lt;&lt; &quot;Error of float accumulator:                &quot; &lt;&lt; float_diff/sum_high_precision &lt;&lt; std::endl;
std::cout &lt;&lt; &quot;Error of float accumulator after sorting:  &quot; &lt;&lt; float_sorted_diff/sum_high_precision &lt;&lt; std::endl;

return 0;

}

I included even higher precision float to confirm, but the double is actually enough here for a "true" answer. I also included the Kahan summation algorithm.

1. Using the Kahan summation completely removes this effect, and the final answer is more accurate.
2. The unsorted sum is smaller than the true answer, and the sorted sum is larger.

Wait what? Larger? I admit I haven't read the floating point spec, but my gut feeling was that adding large_sum + small_term would always produce an underestimate.

Diving into the Kahan summation helps our understanding, we can modify it to just report the summation error, without modifying the values:

template <typename T>
T KahanCheck(const std::vector<T>& numbers) {
    T sum = 0.0;
for (const T&amp; num : numbers) {
    T y = num;
    T t = sum + y;
    T c = (t - sum) - y;
    std::cout &lt;&lt; std::scientific &lt;&lt; std::setprecision(10) &lt;&lt; c &lt;&lt; std::endl;
    sum = t;
}

return sum;

}

Before today, I would have assumed:

1. y can be split into positive high order and low order bits: y = y_high + y_low
2. Adding sum + y = sum + y_high + y_low = sum + y_high as it will truncate and ignore the low order bits of y_low
3. The compensation c = (t - sum) - y = (sum + y_high - sum) - y_high - y_low = -y_low will always be negative.

However, running the KahanCheck on the sorted array, you will find that c can be positive and negative, so clearly this assumption doesn't hold.

As a concrete example, if we consider the following floats (all decimals shown for 32bit float precision):.

sum   = 418.998487293720245361328125000e-05
y     =   9.554042480885982513427734375e-05
t     = 428.552553057670593261718750000e-05 # computed with float
exact = 428.552529774606227874755859375e-05 # computed exactly
                ^t is larger!

we can see that sum t actually over what would be the exact answer. So a floating point sum can be larger, despite losing low order bits.

As such, what I think is happening here is that by sorting the data, we bias the summation towards overestimation by limiting how many low order bits get truncated. I suspect out standard expected behavior of rounding up 0.5 to is what gives is this nudge.

You could compute the c terms from KahanCheck, adding up all the positive and negative contributions, and by sorting, you nudge this one way or the other.

Another thing to learn is that if you want to minimize the numerical inaccuracy Kahan summation way better than a costly sort. You could also consider pairwise recursive summation

---

I have attempted to verify my "rounding up" speculation with many varied experiments, but I have been unable to do so. I think my theory is unfortunately simply wrong. A few more observations:

Experimenting with only the range [0,1) greatly affect things:

1. At [0,100) the errors are reversed; sorting does increase accuracy!
2. At [0.1,1) the errors are the same, sorting or not.

So in my mind any satisfactory explanation would need to account for the distribution [0,1).