执行双浮点除法的正确算法是什么？

Question

I'm following the algorithms provided by this paper by Andrew Thall describing algorithms for performing math using the df64 data type, a pair of 32-bit floating point numbers used to emulate the precision of a 64-bit floating point number. However, there appears to be some inconsistencies (mistakes?) in how they've written their Division and Square Root functions.

This is how the Division function is written in the paper:

float2 df64_div(float2 B, float2 A) {
    float xn = 1.0f / A.x;
    float yn = B.x * xn;
    float diff = (df64_diff(B, df64_mult(A, yn))).x;
    float2 prod = twoProd(xn, diffTerm);

    return df64_add(yn, prodTerm);
}

^{The language used to write this code appears to be Cg, for reference, although you should be able to interpret this code in C++ if you treat float2 as though it's merely an alias for struct float2{float x, y;}; , with some extra syntax to support arithmetic operations directly on the type.}

For reference, these are the headers of the functions being used in this code:

float2 df64_add(float2 a, float2 b);
float2 df64_mult(float2 a, float2 b);
float2 df64_diff(/*Not provided...*/);
float2 twoProd(float a, float b);

So a couple of problems immediately stand out:

• diffTerm and prodTerm are never defined. There's two variables, diff and prod which are defined, but it's not certain that these are the terms that were intended in this code.
• No declaration of df64_diff is provided. Presumably this is meant to support subtraction; but again, this is not clear.
• df64_mult is a function that does not accept a 32-bit float as an argument; it only supports two pairs of 32-bit floats as arguments. It is not clear how the paper expects this function call to compile
• Same also for df64_add , which also only accepts pairs of 32-bit floats as arguments, but is here invoked with the first argument being only a single 32-bit float.

I'm making an educated guess that this is a correct implementation of this code, but because even a correct implementation of this function has unavoidable errors in the computation, I can't tell if it's correct, even if it gives values that "seem" correct:

float2 df64_div(float2 B, float2 A) {
    float xn = 1.0f / A.x;
    float yn = B.x * xn;
    float diff = (df64_diff(B, df64_mult(A, float2(yn, 0)))).x;
    float2 prod = twoProd(xn, diff);

    return df64_add(float2(yn, 0), prod);
}

float2 df64_diff(float2 a, float2 b) {
    return df64_add(a, float2(-b.x, -b.y));
}

So my question is this: is the written implementation of this algorithm as seen in the paper accurate (because it depends on behavior of the Cg language that I'm not aware of?), or isn't it? And regardless, is my interpolation of that code a correct implementation of the division algorithm described in the paper?

Note: my target language is C++, so although the differences between the languages (for this kind of algorithm) are minor, my code is written in C++ and I'm looking for correctness for the C++ language.

Answer 1

Reviewing the Pseudocode Algorithm as written in the book appears to support the C++ implementation of this algorithm, although my unfamiliarity with Cg means I can't prove that that implementation is correct for Cg.

So breaking down these steps into plain english:

1. The function takes two parameters, each of which are [pseudo-]double precision floating point values, and where the second parameter is not equal to 0
2. The variable x _n is assigned the arithmetic reciprocal of the higher order component of the [pseudo-]double divisor, calculated using single precision floating point math
3. The variable y _n is assigned the product of the higher order component of the [pseudo-]double dividend and x _n , calculated using single precision floating point math
4. The product of the [pseudo-]double Divisor and y _n is calculated
   • This is the first tricky part, because the paper doesn't describe an algorithm for [pseudo-]double x single multiplication. We can see in the Cg algorithm that the Cg algorithm clearly maps to this step 1-to-1, but the Cg rules for promoting a scalar value to a vector value are unknown.
   • What we can say, however, is that we do have a function for multiplying a double by a double, and a single can be promoted to a double by padding its lower order component with 0, so we can do that.
5. The difference between the Dividend and the product calculated in step 4 is calculated, and only the higher order component is kept as a single-precision floating point value
   • What makes this tricky is that the paper doesn't describe an algorithm for subtraction. However, it does describe an algorithm for converting a [IEEE754-]double into a [pseudo-]double, and an observation we can make is that negative [IEEE754-]doubles, when converted, have negative values for both its higher order and lower order components. So logically, a [pseudo-]double can be negated by simply negating both of its components. And a negated number added is mathematically equivalent to subtraction, so we can build a subtraction algorithm using this knowledge.
6. The product of x _n and step 5 is performed, preserving the extended precision that would otherwise be lost in a single x single multiplication.
   • The twoProd function exists for exactly this purpose.
7. The sum of step 6 and y _n is calculated
   • Again, we can use the [pseudo-]double addition algorithm if we simply promote y _n to a [pseudo-]double by padding the lower order component with 0
8. The result of step 7 is the returned value

So understanding this algorithm we can map each of these steps directly to the C++ algorithm I wrote:

//(1) Takes two [pseudo-]doubles, returns a [pseudo-]double
float2 df64_div(float2 B, float2 A) {
    //(2) single float divided by single float
    float xn = 1.0f / A.x;
    // (3) single float multiplied by single float
    float yn = B.x * xn;
    //                        (4) double x double multiplication
    //                                       (4a) yn promoted to [pseudo-]double
    //            (5) subtraction                           (5a) only higher order component kept
    float diff = (df64_diff(B, df64_mult(A, float2(yn, 0)))).x;
    // (6) single x single multiplication with extra precision preserved using twoProd
    float2 prod = twoProd(xn, diff);
    // (7) adding higher-order division to lower order division
    //              (7a) yn promoted to [pseudo-]double
    // (8) value is returned
    return df64_add(float2(yn, 0), prod);
}

float2 df64_diff(float2 a, float2 b) {
    //                 (5a) negating both components is a logical negation of the whole number
    return df64_add(a, float2(-b.x, -b.y));
}

From this, we can conclude that this is a correct implementation of the algorithm described in this paper, upheld by some testing I've done to validate that performing these operations in this manner yields results that appear to be correct.

Answer 2

Xirema's answer provides a faithful rendering of Thall's high-radix long-hand division algorithm into C++. Based on fairly extensive testing against a higher-precision reference, I found its maximum relative error to be on the order of 2 ^-45 , provided there are no underflows in intermediate computation.

On platforms that provide a fused multiply-add operation ( FMA ), the following Newton-Raphson-based division algorithm due to Nagai et. al. is likely to be more efficient and achieves identical accuracy in my testing, that is, maximum relative error of 2 ^-45 .

/*
  T. Nagai, H. Yoshida, H. Kuroda, Y. Kanada, "Fast Quadruple Precision 
  Arithmetic Library on Parallel Computer SR11000/J2." In: Proceedings 
  of the 8th International Conference on Computational Science, ICCS '08, 
  Part I, pp. 446-455.
*/
float2 div_df64 (float2 a, float2 b)
{
    float2 t, c;
    float r, s;
    r = 1.0f / b.x;
    t.x = a.x * r;
    s = fmaf (-b.x, t.x, a.x);
    t.x = fmaf (r, s, t.x);
    t.y = fmaf (-b.x, t.x, a.x);
    t.y = a.y + t.y;
    t.y = fmaf (-b.y, t.x, t.y);
    s = r * t.y;
    t.y = fmaf (-b.x, s, t.y);
    t.y = fmaf (r, t.y, s);
    c.x = t.x + t.y;
    c.y = (t.x - c.x) + t.y;
    return c;
}