what is the significance of 3.3121686421112381E-170 ?

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Adam Wozniak
Adam Wozniak on 19 Mar 2019
Edited: Mike Hosea on 20 Mar 2019
using matlab coder, the following is produced. real_T is a double precision floating point.
What is the significance of 3.3121686421112381E-170 ? Why was this specific number chosen, and not some other number?
/* Function for MATLAB Function: '<S1>/MATLAB Function' */
static real_T MyThing_norm(const real_T x[3])
{
real_T y;
real_T scale;
real_T absxk;
real_T t;
scale = 3.3121686421112381E-170;
absxk = fabs(x[0]);
if (absxk > 3.3121686421112381E-170) {
y = 1.0;
scale = absxk;
} else {
t = absxk / 3.3121686421112381E-170;
y = t * t;
}
absxk = fabs(x[1]);
if (absxk > scale) {
t = scale / absxk;
y = y * t * t + 1.0;
scale = absxk;
} else {
t = absxk / scale;
y += t * t;
}
absxk = fabs(x[2]);
if (absxk > scale) {
t = scale / absxk;
y = y * t * t + 1.0;
scale = absxk;
} else {
t = absxk / scale;
y += t * t;
}
return scale * sqrt(y);
}
  3 Comments
Walter Roberson
Walter Roberson on 19 Mar 2019
It looks like it is doing some kind of internal rescaling to prevent underflow. It just isn't obvious to me why 1E-170 is the boundary point, rather than sqrt(realmin) .

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Accepted Answer

Mike Hosea
Mike Hosea on 20 Mar 2019
Edited: Mike Hosea on 20 Mar 2019
Walter is on the right track.
What we see above is essentually a loop-unrolled version of the reference BLAS algorithm of DNRM2. The algorithm, having much longer vectors in mind, was designed to avoid unnecessary overflow and underflow while still making just one pass through the data. Unfortunately, the reference BLAS implementation asks whether abs(x(k)) ~= 0 in the loop. Floating point equality/inequality comparisons are not welcome when the generated code must adhere, say, to the MISRA standard, and calls to NORM are quite common, so we considered how to modify the algorithm without the ~= 0 comparisons.
The number in question is the largest initial value of the scale variable that guarantees that (abs(x(k))/scale)^2 does not underflow and is not denormal for any value of abs(x(k)) > 0, i.e. (abs(x(k))/scale)^2 >= realmin. Hence, the initial value of scale is eps*sqrt(realmin).

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