Small challenge

[J-F Garnier]

Some more clarifications:

(04-25-2023 08:49 PM)J-F Garnier Wrote:  -when subtracting two 15-digit BCD numbers, the larger mantissa is shifted left (moved to 16 digits)

I would better say: subtraction is done on 16 digits, but in any case the result is normalized to 15 digits.
An example of the benefit of the 16-digit subtraction:

Do 1 - 1/3000 in extended precision, that is:
  1.00000000000000
- 0.000333333333333333
Computing with 15 digits only, you would get, after alignement of the mantissa:
  1.00000000000000
- 0.00033333333333
= 0.99966666666667 i.e. only 14 significant digits

But doing the operation on 16 digits (by shifting both mantissa left 1 position):
  1.000000000000000
- 0.000333333333333
= 0.999666666666667 i.e. now 15 significant digits

Quote:-the multiplication of two 15-digit mantissa can provide a mantissa on 15 or 16 digits, for instance 3.000.. x 3.000.. --> 9.000.. (on 15 digits), but 3.000.. x 4.000.. --> 12.000.. (on 16 digits). The result is always normalized on 15 digits but the possible 16-digit mantissa is kept (in D) and sometimes used to improved the accuracy of internal calculations, and this is the case for y^x computed as exp(x*ln(y)).

Note that in the 3^729 case, the term x*ln(y) = 729 x 1.098... doesn't generate a 16-digit mantissa, no extra accuracy here.

Same for the 1e44^10.5 case, the term x*ln(y) is 10.5 x 101.31..
Here are the intermediate results in extended precision on both machines for 1e44^10.5:
                              75c (16 digits)                71b (15 digits)
ln(1e44) =                101.3137440917379        101.313744091738    exact value = 101.3137440917380(10)
10.5*ln(1e44) =        1063.794312963247        1063.79431296324
exp(10.5*ln(1e44)) = 9.999999999977472e461   9.99999999990893e461
rounded to:              9.99999999998e461          9.99999999991e461

The difference comes from the result of the step 10.5*ln(1e44) and the missing 16th digit on the 71b.

J-F