[A83] Re: 83p os

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[A83] Re: 83p os

> > For ease of calculation, lets say we want 9 accurate decimal digits.
> > 10^9 ~ 2^30 (because 1000 = 10^3 ~ 2^10 = 1024). So 9 decimal
> > digits more or less correspond to 30 binary digits, right? (Another
> > way of looking at it: 2^-30 < 10^-9, meaning if the 31st binary digit
> > is wrong, that won't affect the first 9 decimal digits.)
>
> Try it with 1/3.  No matter how many binary digits you have, you'll still
be
> incorrect.

I don't agree.  While, as Henk Poley states, you cannot describe 1/3 even in
decimal form, a certain number of binary digits do give "correct" and
accurate, but of course not infinitely precise answers.  Try some binary
long division.
(monospaced font)

      .0101010101010101010101010101010101010101 ...
    _______________________________________________
11 ) 1.0000000000000000000000000000000000000000 ...
       11
        100
         11
          100
           11
            100
              ...

In the case of 1/3 (1/11 binary) it requires 32 digits (according to my
TI-83 ;) to produce a 10-digit decimal answer of .3333333333.  9 decimal
digits (.333333333) requires 30 binary digits.  Evaluate this yourself:
2^-2 + 2^-4 + 2^-6 + 2^-8 ... 2^-30 the exponent corresponds to the binary
digit right of the point.

2/3, anyone?

       .1010101010101010101010101010101010101010101
    __________________________________________________
11 ) 10.0000000000000000000000000000000000000000000...


1/7?

       .001001001001001001001001001001001001001001001001...
     ______________________________________________________
111 ) 1.000000000000000000000000000000000000000000000000...


Not only correct, but fun (I need a life).


> You need something like BCD to do things accurately.  Binary can
> get you "close enough" for many applications, such as 3D engines on slow
> platforms, but it isn't nearly accurate enough for general numerical work.

Why?  Numbers are numbers, and arithmetic is arithmetic, no matter what base
you use.  IMO, it's just a matter of creating the algorithm for the required
number of digits.  I don't see how it isn't accurate.  Precision of this
math routine is only dependent upon the number of places the calculation is
carried out to.
References: