[Sussex] Challenge

[Geoff Teale]

Steve wrote:
------------
> The output (in bc) of 1.0 / 1010.0 is:
> .0001100110011001100110011001100110011001100110011001100110011001100\
> 11001100110011001100110011001100110011001100110011001100110011001100\
> 11001100110011001100110011001100110011001100110011001100110011001100\
> 11001100110011001100110011001100110011001100110011001100110011001100\
> 11001100110011001100110011

:)

That looks like what I'd expect to see ;)

> This look like a little more than 53 bits to me!  (you did say you
> didn't have bc to hand - install Cygwin as you exit statement to HR
> on your PC).

I installed Cygwin on this machine a while back from a magazine coverdisk,
for some reason this version is missing a lot of stuff (including gcc!) but
has mysql in it.. v.odd.

> But given that:
> 0010.0 * .0001100110011001100...0011 = .0011001100110011001...0110
> and
> 1000.0 * .0001100110011001100...0011 = .1100110011001100110...1001
> 
> Then I agree there must be a fiddle factor in there somewhere :-)


Indeed ;)

> Go 2:
> 
> 0001.0 = 1.0 * 10^0
> 1010.0 = 1.0 * 10^1010
> 0000.1 = 1.0 * 10^-1010
> 
> Therefore when multiplying number one just has to add the exponents
> together:
> 
> 0000.1 * 1010.0 = 1.0 * 1010^(-1010 + 1010)
>                 = 1.0 * 1010^(0)
>                 = 0001.0
> 
> How' that!  Of course the mantissa is always 1.0 and the exponent 
> is an integer itself so this only works for a limited number of
> values (but no fiddle factor).

Good thinking, but that doesn't escape the problem of expressing 0.1 in
binary, it just sidesteps it.  So, whilst it works logically, it isn't in
the spirit of the question - still it shows an interesting way around the
problem (for a limited number of cases) so it carries some merit.

> Mine's still a pint of Horsham Best

We'll see... :)

-- 
GJT
Free Software, Free Society. 
http://www.fsf.org   http://www.gnu.org


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