also sprach Sarad AV <[EMAIL PROTECTED]> [2003.08.16.1321 +0200]:
> f(x)=2x for all x=0,1,2,... ;
> f(x)=2x+1 for all x=0,1,2,...;

You need an extra bit to store which of the two is used. Otherwise,

> f(x)=2x
> 
> =2*5
> =10    for x=5;
> 
> Decimal number 5 can be represented in binary as 101.

... you can't decompress five as you don't know if it's 10 or 11.

and it's exactly this 2^0 bit which your compression kicks off. you
need it, can't do without.

[...]

> If the input programmes are picked truely randomly,then I know 16
> of the programs will halt(i.e 50% of the programs halt).

Look at it differently: you have a 50% chance of guessing whether
a given program from the sack will halt or not. This sheds some
different light onto the issue, doesn't it?

I raise a different question: are there more programs that halt than
programs that don't halt, or the other way around, or are they
equal in number?

> So where is the redundancy in different instances of
> the halting problem? I don't see any redundancy.

It's simple, if I am correct. The redundancy simply makes you care
less about the specific instance you are looking at.

> To represent 32 coins-i need 5 bits of information.
> Since the experiment is truely random-i know half of
> them will be heads,so in this case using 5 bits of
> information,i can determine all the coins that are
> heads and that are tails.

Same deal, unless you are counting pairs, in which case you cannot
distinguish between the members of a pair. You need an extra bit to
tell a head from a tail.

> So-the question is what is the minimum number of bits
> or entropy required to determine which all coins are
> heads and which all coins are tails,is it 5 bits or 6
> bits of information?

With 5 bits, you can count to 31, so you need 6.

Just my two tails.

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