Re: OPPOSITE OF all coin sequences are computable to infinite length ?

From: |-|erc (h_at_r.c)
Date: 01/09/05 

Date: Sun, 9 Jan 2005 16:00:47 +1000

"The Ghost In The Machine" <ewill@sirius.athghost7038suus.net> wrote in
> >> >
> >> > define a TM
> >> > prove it is impossible
> >> > define a number using impossible TM
> >>
> >> The TM for computing Chaitin's Constant is a mutation of
> >> the TM for computing the halting problem.
> >>
> >> [spitsnip]
> >>
> >
> > so you prove the halting value of some programs is unknown
> > then you assign that to a variable, an unknown so to speak?
>
> I'm not proving much of anything here. I'm simply stating
> a result.
>
> But, have it your way. Assume H is a machine with the following
> characteristics.
>
> [1] It always halts.
> [2] If k = Enc1<m> concat ',' concat Enc2<i>, where m is
> a machine descriptor in some UTM formalism and i is
> a tape descriptor, and Enc1<> and Enc2<> are computable
> mappings into H's alphabet (sans ','), then H will
> write a 0 or 1 into the first cell of k
> (the rest of the tape will be munged).
> [3] H will write 0 if m(i) will not halt, 1 if m(i) will.
>
> If I've done this right, I can then construct another
> machine W from H. W does the following.
>
> [a] It looks for the comma, then replaces anything following
> with Enc2<beforecomma ',' aftercomma> , then moves the head
> back to the start of its input.
> [b] It copies H. That is to say, the states of H are isomorphic
> to a proper subset of W's states, and the transitions of H
> are isomorphic to a proper subset of W's transitions.
> [c] If the head is sitting over a 0, it halts and announces success.
> If the head is sitting over a 1, it loops.
>
> W of course has a machine descriptor w. We can construct
> k = Enc1<w> ',' Enc2<w>, and feed it to W.
>
> Case 1: W(k) halts. This means that machine(w) on input w
> does not halt. But machine(w) = W.
>
> Case 2: W(k) loops. This means that machine(w) on input w
> halts. But machine(w) = W.
>
> Therefore H cannot exist.
>
> In this machinery Chaitin's Omega can be construed as that number
> such that the n'th bit is 1 if machine n halts on input Enc2<n>,
> and 0 if machine n does not halt on input Enc2<n>. Very well-defined,
> but not all that computable.

look nancy boy, stop strutting while patting yourself on the head.
whether machine n halts is UNKNOWN, right?

Here's an apple, here's another apple..
NOW, that makes 2 apples, right?

Here's is an unknown value, halt(1000).
Here is another unknown value, halt(2000).

halt(1000) and halt(2000) don't make anything!
2 things that are unknown is just something that is unknown.

They aren't things. Its UNKNOWN.

Here's a set of unknowns { (*& (*(*& ^&(* )(( ^& (()* &**& } encoded so you can't see
Here is another set of unknowns { *&&*(**& (*(**( *&*& ((*)}

They are UNKNOWN.

You can't union them, you can't make them members, you can't do anything with them.
the result ISNT KNOWN. YOU DONT KNOW.

Here's a maths puzzle.

2 x + 3 = 7 x is an_unknown
2x = 4
x = 2 NOW YOU KNOW X

This is not a maths puzzle.

x = halt(1000) x is NOT KNOWN

Sure it *MAY* halt, hence your confusion with it being AN_UNKNOWN.

but its not AN_UNKNOWN.

for some values of H(n), you WILL NEVER KNOW.

the program may halt IN THE FUTURE.
it has no HALT value.

Things that are unknown are not objects.

>
> >
> > X = of Halt(777)
> > an unknown of the unknown. <gag>
> >
> > Y = "there is a god"
> > what is the value of Y?
>
> Y = false. But that's more properly addressed in talk.atheism. :-P
>

only if you can prove it.

"The futures not ours to see".

Now do you see the difference between UNKNOWN and AN_UNKNOWN?

Herc

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