Re: HALTING PROBLEM

In sci.math, tommy1729
<tommy1729@xxxxxxxxx>
wrote
on Wed, 20 Jun 2007 10:18:18 EDT
<4344668.1182349128560.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>:
In sci.math, tommy1729
<tommy1729@xxxxxxxxx>
wrote
on Mon, 18 Jun 2007 19:41:54 EDT
<29855439.1182210147971.JavaMail.jakarta@xxxxxxxxxxxxx
forum.org>:
A halts if B halts

determining if A halts
depends on the halting of B
and B on C
and C on D etc

thats the basic idea of the halting problem and its
socalled unsolvability.

however the idea is WRONG !

1) there is no known algoritm A for wich an
algoritm B C AND D EXISTS !!!!
(in the sence that B C AND D are very different
from A AND EACHOTHER and thus not A repeated n times
in m steps )

2) the proof of the unsolvability is clearly a
selfreference rather than a proof.

3) mathematical proof rather occur like proving
that two opposite things dont add up, rather than
saying this is true if that is true and that is true
if this is true ...

4) iterations cannot be discribed bye totally
different iterations

iteration A can not depend on iteration B unless
there equal (see 1) ) if the iteration cannot be
rewritten in a simpler "closed form"


more promising is

A halts if B does not.

B can then not be tested bye trial , but bye
THEORY.

the idea of taking proofs to tested iterations is
childish and primitive
IN GENERAL NO GREAT MATHEMATICIAN OR PROOF HAS EVER
BEEN DONE LIKE THAT. (with a few exceptions but these
never had an A B AND C AND D , just an A or an A and
B nothing more)

another controversial topic

but i dare you to disproof 1) for any conjecture
that is (not) decidable and no closed form functions
used.

greets
tommy1729

Your logic is way off. First, some formalisms.

[1] Letters such as P denote a program. (For
purposes
of this expose think of it as a function
tion implemented by
a Turing machine. Other methods are also
also possible.)
P(n) is the result of applying P to input n, if
it halts. P is also a string; for every program
gram one
can assign a number or string. (For Turing
ring machines
one might encode the alphabet, number of states,
tape motions, transition matrix, and accepting
ting states
in such a way as to guarantee a unique P for
for every
machine, ignoring problems such as machine
hine equivalence,
which isn't needed for this proof.)

[2] Let k(m,n) be a combining function; this function
combines two strings m and n and spits out a
ut a string
in an injective way. Define two additional
onal functions
if necessary F(k) = m and S(k) = n if k = k(m,n).
One simple function would be to interleave m and
n, with a marker perhaps to indicate end of
d of string
(and maybe doubling the marker to escape it,
it, somewhat
like backslash). Note that F() and S() aren't
en't used
in the proof proper, though they might be
t be internally
used in H().


yes but im intrested in the math not the tapes and writings etc. thats computer stuff.

[3] Let H(i) be our hypothetical halting test, where
i = k(P,n). The result of H(i) will start with
with '1' if P(n)
halts, and '0' if P(n) loops. H() is assumed to
always halt.

i mainly read LET and ASSUMED TO HALT.

they are assumptions , already before it is proven !!!

how "strange"

Read up on reducto ad absurdum sometime.

A -> (contradiction) therefore ~A.


[4] Let L(i) be a machine that loops if the string i
begins with '1',
and halts returning '1' if the string i begins
gins with '0'.


another "let" many definitions and assumpions ,
to eventully proof the definitions and assumptions.

The halting problem is not as simple as it should be, but neither is it
as difficult as some others.



[5] If P(i) = M(N(i)), then one can contemplate
constructing P from M and N using the composition
tion operator. This might be
written P = M o N. For now, we elide the details
ails and simply write
things such as W_h(i) = L(H(k(i,i))), as opposed
osed to attempting
to define composition operators, projectors, and
and such to express
W_h in terms of L, H, k, and some sort of
t of argument duplicator
function (since k cannot actually be implemented
nted in this formalism).

[6] What is the value of H(k(W_h,W_h)), where W_h(i)
= L(H(k(i,i)))?
A: If it starts with a '1', then W_h halts on
input W_h. Since W_h(W_h) is
(W_h) is L(H(k(W_h,W_h)))
and we already know H(k(W_h,W_h)) starts with
rts with a
'1', we evalate L('1') and end up with
up with something
that loops, which means our initial
initial assumption
was false for case A.
B: If it starts with a '0', then W_h loops on
input W_h, and since W_h(W_h) is
(W_h) is L(H(k(W_h,W_h))),
H(k(W_h,W_h)) starts with '0', and L('0')
d L('0') halts,
which means our case B assumption was also
was also false.

Therefore, for every H() I can construct a W_h() that
breaks it.

The main flaw in this proof is k(). I can get around
that by defining
P(x,y,z,...) as being a program (the ',' being a
special designated
separator character), then defining A(x,y,z,...) as a
method to create
a unique string from x,y,z, and additional characters
as necessary.
One can also define E(a,n) to pick the n'th argument,
and work from
there. One can then define H(i) such that i =
A(P,A(args)), or simply
define H(i,A(args)), then construct a W_h() such that
it breaks H().
The details are left to the interested reader.

why so complicated ???

give me an example of a program A wich halts if
the 3n+1 circles somewhere above 50.

/*
1-50 provably stop, though it takes awhile for 27

2->1
3->10->5->16->8->4->2
6->3
7->22->11->34->17->52->26->13->40->20->10
9->28->14->7
12->6
15->46->23->70->35->106->53->160->80->40
18->9
19->58->29->88->44->22
21->64->32->16
24->12
25->76->38->19
27->82->41->124->62->31->94->47->142->71->214->107->322->161->
484->242->121->364->182->91->274->137->412->206->103->310->
155->466->233->700->350->175->526->263->790->395->1186->593->
1780->890->445->1336->668->334->167->502->251->754->377->
1132->566->283->850->425->1276->638->319->958->479->1438->
719->2158->1079->3238->1619->4858->2429->7288->3644->
1822->911->2734->1367->4102->2051->6154->3077->9232->
4616->2308->1154->577->1732->866->433->1300->650->325->
976->488->244->122->61->184->92->46
30->15
33->100->50->25
36->18
37->112->56->28
39->118->59->178->89->268->134->67->202->101->304->152->76
42->21
43->130->65->196->98->49->148->74->37
45->136->68->34
48->24

we assume sufficiently large integers
*/

function collatz50()
{
for(int i = 51; ; i += 2)
{
int t = i;
while(t >= i)
{
if(t mod 2) t div= 2;
else t = 3*t + 1;
if(t == i)
return i;
}
}
}


and a program B that halts/does not halt(choose) if A does halt.

function weirdo(f)
{
while(halts(f()))
;
}

(assuming halt() works)


and a program C that does that for B

Does what for B?

and D for C.

you can add as many veriables , tapes assumptions and heads as you want but you cant even
turn the 3n+1 problem into distinct 4 programs
that halt/halt not depending on 3n+1 and the other ones.


But there's no real self-reference, either way.

yes i just replied to them !

I'm
defining a machine-to-machine mapping: for every
machine H,
I construct a W_h. If H is purported to solve the
halting
problem, W_h breaks H. If H does something silly,
not
involving the halting problem (e.g., it just sits
there and
moves its head), then W_h does even sillier, and
largely
irrelevant, things. But W_h can still be
constructed.

thanks for your effort.

i do appreciate it.

but i still stand with my arguments that there are no programs A B C D for e.g. a simple iteration problem as
3n+1.

The Collatz conjecture is an interesting conundrum, but not all that
relevant to the halting problem except as input.


and therefore i dont believe in the use/proof of the halting problem.

assuming B from A etc is at the core.

and it is wrong.

assuming another "related program" exists is very

"optimistic", even though the halting problem has
a "pessimistic" outcome.

and adding more complicated assumptions, definitions, tapes and variables makes it even more "optimistic"

i see programs as iterations.

therefore my critic.

i hope more people will reply.

Don't count on it; the halting problem was solved by Turing
many many years ago, and is now cut and dried. To contest
Turing's conclusions risks you being labeled ignorant; to
do so with very sloppy setup risks you being labeled a crank.

This is one reason I was trying to be careful during setup.
Evidently I wasn't quite careful enough -- or perhaps
intuitive enough.


greets
tommy1729

--
#191, ewill3@xxxxxxxxxxxxx
Does anyone else remember the 1802?

--
Posted via a free Usenet account from
http://www.teranews.com


1802 ?

http://www.microprocessor.sscc.ru/great/s2.html#1802

It is a very weird little processor.

--
#191, ewill3@xxxxxxxxxxxxx
Linux sucks efficiently, but Windows just blows around
a lot of hot air and vapor.

--
Posted via a free Usenet account from http://www.teranews.com

.

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