Sujet : Two dozen people were simply wrong
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory sci.logicDate : 29. May 2024, 20:31:52
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v37sap$18mfo$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9
User-Agent : Mozilla Thunderbird
On 5/29/2024 1:14 PM, Ben Bacarisse wrote:
Alan Mackenzie <acm@muc.de> writes:
How about a bit of respect? Mike specifically asked you not to cite his
name as a back up for your points. Why do you keep doing it?
He does it to try to rope more people in. It's the same ploy as
insulting people by name. It's hard to ignore being maligned in public
by a fool.
*Thanks for validating my simplified encoding of the Linz*
When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
I really did believe that Ben Bacarisse was lying when I said it.
At the time I was talking about the easily verified fact of the actual
execution trace of fully operational code and everyone was denying the
easily verified facts.
typedef int (*ptr)(); // ptr is pointer to int function in C
00 int H(ptr p, ptr i);
01 int D(ptr p)
02 {
03 int Halt_Status = H(p, p);
04 if (Halt_Status)
05 HERE: goto HERE;
06 return Halt_Status;
07 }
08
09 int main()
10 {
11 H(D,D);
12 return 0;
13 }
It turns out that two dozen people are easily proven wrong when
they claimed that the correct simulation of the input to H(D,D)
is the behavior of int main() { D(D); }
When D is correctly simulated by H using an x86 emulator the only
way that the emulated D can reach its own emulated final state
at line 06 and halt is
(a) The x86 machine code of D is emulated incorrectly
(b) The x86 machine code of D is emulated in the wrong order
*two dozen people were simply wrong*
It now turns out that Richard Damon was not lying when he referred
to the words of Peter Linz.
It did seem ridiculous that the Linz proof merely proved that
a single machine does not get the correct answer to a specific
input. Since Linz actually did use the term "single Turing machine"
I now see that was an honest mistake.
The domain of this problem is to be taken as the set of all
Turing machines and all w; that is, we are looking for a
*single Turing machine* that, given the description of an arbitrary
M and w, will predict whether or not the computation of M applied
to w will halt
-- Copyright 2024 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer