On 6/18/22 8:32 AM, olcott wrote:
On 6/18/2022 7:24 AM, wij wrote:
On Saturday, 18 June 2022 at 20:07:56 UTC+8, olcott wrote:
On 6/18/2022 4:09 AM, Malcolm McLean wrote:
On Friday, 17 June 2022 at 12:29:37 UTC+1, olcott wrote:When H(P,P) correctly determines that P would never reach its "ret"
When a simulating halt decider rejects all inputs as non-haltingYes.
whenever it correctly detects that its correct and complete
simulation
of its input would never reach the final state of this input then all >>>>>> [these] inputs (including pathological inputs) are decided correctly. >>>>>>
*computation that halts* … the Turing machine will halt whenever it >>>>>> enters a final state. (Linz:1990:234)
Linz, Peter 1990. An Introduction to Formal Languages and Automata. >>>>>> Lexington/Toronto: D. C. Heath and Company. (317-320)
#include <stdint.h>
typedef void (*ptr)();
void P(ptr x)
{
if (H(x, x))
HERE: goto HERE;
return;
}
int main()
{
Output("Input_Halts = ", H(P, P));
}
_P()
[00001352](01) 55 push ebp
[00001353](02) 8bec mov ebp,esp
[00001355](03) 8b4508 mov eax,[ebp+08]
[00001358](01) 50 push eax // push P
[00001359](03) 8b4d08 mov ecx,[ebp+08]
[0000135c](01) 51 push ecx // push P
[0000135d](05) e840feffff call 000011a2 // call H
[00001362](03) 83c408 add esp,+08
[00001365](02) 85c0 test eax,eax
[00001367](02) 7402 jz 0000136b
[00001369](02) ebfe jmp 00001369
[0000136b](01) 5d pop ebp
[0000136c](01) c3 ret
Size in bytes:(0027) [0000136c]
(1) It is an easily verified fact that when we assume that H is
only an
x86 emulator that the correctly emulated P never reaches its "ret" >>>>>> instruction it remains stuck in repeated cycles of emulation.
Yes.
(2) It is an easily verified fact that if H has been adapted to
correctly detect (in a finite number of steps) that the correct and >>>>>> complete x86 emulation of its input would never each its "ret"
instruction that H could abort its emulation and return 0 to
report this.
No. Because by changing H from emulator to halt decider, you have
(3) When the halt status criteria is defined as correctly determining >>>>>> whether or not an x86 emulated input would ever reach its "ret"
instruction then it becomes an easily verified fact H(P,P) could
correctly reject its input as non-halting.
changed
P.
instruction (AKA final state) this makes H a halt decider for this
input
by definition:
computation that halts … the Turing machine will halt whenever it
enters
a final state. (Linz:1990:234)
Linz, Peter 1990. An Introduction to Formal Languages and Automata.
Lexington/Toronto: D. C. Heath and Company. (317-320)
--
Copyright 2022 Pete Olcott
"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer
GUR is repeatedly proved:
olcott can never provide a real halting decider except a verbal one.
Thanks olcott.
This one proves that H(P,P) does correctly determine that H(P,P)==0 is
correct:
H(P,P)==0 as a pure function of its inputs is fully operational V2
On 6/17/2022 9:31 PM, olcott wrote:
Except it isn't since H(P,P) returns 0, saying that P(P) is non-halting,
when P(P) will halt when H(P,P) returns 0.
If H(P,P) isn't asking about P(P), then you don't have the right
function P, as the Linz definition of P is that it will ask about that.
On 6/18/2022 7:52 AM, Richard Damon wrote:
On 6/18/22 8:32 AM, olcott wrote:
On 6/18/2022 7:24 AM, wij wrote:
On Saturday, 18 June 2022 at 20:07:56 UTC+8, olcott wrote:
On 6/18/2022 4:09 AM, Malcolm McLean wrote:
On Friday, 17 June 2022 at 12:29:37 UTC+1, olcott wrote:When H(P,P) correctly determines that P would never reach its "ret"
When a simulating halt decider rejects all inputs as non-halting >>>>>>> whenever it correctly detects that its correct and completeYes.
simulation
of its input would never reach the final state of this input then >>>>>>> all
[these] inputs (including pathological inputs) are decided
correctly.
*computation that halts* … the Turing machine will halt whenever it >>>>>>> enters a final state. (Linz:1990:234)
Linz, Peter 1990. An Introduction to Formal Languages and Automata. >>>>>>> Lexington/Toronto: D. C. Heath and Company. (317-320)
#include <stdint.h>
typedef void (*ptr)();
void P(ptr x)
{
if (H(x, x))
HERE: goto HERE;
return;
}
int main()
{
Output("Input_Halts = ", H(P, P));
}
_P()
[00001352](01) 55 push ebp
[00001353](02) 8bec mov ebp,esp
[00001355](03) 8b4508 mov eax,[ebp+08]
[00001358](01) 50 push eax // push P
[00001359](03) 8b4d08 mov ecx,[ebp+08]
[0000135c](01) 51 push ecx // push P
[0000135d](05) e840feffff call 000011a2 // call H
[00001362](03) 83c408 add esp,+08
[00001365](02) 85c0 test eax,eax
[00001367](02) 7402 jz 0000136b
[00001369](02) ebfe jmp 00001369
[0000136b](01) 5d pop ebp
[0000136c](01) c3 ret
Size in bytes:(0027) [0000136c]
(1) It is an easily verified fact that when we assume that H is
only an
x86 emulator that the correctly emulated P never reaches its "ret" >>>>>>> instruction it remains stuck in repeated cycles of emulation.
Yes.
(2) It is an easily verified fact that if H has been adapted to
correctly detect (in a finite number of steps) that the correct and >>>>>>> complete x86 emulation of its input would never each its "ret"
instruction that H could abort its emulation and return 0 to
report this.
No. Because by changing H from emulator to halt decider, you have
(3) When the halt status criteria is defined as correctly
determining
whether or not an x86 emulated input would ever reach its "ret"
instruction then it becomes an easily verified fact H(P,P) could >>>>>>> correctly reject its input as non-halting.
changed
P.
instruction (AKA final state) this makes H a halt decider for this
input
by definition:
computation that halts … the Turing machine will halt whenever it
enters
a final state. (Linz:1990:234)
Linz, Peter 1990. An Introduction to Formal Languages and Automata.
Lexington/Toronto: D. C. Heath and Company. (317-320)
--
Copyright 2022 Pete Olcott
"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer
GUR is repeatedly proved:
olcott can never provide a real halting decider except a verbal one. >>>> Thanks olcott.
This one proves that H(P,P) does correctly determine that H(P,P)==0
is correct:
H(P,P)==0 as a pure function of its inputs is fully operational V2
On 6/17/2022 9:31 PM, olcott wrote:
Except it isn't since H(P,P) returns 0, saying that P(P) is
non-halting, when P(P) will halt when H(P,P) returns 0.
If H(P,P) isn't asking about P(P), then you don't have the right
function P, as the Linz definition of P is that it will ask about that.
(X) H(P,P) correctly determines (in a finite number of steps) that the correct and complete x86 emulation of its input would never reach the
"ret" instruction (AKA final state) of this input.
(Y) computation that halts … the Turing machine will halt whenever it enters a final state. (Linz:1990:234)
(Z) H(P,P) has correctly determined that its input is non-halting.
When we know that X & Y proves Z and we know X & Y then Z is proved.
Failing to agree with this last line is sufficient evidence of dishonesty.
Linz, Peter 1990. An Introduction to Formal Languages and Automata. Lexington/Toronto: D. C. Heath and Company. (317-320)
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