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computers / comp.ai.philosophy / Re: Does the call from P() to H() specify infinite recursion?

SubjectAuthor
* Re: Does the call from P() to H() specify infinite recursion?olcott
`- Re: Does the call from P() to H() specify infinite recursion?olcott

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Subject: Re: Does the call from P() to H() specify infinite recursion?
From: olcott
Newsgroups: comp.theory, sci.logic, sci.math, comp.ai.philosophy
Followup: comp.theory
Date: Thu, 11 Nov 2021 16:31 UTC
References: 1 2
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Subject: Re: Does the call from P() to H() specify infinite recursion?
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From: NoO...@NoWhere.com (olcott)
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On 11/11/2021 10:14 AM, Richard Damon wrote:
On 11/11/21 11:00 AM, olcott wrote:
#define ptr uintptr_t

void P(ptr x)
{
   H(x, x);
}

int H(ptr x, ptr y)
{
   ((void(*)(ptr))x)(y);
   return 1;
}

int main()
{
   H((ptr)P, (ptr)P);
   return 0;
}


Yes.

So all you have proven is that if H unconditionally executes its input you get infinite recursion.

And, H will never return.

If H isn't that function, then the computation of P changes so you have no proof for what the behavior of that machine is.


_P()
[00001a5e](01)  55              push ebp
[00001a5f](02)  8bec            mov ebp,esp
[00001a61](03)  8b4508          mov eax,[ebp+08]
[00001a64](01)  50              push eax        // push P
[00001a65](03)  8b4d08          mov ecx,[ebp+08]
[00001a68](01)  51              push ecx        // push P
[00001a69](05)  e810000000      call 00001a7e   // call H
[00001a6e](03)  83c408          add esp,+08
[00001a71](01)  5d              pop ebp
[00001a72](01)  c3              ret
Size in bytes:(0021) [00001a72]


If H simulates the x86 machine language of its input and sees that its simulated P is calling H with the same parameters that H was called with H can abort its simulation of P and correctly report that P would never reach its final state at 1a72.


--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre minds." Einstein


Subject: Re: Does the call from P() to H() specify infinite recursion?
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Thu, 11 Nov 2021 17:14 UTC
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Subject: Re: Does the call from P() to H() specify infinite recursion?
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On 11/11/2021 10:46 AM, Richard Damon wrote:
On 11/11/21 11:31 AM, olcott wrote:
On 11/11/2021 10:14 AM, Richard Damon wrote:
On 11/11/21 11:00 AM, olcott wrote:
#define ptr uintptr_t

void P(ptr x)
{
   H(x, x);
}

int H(ptr x, ptr y)
{
   ((void(*)(ptr))x)(y);
   return 1;
}

int main()
{
   H((ptr)P, (ptr)P);
   return 0;
}


Yes.

So all you have proven is that if H unconditionally executes its input you get infinite recursion.

And, H will never return.

If H isn't that function, then the computation of P changes so you have no proof for what the behavior of that machine is.


_P()
[00001a5e](01)  55              push ebp
[00001a5f](02)  8bec            mov ebp,esp
[00001a61](03)  8b4508          mov eax,[ebp+08]
[00001a64](01)  50              push eax        // push P
[00001a65](03)  8b4d08          mov ecx,[ebp+08]
[00001a68](01)  51              push ecx        // push P
[00001a69](05)  e810000000      call 00001a7e   // call H
[00001a6e](03)  83c408          add esp,+08
[00001a71](01)  5d              pop ebp
[00001a72](01)  c3              ret
Size in bytes:(0021) [00001a72]


If H simulates the x86 machine language of its input and sees that its simulated P is calling H with the same parameters that H was called with H can abort its simulation of P and correctly report that P would never reach its final state at 1a72.



No it can't.

If H can abort its simulation, then it needs to take into account that H can abort its simulation. PERIOD.

H, in making that conclusion, is PRESUMING that the called H will not abort its simulation, which it is wrong about.

Because I keep repeating these same words dozens of times and you never acknowledge that you have ever seen them I really believe that you actually have attention deficit disorder (ADD):

(a) P only halts if it reaches its final state at 1a72.

(b) If H does not abort its simulation of P then P never reaches its final state at 1a72.

(c) If H aborts its simulation of P then P never reaches its final state as 1a72.

Because P never halts in all possible cases H(P,P)==0 is always correct.




--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre minds." Einstein


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