Re: 195 page execution trace of DDD correctly simulated by HH0

On 2024-06-25 17:29:12 +0000, olcott said:

On 6/25/2024 9:13 AM, Fred. Zwarts wrote:

Op 25.jun.2024 om 15:12 schreef olcott:

On 6/25/2024 7:08 AM, Fred. Zwarts wrote:

Op 24.jun.2024 om 23:04 schreef olcott:

On 6/24/2024 2:36 PM, joes wrote:

Am Mon, 24 Jun 2024 08:48:19 -0500 schrieb olcott:

On 6/24/2024 2:37 AM, Mikko wrote:

On 2024-06-23 13:17:27 +0000, olcott said:

On 6/23/2024 3:22 AM, Mikko wrote:

That code is not from the mentined trace file. In that file _DDD()
is at the addresses 2093..20a4. According to the trace no instruction
at the address is executed (because that address points to the last
byte of a three byte instruction.

 In order to make my examples I must edit the code and this changes the
addresses of some functions.

 Why do you need to make an example when you already have one in the
file mentioned in the subject line?
 

I had to make a few more examples such as HH1(DD,DD)

AFACT HH1 is the same as HH0, right? What happens when HH1 tries to
simulate a function DD1 that only calls HH1?
 

 typedef uint32_t u32;
u32 H(u32 P, u32 I);
 int P(u32 x)
{
   int Halt_Status = H(x, x);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
 int main()
{
   H(P,P);
}
 I am going to have to go through my code and standardize my names.
H(P,P) was the original name. Then I had to make a one parameter
version, a version that is identical to H, except P does not call
it and then versions using different algorithms. People have never
been able to understand the different algorithm.
 typedef void (*ptr)();
typedef int (*ptr2)();
int  HH(ptr2 P, ptr2 I); // used with int D(ptr2 P) that calls HH
int HH1(ptr2 P, ptr2 I); // used with int D(ptr2 P) that calls HH
int  HHH(ptr P);         // used with void DDD() that calls HHH
int HHH1(ptr P);         // used with void DDD() that calls HHH
 *The different algorithm version has been deprecated*
int  H(ptr2 , ptr2 I);  // used with int D(ptr2 P) that calls H
int H1(ptr2 P, ptr2 I); // used with int D(ptr2 P) that calls H
 *It is much easier for people to see the infinite recursion*
*behavior pattern when they see it actually cycle through the*
*same instructions twice*

 Twice is not equal to infinitely. When will you see that?
It is strange that you call that an infinite recursion, when H aborts after two cycles and the simulated H cannot reach its own abort operation, because it is aborted when it had only one more cycle to go.
None of the aborted simulations would cycle more than twice, so infinite recursion is not seen for an H that aborts the simulation of itself.

 typedef void (*ptr)();
int H0(ptr P);
 void DDD()
{
   H0(DDD);
}
 int main()
{
   H0(DDD);
}
 _DDD()
[00002172] 55               push ebp      ; housekeeping
[00002173] 8bec             mov ebp,esp   ; housekeeping
[00002175] 6872210000       push 00002172 ; push DDD
[0000217a] e853f4ffff       call 000015d2 ; call H0(DDD)
[0000217f] 83c404           add esp,+04
[00002182] 5d               pop ebp
[00002183] c3               ret
Size in bytes:(0018) [00002183]
 The call from DDD to H0(DDD) when DDD is correctly emulated
by H0 cannot possibly return.

 Contradictio in terminis. The fact that the simulated H0 does not return shows that the simulation is incorrect.

 void Infinite_Recursion()
{
   Infinite_Recursion();
}
 Ah so you simply *DON'T BELIEVE IN* infinite recursion where a
correct simulating termination analyzer would be required to
abort its simulation to correctly report non-terminating behavior.
That seems quite dumb of you.
 

The simulated H0 does not return, because it is aborted one cycle too soon. One cycle later it would return.

 Complete lack of sufficient software engineering skill.

The relevant area of software engineering is testing. The usual attitude of
software engineers is that a program is accpted when it has been sufficiently
tested and passed all tests. Consequently, an important part of sofware work
is the design of tests.
In the current context the program to be tested is a halting decider. The
essential features are: (1) there is no computation or input that can be
given to the halting decider, (2) for every input the decider halts and
answers either "yes" or "no", and (3) if the computation described by the
input halts then the halting decider answers "yes", otherwise "no".
Software engineers want to have an answet to the questions "What test inpus
shall be given to the halting decider?" and "What are the correct answers
to these test cases?"
--
Mikko