Re: Simulation vs. Execution in the Halting Problem

On 2025-06-01 15:42:52 +0000, Mike Terry said:

On 01/06/2025 05:01, olcott wrote:

On 5/31/2025 10:32 PM, Mike Terry wrote:

On 01/06/2025 02:31, olcott wrote:

On 5/31/2025 7:44 PM, Mike Terry wrote:

On 01/06/2025 01:18, olcott wrote:

[..snip..]

 We cannot do a separate side-by-side execution trace of
HHH(DDD) and HHH1(DDD) because the DDD simulated by HHH1
calls HHH(DDD) as a part of this same simulation.

 Duh!  The DDD simulated by HHH ALSO calls HHH(DDD) as a part of the same simulation.
 They BOTH call HHH(DDD) as part of the simulation.  Duuuuuh....
 I've presented the two traces to you side by side on more than one occasion.  Do you really have no recollection of that?  Your explanation of why we supposedly can't put them side by side is literally gibberish!
 

  From the trace shown below we can see that HHH simulates
DDD one whole execution trace more than HHH1 does.

 Really?  That's not at all what I see - but perhaps you can explain what you're saying.
 Mark on the trace below where you think HHH1's simulation [i.e. the simulation /performed/ by HHH1] starts and ends.  Also mark where you think HHH's simulation starts and ends.
 Then to save me the trouble, try to put them side by side to see if they match up...
  Mike.
 

 I really appreciate your sincere honesty and the great
diligence that you have shown evaluating my work. No
one else on the planet has put nearly the same effort
as you in carefully evaluating the key details of my work.
 

 There is a terminology issue here to resolve.
 If A simulates B and B simulates C, what should a "trace" of A's simulation look like?
a)   it includes both B's and C's instructions, interlaced.
b)   it is just B's instructions.

The main purpose of the emulation is to determine whether the computation
specified by the input halts. For that purpose no traces are needed. They
are there only to enable a check that the algrithm for the determination
works as expected. For that purpose the output should trace both B's
instructions and B's output (which inludes C's instructions) but every
B's output line should have a mark that shows that it is not A's output.
--
Mikko