Re: This function proves that only the outermost HHH examines the execution trace

Op 26.jul.2024 om 22:14 schreef olcott:

On 7/26/2024 2:46 PM, Mike Terry wrote:

On 26/07/2024 16:56, olcott wrote:

This is meant for Mike, Joes and Fred don't have the technical competence to understand it.
>
Richard might be able to understand it yet the fact that he is
stuck in rebuttal mode makes any understanding that he may have
utterly useless.
>
Mike: It seems that HHH has been a pure function of its inputs
and never has provided data downward to its slaves that corrupts
their halt status decision. They don't even make a halt status
decision thus cannot make a corrupted one.

>
Well, the first two claims are literally untrue - outer HHH effectively uses the static mutable data to pass flags to the inner HHH that modify its behaviour.  The Root flag below is derived from the actual static data and causes inner HHH to totally skip its own abort logic!
>
You seem to acknowledge this, but claim it does not matter for various reasons, because whatever mistakes you are making, what finally gets printed out is saying the right thing!
>

 If HHH gets the correct answer in an impure way then it only
counts that it gets it in an impure way if it is impossible
to get in a pure way. This makes it possible for HHH to get
this answer in a pure way:
 Message-ID: <rLmcnQQ3-N_tvH_4nZ2dnZfqnPGdnZ2d@brightview.co.uk>
On 3/1/2024 12:41 PM, Mike Terry wrote:
 >
 > Obviously a simulator has access to the internal state
 > (tape contents etc.) of the simulated machine. No problem there.
 

In a similar way we can create an HHH that is able to reach the return of DDD.
if (Root) then
    simulate 10 levels of recursion before aborting;
else
   simulate two levels before aborting
In this way your claim that HHH cannot possibly reach the end of DDD is proven false.