Sujet : Re: DDD correctly emulated by HHH cannot possibly halt, thinks olcott, but it does.
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theoryDate : 11. Jul 2024, 02:01:53
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v6nau1$24jgn$2@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
User-Agent : Mozilla Thunderbird
On 7/10/2024 7:37 PM, Richard Damon wrote:
On 7/10/24 8:24 PM, olcott wrote:
On 7/10/2024 7:01 PM, Richard Damon wrote:
On 7/10/24 9:41 AM, olcott wrote:
On 7/10/2024 8:27 AM, joes wrote:
Am Tue, 09 Jul 2024 23:19:25 -0500 schrieb olcott:
On 7/9/2024 11:01 PM, joes wrote:
> That means that HHH doesn't return, in particular that it doesn't
> abort.
DDD correctly emulated by any pure function HHH that correctly emulates
1 to ∞ steps of DDD can't make it past the above line of code no matter
what.
That line being the call to itself -> it can't simulate itself.
>
*DDD NEVER HALTS*
DDD ONLY calls HHH...
>
>
void DDD()
{
HHH(DDD);
return;
}
>
DDD correctly emulated by any pure function HHH that
correctly emulates 1 to ∞ lines of DDD can't make it
to the second line of DDD no matter what.
>
>
>
Nope, DDD does if HHH(DDD) returns.
>
>
You have a dead cat in your driveway does not mean that
you have a peanut butter sandwich on your front porch.
It has taken you at least 1000 messages to see that.
>
DDD correctly emulated by any pure function HHH that
correctly emulates 1 to ∞ lines of DDD can't make it
to the second line of DDD no matter what.
WRONG, you don't seem to understand the difference between DDD and HHH's emualtion of it.
Would you bet your immortal soul that DDD simulated
by HHH (as provided above) would terminate normally?
-- Copyright 2024 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer
…