Liste des Groupes | Revenir à theory |
On 5/27/2024 11:37 AM, Richard Damon wrote:Nope, because Linz doesn't try to pass a Template to H, but the machine built from the template H^.On 5/27/24 12:06 PM, olcott wrote:DISHONEST HEAD GAMES. WHEN WE APPLY THIS SAME REASONING TO THEOn 5/27/2024 10:56 AM, Richard Damon wrote:>On 5/27/24 11:43 AM, olcott wrote:>On 5/27/2024 9:58 AM, Richard Damon wrote:>On 5/27/24 10:39 AM, olcott wrote:>
typedef int (*ptr)(); // ptr is pointer to int function in C
00 int H(ptr p, ptr i);
01 int D(ptr p)
02 {
03 int Halt_Status = H(p, p);
04 if (Halt_Status)
05 HERE: goto HERE;
06 return Halt_Status;
07 }
08
09 int main()
10 {
11 H(D,D);
12 return 0;
13 }
>
The above template refers to an infinite set of H/D pairs where D is
correctly simulated by either pure simulator H or pure function H. This
was done because many reviewers used the shell game ploy to endlessly
switch which H/D pair was being referred to.
>
Correct Simulation Defined
This is provided because many reviewers had a different notion of
correct simulation that diverges from this notion.
>
A simulator is an x86 emulator that correctly emulates 1 to N of the
x86 instructions of D in the order specified by the x86 instructions
of D. This may include M recursive emulations of H emulating itself
emulating D.
And how do you apply that to a TEMPLATE that doesn't define what a call H means
*It is completely defined and you are just ignoring this definition*
So, what instruction does the call H in D go to to be simulated?
>
LINZ TEMPLATE YOUR REASONING CALLS THE LINZ TEMPLATE NONSENSE.
Nope, Linz choose A SPECIFIC H out of the set, and gives it a SPECIFIC H^ built from that SPECIFIC H, and then works with that set. There is ZERO logic about infinite sets in the part that shows that THIS H can't get the right answer to THIS input.As a template, there is no fixed H, so no instruction to look at.I AM REFERRING TO THE EXACT SAME SORT OF INFINITE SET
>H correctly simulates 1 to ∞ steps of D with either pure function H>
or pure simulator H. In none of these cases does the correctly simulated
D ever reach its own simulated final state and halt.
>
Do some of these instances of H play a game of poker with themselves
before or after they simulate D? Yes they do because the H/D pairs
are an infinite set.
>
But, how do they correctly simulate something that isn't there?
>
Either they are simulating an INSTANCE of the template, in which case each H is looking at a DIFFERENT instance, and you can't relate one result to the other, or they are trying to simulate the Template, at which point you have the problem that the code to be simulated hasn't been defined, and thus you can't do what you define to do.
THAT THE LINZ TEMPLATE IS REFERRING TO AND YOU KNOW IT.
WHEN EVERY ELEMENT OF AN INFINITE SET HAS THE SAME NON-HALTINGBut they don't.
PROPERTY THEN YOUR SHELL GAME SWITCHEROO IS OBVIOUSLY A HEAD GAME.
IF YOU WOULDN'T HAVE TRIED TO GET AWAY WITH THIS FOR TWO SOLIDBut it is YOU who has been ignoring the definitions of the system, apparently because you never learned them.
YEARS I WOULD NOT BE YELLING.
Les messages affichés proviennent d'usenet.