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

Am Fri, 26 Jul 2024 10:56:55 -0500 schrieb olcott:

This is meant for Mike, Joes and Fred don't have the technical
competence to understand it.

Fuck you too.

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.

The indirectly called HHH's most certainly make a decision.

I could change the code so that the slave instances can and do try to
form a halt status decision on the basis of the execution trace that
they do have.
There is no sense in doing that because they can't possibly have enough
data, they will always be one whole execution trace behind the next
outer instance of HHH.

That's kinda the point.

u32 Decide_Halting_HH(u32**                   Aborted,
                       u32**                   execution_trace,
                       Decoded_Line_Of_Code**  decoded,
                       u32                     code_end,
                       u32                     End_Of_Code, Registers**
                                  master_state, Registers**           
                       slave_state, u32**                   slave_stack,
                       u32                     Root)
{
   u32 aborted_temp = 0;  // 2024-06-05 u32
   Current_Length_Of_Execution_Trace = 0;
   while (aborted_temp == 0) // 2024-06-05 {

     if (Root)  // Master UTM halt decider {

There it is.
Where is this passed from?

   if (aborted_temp == 1) // 2021-01-26 Must be aborted
     return 0;
   return 1;           // 2021-01-26 Need not be aborted
}

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.