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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: This function proves that only the outermost HHH examines the
 execution trace
Date: Fri, 26 Jul 2024 19:24:03 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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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.