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On 6/26/2024 9:39 PM, Richard Damon wrote:But, if it can measure the fact that this is the top level decider, that means that it sees something that it can't know.On 6/26/24 10:06 PM, Mike Terry wrote:Is disabled. It is commented out.On 27/06/2024 02:52, Richard Damon wrote:He published the source code of at least his earlier code, and I suspect he hasn't made major changes to it. I forget it it was a zip file on his server or a Github repository.On 6/26/24 9:30 PM, Mike Terry wrote:>On 27/06/2024 02:15, Mike Terry wrote:>On 27/06/2024 01:42, Richard Damon wrote:>On 6/26/24 8:20 PM, olcott wrote:>On 6/26/2024 6:55 PM, Richard Damon wrote:>On 6/26/24 7:46 PM, olcott wrote:>On 6/26/2024 6:41 PM, Richard Damon wrote:>On 6/26/24 9:42 AM, olcott wrote:>On 6/26/2024 6:02 AM, Richard Damon wrote:>On 6/25/24 11:42 PM, olcott wrote:>>>
That is not the way that it actually works.
That the the way that lies are defined.
Source for you claim?
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Where is you finite set of steps from the truthmakers of the system to that claim?
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_DDD()
[00002172] 55 push ebp ; housekeeping
[00002173] 8bec mov ebp,esp ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call H0(DDD)
[0000217f] 83c404 add esp,+04
[00002182] 5d pop ebp
[00002183] c3 ret
Size in bytes:(0018) [00002183]
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The call from DDD to H0(DDD) when DDD is correctly emulated
by H0 cannot possibly return.
Sure it can. I have shown an H0 that does so.
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I already told you that example does not count.
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I can't keep repeating those details or others
that so far have no idea what an x86 emulator is
will be baffled beyond all hope of comprehension.
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WHy not?
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We have already been over that you know that you cheated.
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Nope, since you didn't put in the rule, and if you had it would have shown that you lied, as if H0 is a pure function then the call to H0 emulated by H0 needs to have the same behaivor as the direct call to H0 by main.
Incidentally, the nonconformance you're referring to is shown explicitly in the "195 page trace" that PO linked to. [I.e. the simulated H does not correctly track the code path of the outer H.]
I suppose I should have made clear, that's not simply due to the simulated H being aborted. There is an instruction in H: [actually, in Init_Halts_HH()]
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[000012e4] 753b jnz 00001321
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and in outer H control proceeds to 000012e6 [i.e. branch not taken],
whilein simulated H control proceeds to 00001321 [i.e. branch taken]
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Mike.
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Would need to look closer at the code, but I bet that the simulated machine is looking into the trace buffer to see if it is simulated or not.
Has PO published the C code for the trace? Anyhow, given that its in Init_Halts_HH(), I expect its a global area being initialised - probably the global trace table.
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In effect, it is misusing static memory just like he says isn't allowed.
Right.
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Mike.
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THe code for Init_Halts_HH() is:
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u32 Init_Halts_HH(u32** Aborted,
u32** execution_trace,
Decoded_Line_Of_Code** decoded,
u32* code_end,
u32 P,
Registers** master_state,
Registers** slave_state,
u32** slave_stack)
{
*decoded = (Decoded_Line_Of_Code*) Allocate(sizeof(Decoded_Line_Of_Code));
*code_end = get_code_end(P);
*master_state = (Registers*) Allocate(sizeof(Registers));
*slave_state = (Registers*) Allocate(sizeof(Registers));
*slave_stack = Allocate(0x10000); // 64k
Output((char*)"New slave_stack at:", (u32)*slave_stack);
if (**execution_trace == 0x90909090)
{
// Global_Recursion_Depth = 0;
**Aborted = 0;
**execution_trace = (u32)Allocate(sizeof(Decoded_Line_Of_Code) * 10000);
Output((char*)"\nBegin Local Halt Decider Simulation "
"Execution Trace Stored at:", **execution_trace);
return 1;
}
return 0;
}
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Note the mention of "Global_Recursion_Depth",
It was only ever used so that humans could see the depth.
It does if it knows that it isn't being simulated, which is knowledge that no simulated machine is allowed to have, as that means the simulation isn't correct. BY DEFINITION.This doesn't have any effect on its computation thus irrelevant.a decider shouldn't beable to know that it isn't the top level decider.
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