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On 5/23/2025 11:16 AM, Richard Damon wrote:Because, by the definition, it doesn't matter, as long as it is "A computation", which means a fully specified sequence of finite deterministic operations.On 5/23/25 12:10 PM, olcott wrote:*They never bothered to say HOW until NOW*On 5/23/2025 2:14 AM, Mikko wrote:>On 2025-05-23 03:31:15 +0000, olcott said:>
>On 5/22/2025 10:23 PM, wij wrote:>On Thu, 2025-05-22 at 21:47 -0500, olcott wrote:>[cut]>
Q: How do computations actually work?
A: Computation is merely step-by-step algorithm.
Nothing says it has to be TM.
>
Do the exercises in textbooks first before any claim of it.
>
int main()
{
DD(); // by what steps can the HHH that DD calls
} // report on the behavior its caller?
If we don't insist that the report be correct:
1. guess
2. tell what was guessed
>
This does not work because all computable functions
that implement termination analyzers must compute
the mapping from their input finite string according
to the behavior that it specifies. In my concrete
examples DDD must be simulated by HHH according to the
rules of the x86 language.
>
_DDD()
[00002172] 55 push ebp ; housekeeping
[00002173] 8bec mov ebp,esp ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404 add esp,+04
[00002182] 5d pop ebp
[00002183] c3 ret
Size in bytes:(0018) [00002183]
>
DDD specifies that it will continue to call HHH
until HHH sees the repeating pattern and aborts
its simulation.
>
Yes, they must compute the mapping, but the rules don't say HOW.
The ultimate definition of a correct simulationRight, and for all the languages we are talking about, it means of exactly the instructions presented, and not stopping until we reach the end.
is a simulation according to the definition of
the simulation language.
One is not free to interpret "push ebp" as "jmp 00002183".Right, nor call HHH as "show the results that HHH will compute". Show me in the intel reference manual for the x86 where it says what you do is correct.
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