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Am Sat, 07 Sep 2024 08:56:02 -0500 schrieb olcott:X86utm is based on a world class x86 emulator thatOn 9/7/2024 3:27 AM, Mikko wrote:Like Sipser said.On 2024-09-06 11:42:48 +0000, olcott said:On 9/6/2024 6:19 AM, Mikko wrote:On 2024-09-05 13:24:20 +0000, olcott said:On 9/5/2024 2:34 AM, Mikko wrote:On 2024-09-03 13:00:50 +0000, olcott said:On 9/3/2024 5:25 AM, Mikko wrote:On 2024-09-02 16:38:03 +0000, olcott said:
>A halt decider is a Turing machine that computes the mappingA halt decider needn't compute the full behaviour, only whether
from its finite string input to the behavior that this finite
string specifies.
that behaviour is finite or infinite.
Your implementation is buggy.I SHOW THE ACTUAL EXECUTION TRACE AND EVERYONE DISAGREES WITH IT.>The directly executed HHH is a decider.If that iis true it means that HHH called by DDD does not return and>New slave_stack at:1038c4 Begin Local Halt Decider Simulation>
Local Halt Decider: Infinite Recursion Detected Simulation Stopped
Hence HHH(DDD)==0 is correct
Nice to see that you don't disagree with what said.
Unvortunately I can't agree with what you say.
HHH terminates, so DDD obviously terminates, too.
DDD emulated by HHH never reaches it final halt state.
therefore is not a ceicder.
If the called HHH behaves differently from the direcly executed HHH
then the DDD is not relevant to classic proofs of the impossibility of
a halting decider.
If you can't show encoding rules that permit the encoidng of the
behaviour of the directly executed DDD to HHH then HHH is not a halting
decider.
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