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On 6/6/2024 10:07 AM, Mikko wrote:This time I didn't fail to notice that you did not answer my question.On 2024-06-06 13:41:05 +0000, olcott said:
On 6/6/2024 4:02 AM, Mikko wrote:What makes you think that I failed to notice the correctnessOn 2024-06-05 13:59:53 +0000, olcott said:*Your failure to understand the correctness of these words is noted*
On 6/5/2024 3:11 AM, Fred. Zwarts wrote:Be specific: the test of HH workeed correcty when itOp 05.jun.2024 om 04:05 schreef olcott:Notice the subject line of this thread.On 6/4/2024 8:48 PM, Richard Damon wrote:But that has nothing to do with your simulation result.On 6/4/24 1:40 PM, olcott wrote:Turing Machine descriptions specify behavior to UTMs.On 6/4/2024 3:28 AM, Mikko wrote:But strings don't HAVE "Behavior", they only represent things that do.On 2024-06-03 18:14:39 +0000, olcott said:On 6/3/2024 9:53 PM, Richard Damon wrote:
On 6/3/2024 9:27 AM, Mikko wrote:That does not restrict what a problem statement can specify.On 2024-06-03 12:20:01 +0000, olcott said:int sum(int x, int y) { return x + y; }
On 6/3/2024 4:42 AM, Ben Bacarisse wrote:That does not restrict what a problem statement can specify.Mike Terry <news.dead.person.stones@darjeeling.plus.com> writes:Deciders only compute the mapping *from their inputs* to their own
PO's D(D) halts, as illustrated in various traces that have been posted here.He has quite explicitly stated that false (0) is the correct result for
PO's H(D,D) returns 0 : [NOT halting] also as illustrated in various traces.
i.e. exactly as the Linz proof claims. PO has acknowledged both these
results. Same for the HH/DD variants.
You might imagine that's the end of the matter - PO failed. :)
That's right, but PO just carries on anyway!
H(D,D) "even though D(D) halts". I am mystified why anyone continues to
discuss the matter until he equally explicitly repudiates that claim.
accept or reject state.
If the computed mapping differs from the specified one the
decider does not solve the problem.
sum(2,3) cannot return the sum of 5 + 6.
If the mapping computed by sum differs from the specified one
the program sum does not solve the problem.
> Because you keep on mentioning about DD Halting,
> which IS about the direct execution of DD
Only when one contradicts the definition of a decider that must
compute the mapping FROM ITS INPUTS BASED ON THE ACTUAL BEHAVIOR
OF THESE INPUTS (as measured by DD correctly simulated by HH).
And, for a Halt decider, that thing they represent is the program, whose direct execution specifies the proper behavior of the input.When H is asked about the behavior of a Machine that is programmed
The DEFINITON IS NOT "as measured by DD correctly simulated by HH", as deciders, by their definiton, are trying to compute the mapping of their input according to a defined function, which is a function of just that input. Since that function doesn't know which "H' is going to try to decide on it, it can't change its answer based on which H we ask.
Proper Deciders can not be asked "Subjective" questions, unless we SPECIFICALLY define the mapping to include the decider as one of the inputs, and at that point, the question actually ceases to be subjective, as it becomes, what should THAT H say about this input, which is back to an objective agian (since machines are deterministic, so the definition of H tells us what H will answer to that question).
When we go ahead and contradict this definition then theNope, YOU are wrong, because you
*HALTING PROBLEM IS STILL WRONG IN A DIFFERENT WAY*
When D is defined to do the opposite of whatever yes/noNope, Just shows how small your mind is.
an answer that H provides then the counter-example input
is precisely isomorphic to the question:
Is this sentence: "This sentence is not true." true or false?
Thus that question and the HP question are both incorrect
because both yes and no are the wrong answer.
Proven elsewhere.,
The theory of computation may be ignorant of the details ofBut the question it asks is an OBJECTIVE question that doesn't depend on who it is asked of.
how the context of who is asked a question changes the meaning
of this question, none-the-less this cannot be ignored.
It is and remains incorrect for the theory of computation
to ignore this.
to do the opposite of whatever it says then the context that it is H
that is being asked is an inherent aspect of the meaning of this
question and cannot be correctly ignored.
That HH is being asked an incorrect question is the second
way that the Halting Problem is wrong.
Your simulation does not even reach the part that contradict its result.*I was surprised that this worked correctly:
Your decider even diagnoses programs as non-halting when they do not contradict the result of the decider, as in:
typedef int (*ptr)(); // ptr is pointer to int function in C
int H(ptr p, ptr i);
int main()
{
H(main, 0);
}
It is clear that main does not programmed to do the opposite of what H says.
correctrly demonstrated that HH woked incorrectly.
HH did not work correctly when it, just before main
halted, said that main will not
here are the details*Here main reports that HH said that main will not halt.
int main()
{
Output("Input_Halts = ", HH(main,(ptr)0));
}
machine stack stack machine assembly
address address data code language
======== ======== ======== ========= =============
[00001e42][00103375][00000000] 55 push ebp ; begin main
[00001e43][00103375][00000000] 8bec mov ebp,esp
[00001e45][00103371][00000000] 6a00 push +00
[00001e47][0010336d][00001e42] 68421e0000 push 00001e42 ; push main
[00001e4c][00103369][00001e51] e831f5ffff call 00001382 ; call HH
New slave_stack at:103419
Begin Local Halt Decider Simulation Execution Trace Stored at:113421
[00001e42][0011340d][00113411] 55 push ebp ; begin main
[00001e43][0011340d][00113411] 8bec mov ebp,esp
[00001e45][00113409][00000000] 6a00 push +00
[00001e47][00113405][00001e42] 68421e0000 push 00001e42 ; push main
[00001e4c][00113401][00001e51] e831f5ffff call 00001382 ; call HH
New slave_stack at:14de41
[00001e42][0015de35][0015de39] 55 push ebp ; begin main
[00001e43][0015de35][0015de39] 8bec mov ebp,esp
[00001e45][0015de31][00000000] 6a00 push +00
[00001e47][0015de2d][00001e42] 68421e0000 push 00001e42 ; push main
[00001e4c][0015de29][00001e51] e831f5ffff call 00001382 ; call HH
Local Halt Decider: Infinite Recursion Detected Simulation Stopped
[00001e51][00103375][00000000] 83c408 add esp,+08
[00001e54][00103371][00000000] 50 push eax
[00001e55][0010336d][00000743] 6843070000 push 00000743
[00001e5a][0010336d][00000743] e843e9ffff call 000007a2
Input_Halts = 0
[00001e5f][00103375][00000000] 83c408 add esp,+08Here main halts.
[00001e62][00103375][00000000] eb79 jmp 00001edd
[00001edd][00103375][00000000] 33c0 xor eax,eax
[00001edf][00103379][00000018] 5d pop ebp
[00001ee0][0010337d][00000000] c3 ret ; end main
Number of Instructions Executed(12311) == 184 PagesThat main halts does not prove that HH returns 0 but that doesn't
So main() does halt at its final state at [00001ee0] which proves
that the directly executed HH(main,(ptr)0) called by main() halts
and returns 0;
mater as that was already proven right after HH returned 0.
*Your failure to understand the correctness of these words is noted*
*Your failure to understand the correctness of these words is noted*
of my words? Or is "those" intended to point to something else?
This time I did notice that you attempt a straw-man deception.In other words you can show exactly how DD correctly<Professor Sipser agreed>But did not agree with your claim that there is an error in his proof.
If simulating halt decider H correctly simulates its input D
until H correctly determines that its simulated D would never
stop running unless aborted then
H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
</Professor Sipser agreed>
simulated by HH can
*stop running without having its simulation aborted*I say that you are incorrect about that and about somemething else, too.
I say that you are least incorrect until you prove otherwise.
_DD()--
[00001e12] 55 push ebp
[00001e13] 8bec mov ebp,esp
[00001e15] 51 push ecx
[00001e16] 8b4508 mov eax,[ebp+08]
[00001e19] 50 push eax ; push DD
[00001e1a] 8b4d08 mov ecx,[ebp+08]
[00001e1d] 51 push ecx ; push DD
[00001e1e] e85ff5ffff call 00001382 ; call HH
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