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On 4/24/25 7:52 PM, olcott wrote:I am not the one that stupid repeats that HHH is undefined.On 4/24/2025 6:26 PM, Richard Damon wrote:And you are stuck having to lie about what HHH does,On 4/24/25 5:19 PM, olcott wrote:>On 4/24/2025 5:53 AM, Richard Damon wrote:>On 4/23/25 11:40 PM, olcott wrote:>On 4/23/2025 10:34 PM, olcott wrote:>On 4/23/2025 7:31 PM, Mike Terry wrote:>On 23/04/2025 16:38, olcott wrote:>On 4/23/2025 10:28 AM, Mike Terry wrote:>On 23/04/2025 10:02, Fred. Zwarts wrote:>Op 22.apr.2025 om 21:50 schreef olcott:>On 4/22/2025 2:30 PM, Fred. Zwarts wrote:You never showed a proof. You only repeated a dream. You are dreaming many years without any logic. You failed to show the first state change where the direct execution is different from the simulation. You only showed an erroneous HHH that fails to reach the end of the simulation of a halting program.Op 22.apr.2025 om 21:14 schreef olcott:>On 4/22/2025 1:10 PM, Fred. Zwarts wrote:Therefore HHH should report on the actual input, the finite string that describes a halting program. Not on the hypothetical input that does not halt, because it is based on a hypothetical HHH that does not abort.Op 22.apr.2025 om 18:38 schreef olcott:>>And it has been proven that no finite string transformations are possible that report the halting behaviour for all inputs that specify a correct program.
a function is computable if there exists an algorithm
that can do the job of the function, i.e. given an input
of the function domain it can return the corresponding output.
https://en.wikipedia.org/wiki/Computable_function
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On Turing Machines inputs <are> finite strings, and
finite string transformation rules <are> applied to
these finite strings to derive corresponding outputs.
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int sum(int x, int y) { return x + y; }
Only when people stupid assume the same thing as
sum(3,2) should return the sum of 5 + 3.
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Why do you maintain that HHH should process the hypothetical input instead of the actual input.
Do you really believe that 3+2 equals 5+3?
I have proven that the directly executed DD and DD
emulated by HHH according to the semantics of the
x86 language have a different set of state changes
many hundreds of times for several years.
Worse than this, on more than one occasion I've actually posted traces of computation DDD(DDD) executed directly and simulated by HHH side by side. Both traces were of course / identical/, up to the point where HHH stops simulating.
*Factually incorrect* (You are usually very careful about these things)
The call to HHH(DD) from the directly executed DD returns.
The call to HHH(DD) from DD emulated by HHH cannot possibly return.
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...because HHH stops simulating before reaching that step in the computation. Note that I said
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MT: Both traces were of course /identical/,
*up to the point where HHH stops simulating*
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So I was factually correct.
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Mike.
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It *is not* up to the point where HHH stops simulating.
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It is up to the point where the simulated versus directly
executed calls HHH(DD).
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This call immediately from the directly executed DD and
cannot possibility return from DD emulated by HHH according
to the finite string transformation rules of the x86 language.
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According to the finite string transformation rules of the x86 language.
The call from the directly executed DD to HHH(DD) immediately returns.
The call from DD emulated by HHH to HHH(DD) cannot possibility return.
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According to the rules of the x86 language, your provided input is invalid as it references code outside the input. PERIOD.
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*Repetition seems to help you overcome your ADD*
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I have told you that the whole Halt.obj is in scope
for every function in Halt.c several times now.
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And thus there is only every that ONE HHH, so HHH *NEVER* correctly emulates it input,
*At least you will quit STUPIDLY saying that HHH is undefined*
*At least you will quit STUPIDLY saying that HHH is undefined*
*At least you will quit STUPIDLY saying that HHH is undefined*
*At least you will quit STUPIDLY saying that HHH is undefined*
*At least you will quit STUPIDLY saying that HHH is undefined*
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