Sujet : Re: Analysis of Richard Damon’s Responses to Flibble
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theoryDate : 18. May 2025, 21:18:04
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <f7ac56997ec57e98f9002a12042e482f367e6092@i2pn2.org>
References : 1 2 3 4 5 6 7
User-Agent : Mozilla Thunderbird
On 5/18/25 4:09 PM, Mr Flibble wrote:
On Sun, 18 May 2025 16:03:13 -0400, Richard Damon wrote:
On 5/18/25 3:58 PM, Mr Flibble wrote:
On Sun, 18 May 2025 15:49:33 -0400, Richard Damon wrote:
>
On 5/18/25 3:45 PM, Mr Flibble wrote:
On Sun, 18 May 2025 15:19:38 -0400, Richard Damon wrote:
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On 5/18/25 1:07 PM, Mr Flibble wrote:
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4. Stack Overflow as a Semantic Signal
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Damon argues that stack overflow represents a failed computation:
"...it just got the wrong answer."
>
Flibble’s view is different:
- A stack overflow (or crash) isn’t failure.
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Sure it is. A program that fails to complete and give the correct
answer has just failed to give an answer.
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If you want to define "stack overflow" as an "I don't know" result,
fine, but first you have to define that this is a "valid" result.
>
No it isn't. Why? Because the stack overflow a property of the
simulation environment (the fact that the SHD has finite resources)
and NOT a property of the program, P, being analysed per se. P is
NOT halting, it is the SHD that is halting due to the detection of
infinite recursion on the part of P. It is perfectly valid for the
SHD to treat this as NON- HALTING as far as P is concerned.
>
/Flibble
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No, it is a property of the decider. If your "environment" is
inadiquite, it just shows you aren't using a proper environment.
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The SHD and the simulation environment are on in the same.
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And thus a failure of the environment is a failure of the SHD.
Not at all, the SHD can abort the simulation if it would result in stack
overflow due to infinite recursion and then return a result of NON-HALTING.
/Flibble
Yes, *IF* it can show that the simulation WOULD be infinite for that exact input.
Remember, to simulate it, it must be a complete program, that includes all its defined code.
Thus the "pathological" program, since it is built on a specific decider, has fixed behavior, as does that specific decider.
SO, if the SHD aborts and returns an answer, the the correct simulation of the "pathological" program will have its decider do exactly the same thing.