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On 10/5/2024 3:20 AM, Mikko wrote:The former is the behaviour of DDD. The latter is someting else thoughOn 2024-10-04 14:54:06 +0000, olcott said:When DDD is executed before HHH its behavior its different
On 10/4/2024 4:05 AM, Mikko wrote:DDD does not "rely" on anything. It simply is a specification ofOn 2024-10-03 21:01:51 +0000, olcott said:Directly executed DDD can rely on having its second invocation
On 10/3/2024 7:15 AM, Mikko wrote:So the behaviour of DDD "correctly" emulated is not the behaviourOn 2024-10-03 02:09:39 +0000, olcott said:The behavior of the directly executed DDD is essentially the
On 10/2/2024 5:48 PM, Richard Damon wrote:The behaviour of DDD is what DDD does if executed. As DDD takes no inputOn 10/2/24 10:39 AM, olcott wrote:You guys try to pretend that the pathological relationship betweenOn 10/2/2024 6:08 AM, Richard Damon wrote:Try asking itOn 10/1/24 7:26 PM, olcott wrote:In the case of their evaluation of my work they are correct.On 10/1/2024 12:58 PM, joes wrote:No, it will tell you something that matches the words you told it.Am Tue, 01 Oct 2024 12:31:41 -0500 schrieb olcott:You feed it your objections.On 10/1/2024 8:09 AM, joes wrote:q=Termination+Analyzer+H+is+Not+Fooled+by+Pathological+Input+D&sca_esv=889093c5cb21af9e&sca_upv=1&source=hp&ei=Muf7ZpOyMZHfwN4PwYL2gAc&iflsig=AL9hbdgAAAAAZvv1Qg04jNg2ze170z3a8BSGu8pA29Fj&ved=0ahUKEwiTk7zkk-2IAxWRL9AFHUGBHXAQ4dUDCBg&uact=5&oq=Termination+Analyzer+H+is+Not+Fooled+by+Pathological+Input+D&gs_lp=Egdnd3Mtd2l6IjxUZXJtaW5hdGlvbiBBbmFseXplciBIIGlzIE5vdCBGb29sZWQgYnkgUGF0aG9sb2dpY2FsIElucHV0IERIAFAAWABwAHgAkAEAmAEAoAEAqgEAuAEDyAEA-Am Tue, 01 Oct 2024 07:39:18 -0500 schrieb olcott:On 10/1/2024 7:19 AM, olcott wrote:https://www.google.com/search?
AEC-AEBmAIAoAIAmAMAkgcAoAcA&sclient=gws-wizhttps://chatgpt.com/ share/66fbec5c-7b10-8011-9ce6-3c26424cb21cIt sounds like it’s trained on your spam. LLMs don’t know anything
anyway.I asked: „What happens when HHH tries to simulate itself?”
ChatGPT: [my comments in brackets]### 1. **Self-Invocation in HHH**
If `HHH` calls itself (i.e., `HHH(HHH)`), it creates an infinite loop
of calls unless there is some form of termination condition built into
`HHH`.
Without such a condition, `HHH` would never return, resulting in non-
termination.
[So not a decider.]
### 2. **Halting Criteria**
In a well-designed termination analyzer, `HHH` would need to have
specific logic to prevent endless recursion:I’ll leave that to you.[Not present.]*You can continue this conversation with ChatGPT*
### 3. **Practical Consequences**
If `HHH` lacks the ability to handle self-referential calls correctly:
- **Stack Overflow**: Continuous self-invocation will eventually
exhaust the stack memory, leading to a stack overflow error.
[Not considered in theory.]
- **Incorrect Results**: If not designed carefully, it may lead to an
incorrect conclusion about halting, misrepresenting itself as either
halting or non-halting.
### 4. **Best Practices for Termination Analyzers**
To avoid these pitfalls, a termination analyzer should implement:
- **Guard Clauses**: Checks at the beginning of the function to prevent
unnecessary self-simulation.
- **State Management**: A way to track the functions being analyzed,
allowing `HHH` to recognize if it's being invoked with itself and act
accordingly.
### Conclusion In summary, if `HHH` tries to simulate itself without
proper safeguards,
it risks falling into an infinite recursion or causing a stack
overflow. A robust design would include checks to prevent
self-invocation, ensuring the termination analyzer can handle all
cases, including its own function,
gracefully.
Click on the link and see how it answers this question:You should feed it our objections.
Is H a Halt Decider for D?
If you believe in it only when you prompt it, it is not suited as an
authority (fallacious anyway).
It will tell you how and why you are wrong.
You don't seem to understand what Large Language Models are.
You seem to forget that LLM know nothing of the "truth", only what matches their training data.
They are know to be liars, just like you.
*Try this yourself*
https://chatgpt.com/share/66fbec5c-7b10-8011-9ce6-3c26424cb21c
Does HHH have to abort its emulation of DDD to prevent the infinite execution of DDD?
Why is it ok for HHH to say that DDD doesn't halt when it will when run?
HHH and DDD does not make any difference to the behavior of DDD
knowing full well that it does make a difference.
its behavour is always the same. What does "make a difference" mean
in this context?
behavior of what would otherwise be infinite recursion except
that the second recursive call has already been aborted.
This is not the same as the behavior of DDD correctly emulated
by the same emulator that it calls where HHH cannot rely on DDD
being aborted by any other process than itself.
of DDD.
aborted by HHH is different behavior than DDD emulated by HHH
according to the behavior specified by x86 the machine code of
DDD and HHH where HHH cannot rely on DDD having already been
aborted by anything else.
a behaviour.
than when DDD is emulated by HHH. This is easy to see by
anyone that is not totally clueless about the x86 language.
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