On 4/18/2021 9:13 PM, Richard Damon wrote:
> On 4/18/21 9:27 PM, olcott wrote:
>> On 4/18/2021 6:35 PM, Richard Damon wrote:
>>> On 4/18/21 7:16 PM, olcott wrote:
>>>> On 4/18/2021 4:58 PM, Richard Damon wrote:
>>>>> On 4/18/21 5:43 PM, olcott wrote:
>>>>>> On 4/18/2021 4:37 PM, Richard Damon wrote:
>>>>>>> On 4/18/21 4:17 PM, olcott wrote:
>>>>>>>> On 4/18/2021 2:59 PM, Richard Damon wrote:
>>>>>>>>> On 4/18/21 3:43 PM, olcott wrote:
>>>>>>>>>> On 4/18/2021 2:36 PM, Richard Damon wrote:
>>>>>>>>>>> The problem here is you are ignoring the definition of
>>>>>>>>>>> Computations.
>>>>>>>>>>
>>>>>>>>>> This is not a problem at all because I am only referring to the
>>>>>>>>>> simple
>>>>>>>>>> fact that my C code does do what I claim that it does do.
>>>>>>>>>
>>>>>>>>> No, you CLAIM that you are proving an error in Theories about
>>>>>>>>> Computation Theory. If you don't follow the rules of Computation
>>>>>>>>> Theory
>>>>>>>>> that is not true.
>>>>>>>>>>
>>>>>>>>>> The commercial purpose of a halt decider is to detect programs
>>>>>>>>>> that
>>>>>>>>>> would not otherwise halt and abort these programs to prevent
>>>>>>>>>> things
>>>>>>>>>> such
>>>>>>>>>> as a denial-of-service attack.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> But that is not the purpose of the Halting Theory of Conventional
>>>>>>>>> Computaiton Theory.
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>> My C code proves that the conventional halting problem
>>>>>>>>>> undecidability
>>>>>>>>>> proof counter-example does not prevent such a denial-of-service
>>>>>>>>>> attack
>>>>>>>>>> program from being created.
>>>>>>>>>
>>>>>>>>> The Conventional Halting Problem never claimed to prevent DOS
>>>>>>>>> attacks.
>>>>>>>>>
>>>>>>>>> It claimed that we can not always predict if a given potentially
>>>>>>>>> long
>>>>>>>>> running computation will eventually end.
>>>>>>>>>
>>>>>>>>
>>>>>>>> That a program X can correctly decide whether or not another
>>>>>>>> program Y
>>>>>>>> is calling X in what would otherwise be infinite recursion <is>
>>>>>>>> computationally equivalent to a halt decider whether you understand
>>>>>>>> this
>>>>>>>> or not.
>>>>>>>>
>>>>>>>>
>>>>>>>
>>>>>>> Wrong. The fact that when Y() calls X() that X() will decide to limit
>>>>>>> the recursion makes Y() not have infinite-recursion.
>>>>>>>
>>>>>>> The fact that the outer X() can't deduce this properly is a
>>>>>>> failure of
>>>>>>> X().
>>>>>>>
>>>>>>
>>>>>> Yes of source you say very stupid things like this when you merely
>>>>>> glance at a few of my words before artificicially contriving a what
>>>>>> might appear to be a rebuttal to someone that is hardly paying any
>>>>>> attention at all.
>>>>>>
>>>>>> what would otherwise be infinite recursion
>>>>>> what would otherwise be infinite recursion
>>>>>> what would otherwise be infinite recursion
>>>>>> what would otherwise be infinite recursion
>>>>>
>>>>>
>>>>> We don't care about 'otherwise...'
>>>>
>>>> "Otherwise" is an integral part of the criteria, if you don't care about
>>>> otherwise then you are simply ignoring my criteria rather than
>>>> critiquing it.
>>>>
>>>> I will try to formalize it a little better:
>>>> Does X() stop simulating Y() on the basis of:
>>>>
>>>> (1) Function X() is called twice in sequence from the same machine
>>>> address of Y().
>>>> (2) With the same parameters to X().
>>>> (3) With no conditional branch or indexed jump instructions in Y().
>>>> (4) With no function call returns from X().
>>>>
>>>
>>> It becomes a matter of definition, and because of the context of your
>>> comments that context is fairly obvious, especially since it is present
>>> upthread.
>> void X(u32 P, i32 I)
>> {
>> Simulate(P, I);
>> }
>>
>> void Y(u32 P)
>> {
>> X(P, P);
>> }
>>
>> int main()
>> {
>> X((u32)Y, (u32)Y);
>> }
>>
>> When X() simulates the machine language of Y() and examines resulting
>> execution trace after simulating each machine instruction of Y() then
>> X() can apply the criteria provided in the above to stop simulating Y()
>> on the basis that Y() would otherwise cause X() have infinite execution.
>>
>> Do you agree with this?
>>
>>
>
> With the code provided here, X has no conditions at all,
No try again you are not paying close enough attention, it never says
that X() has no conditional code.