Re: The truth about the halting problem counter-example input

On 7/3/2026 11:11 PM, olcott wrote:
> On 7/3/2026 9:43 PM, dbush wrote:
>> On 7/3/2026 10:37 PM, olcott wrote:
>>> On 7/3/2026 9:19 PM, dbush wrote:
>>>> On 7/3/2026 10:05 PM, olcott wrote:
>>>>> On 7/3/2026 8:58 PM, dbush wrote:
>>>>>> On 7/3/2026 9:52 PM, olcott wrote:
>>>>>>> On 7/3/2026 5:51 PM, André G. Isaak wrote:
>>>>>>>> On 2026-07-03 16:37, olcott wrote:
>>>>>>>>> On 7/3/2026 1:47 PM, André G. Isaak wrote:
>>>>>>>>>> On 2026-07-03 12:36, olcott wrote:
>>>>>>>>>>> On 7/3/2026 1:18 PM, dbush wrote:
>>>>>>>>>>
>>>>>>>>>>>> If an algorithm takes an input and produces an output, that 
>>>>>>>>>>>> is by definition a mapping. 
>>>>>>>>>>> That only proves that the definition is incoherent.
>>>>>>>>>>> The coherent way that it actually works is that
>>>>>>>>>>> inputs are transformed into outputs by applying
>>>>>>>>>>> finite string transformation rules to inputs to
>>>>>>>>>>> derive outputs.
>>>>>>>>>>
>>>>>>>>>> Apparently you don't understand the difference between a 
>>>>>>>>>> mapping and an algorithm. They are two different things.
>>>>>>>>>>
>>>>>>>>>> André
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> A function that ignores its input and only returns 0
>>>>>>>>> is not any sort of halt function.
>>>>>>>>
>>>>>>>> He was defining 'mapping', not 'halt function'.
>>>>>>>>
>>>>>>>> André
>>>>>>>>
>>>>>>>
>>>>>>> A actual halt function must compute 
>>>>>> The mathematical halting function:
>>>>>>
>>>>>
>>>>> When you actually implement this concretely 
>>>>
>>>> We find that it is not possible, as Linz and others have proved.
>>>>
>>>
>>> Impossible requirements are incorrect requirements.
>>>
>>
>> Nope.  Requirements are requirements for a reason.  If they can't be 
>> satisfied, then that's just the way it is.
>>
> 
> The halting problem requires a decider that correctly reports the halt 
> status of an input that does the opposite of whatever it reports. 

One such example is shown below where algorithm H1 does not correctly 
report the halt status of algorithm D1 which is built using the 
counter-example template.


void D(ptr *I)
{
      // algorithm D1; input: I
      ptr *X = D;
      ptr *Y = I;
      int result;
      {
          // algorithm H1; inputs: X,Y
          result = 0 + (X-X) + (Y-Y);
      }
      if (result == 1) {
          while (1);
      }
}

int H(ptr *X, ptr *Y)
{
      int result;
      {
          // algorithm H1; inputs: X,Y
          result = 0 + (X-X) + (Y-Y);
      }
      return result;
}