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We know that the problem “Does this Turing machine take at least this finite number of steps on that input?” is decidable, because it will always answer yes or no, where it will say yes if the machine reaches the given number of steps and no if it halts before that.

Now here is my doubt: if it halts before reaching those many steps — i.e. the input either (1) got accepted or (2) got rejected or maybe (3)if it doesn’t halt but rather goes into an infinite loop — then, when we are in case (3), how can we be sure that it will always be in that loop? What I mean to say is that if it doesn't run forever but comes out of the loop at some point of time then it might cross the asked number of steps and the decision can be made now which was earlier not possible. If so, then how can we conclude that it's decidable when we know that being stuck in a loop we won’t be able to say anything about the outcome?

Ananya Nayak
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  • What does “the input got into a loop” mean, according to you? If the machine has halted, then it cannot be stuck in a loop. That’s almost by definition… – Maëlan Oct 19 '21 at 15:06
  • @Maëlan Yes, I think I wrote that by mistake. What I actually meant to say was that if on some input the machine gets into an infinite loop at some state (because it’s not a halting Turing machine )which is occurring before the given number of steps then how can we be sure that it will never come out of that loop ? Because if it comes out of the loop then it might cover the given steps and hence it becomes undecidable. So why aren’t we considering that as an option? – Ananya Nayak Oct 19 '21 at 19:09

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(I already more or less answered your question when I edited it.)

The thing is, the decision system (a Turing machine, an algorithm or any other equivalent formalism) that takes as inputs a Turing machine M, a number N and a value X, and returns yes or no, has total control over how it executes M on X. It simulates it step by step. So it can run one step of M(X), increment an instruction counter, compare it to N and, as soon as the given number of steps is reached, it stops and returns yes. At that point, there is no need that the simulated machine M be in a final state, and actually the full computation M(X) could very well diverge. We don’t care, because we only run the first N steps.

Maëlan
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  • Yes I got it .I was actually getting confused in some similar kind of problems and that’s why I couldn’t understand that how simple it actually is but now I am clear with the process. Thanks again for the explanation. – Ananya Nayak Oct 20 '21 at 09:09
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Most likely the "conditional structures where not being debuged/developed enough so that multiple conditions often conflicted each other..the error reporting where not as definitive, so it where used semi abstract notions as "decidable" and "undecidable"

as a semi example i writen years ago in vbs a "64 bit rom memory" simulator, as i tried to manage the memory cells, where i/o read/write locations where atributed , using manny formulas and conditions to set conversions from decimal to binary and all the operations, indexing, etc.

I had allso run into bugs becouse that the conditons where not perfect.Why? becouse the program had some unresolved somewhat arbitrary results that could had ended up in :

print.debug "decidable"

On Error Resume h

h:

print.debug "undecidable"

this was a example with a clear scope and with a debatable result.

to resume to your question : > "so how do we conclude that it's decidable??"

wikipedia : The Turing machine was invented in 1936 by Alan Turing, who called it an "a-machine" (automatic machine). With this model, Turing was able to answer two questions in the negative:

Does a machine exist that can determine whether any arbitrary machine on its tape is "circular" (e.g., freezes, or fails to continue its computational task)?

Does a machine exist that can determine whether any arbitrary machine on its tape ever prints a given symbol?

Thus by providing a mathematical description of a very simple device capable of arbitrary computations, he was able to prove properties of computation in general—and in particular, the uncomputability of the ('decision problem').

Cris
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