The problem {x ∣ Tx(x)=1} is undecidable

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In summary, the problem $B$ of proving that the problem $\{x\mid T_x(x)=1\}$ is undecidable can be translated to the problem $A$ of constructing a computable function $f$ such that $x\in A\Leftrightarrow f(x)\in B$. However, to find the function, the original program $M$ must be modified so that it counts the number of steps it takes.
  • #1
mathmari
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Hey! :eek:

I want to prove that the problem $\{x\mid T_x(x)=1\}$ is undecidable.

Let $A=\text{HALT}=\{x\in \mathbb{N}_0\mid T_x(x)\downarrow\}$ and $B$ the above problem.

We want to construct a "translation" of $A$ to $B$, i.e., we want to construct a computable function $f$ such that $x\in A\Leftrightarrow f(x)\in B$.

Could you give me some hints how we can find that function? (Wondering)
 
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  • #2
mathmari said:
I want to prove that the problem $\{x\mid T_x(x)=1\}$ is undecidable.
Could you remind what $T_x(y)$ is? I know about Kleene's T predicate, but it has three arguments.
 
  • #3
Evgeny.Makarov said:
Could you remind what $T_x(y)$ is? I know about Kleene's T predicate, but it has three arguments.

$T_x(y)$ is the Turing machine with code $x$ and input $y$.
 
  • #4
Given a machine, $f$ should change it so that if it halts, it outputs 1.
 
  • #5
Evgeny.Makarov said:
Given a machine, $f$ should change it so that if it halts, it outputs 1.

But how could we define the function so that it does that? (Wondering)
 
  • #6
A machine halts when it enters the accepting or rejecting state. Replace those states with ordinary state $q_0$ throughout the program (set of instructions) and add a new block of instructions so that when the machine is in $q_0$, it erases the content of the tape and writes 1 on it.

Now there is a question how to erase the content of the tape. How does the machine know that beyond a certain position all cells contain the blank symbol? The original program $M$ can be modified so that it counts the number of steps it takes. That is, the original program is not simply run as a subroutine, but rather interpreted, and the number of steps is counted. Then the target program (i.e., the output of $f(\lceil M\rceil)$ erases as many cells as the number of steps taken by $M$ and then writes 1. If $M$ does not stop, $f(\lceil M\rceil)$ does not stop either. This all assumes I understood correctly that $T_x(x)=1$ means that the machine with code $x$ and input $x$ outputs 1.

I recommend reading section 5.3 "Mapping Reducibility" in the Sipser's book to see more examples of reductions.
 

FAQ: The problem {x ∣ Tx(x)=1} is undecidable

What does it mean for a problem to be undecidable?

When a problem is undecidable, it means that there is no algorithm or method that can determine a definitive answer for every instance of the problem. This means that there may be some instances of the problem that can be solved, but there will always be some instances that cannot be solved.

What is the problem {x ∣ Tx(x)=1}?

The problem {x ∣ Tx(x)=1} is also known as the halting problem. It asks whether a given algorithm will halt or loop indefinitely when given a specific input. In other words, it is trying to determine if there is a way to predict the behavior of every possible algorithm.

Why is the problem {x ∣ Tx(x)=1} important?

The problem {x ∣ Tx(x)=1} is important because it proves the existence of undecidable problems in computer science. It also has significant implications for artificial intelligence and the limits of computation.

Can any undecidable problems be solved?

Some undecidable problems can be solved in specific cases, but there is no general algorithm that can solve all instances of an undecidable problem. This means that even when a solution is found, it may not apply to all instances of the problem.

How does the undecidability of this problem impact computer science?

The undecidability of this problem has had a profound impact on computer science, specifically in the study of algorithms and the limits of computation. It has also led to the development of alternative methods for tackling complex problems that cannot be solved using traditional algorithms.

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