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

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Discussion Overview

The discussion centers around the undecidability of the problem $\{x \mid T_x(x) = 1\}$, exploring the relationship between this problem and the Halting problem. Participants are seeking to construct a computable function that translates instances of the Halting problem to this new problem.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant proposes proving the undecidability of $\{x \mid T_x(x) = 1\}$ by constructing a computable function that translates instances of the Halting problem to this problem.
  • Another participant seeks clarification on the definition of $T_x(y)$, noting familiarity with Kleene's T predicate but mentioning it has three arguments.
  • A clarification is provided that $T_x(y)$ refers to the Turing machine with code $x$ and input $y$.
  • One participant suggests modifying a Turing machine so that if it halts, it outputs 1, but questions how to define this function effectively.
  • A more detailed proposal is made to modify the machine's states and add instructions to erase the tape content and write 1, while also addressing how to count the number of steps taken by the original program.
  • A recommendation is made to read a specific section in Sipser's book for further examples of reductions related to this topic.

Areas of Agreement / Disagreement

Participants are generally exploring the same problem but have not reached a consensus on the specific construction of the computable function or the details of the proposed modifications to the Turing machine.

Contextual Notes

There are unresolved questions regarding the implementation details of the proposed modifications to the Turing machine, particularly in how to effectively erase tape content and count steps.

mathmari
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Hey! :o

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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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.
 
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 $y$.
 
Given a machine, $f$ should change it so that if it halts, it outputs 1.
 
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)
 
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.
 

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