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I was wondering how you would go about making a two-tape Turing machine in linear time that starts with input 𝑥#𝑦 on Tape 1, where 𝑥 and 𝑦 are binary integers that might have leading 0's. It would accept the input if 𝑥 and 𝑦 have the same value such as 001101 has the same value as 1101. Would it be a good idea to replace the 1s and 0s with something like n and m to read the inputs and then if the turing machine gets to the # symbol it goes to the next binary string. Then if the n/m are the same it would accept the input and if they are different it would reject it? How would this be implemented?

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Here is a sketch: Put the machine in a state where the first tape contains $x$ with no leading zeros, and the second tape contains $y$ with no leading zeros, then check that the contents of two tapes are equal.

To reach such state, assume that both $x$ an $y$ contain at least one $1$ (this can be easily checked). Then, scan the first (input) tape deleting all leading zeros. Continue scanning the tape without changing it until you find $\#$. Delete $\#$, then "move" $y$ to the second tape ignoring the leading zeros (write blanks on the first tape and the significant digits of $y$ to the second tape). Reset both heads to the first non-zero character of the respective tapes.

Interesting there exists no subquadratic Turing machine that solves your problem using only one tape.

Suppose such a Turing machine existed $T$, then you could use it to build the following Turing machine that decides the language $L$ containing all strings of the form $x' 2^n y'$ where $x',y' \in \{0,1\}^\ell$, $x'=y'$, and $\ell \ge 2$.

To do so you can:

  • Check that the string is of the form $\{0,1\}^*2^* \{0,1\}^*$ (if not, reject). This can be done in linear time.
  • Check that the string is of the form $\{0,1\}^\ell 2^\ell \{0,1\}^\ell$ with $\ell \ge 2$. This can be done in time $O(n \log n)$, where $n$ is the length of the input.
  • Prepend a $1$ to $x$
  • Replace all $2$s except the first and the last one with $0$.
  • Replace the first $2$ with $\#$ and the last $2$ with $1$.
  • Run $T$ on the current tape contents (which are now of the form $1x'\#00\dots 01y')$.

The overall time spent is $O(n \log n)$, where $n$ is the length of the input. However $L$ cannot be decided by any Turing machine with running time $o(n^2)$. See pages 566-569 of One-Tape, Off-Line Turing Machine Computations.

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  • $\begingroup$ Good analysis. However: the question mentions two tapes, for which the problem becomes much simpler. $\endgroup$ Nov 13, 2023 at 10:47
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    $\begingroup$ @HendrikJan, agreed. The first part of my answer addresses two-tape Turing machines. I went on a tangent researching the problem for single-tape Turing machines (because I had originally misread the question) and I thought I'd also discuss that case. $\endgroup$
    – Steven
    Nov 13, 2023 at 10:49
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    $\begingroup$ My bad! I should actually read the full answer before commenting. $\endgroup$ Nov 13, 2023 at 10:51

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