## How should we define the behavior of a Turing machine where the head tries to move left from the leftmost tape position?

If we have a Turing machine in a model with a tape that is infinite only to the right and assume at some point the head tries to move left from the leftmost position.

How should we define the behavior in such a case? Is a machine doing so for some input not a valid Turing machine? And if so, how can we make sure when we define a Turing machine that this situation can’t occur for any input?

I’ve read some sources about Turing machines though couldn’t find the answer to this specific case, and I see no reason why this case won’t happen for an arbitrary Turing machine and some input.

## Turing machine with a finite tape after the input word ends

How can I prove that for a natural number K, a language that accepted by a Turing machine with K cells after the input word ends, belongs to R (which R is the set of languages that there is Turing machine that accept them while for each input the run is finite)?

## Probabilistic Turing machine – Probability that the head has moved k steps to the right on the work tape

I have a PTM with following transition:

$$\delta(Z_0, \square , 0) = \delta(Z_0, \square , L, R)$$,

$$\delta(Z_0, \square , 1) = \delta(Z_0, \square , R, R)$$

Suppose that this PTM executes n steps. What is the probability that the head has moved k steps to the right on the work tape (in total, i.e., k is the difference between moves to the right and moves to the left) ?

## Turing machine on input w tries to move its head past the left end of the tape

Consider the language

$$L = \{ \langle M,w \rangle \mid \text{ M on input w tries to move its head past the left end of the tape}\}.$$

Prove whether L is decidable or not.

I tried to prove it as undecidable through reduction method but could’nt reduce the language halt to L inorder to prove that it is undecidable.

## Is it decidable whether Turing Machine never scans any tape cell more than once when started with given string

The problem:

Is it decidable that the set of pairs $$(M,w)$$ such that TM $$M$$, started with input $$w$$, never scans any tape cell more than once.

How can I easily prove above to be decidable. I found following proof confusing:

How is $$l+m$$ is upper bound on number of steps? I feel we should be doing at least $$l\times 𝑄\times \Gamma\times\{𝐿,𝑅\}+1$$ steps ($$Q$$ being number of states,$$\Gamma$$ being set of tape alphabet, $$l$$ is string length, $$L$$ and $$R$$ are head movement directions).

## Is it decidable “Given a TM M, whether M ever writes a non blank symbol when started on the empty tape.”

I came across below problem in this pdf:

Given a TM M, whether M ever writes a non blank symbol when started on the empty tape.

Solution given is as follows:

Let the machine only writes blank symbol. Then its number of configurations in the com computation on w is q × 2, where q is the number of states of M; the factor 2 is for the choices re. the direction of heads movement; there is no factor for the written symbol because that is always blank. So the problem is decidable, decided by the following machine: input (M,w), run M on w for q × 2 steps; if it M ever writes a non blank symbol, stop with yes answer; if M never writes a non blank symbol, stop with no answer

Doubts:

Q1. How be sure all q x 2 configurations will happen while running q x 2 steps on w? Some configuration may get repeated in q x 2 steps.

Q2. Question says “when started on the empty state”, but the answer tried to simulate TM on non empty string w. How does it makes sense?

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## Turing machine with k-tape, tape of output

Consider the Turing Machine with input alphabet {a,b} that computes the function:

$$f(w, v) = \left\{\begin{matrix} w & if length(w) > length(v)\ v & otherwise \end{matrix}\right.$$

I wish to use a TM of 2 tapes where in the tape-1 there is the input string, i.e. $$*w*v*$$. I have not clear, reading several books, about the output tape when I use a k-tape TM. For example, the output tape for a TM of 2 tapes is the input tape (w or v) or the 2-tape? If I use 3-tapes: 1-input, 2-working, 3-output is correct?

## How to compute within a Multi tape turing machine [a binary substraction between 2 numbers and result on 3rd tape]?

I’m trying to compute a Multi-tape turing machine (of 3 tapes) that follows this statement.

“A binary subtraction operation between the first and second tape printing the result on the third tape”

I’ve been trying to do it in JFlap but I reached a halt point being pretty confused in this topic overall.

Any help regarding this MTTM would be really appreciated.

## Difference between multi-tape Turing machine and single tape machine

A beginner’s question about “fine-grained” computational power.
Let $$M_k$$ be a $$k$$-tapes turing machine, and let $$M$$ be a single tape turing machine. We know that $$M_k$$ and $$M$$ both have the same “computable power”. In addition, one can simulate $$M_k$$ on $$M$$ in a way that every computation which takes $$O(t(n))$$ on $$M_k$$ will take $$O(t(n) \log(t(n))$$ on $$M$$.
Here is my question:
Is there a language $$L$$ such that $$L$$ can be decided in $$O(n)$$ time in $$k$$-tape Turing machine (for fixed $$k$$, say 2), but can’t be decided in $$O(n)$$ time in a single tape machine? (every single tape machine which decides $$L$$ needs $$\Omega(n \log n)$$ time).
In addition, are there any examples of two computational models (classical, not the quantum model) with the same computable power, but with fine-grained differences in their running time? (I guess that major changes in running time would contradict the extended Church-Turing thesis, which is less likely).