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.

Maximum characters in a deterministic Turing machine

Assume we have a deterministic Turing machine $ M = (q_s, q_a, q_r, \Sigma, \Gamma, \delta, Q, b)$ where $ q_s,q_a,q_r$ are the (unique) starting state, accept state and reject state respectively, $ Q$ the set of non-final states, $ \Sigma$ the input alphabet, $ \Gamma$ the tape alphabet, $ \delta$ the transition function and $ b \in \Gamma$ the blank symbol.

How many characters can fit in $ \Gamma$ , as a function of $ |\Sigma|, |Q|$ such that for each $ c \in \Gamma$ , $ \delta$ will be defined by it for some state and character?

How to proof that Turing machine that can move right only a limit number of steps is not equal to normal Turing machine

I need to prove that a Turing machine that can move only k steps on the tape after the last latter of the input word is not equal to a normal Turning machine.

My idea is that given a finite input with a finite alphabet the limited machine can write only a finite number of “outputs” on the tape while a normal Turing machine has infinite tape so it can write infinite “outputs” but I have no idea how to make it a formal proof.

any help will be appreciated.

Question on the decidable of Turing Machine

I am a bit confused about using the subset of the turning machine to prove the desirability of the turning machine.

If I have a Turing Machine M and we have already know M has a single halting state. If we already the machine takes a string was the input and reaches to the state q0, is it possible to prove that M is decidable by considering constructing a new Turing Machine that halts on q0?

Thanks a lot!

Is solving a quadratic equation using Turing machine impossible?

I’ve just started Algorithms at university. There’s a task to write an algorithm for a Turing machine to solve quadratic equations. The task doesn’t specify if it’s x^2+bx+c or ax^2+bx+c. I’ve searched whole bunch of information over Russian and English Internet.

I did find articles, which say it’s not possible because we’ve got real numbers A, B, C. Please confirm if that’s true. I may not get it correct.. But I think that’s impossible. I still don’t know how to prove my thoughts.

Thanks in advance!

All problems about Turing machines that involve only the language that the TM accepts are undecidable

I came across the below statement in the classic text “Introduction to Automata Theory, Languages, and Computation” by Hopcroft, Ullman, Motwani.

All problems about Turing machines that involve only the language that the TM accepts are undecidable

They say that the above theorem is per Rice theorem which states that:

“Every nontrivial property of the RE languages is undecidable.”

How are these two statements equivalent? The former deals only problems while the later deals with non trivial property.

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) ?