The paper “Infinite Time Turing Machines” contains the following information:
At each step of computation, the head reads the cell values which it overlies, reflects on its state, consults the program about what should be done in such a situation and then carries out the instructions: it writes a $0$ or $1$ on (any of) the tapes, moves left or right and switches to a new state accordingly. This procedure determines the configuration of the machine at stage $\alpha + 1$, given the configuration at stage $\alpha$, for any $\alpha$. It remains to somehow take a limit of these computations in order to identify the configuration of the machine at stage $\omega$ and, more generally, at limit ordinal stages in the supertask computation.
The ordinal $\omega$ is also clockable, since a machine can be programmed, for example, to move the head always to the right, until a limit state is obtained, and then halt.
If a machine $M$ encodes some ordinal, then $M$ can operate as follows: given an input $N$, this machine converts $N$ to a pair of natural numbers, obtains the relation between these two numbers, writes the corresponding bit to the output tape, then generates the next number ($N+1$) and repeats the procedure. But what I don’t understand is how exactly can we determine whether $M$ reaches a limit stage? There is no upper bound for $N$. And if $\omega$ is an abstract ordinal not bounded from above by some finite number, then how can we technically know that at some particular moment, a particular machine has $\omega$ steps done and should be put into a limit state?
I have also read this article. It contains some descriptions, but for example, in the following description:
Here is brief look of how machine works:
1. We know $0=\langle 0, 0\rangle$ doesn't belong to $\omega$, so we "hard-code" it into machine.
8. After $\omega$ steps halt.
I don't understand why $0=\langle 0, 0\rangle$ doesn't belong to $\omega$ and how exactly can a machine halt after $\omega$ steps.
My mistake was that I was trying to imagine an ITTM computation as a physically possible process, instead of a purely mathematical construction. But all I needed to know is that at any stage, for any cell $C_i$, there exists only one, mathematically predetermined value for a symbol that may appear on $C_i$ before the machine reads it! There are no special moments of computation when the LIMIT transition is required to occur, because limit stages are mere abstractions implying that a non-halting computation has reached its limit and all cells are ready. Is my understanding correct?
For example, let $R$ denote a particular representation of a real number that encodes the following wellorder: $$0 \prec 2 \prec 4 \prec \ldots \prec 1 \prec 3 \prec 5 \prec \ldots$$
Let $M_1$ denote a standard (non-halting) Turing machine that generates $R$ on the output tape.
Consider an ITTM $M_2$ that is identical to $M_1$, but with the following addition: as soon as the machine enters the limit stage, it halts.
Question: can we assume that a real number $R$ encoding the ordinal $\omega + \omega$ is the final output of $M_2$ (so the ordinal $\omega + \omega$ is writable by $M_2$), but $M_2$ will halt at time $\omega$ (and clock the ordinal $\omega$)?