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A polynomially bounded Turing machine is the one which, on input $w$, uses no more than $f(|w|)$ cells on its tape, where $f$ is a polynomial. For this problem halting is decidable.

I do not understand why is halting decidable for this problem because $f$ is unknown to us.

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You don't need to know $f$. It is sufficient to use the fact that there are only finite possible configurations.

Given an input $\langle \langle M\rangle,w\rangle$ where $M$ is a polynomially bounded TM, simulate $M$ on $w$ and record its configuration at every step. Whenever we find a repeated configuration, we can assert that $M$ does not halt on $w$.

Note if $M$ does not halt on $w$, it must experience a repeated configuration, so the process described above will always halt. Therefore this process is the decider for the halting problem.

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