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I have a question described as follows from the Sipser's TOC textbook:

Let $\Gamma=\{0,1, \sqcup\}$ be the tape for all TMs in this problem. Define the busy beaver function $BB: N \rightarrow N$ as follows. For each value of $k$, consider all $k$-state TMs that halt when started with a blank tape. Let $BB(k)$ be the maximum number of $1$s that remain on the tape among all of these machines. Show that $BB$ is not a computable function.

I want to prove $BB$ is not computable by reducing $A_{TM}$ to $BB$. I found a similar question but proven using a different approach. My thought is given $M$, I know the number of states $|Q|$, let's denote it as $k$. Hence, $BB(k)$ tells me the maximum number of $1$s that remain on the tape among all $k$-state machines that halt when started with a blank tape.

If I construct a decider for $A_{TM}$, is $BB(k)$ an upper bound of steps that $M(w)$ could possibly run if it were to halt?

jamm
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As noted in the comments, no $BB(k)$ is not an upper bound on the number of steps for any halting computation on a machine with $k$ states. You can do the trick given in this answer to the same question you linked though, which I'll paraphrase:

Given a TM $M$ and input $w$, you can compute a TM $M'$ that, on a blank input, simulates $M$ on $w$, keeping track of how many steps, and if and when that computation finishes after some number of steps $t$, it cleans up and writes $t$ 1's to its tape and halts. Then, you can compute $k$, the number of states in $M'$, and then $BB(k).$ Then you can actually simulate $M$ on $w$ accept if that computation accepts within $BB(k)$ steps, otherwise reject. The machine $M$ running on $w$ cannot halt after more than $BB(k)$ steps, since if it did, $M'$ would be a $k$-state machine that on a blank input, halts with more than $BB(k)$ ones on the tape, contradicting the definition of $BB.$

spaceisdarkgreen
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