Problem 3

For $n\equiv N'\pmod2$, use the heights after $B_n$ and define $T(n)=(h_1-h_2,h_2-h_3,\ldots,h_{L-1}-h_L,a_n)$. We spell out the two-step transition. Put $q=h_{a_n}$; by Step 4, $q=a_{n+1}>M$. The next block is placed in tower $q$. Since every tower with label greater than $M$ has height at most $M$, after this placement the next small term is $a_{n+2}=\#\{i:h_i\ge q\}$, where only $i=1,\ldots,L$ can contribute. The value of this count is determined by the relative heights, because $q$ is one of the heights $h_1,\ldots,h_L$. Then $B_{n+2}$ increments tower $a_{n+2}$ by one, so all successive differences and the last coordinate are determined from $T(n)$. To see that the state set is finite, Step 2 gives $h_{k+1}\le h_k+C$; we also claim $h_k\le h_{k+1}+C(L-1)$ for $k<L$. If this failed just after tower $k$ was updated, then $h_k>h_{k+1}+C(L-1)$. Repeatedly using $h_j\ge h_{j+1}-C$ shows that every one of $h_1,\ldots,h_k$ is larger than every one of $h_{k+1},\ldots,h_L$. Let $q=h_k$; then exactly the first $k$ towers have height at least $q$, so the next small term is $k$, and the same two-step transition updates tower $k$ forever. This leaves towers $k+1,\ldots,L$ bounded, contradicting their definition. Hence all adjacent differences lie in a fixed finite interval, and with $a_n\in\{1,\ldots,L\}$, only finitely many states $T(n)$ occur.