Where did this problem come up, and what have you tried?

You can parameterize the solutions to $\pi K = 0$ by Gaussian elimination, then eliminate some parameters using the zero row sum constraint on $K$. Generically this should give you $n(n-2)$ parameters. Are you looking for something better than that?

Welcome to MathOverflow! It might be helpful if you provided definitions of key terms for people who can't access the article or don't feel like looking it up. You can assume that people know what a Hamiltonian graph is, and probably a cycle decomposition, but e.g. I don't know what it means in this context for a cycle decomposition to be compatible.

Slightly subtle aspect of the simple connectedness issue (because this caused me to doubt myself after I posted that comment): $U$ isn't a Hermann ring because it's simply connected in the Riemann sphere $\mathbb{C} \cup \{ \infty \}$, and it's not a Siegel disk because it's not simply connected in the plane $\mathbb{C}$.

The complement of $T$ is simply connected in the Riemann sphere, so the Fatou set of any map $f$ with $J(f) = T$ has a single component $U$, and the classification of Fatou components tells us what $U$ could be like. $U$ is neither a Hermann ring (because it's simply connected) nor a Siegel disk (because it's clearly not the image of the unit disk under an analytic map). So either $U$ is parabolic or it contains a single attracting fixed point. Have you investigated what either of these possibilities would imply about $f$?

Sorry I never replied to this answer, but I cited it (just for context, not in the proof of any results) in a paper that is now in ETDS! "Encoding subshifts through sliding block codes", FirstView sometime last year, arXiv:2210.08150.

If you don't mind, I'm just going to email you -- I'm interested in understanding exactly what is going on with the various formulations of your question, and I think there are enough details to spell out that it's best not to try to get them right through successive modifications to this answer.

Correction to my response to the first question (sorry for the multiple comments): instead of "you're really computing the probability of accepting assuming it's all 0's to the left", I should have said that you're deciding whether $n \in S$, assuming it's all $0$'s to the left. So you accept or reject based on partial information ($T$ digits) about the infinite string, and the probability of accepting after $T$ digits happens to approximate the density of $S$ among $T$-digit integers.

Second question: yes, easy to add in. The limiting acceptance probability is the sum of the probabilities of the accepting sinks. If most strings have hit a sink already at $T$ digits, then the fraction that have reached a given sink is going to be close to the limiting probability of that sink.

First question: by reading from the big end, I guess you mean that you fix $T \approx \log_b N$ then compute $|S \cap [1,N]|$ by looking at the integers with $T$ digits? The difference, digging into the weeds a bit, is that you're then no longer looking at successive refinements of partitions/algebras on the space of left-infinite strings $\{0, 1, \dots, b-1 \}^{\mathbb{N}}$. It occurs to me that maybe what I need to say in my second para is that when you halt after $T$ digits from the right, you're really computing the probability of accepting assuming it's all $0$'s to the left.

I don't quite understand the model. Is the point that the DFA only accepts or rejects at a sink? If so, this seriously restricts the class of automatic sets/languages you can recognize --- it's equivalent to the language being a suffix code.

You know, I tried that in the singleton case, and I don't think it helps there, at least when $C$ is not much larger than $(n \log n)^n$, but when $C$ can be bigger and the $|E_j|$'s can be bigger, too, it looks like it may help. I'll either revise or answer the question if I get anywhere with this. Thanks!