DFA with limited states
Constructing a DFA

There is an algorithm to construct a minimal DFA for a regular language. In this case, it isn't too hard to come up with a DFA with $z(z+1)/2$ states.

Observe that $L_z = \{\epsilon, abc, aabbcc, \ldots, a^{z-1}b^{z-1}c^{z-1}\}$. Once you know the number of $a$'s at the beginning of the word, you know how many $b$'s and $c$'s to expect. Here's a simple DFA built from this observation (with $z' = z-1$):


[source]

Following the usual convention, if there is no transition for a particular letter from a particular node, the automaton goes to a sink state.

This automaton has $n$ states reached after reading $0$ to $z-1$ letters a ($k,0,0$ for $0\le k\le z-1$), and branches of length $2k$ for $0 \le k \le z-1$ to read the right number of b's and c's ($k,i,0$ and $k,k,i$ for $1 \le i \le k \le z-1$). That's a total of $z + \sum_{k=0}^{z-1} (2k) = z^2$ states, plus one to account for the implicit sink state.

It's possible to compress this automaton a bit.


   Notice that every side branch finishes by counting the c's. Instead of tracking how many c's have already been seen, think in terms of how many c's are left to read. You can merge all the states of the form $k,k,k-n$, for all $k \ge 1$, into a state $-,n$ from which $n$ occurrences of c must be read. In particular, there will be only two final states: one for the empty word, and one after reading the right number of c's. In this automaton, the $k$th side branch has $k$ edges to read the b's (so $k-1$ states, as the start state belongs to the a trunk, and the end state belongs to the c collector), and there are $z-1$ transitions to count the c's (so $z$ states, corresponding to $0$ through $z-1$ c's left).


The total number of states in this automaton (including the sink state) is $z + \sum_{k=1}^{z-1} (k-1) + (z-1) + 1 = z(z+1)/2 + 1$.

Constructing an NFA

In other words, $n_z$ is the number of states in a minimal NFA that recognizes $L_z$. If you have a DFA with $N$ states, that's an NFA with $N$ states. So $n_z \le z(z+1)/2+1$. For most languages, a minimal NFA is smaller than a minimal DFA: nondeterminism adds expressive power in terms of how much you can accomplish with a number of states.

In this particular case, it happens that the minimal DFA above is also a minimal NFA (if I didn't make a mistake). Here's an informal proof.


   Consider the states in the middle of reading the b's. When reading $a^k b^i | b^{k-i} c^k$, where $|$ represents the current position, the state must account for both $k$ and $i$, to remember both the number of seen a's (which is necessary, because we'll need to count the c's later) and the number of seen b's (or, equivalently, the number of b's still required). This requires distinct states for every $(i,k)$ such that $1 \le i \lt k \le z-1$. In addition, there must be $z-1$ states to count the a's at the beginning, and $z-1$ states to count the c's at the beginning, plus a sink state. That's a total of $z(z+1)/2+1$ states.