= Solution
For $0\leq j\leq9$, consider the ten prefixes
$$
u_j=b^j.
$$
If $0\leq i<j\leq9$, append
$$
z_j=b^{11-j}a.
$$
Then
$$
u_i z_j=b^{i+11-j}a\in L,
$$
because $i+11-j\leq10$, whereas
$$
u_jz_j=b^{11}a\notin L.
$$
Indeed, in a word ending in one $a$, the prefix before the final positive run of $a$'s must have length at most ten. Hence every pair $u_i,u_j$ is distinguished by some suffix. They occupy ten distinct <Myhill-Nerode equivalence> classes, so the <minimal deterministic finite automaton> for $L$ has at least ten states.
Back to article page
