We prove that two-way nondeterministic finite automata cannot be complemented with a polynomial number of states independent of the alphabet. For each n ≥ 4 we construct an n-state automaton over a finite alphabet whose complement requires at least $\tfrac12 2^{\lfloor(n-4)/127\rfloor}-1$ states.

One-way liveness on h points accepts a word of binary relations when their ordered product is nonempty. For every h ≥ 2, it has a nondeterministic automaton with $h+3$ states and no left moves, whereas every equivalent s-state two-way deterministic automaton satisfies $4(s+2)^2\ge2^{\lfloor(h-2)/31\rfloor}$. Partial transition rules, stay moves, and nonaccepting infinite computations are allowed. The alphabets are finite and grow with h, so the result rules out an alphabet-independent polynomial state bound for deterministic two-way simulation, already for one-way nondeterministic sources.