Question

Show that ${\mathbf{E}}_{\mathbf{DFA}}$is NL-complete.

Short answer

If there is a path to an accept state, it must include at least n stages. $\mathrm{M}$ does not accept any string if there is no path to an accept state within n steps, therefore M∈ $M\in E_{DFA}$.The string is not accepting path then $\mathrm{l}\left(\mathrm{m}\right)\ne \mathrm{\varphi }$And ${\mathrm{m\xi E}}_{\mathrm{DFA}}$.

Step-by-step solution

Step-1: NL complete

NL-complete is a complexity class in computational complexity theory that includes all languages that are complete for NL, a class of decision problems that can be handled by a nondeterministic Turing machine with a logarithmic amount of memory space.

Step-2: EDFA  is NL complete

The complexity of the task is referred to as a function of the input size. DFA is NL-complete because it recognises the language of finite states.

${\mathrm{E}}_{\mathrm{DFA}}$is context-free since it is regular.${\mathrm{E}}_{\mathrm{DFA}}$is in NL. Because a DFA with n states can start at the beginning state and nondeterministically go throughMstates for at most n steps, saving just the current state, it can start at the initial state and nondeterministically traverse through M states for at most n steps.