Question

${\mathbf{BOTH}}_{\mathbf{NFA}}\mathbf{=}\mathbf{\left\{}\mathbf{(}\mathbf{M}\mathbf{1}\mathbf{,}\mathbf{M}\mathbf{2}\mathbf{)}\mathbf{|}{\mathbf{M}}_{\mathbf{1}}{\mathbf{andM}}_{\mathbf{2}}\mathbf{are}\mathbf{NFAs}\mathbf{where}\mathbf{L}\mathbf{\left(}{\mathbf{M}}_{\mathbf{1}}\mathbf{\right)}\mathbf{\cap }\mathbf{L}\mathbf{\left(}{\mathbf{M}}_{\mathbf{2}}\mathbf{\right)}\mathbf{\ne }\mathbf{\varphi }\mathbf{\right\}}$.

Show that ${\mathbf{BOTH}}_{\mathbf{NFA}}$ is NL-complete.

Short answer

Its reduction occurs in polynomial time as well. So it's NP Difficult. As a result, ${\mathrm{BOTH}}_{\mathrm{NFA}}$is NL-complete.

Step-by-step solution

Step-1: NP complete

There are two conditions that must be met in order to prove that the problem is NP complete.

1. The problem is in NP.
2. The problem is NP-hard.

Step-2: BOTHNFA is NL complete

M=" on input (M1, M2), so that they are both NFAs.

1. Mark the initials state of both NFAs using markers.
2. Repeat the procedure for $2^{\mathrm{q}1*\mathrm{q}2}$time.
3. Change the position of the marker in both stages in a nondeterministic manner to simulate symbol reading.
4. Accept if a string exists in stages 2 and 3, else reject."

This occurs in a logarithmic space. So ${\mathrm{BOTH}}_{\mathrm{NFA}}\in \mathrm{NL}$.