Deciding emptiness of intersection of regular languages in subquadratic time
Simple answer: If there does exist a more efficient algorithm that runs in $O(n^{\delta})$ time for some $\delta &lt; 2$, then the strong exponential time hypothesis would be refuted.



We will prove a stronger theorem and then the simple answer will follow.

Theorem: If we can solve the intersection non-emptiness problem for two DFA's in $O(n^{\delta})$ time, then any problem that's non-deterministically solvable using only n bits of memory is deterministically solvable in $poly(n)\cdot2^{(\delta n/2)}$ time.

Justification: Suppose that we can solve intersection non-emptiness for two DFA's in $O(n^{\delta})$ time.  Let a non-deterministic Turing machine M with a read only input tape and a read/write binary work tape be given.  Let an input string x of length n be given.  Suppose that M doesn't access more than n bits of memory on the binary work tape.

A computation of M on input x can be represented by a finite list of configurations.  Each configuration consists of a state, a position on the input tape, a position on the work tape, and up to n bits of memory that represent the work tape.

Now, consider that the work tape was split in half.  In other words, we have a left section of $\frac{n}{2}$ cells and a right section of $\frac{n}{2}$ cells.  Each configuration can be broken up into a left piece and a right piece.  The left piece consists of the state, the position on the input tape, the position on the work tape, and the $\frac{n}{2}$ bits from the left section.  The right piece consists of the state, the position on the input tape, the position on the work tape, and the $\frac{n}{2}$ bits from the right section.

Now, we build a DFA $D_1$ whose states are left pieces and a DFA $D_2$ whose states are right pieces.  The alphabet characters are instructions that say which state to go to, how the tape heads should move, and how the work tape's active cell should be manipulated.

The idea is that $D_1$ and $D_2$ read in a list of instructions corresponding to a computation of M on input x and together verify that it is valid and accepting.  Both $D_1$ and $D_2$ will always agree on where the tape heads are because that information is included in their input characters.  Therefore, we can have $D_1$ verify that the instruction is appropriate when the work tape position is in the left piece and $D_2$ verify when in the right piece.

In total, there are at most $poly(n) \cdot 2^{n/2}$ states for each DFA and at most $poly(n)$ distinct alphabet characters.

By the initial assumption, it follows that we can solve intersection non-emptiness for the two DFA's in $poly(n) \cdot 2^{(\delta n /2)}$ time.

You might find this helpful: https://rjlipton.wordpress.com/2009/08/17/on-the-intersection-of-finite-automata/



CNF-SAT is solvable using $k+O(\log(n))$ bits of memory where k is the number of variables.  The preceding construction can be used to show that if we can solve intersection non-emptiness for two DFA's in $O(n^{\delta})$ time, then we can solve CNF-SAT in $poly(n) \cdot 2^{(\delta k/2)}$ time.  Therefore, the simple answer holds.

Comments, corrections, suggestions, and questions are welcomed.  :)