Question

In calculate if two arbitrary regular expressions have any overlapping solutions (assuming it’s possible).

For example these two regular expressions can be shown to have no intersections by brute force because the two solution sets are calculable because it’s finite.

^1(11){0,1000}$ ∩     ^(11){0,1000}$        = {}
{1,111, ..., ..111} ∩ {11,1111, ..., ...11} = {}
{}                                          = {}

But replacing the {0,1000} by * remove the possibility for a brute force solution, so a smarter algorithm must be created.

^1(11)*$ ∩ ^(11)*$ = {}
{1,^1(11)*$} ∩ {^(11)*$} = {}
{1,^1(11)*$} ∩ {11,^11(11)*$} = {}
{1,111,^111(11)*$} ∩ {11,^(11)*$} = {}
.....

In another similar question one answer was to calculate the intersection regex. Is that possible to do? If so how would one write an algorithm to do such a thing?

I think this problem might be domain of the halting problem.

EDIT:

I’ve used the accepted solution to create the DFAs for the example problem. It’s fairly easy to see how you can use a BFS or DFS on the graph of states for M_3 to determine if a final state from M_3 is reachable.

Answer 1

It is not in the domain of the halting problem; deciding whether the intersection of regular languages is empty or not can be solved as follows:

1. Construct a DFA M1 for the first language.
2. Construct a DFA M2 for the second language. Hint: Kleene’s Theorem and Power Set machine construction
3. Construct a DFA M3 for M1 intersect M2. Hint: Cartesian Product Machine construction
4. Determine whether L(M3) is empty. Hint: If M3 has n states, and M3 doesn’t accept any strings of length no greater than n, then L(M3) is empty… why?

Each of those things can be algorithmically done and/or checked. Also, naturally, once you have a DFA recognizing the intersection of your languages, you can construct a regex to match the language. And if you start out with a regex, you can make a DFA. This is definitely computable.

EDIT:

So to build a Cartesian Product Machine, you need two DFAs. Let M1 = (E, q0, Q1, A1, f1) and M2 = (E, q0′, Q2, A2, f2). In both cases, E is the input alphabet, q0 is the start state, Q is the set of all states, A is the set of accepting states, and f is the transition function. Construct M3 where…

1. E3 = E
2. Q3 = Q1 x Q2 (ordered pairs)
3. q0” = (q0, q0′)
4. A3 = {(x, y) | x in A1 and y in A2}
5. f3(s, (x, y)) = (f1(s, x), f2(s, y))

Provided I didn’t make any mistakes, L(M3) = L(M1) intersect L(M2). Neat, huh?