Question

I am to construct a DFA from the intersection of two simpler DFAs. The first simpler DFA recognizes languages of all strings that have at least three 0s, and the second simpler language DFA recognizes languages of strings of at most two 1s. The alphabet is (0,1). I’m not sure how to construct a larger DFA combining the two. Thanks!

Answer 1

Here’s a general idea:

The most straightforward way to do this is to have different paths for counting your 0s that are based on the number of 1s you’ve seen, such that they are “parallel” to each other. Move from one layer of the path to the next any time you see a 1, and then move from the last layer to a trap state if you see a third 1. Depending on the exact nature of the assignment you might be able to condense this, but once you have a basic layout you can determine that. Typically you can combine states from the first DFA with states in the second DFA to produce a smaller end result.

Here’s a more mathematical explanation:

Constructing automata for the
intersection operation.
Assume we are
given two DFA M1 = (S1, q(1) 0 , T1,
F1) and M2 = (S2, q(2) 0 , T2, F2).
These two DFA recognize languages L1 =
L(M1) and L2 = L(M2). We want to
design a DFA M= (S, q0, T, F) that
recognizes the intersection L1 ∩L2. We
use the idea of constructing the DFA
for the union of languages. Given an
input w, we run M1 and M2 on w
simultaneously as we explained for the
union operation. Once we finish the
runs of M1 and of M2 on w, we look at
the resulting end states of these two
runs. If both end states are accepting
then we accept w, otherwise we reject
w.

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When constructing the new transition function, the easy way to think of it is by using pairs of states. For example, consider the following DFAs:

Now, we can start combining these by traversing both DFAs at the same time. For example, both start at state 1. Now what happens if we see an a as input? Well, DFA1 will go from 1->2, and DFA2 will go from 1->3. When combining, then, we can say that the intersection will go from state “1,1” (both DFAs are in state 1) to state “2,3”. State 2 is an accept state in DFA1 and state 3 is an accept state in DFA2, so state “2,3” is an accept state in our new DFA3. We can repeat this for all states/transitions and end up with:

Does that make sense?

Reference: Images found in this assignment from Cornell University.