Intersection process of two DFAs (original) (raw)

Last Updated : 12 Jul, 2025

Let's understand the intersection of two DFA with an example.
Designing a DFA for the set of string over {0, 1} such that it ends with 01 and has even number of 1's.
There two desired language will be formed:

L1= {01, 001, 101, 0101, 1001, 1101, ....}
L2= {11, 011, 101, 110, 0011, 1100, .....}

L = L1 and L2 = L1 ∩ L2

**State Transition Diagram for the language L 1 :
This is a DFA for language L1

It accepts all the string that accept 01 at end.

**State Transition Diagram for the language L 2 :
This is a DFA for language L2

It accepts all the string that accept with even number of 1's.

**State Transition Diagram of L 1 ∩ L 2 :
Intersection of L1 and L2 can be explained by language that a string over {0, 1} accept such that it ends with 01 and has even number of 1's.

L = L1 ∩ L2
= {1001, 0101, 01001, 10001, ....}

Thus as we see that L1 and L2 have been combined through intersection process and this final DFA accept all the language that has even number of 1's and is ending with 01.