DFA to Regular Expression-
The two popular methods for converting a DFA to its regular expression are-
- Arden’s Method
- State Elimination Method
In this article, we will discuss State Elimination Method.
State Elimination Method-
This method involves the following steps in finding the regular expression for any given DFA-
Step-01:
Thumb RuleThe initial state of the DFA must not have any incoming edge. |
- If there exists any incoming edge to the initial state, then create a new initial state having no incoming edge to it.
Example-
Step-02:
Thumb RuleThere must exist only one final state in the DFA. |
- If there exists multiple final states in the DFA, then convert all the final states into non-final states and create a new single final state.
Example-
Step-03:
Thumb RuleThe final state of the DFA must not have any outgoing edge. |
- If there exists any outgoing edge from the final state, then create a new final state having no outgoing edge from it.
Example-
Step-04:
- Eliminate all the intermediate states one by one.
- These states may be eliminated in any order.
In the end,
- Only an initial state going to the final state will be left.
- The cost of this transition is the required regular expression.
NOTEThe state elimination method can be applied to any finite automata. (NFA, ∈-NFA, DFA etc) |
Also Read- Construction of DFA
PRACTICE PROBLEMS BASED ON CONVERTING DFA TO REGULAR EXPRESSION-
Problem-01:
Find regular expression for the following DFA-
Solution-
Step-01:
- Initial state A has an incoming edge.
- So, we create a new initial state q_{i}.
The resulting DFA is-
Step-02:
- Final state B has an outgoing edge.
- So, we create a new final state q_{f.}
The resulting DFA is-
Step-03:
Now, we start eliminating the intermediate states.
First, let us eliminate state A.
- There is a path going from state q_{i} to state B via state A.
- So, after eliminating state A, we put a direct path from state q_{i} to state B having cost ∈.0 = 0
- There is a loop on state B using state A.
- So, after eliminating state A, we put a direct loop on state B having cost 1.0 = 10.
Eliminating state A, we get-
Step-04:
Now, let us eliminate state B.
- There is a path going from state q_{i} to state q_{f} via state B.
- So, after eliminating state B, we put a direct path from state q_{i} to state q_{f} having cost 0.(10)*.∈ = 0(10)*
Eliminating state B, we get-
From here,
Regular Expression = 0(10)* |
NOTE-
In the above question,
- If we first eliminate state B and then state A, then regular expression would be = (01)*0.
- This is also the same and correct.
Problem-02:
Find regular expression for the following DFA-
Solution-
Step-01:
- There exist multiple final states.
- So, we convert them into a single final state.
The resulting DFA is-
Step-02:
Now, we start eliminating the intermediate states.
First, let us eliminate state q_{4}.
- There is a path going from state q_{2} to state q_{f} via state q_{4}.
- So, after eliminating state q_{4} , we put a direct path from state q_{2} to state q_{f} having cost b.∈ = b.
Step-03:
Now, let us eliminate state q_{3}.
- There is a path going from state q_{2} to state q_{f} via state q_{3}.
- So, after eliminating state q_{3} , we put a direct path from state q_{2} to state q_{f} having cost c.∈ = c.
Step-04:
Now, let us eliminate state q_{5}.
- There is a path going from state q_{2} to state q_{f} via state q_{5}.
- So, after eliminating state q_{5} , we put a direct path from state q_{2} to state q_{f} having cost d.∈ = d.
Step-05:
Now, let us eliminate state q_{2}.
- There is a path going from state q_{1} to state q_{f} via state q_{2}.
- So, after eliminating state q_{2} , we put a direct path from state q_{1} to state q_{f} having cost a.(b+c+d).
From here,
Regular Expression = a(b+c+d) |
Problem-03:
Find regular expression for the following DFA-
Solution-
Step-01:
- Initial state q_{1} has an incoming edge.
- So, we create a new initial state q_{i}.
The resulting DFA is-
Step-02:
- Final state q_{2} has an outgoing edge.
- So, we create a new final state q_{f}.
The resulting DFA is-
Step-03:
Now, we start eliminating the intermediate states.
First, let us eliminate state q_{1}.
- There is a path going from state q_{i} to state q_{2} via state q_{1} .
- So, after eliminating state q_{1}, we put a direct path from state q_{i} to state q_{2} having cost ∈.c*.a = c*a
- There is a loop on state q_{2} using state q_{1} .
- So, after eliminating state q_{1} , we put a direct loop on state q_{2} having cost b.c*.a = bc*a
Eliminating state q_{1}, we get-
Step-04:
Now, let us eliminate state q_{2}.
- There is a path going from state q_{i} to state q_{f} via state q_{2} .
- So, after eliminating state q_{2}, we put a direct path from state q_{i} to state q_{f} having cost c*a(d+bc*a)*∈ = c*a(d+bc*a)*
Eliminating state q_{2}, we get-
From here,
Regular Expression = c*a(d+bc*a)* |
Problem-04:
Find regular expression for the following DFA-
Solution-
Step-01:
- State D is a dead state as it does not reach to any final state.
- So, we eliminate state D and its associated edges.
The resulting DFA is-
Step-02:
- Initial state A has an incoming edge (self loop).
- So, we create a new initial state q_{i}.
The resulting DFA is-
Step-03:
- There exist multiple final states.
- So, we convert them into a single final state.
The resulting DFA is-
Step-04:
Now, we start eliminating the intermediate states.
First, let us eliminate state C.
- There is a path going from state B to state q_{f} via state C.
- So, after eliminating state C, we put a direct path from state B to state q_{f} having cost b.b*.∈ = bb*
Eliminating state C, we get-
Step-05:
Now, let us eliminate state B.
- There is a path going from state A to state q_{f} via state B.
- So, after eliminating state B, we put a direct path from state A to state q_{f} having cost a.a*.(bb*+∈) = aa*(bb*+∈)
Eliminating state B, we get-
Step-06:
Now, let us eliminate state A.
- There is a path going from state q_{i} to state q_{f} via state A.
- So, after eliminating state A, we put a direct path from state q_{i} to state q_{f} having cost ∈.b*.(aa*(bb*+∈)+∈) = b*(aa*(bb*+∈)+∈)
Eliminating state A, we get-
From here,
Regular Expression = b*(aa*(bb*+∈)+∈) |
We know, bb* + ∈ = b*
So, we can also write-
Regular Expression = b*(aa*b*+∈) |
Problem-05:
Find regular expression for the following DFA-
Solution-
Step-01:
- Since initial state A has an incoming edge, so we create a new initial state q_{i}.
- Since final state A has an outgoing edge, so we create a new final state q_{f}.
The resulting DFA is-
Step-02:
Now, we start eliminating the intermediate states.
First, let us eliminate state B.
- There is a path going from state C to state A via state B.
- So, after eliminating state B, we put a direct path from state C to state A having cost b.b = bb.
- There is a loop on state A using state B.
- So, after eliminating state B, we put a direct loop on state A having cost a.b = ab.
Eliminating state B, we get-
Step-03:
Now, let us eliminate state C.
- There is a loop on state A using state C.
- So, after eliminating state C, we put a direct loop on state A having cost b.(a+bb) = b(a+bb)
Eliminating state C, we get-
Step-04:
Now, let us eliminate state A.
- There is a path going from state q_{i} to state q_{f} via state A.
- So, after eliminating state A, we put a direct path from state q_{i} to state q_{f} having cost ∈.(ab + b(a+bb))*∈ = (ab + b(a+bb))*
Eliminating state A, we get-
From here,
Regular Expression = (ab + b(a+bb))* |
Problem-06:
Find regular expression for the following DFA-
Solution-
- State B is a dead state as it does not reach to the final state.
- So, we eliminate state B and its associated edges.
The resulting DFA is-
From here,
Regular Expression = a |
Problem-07:
Find regular expression for the following DFA-
Solution-
Step-01:
- There exist multiple final states.
- So, we create a new single final state.
The resulting DFA is-
Step-02:
Now, we start eliminating the intermediate states.
First, let us eliminate state B.
- There is a path going from state A to state q_{f} via state B.
- So, after eliminating state B, we put a direct path from state A to state q_{f} having cost a.a*.∈ = aa*.
Eliminating state B, we get-
Step-03:
Now, let us eliminate state C.
- There is a path going from state A to state q_{f} via state C.
- So, after eliminating state C, we put a direct path from state A to state q_{f} having cost b.a*.∈ = ba*.
Eliminating state C, we get-
From here,
Regular Expression = aa* + ba* |
Also Read- Minimization of DFA
To gain better understanding about Converting DFA to Regular Expression,
Next Article- Arden’s Theorem
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