- Moore & Mealy Machines
- DFA Minimization
- NDFA to DFA Conversion
- Non-deterministic Finite Automaton
- Deterministic Finite Automaton
- Automata Theory Introduction
- Automata Theory - Home
Classification of Grammars
Regular Grammar
- DFA Complement
- Pumping Lemma for Regular Grammar
- Constructing FA from RE
- Regular Sets
- Regular Expressions
Context-Free Grammars
- Pumping Lemma for CFG
- Greibach Normal Form
- Chomsky Normal Form
- CFG Simplification
- CFL Closure Properties
- Ambiguity in Grammar
- Context-Free Grammar Introduction
Pushdown Automata
- PDA & Parsing
- PDA & Context Free Grammar
- Pushdown Automata Acceptance
- Pushdown Automata Introduction
Turing Machine
- Linear Bounded Automata
- Semi-Infinite Tape Turing Machine
- Non-Deterministic Turing Machine
- Multi-Track Turing Machine
- Multi-tape Turing Machine
- Accepted & Decided Language
- Turing Machine Introduction
Decidability
- Post Correspondence Problem
- Rice Theorem
- Turing Machine Halting Problem
- Undecidable Language
- Language Decidability
Automata Theory Useful Resources
Selected Reading
- Who is Who
- Computer Glossary
- HR Interview Questions
- Effective Resume Writing
- Questions and Answers
- UPSC IAS Exams Notes
NDFA to DFA Conversion
Problem Statement
Let X = (Qx, ∑, δx, q0, Fx) be an NDFA which accepts the language L(X). We have to design an equivalent DFA Y = (Qy, ∑, δy, q0, Fy) such that L(Y) = L(X). The following procedure converts the NDFA to its equivalent DFA −
Algorithm
Input − An NDFA
Output − An equivalent DFA
Step 1 − Create state table from the given NDFA.
Step 2 − Create a blank state table under possible input alphabets for the equivalent DFA.
Step 3 − Mark the start state of the DFA by q0 (Same as the NDFA).
Step 4 − Find out the combination of States {Q0, Q1,... , Qn} for each possible input alphabet.
Step 5 − Each time we generate a new DFA state under the input alphabet columns, we have to apply step 4 again, otherwise go to step 6.
Step 6 − The states which contain any of the final states of the NDFA are the final states of the equivalent DFA.
Example
Let us consider the NDFA shown in the figure below.
q | δ(q,0) | δ(q,1) |
---|---|---|
a | {a,b,c,d,e} | {d,e} |
b | {c} | {e} |
c | ∅ | {b} |
d | {e} | ∅ |
e | ∅ | ∅ |
Using the above algorithm, we find its equivalent DFA. The state table of the DFA is shown in below.
q | δ(q,0) | δ(q,1) |
---|---|---|
[a] | [a,b,c,d,e] | [d,e] |
[a,b,c,d,e] | [a,b,c,d,e] | [b,d,e] |
[d,e] | [e] | ∅ |
[b,d,e] | [c,e] | [e] |
[e] | ∅ | ∅ |
[c, e] | ∅ | [b] |
[b] | [c] | [e] |
[c] | ∅ | [b] |
The state diagram of the DFA is as follows −
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