- Automata Theory Tutorial
- Automata Theory - Home
- Automata Theory Introduction
- Deterministic Finite Automaton
- Non-deterministic Finite Automaton
- NDFA to DFA Conversion
- DFA Minimization
- Moore & Mealy Machines

- Classification of Grammars
- Introduction to Grammars
- Language Generated by Grammars
- Chomsky Grammar Classification

- Regular Grammar
- Regular Expressions
- Regular Sets
- Arden's Theorem
- Constructing FA from RE
- Pumping Lemma for Regular Grammar
- DFA Complement

- Context-Free Grammars
- Context-Free Grammar Introduction
- Ambiguity in Grammar
- CFL Closure Properties
- CFG Simplification
- Chomsky Normal Form
- Greibach Normal Form
- Pumping Lemma for CFG

- Pushdown Automata
- Pushdown Automata Introduction
- Pushdown Automata Acceptance
- PDA & Context Free Grammar
- PDA & Parsing

- Turing Machine
- Turing Machine Introduction
- Accepted & Decided Language
- Multi-tape Turing Machine
- Multi-Track Turing Machine
- Non-Deterministic Turing Machine
- Semi-Infinite Tape Turing Machine
- Linear Bounded Automata

- Decidability
- Language Decidability
- Undecidable Language
- Turing Machine Halting Problem
- Rice Theorem
- Post Correspondence Problem

- Automata Theory Useful Resources
- Automata Theory - Quick Guide
- Automata Theory - Useful Resources
- Automata Theory - Discussion

A TM accepts a language if it enters into a final state for any input string w. A language is recursively enumerable (generated by Type-0 grammar) if it is accepted by a Turing machine.

A TM decides a language if it accepts it and enters into a rejecting state for any input not in the language. A language is recursive if it is decided by a Turing machine.

There may be some cases where a TM does not stop. Such TM accepts the language, but it does not decide it.

The basic guidelines of designing a Turing machine have been explained below with the help of a couple of examples.

Design a TM to recognize all strings consisting of an odd number of α’s.

*Solution*

The Turing machine **M** can be constructed by the following moves −

Let

**q**be the initial state._{1}If

**M**is in**q**; on scanning α, it enters the state_{1}**q**and writes_{2}**B**(blank).If

**M**is in**q**; on scanning α, it enters the state_{2}**q**and writes_{1}**B**(blank).From the above moves, we can see that

**M**enters the state**q**if it scans an even number of α’s, and it enters the state_{1}**q**if it scans an odd number of α’s. Hence_{2}**q**is the only accepting state._{2}

Hence,

M = {{q_{1}, q_{2}}, {1}, {1, B}, δ, q_{1}, B, {q_{2}}}

where δ is given by −

Tape alphabet symbol | Present State ‘q_{1}’ |
Present State ‘q_{2}’ |
---|---|---|

α | BRq_{2} |
BRq_{1} |

Design a Turing Machine that reads a string representing a binary number and erases all leading 0’s in the string. However, if the string comprises of only 0’s, it keeps one 0.

*Solution*

Let us assume that the input string is terminated by a blank symbol, B, at each end of the string.

The Turing Machine, **M**, can be constructed by the following moves −

Let

**q**be the initial state._{0}If

**M**is in**q**, on reading 0, it moves right, enters the state_{0}**q**and erases 0. On reading 1, it enters the state_{1}**q**and moves right._{2}If

**M**is in**q**, on reading 0, it moves right and erases 0, i.e., it replaces 0’s by B’s. On reaching the leftmost 1, it enters_{1}**q**and moves right. If it reaches B, i.e., the string comprises of only 0’s, it moves left and enters the state_{2}**q**._{3}If

**M**is in**q**, on reading either 0 or 1, it moves right. On reaching B, it moves left and enters the state_{2}**q**. This validates that the string comprises only of 0’s and 1’s._{4}If

**M**is in**q**, it replaces B by 0, moves left and reaches the final state_{3}**q**._{f}If

**M**is in**q**, on reading either 0 or 1, it moves left. On reaching the beginning of the string, i.e., when it reads B, it reaches the final state_{4}**q**._{f}

Hence,

M = {{q_{0}, q_{1}, q_{2}, q_{3}, q_{4}, q_{f}}, {0,1, B}, {1, B}, δ, q_{0}, B, {q_{f}}}

where δ is given by −

Tape alphabet symbol | Present State ‘q_{0}’ |
Present State ‘q_{1}’ |
Present State ‘q_{2}’ |
Present State ‘q_{3}’ |
Present State ‘q_{4}’ |
---|---|---|---|---|---|

0 | BRq_{1} |
BRq_{1} |
ORq_{2} |
- | OLq_{4} |

1 | 1Rq_{2} |
1Rq_{2} |
1Rq_{2} |
- | 1Lq_{4} |

B | BRq_{1} |
BLq_{3} |
BLq_{4} |
OLq_{f} |
BRq_{f} |

Advertisements