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Lecture No. 17

Dated: 02-11-2024

Method 3

In a non deterministic finite automaton,1 for a letter, there may be no transition at all or multiple transitions.
The corresponding finite automaton2 can be represented as having empty state and having combination of states.

Example

Consider the non deterministic finite automaton1 which accepts the language containing the string3 \(bb\).
CS402_e_17_1.svg

Old States New Starts After Reading
a b
\(z_1^- \equiv x_1\) \(x_1 \equiv z_1\) \((x_1, x_2) \equiv z_2\)
\(z_2 \equiv (x_1, x_2)\) \((x_1, \varnothing) \equiv x_1 \equiv z_1\) \((x_1, x_2, x_3) \equiv z_3\)
\(z_3^+ \equiv (x_1, x_2, x_3)\) \((x_1, x_3) \equiv z_4\) \((x_1, x_2, x_3) \equiv z_3\)
\(z_4^+ \equiv (x_1, x_3)\) \((x_1, x_3) \equiv z_4\) \((x_1, x_2, x_3) \equiv z_3\)

CS402_e_17_2.svg

Non Deterministic Finite Automaton1 And Kleene's Theorem4

The non deterministic finite automaton1 can help proving the 3rd part of kleene's theorem4
There are 2 methods for this

Method 1

Example

\[L_1 = \{a\}, L_2=\{b\}, L_3=\{\Lambda\}\]

\(NFA_1\)

CS402_e_17_3.svg

\(NFA_2\)

CS402_e_17_4.svg

\(NFA_3\)

CS402_e_17_5.svg

\(FA_1\)

CS402_e_17_6.svg

\(FA_2\)

CS402_e_17_7.svg

\(FA_3\)

CS402_e_17_8.svg

Method 2

Example

\(FA_1\)

CS402_e_17_9.svg

\(FA_2\)

CS402_e_17_10.svg

Non Deterministic Finite Automaton1 Corresponding to \(FA_1 \cup FA_2\)

CS402_e_17_11.svg

References


  1. Read more about non deterministic finite automaton

  2. Read more about finite automaton

  3. Read more about strings

  4. Read more about kleene's theorems