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 automaton
2 can be represented as having empty state
and having combination of states
.
Example
Consider the non deterministic finite automaton
1 which accepts the language
containing the string
3 \(bb\).
Old States | New Starts After Reading | |
---|---|---|
a | b | |
\(z_1^- \equiv x_1\) | \(x_1 \equiv z_1\) | \((x_1, x_2) \equiv z_2\) |
\(z_2 \equiv (x_1, x_2)\) | \((x_1, \varnothing) \equiv x_1 \equiv z_1\) | \((x_1, x_2, x_3) \equiv z_3\) |
\(z_3^+ \equiv (x_1, x_2, x_3)\) | \((x_1, x_3) \equiv z_4\) | \((x_1, x_2, x_3) \equiv z_3\) |
\(z_4^+ \equiv (x_1, x_3)\) | \((x_1, x_3) \equiv z_4\) | \((x_1, x_2, x_3) \equiv z_3\) |
Non Deterministic Finite Automaton
1 And Kleene's Theorem
4
The non deterministic finite automaton
1 can help proving the 3rd part of kleene's theorem
4
There are 2 methods for this
Method 1
Example
\[L_1 = \{a\}, L_2=\{b\}, L_3=\{\Lambda\}\]
\(NFA_1\)
\(NFA_2\)
\(NFA_3\)
\(FA_1\)
\(FA_2\)
\(FA_3\)
Method 2
Example
\(FA_1\)
\(FA_2\)
Non Deterministic Finite Automaton
1 Corresponding to \(FA_1 \cup FA_2\)
References
-
Read more about finite automaton. ↩
-
Read more about kleene's theorems. ↩↩