Skip to main content
Computer Science

Return to Answer

Avoid confusion due to FSM vs FSA
Source Link
Tobias
Tobias
  • 325
  • 2
  • 8

Finite state machinesautomatons recognize regular languages. The advantage of regular languages is that it is usually trivial to express them formally.

FSA #1

$$L(M_1) = \left\{ a^kb^lc^m | k > 0, l \geq 0, m \in \{ 0, 1 \} \right\}$$

FSA #2

$$L(M_2) = \left\{ a^kb^lc^m | k > 0, l > 0, m \in \{ 0, 1 \} \right\}$$

Finite state machines recognize regular languages. The advantage of regular languages is that it is usually trivial to express them formally.

FSA #1

$$L(M_1) = \left\{ a^kb^lc^m | k > 0, l \geq 0, m \in \{ 0, 1 \} \right\}$$

FSA #2

$$L(M_2) = \left\{ a^kb^lc^m | k > 0, l > 0, m \in \{ 0, 1 \} \right\}$$

Finite state automatons recognize regular languages. The advantage of regular languages is that it is usually trivial to express them formally.

FSA #1

$$L(M_1) = \left\{ a^kb^lc^m | k > 0, l \geq 0, m \in \{ 0, 1 \} \right\}$$

FSA #2

$$L(M_2) = \left\{ a^kb^lc^m | k > 0, l > 0, m \in \{ 0, 1 \} \right\}$$

Source Link
Tobias
Tobias
  • 325
  • 2
  • 8

Finite state machines recognize regular languages. The advantage of regular languages is that it is usually trivial to express them formally.

FSA #1

$$L(M_1) = \left\{ a^kb^lc^m | k > 0, l \geq 0, m \in \{ 0, 1 \} \right\}$$

FSA #2

$$L(M_2) = \left\{ a^kb^lc^m | k > 0, l > 0, m \in \{ 0, 1 \} \right\}$$