Introduction to Queue Automata

We already know about Finite Automata which can be used to accept regular languages and Pushdown Automata that can be used to recognize Context Free Languages.

Queue Automata(QDA) is a non-deterministic automata that is similar to Pushdown Automata but has a queue instead of a stack which helps Queue automata to recognize languages beyond Context Free Languages.

A QDA is a 6 – tuple $M = (Q, \sigma, \Gamma, \delta, q_0, F)$

Where

Q is the set of finite states.
$\bf{\sigma}$ is the set of finite input alphabets.
$\bf{\Gamma}$ is the set of finite queue alphabets.
$\delta : Q \times \sigma_\epsilon \times \Gamma_\epsilon \rightarrow P(Q \times \Gamma_\epsilon )$ .
$q_0 \in Q$ is the start state.
F $\subseteq$ Q is the set of accept states.

Acceptance of a string

A QDA $M = (Q, \sigma, \Gamma, \delta, q_0, F)$ accepts input $w$ if $w$ can be written as $w= w_1w_2w_3 ... . w_m$ , where each $w_i \in$ $\sigma_\epsilon$ and there are states $r_0, r_1, r_2, ... . ., r_m \in Q$ and strings $s_0, s_1, s_2, ... . ., s_m \in \Gamma^*$ exist, such that they satisfy the following conditions:

$r_0 = q_0$ and $s_0 = \epsilon$ .
For $0\leq i \leq m-1$ $( r_{i+1}, b) = \delta(r_i, w_{i+1}, a), where\, a, b \in \Gamma_\epsilon$ and $s_i = ta$ and $s_{i+1} = bt$ and $t \in \Gamma^*$
$r_m \in F$

Example:
Define the queue automata for language ${a^nb^n | n \geq 0}$

Solution:
Q = {q0, q1, q2, q3} and $\sigma$ ={a, b} and $\Gamma$ = {a, b, $}
And the transition functions are given by:
$\delta(q0, a, \epsilon)={(q0, a)}$
$\delta(q0, \epsilon, \epsilon)={(q1, \$)}$
$\delta(q1, \epsilon, a)={(q2, \epsilon)}$
$\delta(q2, \epsilon, a)={(q2, a)}$
$\delta(q2, b, \$)={(q1, \$)}$
$\delta(q1, \epsilon, \$)={(q3, \$)}$

Let us see how this automata works for aabb.

Row	State	Input	Transition function	Queue(Input from left)	State after move
1	q0	aabb	δ(q0, a, ε)={(q0, a)}	a	q0
2	q0	aabb	δ(q0, a, ε)={(q0, a)}	aa	q0
3	q0	ε	δ(q0, ε, ε)={(q1, $)}	$aa	q1
4	q1	ε	δ(q1, ε, a)={(q2, ε)}	$a	q2
5	q2	ε	δ(q2, ε, a)={(q2, a)}	a$	q2
6	q2	aabb	δ(q2, b, $)={(q1, $)}	$a	q1
7	q1	ε	δ(q1, ε, a)={(q2, ε)}	$	q2
8	q2	aabb	δ(q2, b, $)={(q1, $)}	$	q1
9	q1	ε	δ(q1, ε, $)={(q3, $)}	$	q3

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