Introduction to Queue Automata

Last Updated : 08 Jul, 2020

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.

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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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