No A B C 4. Consider the following PDAs. Match them with the suitable definition of the languages and grammars. Write your answer in the blank column of the table below: Matched PDA Matched PDA Pushdown Automata Language Grammar (No.) (No.) (a,ɛ/X) (b,X/ɛ) (c,ɛ/E) S→ as | aB L = anbmc | n > 0, m >= 0 B→ bBc | A qo (a,ɛ/X) (c,X/ɛ) SAB L = amb c | n >= 0, m> 0 A → aAb | ab BcB | C qo L = a b cm | n, m > 0 S→ aSc | aBc B→ bB | A (b,X/E) (ɛ,ɛ/E) (a,ɛ/ɛ) 91 (c,ɛ/E) (b,ɛ/ɛ) 91 (c,X/E). (a,ɛ/ɛ) $(6.2/6), (b,ɛ/X) (E,ɛ/E) (c,X/ɛ)

No A B C 4. Consider the following PDAs. Match them with the suitable definition of the languages and grammars. Write your answer in the blank column of the table below: Matched PDA Matched PDA Pushdown Automata Language Grammar (No.) (No.) (a,ɛ/X) (b,X/ɛ) (c,ɛ/E) S→ as | aB L = anbmc | n > 0, m >= 0 B→ bBc | A qo (a,ɛ/X) (c,X/ɛ) SAB L = amb c | n >= 0, m> 0 A → aAb | ab BcB | C qo L = a b cm | n, m > 0 S→ aSc | aBc B→ bB | A (b,X/E) (ɛ,ɛ/E) (a,ɛ/ɛ) 91 (c,ɛ/E) (b,ɛ/ɛ) 91 (c,X/E). (a,ɛ/ɛ) $(6.2/6), (b,ɛ/X) (E,ɛ/E) (c,X/ɛ)

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Concept explainers

Intelligent Machines

Machines that use artificial intelligence to learn from the environment and past experiences to perform better in the presence of variability and uncertainty are called intelligent machines. Intelligence develops according to environment and experience. It is human-like behavior. When the computer system or any programmed machine learns from the inputs and corresponding outputs, it can be called an intelligent machine.

Question

No
4.
Consider the following PDAs. Match them with the suitable definition of the
languages and grammars. Write your answer in the blank column of the table below:
Matched
PDA
Matched
PDA
Pushdown Automata
Language
Grammar
(No.)
(No.)
(a,ɛ/X)
(b,X/ɛ) (c,ɛ/ɛ)
L = anbmc | n > 0, m >= 0
SaS | aB
BbBc | A
(c,ɛ/ɛ)
go
91
92
(a,ɛ/X)
(c,X/ɛ)
S→ AB
L = ambncn | n >= 0, m > 0
A → aAb | ab
B→ CB | C
L = a b cm | n, m > 0
SaSc | aBc
B→ bB | A
(b,X/ɛ)
8
qo
9⁰
(ɛ,ɛ/ɛ)
(a,ɛ/ɛ)
(b,ɛ/ɛ)
91
(a,ɛ/ɛ)
91
(c,X/ɛ),
(ε,ɛ/ε)
(b,ɛ/X)
92
(ε,ɛ/E)
(c,X/ɛ)
93

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