# CS 101 Introduction to Computing Assignment # 01

Internet και Εφαρμογές Web

21 Οκτ 2013 (πριν από 4 χρόνια και 6 μήνες)

75 εμφανίσεις

CS
101

Introduction to Computing

Assignment
#

01

Your assignment must be uploaded / submitted before or on

____
20
-
10
-
2009
__________.

Please view the assignment submission process document provided to you by the Virtual Universi
ty to

Rules for Marking

Please note that your assignment will not be graded if:

It is submitted after due date

The file you uploaded does not open

The file you uploaded is copied from some one else

It is in some format other than
.
d
oc

Objective

The assignment has been designed to enable you:

To learn about the concept of Semantic Web

To learn about the decimal to binary conversion and vice versa.

To know about the truth table.

Question No. 1

As you have studied basic concept of “Se
mantic Web” in Lecture #3

You are required to read carefully the below listed definition of semantic web

“The semantic Web is an extension of the current Web in which
information is given well
-
defined
meaning,

better enabling computers

and
people to wor
k in cooperation.”

The Web as a whole can be made more intelligent and perhaps even intuitive about how to serve a user's needs.
they have little ability to select the pages that a user really wants or needs.

The c
ontext understanding programs
can select
ively find what users want.

I need to know a very brief and precise description of the above highlighted
words in a
definition in your
own words in a single paragraph.
(Marks

5
)

Question No. 2

Convert the following binary and decimal number into their respective number system.

i) 396 = (
110001100

)
2

ii) 101011 = ( 44

)
10

(

Marks 5)

Question No. 3

Prove right hand side equal to left hand site by using truth table

((A
+B.C).(A.C+B))’ =A’.B’+ A’.C’+B’.C’

(

Marks 5)

(NOTE:
Fill the given table

and make sure the sequence of input remains same as it is given in the
table below)

A

B

C

((A+B.C) .(A.C+B))’

A’.B’+ A’.C’+B’.C’

0

0

0

0

1

0

0

1

0

1

0

1

0

0

1

0

1

1

1

0

1

0

0

0

1

1

0

1

1

0

1

1

0

0

0

1

1

1

1

0