CS201 Solved Progmming Assignment Spring 2011 18 april

#include<iostream.h>
using std::cout;
using std::endl;
main()
{
// declaration of variables..........
char model,name[15], car#[15];
int days, amount, rentalAmount=0;
// little description and prompt the user user to type the input..
cout" Please provide the customer Name: " endl;
cin>> name;
cout" Please provide the Car Model Description. " endl;
cout"\t Enter 'A' for Model 2009. " endl;
cout"\t Enter 'B' for Model 2010. " endl;
cout"\t Enter 'C' for Model 2011. " endl;
cin>> model;
cout " Please provide the following information : " endl;
cout" Car No. : " endl;
cin>> car#;
cout" Number of day's : " endl;
cin>> days;
if (model == 'A')
{
amount = 5000;
model = 2009;
rentalAmount = (amount * days);
cout"======================================" endl;
cout"Customer Name :" name;
cout"\n""Car Model :" model ;
cout"\n""Number of days :" days;
cout"\n""Your Rental Amount is :" rentalAmount ;
cout"\n""======================================" endl;
break;
}
else if (model == 'B')
{
amount = 8000;
model = 2010;
rentalAmount = (amount * days);
cout"======================================" endl;
cout"Customer Name :" name;
cout"\n""Car Model :" model ;
cout"\n""Number of days :" days;
cout"\n""Your Rental Amount is :" rentalAmount ;
cout"\n""======================================" endl;
break;
}
else if (model == 'C')
{
amount = 10000;
model = 2011;
rentalAmount = (amount * days);
cout"======================================" endl;
cout"Customer Name :" name;
cout"\n""Car Model :" model ;
cout"\n""Number of days :" days;
cout"\n""Your Rental Amount is :" rentalAmount ;
cout"\n""======================================" endl;
break;
}
}

CS402 Theory of Automata Solved First Assignment spring 2011

CS402 assignment no 1 Spring 2011 Idea solution
Question No.1
RECURSIVE DEFINITION
a. Give recursive definition of language defined over alphabet Σ = {a,
b}, having all
strings STARTING WITH aa OR ENDING WITH bb
b. Give recursive definition of language defined over alphabet Σ = {a,
b}, having all
strings with length MULTIPLE OF 2
c. Give recursive definition of language defined over alphabet Σ = {a,
b}, having all
strings NOT ENDING with aa or bb
d. Give recursive definition of language defined over alphabet Σ = {a, b}, NOT
HAVING ab at any place.
e. Give recursive definition of ODD PALINDROME (PALINDROME WITH
ODD STRINGS ONLY) defined over alphabet Σ = {a, b}
Solution:

Recursive definition of languages
The following three steps are used in recursive definition
Some basic words are specified in the language.
Rules for constructing more words are defined in the language.
No strings except those constructed in above, are allowed to be in the language.
Examples
Defining language of INTEGER
Step 1: 1 is in INTEGER.
Step 2: If x is in INTEGER then x+1 and x-1 are also in INTEGER.
Step 3: No strings except those constructed in above, are allowed to
be in INTEGER.
Defining language of EVEN
Step 1: 2 is in EVEN.
Theory of Automata (CS402)
(c) Copyright Virtual University of Pakistan
11
Step 2: If x is in EVEN then x+2 and x-2 are also in EVEN.
Step 3: No strings except those constructed in above, are allowed to be in EVEN.
Defining the language factorial
Step 1: As 0!=1, so 1 is in factorial.
Step 2: n!=n*(n-1)! is in factorial.
Step 3: No strings except those constructed in above, are allowed to
be in factorial.
Defining the language PALINDROME, defined over Σ = {a,b}
Step 1: a and b are in PALINDROME
Step 2: if x is palindrome, then s(x)Rev(s) and xx will also be
palindrome, where s belongs to Σ*
Step 3: No strings except those constructed in above, are allowed to
be in palindrome
Defining the language {anbn }, n=1,2,3,... , of strings defined over Σ={a,b}
Step 1: ab is in {anbn}
Step 2: if x is in {anbn}, then axb is in {anbn}
Step 3: No strings except those constructed in above, are allowed to
be in {anbn}
Defining the language L, of strings ending in a , defined over Σ={a,b}
Step 1: a is in L
Step 2: if x is in L then s(x) is also in L, where s belongs to Σ*
Step 3: No strings except those constructed in above, are allowed to be in L
Defining the language L, of strings beginning and ending in same
letters , defined over Σ={a, b}
Step 1: a and b are in L
Step 2: (a)s(a) and (b)s(b) are also in L, where s belongs to Σ*
Step 3: No strings except those constructed in above, are allowed to be in L
Defining the language L, of strings containing aa or bb , defined over Σ={a, b}
Step 1: aa and bb are in L
Step 2: s(aa)s and s(bb)s are also in L, where s belongs to Σ*
Step 3: No strings except those constructed in above, are allowed to be in L
Defining the language L, of strings containing exactly one a, defined
over Σ={a, b}
Step 1: a is in L
Step 2: s(aa)s is also in L, where s belongs to b*
Step 3: No strings except those constructed in above, are allowed to be in L

Question No.2
REGAULAR EXPRESSIONS
Give Regular Expression for each of the following language defined
over alphabet Σ =
{a, b}
a. Language having all strings STARTING AND ENDING WITH ab
b. Language of strings NOT having bb OR aa at any place
c. Language of all strings NOT HAVING aab in start
d. Language of all strings NOT HAVING aab in end
e. Language of all strings HAVING count of b's multiple of 2 [No restriction on
count of a]
Solution:

Regular Expression
As discussed earlier that a* generates Λ, a, aa, aaa, ... and a+
generates a, aa, aaa, aaaa, ..., so the language L1
= {Λ, a, aa, aaa, ...} and L2 = {a, aa, aaa, aaaa, ...} can simply be
expressed by a* and a+, respectively.
a* and a+ are called the regular expressions (RE) for L1 and L2 respectively.
Note a+, aa* and a*a generate L2.
Recursive definition of Regular Expression(RE)
Step 1: Every letter of Σ including Λ is a regular expression.
Step 2: If r1 and r2 are regular expressions then
(r1)
r1 r2
r1 + r2 and
r1*
are also regular expressions.
Step 3: Nothing else is a regular expression.
Method 3 (Regular Expressions)
Consider the language L={Λ, x, xx, xxx,...} of strings, defined over Σ = {x}.
We can write this language as the Kleene star closure of alphabet Σ or
L=Σ*={x}* .
This language can also be expressed by the regular expression x*.
Similarly the language L={x, xx, xxx,...}, defined over Σ = {x}, can be
expressed by the regular expression x+.
Now consider another language L, consisting of all possible strings,
defined over Σ = {a, b}. This language can
also be expressed by the regular expression (a + b)*.
Now consider another language L, of strings having exactly one a,
defined over Σ = {a, b}, then it's regular
expression may be b*ab*.
Now consider another language L, of even length, defined over Σ = {a,
b}, then it's regular expression may be
((a+b)(a+b))*.
Now consider another language L, of odd length, defined over Σ = {a,
b}, then it's regular expression may be
(a+b)((a+b)(a+b))* or ((a+b)(a+b))*(a+b).
Remark
It may be noted that a language may be expressed by more than one
regular expression, while given a regular
expression there exist a unique language generated by that regular expression.
Example
Consider the language, defined over
Σ = {a , b} of words having at least one a, may be expressed by a
regular expression (a+b)*a(a+b)*.
Consider the language, defined over Σ = {a, b} of words having at
least one a and one b, may be expressed by a
regular expression (a+b)*a(a+b)*b(a+b)*+ (a+b)*b(a+b)*a(a+b)*.
Consider the language, defined over Σ ={a, b}, of words starting with
double a and ending in double b then its
regular expression may be aa(a+b)*bb
Consider the language, defined over Σ ={a, b} of words starting with a
and ending in b OR
starting with b and ending in a, then its regular expression may be
a(a+b)*b+b(a+b)*a
Theory of Automata (CS402)
(c) Copyright Virtual University of Pakistan
13
Regular expression of EVEN-EVEN language, Difference between a* + b*
and (a+b)*, Equivalent regular
expressions; sum, product and closure of regular expressions; regular
languages, finite languages are regular,
introduction to finite automaton, definition of FA, transition table,
transition diagram
An important example
The Language EVEN-EVEN
Language of strings, defined over Σ={a, b} having even number of a's
and even number of b's. i.e.
EVEN-EVEN = {Λ, aa, bb, aaaa,aabb,abab, abba, baab, baba, bbaa,
bbbb,...}, its regular expression can be
written as (aa+bb+(ab+ba)(aa+bb)*(ab+ba))*
Note
It is important to be clear about the difference of the following
regular expressions
r1 = a*+b*
r2 = (a+b)*
Here r1 does not generate any string of concatenation of a and b,
while r2 generates such strings.
Equivalent Regular Expressions
Definition
Two regular expressions are said to be equivalent if they generate the
same language.
Example
Consider the following regular expressions
r1 = (a + b)* (aa + bb)
r2 = (a + b)*aa + ( a + b)*bb then both regular expressions define the
language of strings ending in aa or bb.
Note
If r1 = (aa + bb) and r2 = ( a + b) then
r1+r2 = (aa + bb) + (a + b)
r1r2 = (aa + bb) (a + b)
= (aaa + aab + bba + bbb)
(r1)* = (aa + bb)*

Question No.3
FINITE AUTOMATA
Give Finite Automata for each of the following language defined over
alphabet Σ = {a, b}
a. Language having all strings with alternating a's and b's , some
example strings
are ababab... or bababa...
b. Language having all strings NOT containing aa at any place
c. Language of all strings NOT STARTING with bb
d. Language of all strings STARTING WITH bba
e. Language having all strings NOT having even no of a's and b's
Solution:

CS403 Solved Assignment Spring 2011 Assignment 1

CS 403 (Database Management System)

Assignment # 01

Q#01:
State the significance of DFDs in the process of Database designing. (5)
Solution:
DBMS Development:

There are some tools used to describe a design process. These also
help the designers of database system to mutually agree on a specific
design.
If there is no standard tool for designing a specific DBMS then every
designer will have his own design for developing that specific design.
This can create misunderstanding or conflict among the designs for
that specific DBMS.
So, DFD is that standard tool that can be used to resolve this
conflict among different designs and to get utmost the same result by
different designers.

Significance of DFDs in the process of Database development:

DFD are used to design data centric applications. They follow the data
through entry, storage, retrieval transformations and output. Storage
and retrieval are where the database comes in. You can use the DFD to
design the schema for the most efficient use of the data in the
context of the application. Of course this is simplified but this is
the general idea.
• DFDs are most important and common tools for designing a database system.
• It presents a system diagrammatically and expresses details of the
different DFD levels.
• It shows the flow of data among various processes / levels.
• These are very simple. DFDs hide the complexities of the system.
• These are easy to understand and create links among different levels
and process of the information flow.
DFDs represent the functions a system performs hierarchically,
starting with the highest-level functions and moving through various
layers or levels of sub-functions. As a modeling technique, DFDs are
useful for performing a structured analysis of software problems,
allowing developers to spot and pinpoint issues in software
development.
It is, therefore, important and common because of its simple graphical
representation. It has various symbols that are simple and easy to
understand. These symbols make the DFD more elegant. They're often
used to model software and to show the flow of data in a system.

Following are some common components of DFDs:
There are some components of DFDs which enhances the importance of it
in a system:
Dataflow symbol_ its purpose in DFDs is to express the flow of
information form one entity to another.
Data Store_ data store is for the storage of data. It is for the
permanent store of data in the system. It is represented by a
rectangle shape open on right width and left one is drawn with double
line.
Processes_ processes are expressed with ovals or rounded rectangles.
When data is transformed from one form into a processed form.
DFD Process_ these are numbered for showing the existence of DFDs at a
certain level in the system.
External Entities_ these are interacting with the system in two
different ways. They may be either receiving data from the system or
may be producing data for the system.
Collector_ it is used to show the convergence of data to a single point.
Separator_ this symbol is used for separating data from a single
source to multiple sinks.
Ring Sum Operator_ this operator is used when data from a single
source process flows to one of the required sinks.
AND Operator_ this operator is used when data from a single source
process must flow to all of the required sinks.

Q#02:

Draw a Context and level 0 Data Flow Diagrams of mobile SIM
registration System. Guidelines are given below. (3+7)

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cs401 assignment no 1 spring 2011 idea solution

*
On Sun, Apr 17, 2011 at 11:18 AM, afaaq <afaaqtariq233@gmail.com> wrote:

> *
>
> cs401 assignment no 1 spring 2011 idea solution
>
>
>
> Suppose your computer has a processor with 24-bit address lines. What is
> maximum amount of memory that can be attached in your system? (Show the
> step(s) for calculation of maximum addressable memory) (2.5 marks)
>
>
>
> b)
>
> How many address bits are required for accessing 1GB RAM? (Show the step(s)
> for calculation of required address bits) (2.5 marks)
>
>
>
> Question No. 2:
>
> What are the contents of memory locations 0151, 0152, 0153, ………….,0158 if
> 0151 is starting address for Label1. (1 mark for each location)
>
>
>
> Label1: dw 8494
>
> db 42
>
> dw 54
>
> dw 7500
>
> db 01
>
>
>
>
>
> Question No. 3:
>
>
>
> a)
>
> Calculate physical address using the following segment offset pairs. (1
> mark each)
>
>
>
>
>
> 00EA:02A4
>
> 0100:AA23
>
> D3B8:F222
>
> 00A0:1234
>
> 8FEf:0FFF
>
>
>
> b)
>
> What is effective address generated by the following instructions? Every
> instruction is independent of others. Initially bx = 0x0101, bp=0x0222,
> si=0x1234, var1=0x1771
>
> (1 mark each)
>
>
>
> mov ax, [bx+si]
>
> mov ax, [bx+100] (100 is in decimal)
>
> mov ax, [bp+si]
>
> mov ax, [var1+bp]
>
> mov ax, [si+var1]
> *
> *
>
> Remember Me In Your Prayers
> Best regard's
> Muhammad Afaaq
> MBA 4th (Final Semester) Finance Group
> Afaaq_Tariq@yahoo.com
> Islamabad
> For latest assignments solved quizzes files gdb solve n unsolved past
> papers Come join us in http://vugoogle.com
>
> http://www.alliswell.com.pk/
> http://groups.google.com/group/vustudymania
>
> 0346-5329264
>
> If u like me than raise your hand with me
> If not then raise ur standard
> That's about me … !
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>
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