CUPLEX2B

Write a Client-Server socket program to implement “TCP client server” architecture.

2022-02-25T12:17:00.009+05:30

Write a Client-Server socket program to implement “TCP client server” architecture.

CODE for server side in c++:


#include < bits/stdc++.h >
#include < sys/socket.h >
#include < netinet/in.h >
using namespace std;
#define PORT 8000

int main()
{
    int server_fd;
    int nRet=0;
    struct sockaddr_in address;
    int opt = 1;
    int addrlen = sizeof(address);
    char buffer[1024] = {0};
    // Socket creation
    server_fd=socket(AF_INET, SOCK_STREAM,0);
    if (server_fd == 0)
    {
        cout << "Socket creation failed" << endl;
        exit(EXIT_FAILURE);
    }
    /*Initialize environment for sockaddr structure*/
    address.sin_family = AF_INET;
    address.sin_addr.s_addr = INADDR_ANY;
    address.sin_port = htons( PORT );
    // Forcefully attaching socket to the port 8080
    nRet=setsockopt(server_fd, SOL_SOCKET, SO_REUSEADDR | SO_REUSEPORT,&opt, sizeof(opt));
    if (nRet)
    {
        cout << "setsockopt Failed" << endl;
        exit(EXIT_FAILURE);
    }
    /*Bind the socket to local port*/
    nRet=bind(server_fd,(struct sockaddr *)&address,sizeof(address));
    if (nRet < 0)
    {
        cout << "Failed to bind"  << endl;
        exit(EXIT_FAILURE);
    }
    //listen request from client
    nRet=listen(server_fd,5);
    if (nRet < 0)
    {
        cout << "Failed to listen" << endl;
        exit(EXIT_FAILURE);
    }
    //accept and process request
    int new_socket=accept(server_fd,(struct sockaddr *)&address,(socklen_t*)&addrlen);
    if (new_socket < 0)
    {
        cout << "Failed to accept request" << endl;
        exit(EXIT_FAILURE);
    }
    //int nread = read( new_socket , buffer, 1024);
    send(new_socket ,"Connection accepted !! Hello from server" , 41 , 0 );
    cout << "Connection successfull" << endl;
    return 0;
}

CODE for client side in c++:


#include <bits/stdc++.h >
#include <sys/socket.h >
#include < arpa/inet.h >
#include < unistd.h >
#define PORT 8080

int main()

{
    int sock = 0, valread;
    struct sockaddr_in serv_addr;
    char buffer[1024] = {0};
    if ((sock = socket(AF_INET, SOCK_STREAM, 0)) < 0)
    {
        printf("\n Socket creation error \n");
        return -1;
    }
    serv_addr.sin_family = AF_INET;
    serv_addr.sin_port = htons(PORT);
    // Convert IPv4 and IPv6 addresses from text to binary form
    if(inet_pton(AF_INET, "127.0.0.1", &serv_addr.sin_addr) < =0) 
    {
        printf("\nInvalid address/ Address not supported \n");
        return -1;
    }
    if (connect(sock, (struct sockaddr *)&serv_addr, sizeof(serv_addr)) < 0)
    {
        printf("\nConnection Failed \n");
        return -1;
    }
    valread = read( sock , buffer, 1024);
    printf("%s\n",buffer );
    return 0;

}

Use the following customer table and write the SQL query to:

2022-01-01T18:28:00.005+05:30

Please feel free to suggest any edits....

Use the following customer table and write the SQL query to:

i) Find any values that start with "a".

ii) Find any values that end with "a".

iii) Find any values that have "or" in any position.

iv) Find any values that have "r" in the second position.

v) Find any values that start with "a" and are at least 2 characters in Length.

vi) Find any values that start with "a" and end with "e".


create table customers(
  customerId int,
  customerName varchar(255),
  contactName varchar(255),
  address varchar(255),
  city varchar(255),
  postalCode varchar(255),
  country varchar(255)
  );
insert into customers values (1,"alfreds futterkiste","maria anders","obere str. 57","berlin","12209","germany"),
  (2,"ana trujillo emparedadosy helados","ana trujillo","avda. de la constitucion 2222","mexico d.f.","5021","mexico"),
  (3,"antonio moreno taqueria","antonio moreno","mataderos 2312","mexico d.f.","5023","mexico"),
  (4,"around the horn","thomas hardy","120 hanover sq.","london","wa11dp","uk"),
  (5,"berglunds snabbkop","christina berglund","berguvsvagen 8","lulea","s-95822","sweden"),
  (6,"blauer see delikatessen","hanna moos","forsterstr. 57","mannheim","68306","germany"),
  (7,"blondel pere et fils","frederique citeaux","24, place kleber","strasbourg","67000","france"),
  (8,"bolido comidas preparadas","martin sommer","c/araquil,67","madrid","28023","spain"),
  (9,"bon app'","laurence lebihans","12,rue des bouchers","marseille","13008","france"),
  (10,"bottom-dollar marketse","elizabeth lincoln","23 tsawassen blyd","tsawssen","t2f8m4","canada"),
  (11,"b's beverages","victoria ashworth","fauntleroy circus","london","ec25nt","uk"),
  (12,"cactus comidas parallevar","patricio simpson","cerrito 333","buenos aires","1010","argentina"),
  (13, "centro comercial moctezuma","francisco chang","sierras de granada 9993","mexico d.f.","5022","mexico"),
  (14,"chop-suey chinese","yang wang","haupstr. 29","bern","3012","switzerland"),
  (15,"comercio mineiro","pedro afonso","av. dos lusiadas,23","sao paulo","05432043","brazil"),
  (16,"consolidated holdings","elizabeth brown","berkeley gardens 12 brewery","london","wx16lt","uk"),
  (17,"drachenblut delikatessend","sven ottlieb","waiserweg 21","aachen","52066","germany"),
  (18,"du monde entier","janine labrune","67, rue des cinquante otages","nantes","44000","france"),
  (19,"eastern connection","ann devon","35 king george","london","wx36fw","uk"),
  (20,"ernst handel","roland mendel","kirchgasse 6","graz","8010","austria");

i) Find any values that start with "a".


select * from customers where customerName like 'a%';

ii) Find any values that end with "a".


select * from customers where customerName like '%a';

iii) Find any values that have "or" in any position.


select * from customers where customerName like '%or%';

iv) Find any values that have "r" in the second position.


select * from customers where customerName like '_r%';

v) Find any values that start with "a" and are at least 2 characters in Length.


select * from customers where customerName like 'a_%';

vi) Find any values that start with "a" and end with "e".


select * from customers where customerName like 'a%e';

Use the following Product table and write the SQL query to find the count , sum and average price.

2022-01-01T18:19:00.003+05:30

Please feel free to suggest any edits....

Use the following Product table and write the SQL query to find the count , sum and average price.


create table product(
  productId int,
  productName varchar(255),
  supplierId int,
  categoryId int,
  unit varchar(255),
  price double
  );
insert into product values
  (1,"chais",1,1,"10boxes*20bags",18),
  (2,"chang",1,1,"24-12 oz bottles",19),
  (3,"aniseed syrap",1,2,"12-550ml bottles",10),
  (4,"chef anton's cajon seasoning",2,2,"48-6 oz jars",22),
  (5,"chef anton's gumbo mix",2,2,"36 boxes",21.35);
select COUNT(price), SUM(price), AVG(price) from product;

Use the following customer table and write the SQL query to select the first three records from the "Customers" table.

2022-01-01T18:12:00.007+05:30

Use the following customer table and write the SQL query to select the first three records from the "Customers" table.


create table customers(
customerId int,
customerName varchar(255),
contactName varchar(255),
address varchar(255),
city varchar(255),
postalCode varchar(255),
country varchar(255));

insert into customers values (1,"alfreds futterkiste","maria anders","obere str. 57","berlin","12209","germany"),
(2,"ana trujillo emparedadosy helados","ana trujillo","avda. de la constitucion 2222","mexico d.f.","5021","mexico"),
(3,"antonio moreno taqueria","antonio moreno","mataderos 2312","mexico d.f.","5023","mexico"),
(4,"around the horn","thomas hardy","120 hanover sq.","london","wa11dp","uk"),
(5,"berglunds snabbkop","christina berglund","berguvsvagen 8","lulea","s-95822","sweden"),
(6,"blauer see delikatessen","hanna moos","forsterstr. 57","mannheim","68306","germany"),
(7,"blondel pere et fils","frederique citeaux","24, place kleber","strasbourg","67000","france"),
(8,"bolido comidas preparadas","martin sommer","c/araquil,67","madrid","28023","spain"),
(9,"bon app'","laurence lebihans","12,rue des bouchers","marseille","13008","france"),
(10,"bottom-dollar marketse","elizabeth lincoln","23 tsawassen blyd","tsawssen","t2f8m4","canada"),
(11,"b's beverages","victoria ashworth","fauntleroy circus","london","ec25nt","uk"),
(12,"cactus comidas parallevar","patricio simpson","cerrito 333","buenos aires","1010","argentina"),
(13, "centro comercial moctezuma","francisco chang","sierras de granada 9993","mexico d.f.","5022","mexico"),
(14,"chop-suey chinese","yang wang","haupstr. 29","bern","3012","switzerland"),
(15,"comercio mineiro","pedro afonso","av. dos lusiadas,23","sao paulo","05432043","brazil"),
(16,"consolidated holdings","elizabeth brown","berkeley gardens 12 brewery","london","wx16lt","uk"),
(17,"drachenblut delikatessend","sven ottlieb","waiserweg 21","aachen","52066","germany"),
(18,"du monde entier","janine labrune","67, rue des cinquante otages","nantes","44000","france"),
(19,"eastern connection","ann devon","35 king george","london","wx36fw","uk"),
(20,"ernst handel","roland mendel","kirchgasse 6","graz","8010","austria");

select* from customers order by customerId asc limit 3;

Use the following customer table and write the SQL query to select all fields from "Customers" where the country is NOT "Germany" and NOT "USA".

2022-01-01T18:04:00.004+05:30

Use the following customer table and write the SQL query to select all fields from "Customers" where the country is NOT "Germany" and NOT "USA".


create table customers(
  customerId int,
  customerName varchar(255),
  contactName varchar(255),
  address varchar(255),
  city varchar(255),
  postalCode varchar(255),
  country varchar(255)
  );
insert into customers values (1,"alfreds futterkiste","maria anders","obere str. 57","berlin","12209","germany"),
  (2,"ana trujillo emparedadosy helados","ana trujillo","avda. de la constitucion 2222","mexico d.f.","5021","mexico"),
  (3,"antonio moreno taqueria","antonio moreno","mataderos 2312","mexico d.f.","5023","mexico"),
  (4,"around the horn","thomas hardy","120 hanover sq.","london","wa11dp","uk"),
  (5,"berglunds snabbkop","christina berglund","berguvsvagen 8","lulea","s-95822","sweden"),
  (6,"blauer see delikatessen","hanna moos","forsterstr. 57","mannheim","68306","germany"),
  (7,"blondel pere et fils","frederique citeaux","24, place kleber","strasbourg","67000","france"),
  (8,"bolido comidas preparadas","martin sommer","c/araquil,67","madrid","28023","spain"),
  (9,"bon app'","laurence lebihans","12,rue des bouchers","marseille","13008","france"),
  (10,"bottom-dollar marketse","elizabeth lincoln","23 tsawassen blyd","tsawssen","t2f8m4","canada"),
  (11,"b's beverages","victoria ashworth","fauntleroy circus","london","ec25nt","uk"),
  (12,"cactus comidas parallevar","patricio simpson","cerrito 333","buenos aires","1010","argentina"),
  (13, "centro comercial moctezuma","francisco chang","sierras de granada 9993","mexico d.f.","5022","mexico"),
  (14,"chop-suey chinese","yang wang","haupstr. 29","bern","3012","switzerland"),
  (15,"comercio mineiro","pedro afonso","av. dos lusiadas,23","sao paulo","05432043","brazil"),
  (16,"consolidated holdings","elizabeth brown","berkeley gardens 12 brewery","london","wx16lt","uk"),
  (17,"drachenblut delikatessend","sven ottlieb","waiserweg 21","aachen","52066","germany"),
  (18,"du monde entier","janine labrune","67, rue des cinquante otages","nantes","44000","france"),
  (19,"eastern connection","ann devon","35 king george","london","wx36fw","uk"),
  (20,"ernst handel","roland mendel","kirchgasse 6","graz","8010","austria");

select * from customers where not country="germany" and not country="usa";

Use the following customer table and write the SQL query to select all fields from "Customers" where the country is "Germany" AND the city must be "Berlin" OR "München".

2022-01-01T17:59:00.007+05:30

1. Use the following customer table and write the SQL query to select all fields from "Customers" where the country is "Germany" AND the city must be "Berlin" OR "München".


  customerId int,
  customerName varchar(255),
  contactName varchar(255),
  address varchar(255),
  city varchar(255),
  postalCode varchar(255),
  country varchar(255)
  );
insert into customers values (1,"alfreds futterkiste","maria anders","obere str. 57","berlin","12209","germany"),
  (2,"ana trujillo emparedadosy helados","ana trujillo","avda. de la constitucion 2222","mexico d.f.","5021","mexico"),
  (3,"antonio moreno taqueria","antonio moreno","mataderos 2312","mexico d.f.","5023","mexico"),
  (4,"around the horn","thomas hardy","120 hanover sq.","london","wa11dp","uk"),
  (5,"berglunds snabbkop","christina berglund","berguvsvagen 8","lulea","s-95822","sweden"),
  (6,"blauer see delikatessen","hanna moos","forsterstr. 57","mannheim","68306","germany"),
  (7,"blondel pere et fils","frederique citeaux","24, place kleber","strasbourg","67000","france"),
  (8,"bolido comidas preparadas","martin sommer","c/araquil,67","madrid","28023","spain"),
  (9,"bon app'","laurence lebihans","12,rue des bouchers","marseille","13008","france"),
  (10,"bottom-dollar marketse","elizabeth lincoln","23 tsawassen blyd","tsawssen","t2f8m4","canada"),
  (11,"b's beverages","victoria ashworth","fauntleroy circus","london","ec25nt","uk"),
  (12,"cactus comidas parallevar","patricio simpson","cerrito 333","buenos aires","1010","argentina"),
  (13, "centro comercial moctezuma","francisco chang","sierras de granada 9993","mexico d.f.","5022","mexico"),
  (14,"chop-suey chinese","yang wang","haupstr. 29","bern","3012","switzerland"),
  (15,"comercio mineiro","pedro afonso","av. dos lusiadas,23","sao paulo","05432043","brazil"),
  (16,"consolidated holdings","elizabeth brown","berkeley gardens 12 brewery","london","wx16lt","uk"),
  (17,"drachenblut delikatessend","sven ottlieb","waiserweg 21","aachen","52066","germany"),
  (18,"du monde entier","janine labrune","67, rue des cinquante otages","nantes","44000","france"),
  (19,"eastern connection","ann devon","35 king george","london","wx36fw","uk"),
  (20,"ernst handel","roland mendel","kirchgasse 6","graz","8010","austria");

select* from customers where country="germany" and (city="berlin" or city="munchen");

Write an “Echo Client” and “Echo Server” using UDP to estimate the round trip Time from client to the server. The server should be such that it can accept Multiple connections at any given time

2021-11-25T20:14:00.003+05:30

Write an “Echo Client” and “Echo Server” using UDP in Cto estimate the round trip Time from client to the server. The server should be such that it can accept Multiple connections at any given time.

CODE FOR SERVER:


#include < stdio.h >
#include < sys/types.h >
#include < sys/socket.h >
#include < netinet/in.h >
#include < unistd.h >
#include < netdb.h >
#include < string.h >
#include < arpa/inet.h >
#define MAXLINE 1024

int main()

{
  int sockfd;
  int n;
  socklen_t len;
  char msg[1024];
  struct sockaddr_in servaddr,cliaddr;
  sockfd=socket(AF_INET,SOCK_DGRAM,0);
  bzero(&servaddr,sizeof(servaddr));
  servaddr.sin_family=AF_INET;
  servaddr.sin_addr.s_addr=INADDR_ANY;
  servaddr.sin_port=htons(5035);
  bind(sockfd,(struct sockaddr*)&servaddr,sizeof(servaddr));
  printf("\n\n Successfully Binded");
  printf("\n\n Listening...");
  for(;;)

  {
       printf("\n ");
       len=sizeof(cliaddr);
       n=recvfrom(sockfd,msg,MAXLINE,0,(struct sockaddr*)&cliaddr,&len);
       printf("\n Client's Message : %s\n",msg);
       if(n < 6)
          perror("send error");
       sendto(sockfd,msg,n,0,(struct sockaddr*)&cliaddr,len);
  }
return 0;

}

CODE FOR CLIENT


#include < stdio.h >
#include < stdlib.h >
#include < sys/types.h >
#include < sys/socket.h >
#include < netinet/in.h >
#include < string.h >
#include < arpa/inet.h >
#include < time.h >
#define MAXLINE 1024
int main()
{
  int sockfd;
  int n;
  socklen_t len;
  char sendline[1024],recvline[1024];
  struct sockaddr_in servaddr;
  memset(sendline,' ',sizeof(sendline));
  printf("\n Enter the message : ");
  gets(sendline);
  sockfd=socket(AF_INET,SOCK_DGRAM,0);
  bzero(&servaddr,sizeof(servaddr));
  servaddr.sin_family=AF_INET;
  servaddr.sin_addr.s_addr=inet_addr("127.0.0.1");
  servaddr.sin_port=htons(5035);
  len=sizeof(servaddr);
  clock_t t;
  t=clock();
  sendto(sockfd,sendline,MAXLINE,0,(struct sockaddr*)&servaddr,len);
  n=recvfrom(sockfd,recvline,MAXLINE,0,NULL,NULL);
  t=clock()-t;
  double time_taken = ((double)t)/CLOCKS_PER_SEC; 
  recvline[n]=0;
  printf("\n Server's Echo : %s\n",recvline);
  printf("\n Round trip time=%f sec\n\n",time_taken);
return 0;

}

Consider the E-R diagram in the following Figure , which models an online bookstore.a. List the entity sets and their primary keys.b. Suppose the bookstore adds music cassettes and compact disks to its collection. The same music item may be present in cassette or compact disk format, with differing prices. Extend the E-R diagram to model this addition, ignoring the effect on shopping baskets.c. Now extend the E-R diagram, using generalization, to model the case where a shopping basket may contain any combination of books, music cassettes, or compact disks.

2021-11-25T11:30:00.000+05:30

Consider the E-R diagram in the following Figure , which models an online bookstore.

a. List the entity sets and their primary keys.

Solution:

The entity sets are:

i) Author ( name, address, URL)

ii) Book(year, title, year, ISBN)

iii) Warehouse(code, phone, address)

iv) Publisher(name, address, URL, phone)

v) Shopping Basket(basket ID)

vi) Customer(name, address, phone, email)

b. Suppose the bookstore adds music cassettes and compact disks to its collection. The same music item may be present in cassette or compact disk format, with differing prices. Extend the E-R diagram to model this addition, ignoring the effect on shopping baskets.

Solution:

c. Now extend the E-R diagram, using generalization, to model the case where a shopping basket may contain any combination of books, music cassettes, or compact disks.

Solution:

Data communication and networking by Forouzan chapter 19 solution

2021-11-13T20:39:00.003+05:30

11. What is the address space in each of the following systems?

a. A system with 8-bit addresses

=> 2^8

b. A system with 16-bit addresses

=>2^16

c. A system with 64-bit addresses

=>2^64

12. An address space has a total of 1024 addresses. How many bits are needed to represent an address.

solution:

Let n bits are needed to represent an address.

so, 2^n = 1024

=>n= 9

13. An address space uses the three symbols 0, 1, and 2 to represent addresses. If each address is made of 10 symbols, how many addresses are available in this system?

Solution:

Since, no of symbols used= 3

no of symbols a address consists of =10

So, addresses available= 3^10

14. Change the following IP addresses from dotted-decimal notation to binary notation.

a. 114.34.2.8

=> 01011110 00100010 00000010 00001000

b. 129.14.6.8

=> 10000001 00001110 00000110 00001000

c. 208.34.54.12

=> 11010000 00100010 00110110 00001100

d. 238.34.2.1

=> 11101110 00100010 00000010 00000001

15. Change the following IP addresses from binary notation to dotted-decimal notation.

a. 01111111 11110000 01100111 01111101

=> 127.240.103.125

b. 10101111 11000000 11111000 00011101

=>175.192.248.29

c. 11011111 10110000 00011111 01011101

=>223.176.31.129

d. 11101111 11110111 11000111 00011101

=> 239.247.199.29

16. Find the class of the following IP addresses.

a. 208.34.54.12

=> class C

b. 238.34.2.1

=> class D

c. 114.34.2.8

=> class A

d. 129.14.6.8

=> class B

17. Find the class of the following IP addresses.

a. 11110111 11110011 10000111 11011101

=> class E

b. 10101111 11000000 11110000 00011101

=> class B

c. 11011111 10110000 00011111 01011101

=> class C

d. 11101111 11110111 11000111 00011101

=> class D

18. Find the netid and the hostid of the following IP addresses.

a. 114.34.2.8

=> Netid=114

Hostid=34.2.8

b. 132.56.8.6

=> Netid=132.56

Hostid=8.6

c. 208.34.54.12

=> Netid=208.34.54

Hostid=12

19. In a block of addresses, we know the IP address of one host is 25.34.12.56/16. What are the first address (network address) and the last address (limited broadcast address) in this block?

Solution:

The IP address can be written as: 00011001 00100110 00001100 00111000

To obtain the first address, we will set the rightmost 32-16 bits to 0

The first address will be-

00011001 00100110 00000000 00000000

i.e. 25.34.0.0

To obtain the last address, we will set the rightmost 16 bits to 1.

So, the last address will be-

00011001 00100110 11111111 11111111

i.e. 25.34.255.255

20. In a block of addresses, we know the IP address of one host is 182.44.82.16/26. What are the first address (network address) and the last address in this block?

Solution:

The IP address can be written as-

10110110 00101100 01010010 00010000

The first address can be obtained by replacing rightmost 32-26=6 bits with 0

So, the first address will be

10110110 00101100 01010010 00000000

i.e. 182.44.82.0

The last address can be obtained by replacing last rightmost 6 bits by 1

So, the last address will be

10110110 00101100 01010010 00111111

i.e. 182.44.82.63

21. An organization is granted the block 16.0.0.0/8. The administrator wants to create 500 fixed-length subnets.

a. Find the subnet mask.

b. Find the number of addresses in each subnet.

c. Find the first and last addresses in subnet 1.

d. Find the first and last addresses in subnet 500

Solution:

Total number of addresses in the block will be:

2^(32-8)

=2^24

=16,777,216

So, each subnet will have (16,777,216)%500=33,554.432

The nearest power of 2 is 32,768

Let, mask of each subnet be n, then

2^(32-n)=32,768

=>2^(32-n)=2^15

=>32-n=15

=>n=17

So, subnet mask is 17

The number of address in each subnet

= 2^(32-subnet mask)

=2^(32-17)

=2^15

=32,768

The first address in first subnet is 16.0.0.0

which can be written

00010000 00000000 00000000 00000000

To obtain the last address of first subnet, we will replace the rightmost (32-17)=15 bits with 1

So, the last address will be

00010000 00000000 01111111 11111111

i.e. 16.0.127.255

What is Peterson's solution?

2021-09-27T11:51:00.000+05:30

As discussed in the post, What is Process Synchronization?, we came to know what is process synchronization and what is critical section problem.

Peterson's solution is a software based solution to the critical section problem.

{

flag[i]=true;

turn=j;

while (flag[j] && turn==j);

CRITICAL SECTION

flag[i]=false;

REMAINDER SECTION

} while(true);

The above code is the basic structure of a process in Peterson's solution. The variable turn indicates whose turn it is to enter the critical section and flag[i] is used to indicate if a process is ready to enter it's critical section.

In the entry section before entering critical section, process Pi sets the flag[i] value to true indicating that Pi is ready to enter it's critical section and then set the turn value to j asserting if any other process wants to enter the critical section they can do so to ensure progress. Then it execute the while loop until flag[j] is true and turn is j i.e. other process is using the critical section. After that if any of the above two conditions in the while loop becomes false, it can enter the critical section. In the exit section, it sets flag[i] to false indicating that it is leaving critical section.

If both processes try to enter the critical section at the same time, turn value will be set to i and j apparently at the same time and the first turn value will be overwritten immediately, the one which will make change in the turn value first will get to enter the critical section.

Peterson's solution preserves all the conditions mutual exclusion, progress and bounded waiting.

When both the processes are ready to enter critical section i.e. flag[i]==flag[j]==true, both cannot enter the critical section at the same time because only one while loop will get executed as the value of turn can be either i or j not both at the same time. Hence ensures mutual exclusion.

Process Pi can be prevented from entering critical section if it is stuck in the while loop i.e. when turn==j and flag[j]==true. But if Pj is not ready to enter the critical section, then flag[j]=false and then Pi can enter the critical section. If flag[j]=false, then turn value must be i and in the while loop, Pi cannot change the turn value, that's why Pi will get to enter the critical section after at most one entry of Pj ensuring both progress and bounded waiting.

IMPORTANT POINTS:

Peterson's solution is restricted to two processes that alternate executions between their critical section and remainder section.
The shared data by the two processes P0 and P1 in Peterson's solution are:

int turn;

boolean flag[2];

flag[i] indicates that process Pi is ready to enter the critical section and turn =j makes sure that if the other process is also ready to enter the critical section, then it can enter before i ensuring mutual exclusion.

What is Process Synchronization?

2021-09-23T14:32:00.000+05:30

WHY DO WE REQUIRE PROCESS SYNCHRONIZATION?

In Multiprogramming, there are two kinds of processes. One is independent processes and the another one is cooperative processes. Independent processes are independent of each other, but cooperative processes can affect or affected by other processes executing in the system. They share logical address space or data through files or messages. Concurrent access to the shared data by the cooperating processes can result into data inconsistency.

For example, we can take race condition when several processes access and manipulate the same data concurrently and the outcome varies with the particular order in which access takes place.

Consider the following processes.

initially, count=5

P1 P2

count++ count--;

The increment or decrement of count inside the processor happens as follows:

P1 P 2

R1<=count R2<=count

R1<=count+1 R2<=count-1

count<=R1 count<=R2

Consider the following sequence of the steps,

R1<=count R1=5

R1<=R1+1 R1=6

R2<=count R2=5

R2<=R2-1 R2=4

count<=R2 count=4

count<=R1 count=6

The final output will be either 4 or 6 depending on which process will execute it's final step last. But, You can see that the actual answer should be 5. Thus, both the processes are racing with each other to manipulate the final result. This is known as race condition.

To avoid these kind of situations, we require Process Synchronization.

THE CRITICAL SECTION PROBLEM:

The critical section is the part of code of the cooperative processes which involves changing common variables or accessing shared resources.

If one process is executing in its critical section, no other process is allowed to enter into critical section. There are some protocols designed regarding critical section which every process must follow. Each process must request permission to enter it's critical section. The section of code implementing this request is called entry section. After passing through entry section, a process enters the critical section and then followed by a exit section. The remaining code of the process is known as remainder section.

There are several solutions are designed for this critical solution problem. But every solution must satisfy the following requirements.

1) Mutual exclusion: If one process is executing in it's critical section, then no other process can enter it's critical section.

2) Progress: If no process is executing in it's critical section, and some processes want to enter in their critical section, then only those who are not in their remainder section can participate to decide who will enter the critical section next.

3) Bounded Waiting: There will be a limit on the umber of times that other processes are allowed to enter their critical section when a process made a request to enter it's critical section and before that request is granted.

SOLUTIONS OF CRITICAL SECTION PROBLEM:

1) Peterson's solution.

2) Mutex Locks.

3) Test and set

2) compare and swap.

4) Semaphores.

How to create a .sh file in linux?

2021-09-22T19:48:00.002+05:30

STEPS TO CREATE A .SH FILE IN LINUX:

Sorry for the poor quality of the images, we will update it very soon :)

STEP 1: Open a folder using cd command : cd ~/Desktop/

Use vi command to create a .sh file : vi tutorial.sh

STEP 2:

The screen will look like this:

Type i

STEP 3:

Now, You can write whatever You want to write inside .sh file

STEP 4:

Press Esc key

STEP 5:

Enter :wq

After entering, it will look like this

STEP 6:

Now make the .sh file executable : chmod +x tutorial.sh

Execute the file : ./tutorial.sh

What are the different types of Network Topology in computer networking?

2021-09-06T21:23:00.003+05:30

Network Topology in computer networking explains how different devices or nodes are connected physically to form a network and the logical flow of information in the network. It is the geometric representation of the relationship of all the links and nodes to one another.

MESH TOPOLOGY:

In Mesh Topology, every device has point-to-point unique connections(links) with every other device.

If there are n nodes, then one node will have (n-1) links.
n nodes should have n*(n-1) total number of links. But since one link connects two devices, so the number of links for n nodes will be n*(n-1)/2.

ADVANTAGES OF MESH TOPOLOGY:

Each connection can carry its own data, so there is no traffic problem that occurs when links are shared by multiple devices.
Failure to one link doesn't affect others as all the connections are independent.
Every message is transferred through dedicated lines, only the intended recipient can see the messages, so there is no issue of privacy or security.
Fault isolation and fault identification become much easier.

DISADVANTAGES OF MESH TOPOLOGY:

Installation and reconfiguration are difficult as every device is connected to every other device through unique connections.
Large number of I/O ports and cabling is required.
Takes lots of physical space.
Hardware required to connect each link can be expensive.

STAR TOPOLOGY

In star topology, there is a central controller called a hub to which every other device is connected point to point. There is no direct link between the devices. One device can send a message to another via the hub.

If there are n nodes or devices, then there will be n links

ADVANTAGES OF STAR TOPOLOGY:

Easy to install and reconfigure as compared to mesh topology.
Less expensive than mesh topology as each device or node requires only one link and one I/O port to connect to other devices.
If one link fails, then no other link is affected.

DISADVANTAGES OF STAR TOPOLOGY:

If the hub fails, then the whole system goes dead as all other devices are dependent on the hub.
Although the number of cables is lesser than mesh topology, the number is still high than some other topologies.

BUS TOPOLOGY:

In bus topology, there is one long cable to which all the devices are connected. The cable acts as a backbone of the network. Nodes are connected to the bus cable by droplines and tap. Droplines are the links that connect the main bus cable and each device.

There will always be one bus cable irrespective of the number of nodes or devices.

ADVANTAGES OF BUS TOPOLOGY:

Easy installation.
Lesser number of cables is required than star and mesh topology.

DISADVANTAGES OF BUS TOPOLOGY:

Signals become weaker to the farthest point of the bus cable from the starting point as some energy is transformed into heat.
Difficult reconnection and fault isolation.
A fault or break in the bus cable disables the entire system by disabling all transmissions.

RING TOPOLOGY:

In ring topology, the devices are connected as a ring. Each device has point-to-point connection to only the two devices on either side of it. The signal passes through one device to another in one direction to the destination.

ADVANTAGES OF RING TOPOLOGY:

Easy to install and reconfigure as each device is connected to only its immediate neighbors.
Easy fault isolation.

DISADVANTAGES OF RING TOPOLOGY:

Break in one ring can disable the entire network.

HYBRID TOPOLOGY:

Hybrid topology is obtained from the combination of the above-mentioned topologies. Hybrid topologies have more advantages than the above topologies.

What is the output of the following code- fork() system call

2021-06-16T19:02:00.000+05:30

What is the output of the following code- fork() system call

#include <stdio.h>

#include<unistd.h>

int main()

{

if(fork() && fork())

fork();

printf("Hi ");

return 0;

}

OUTPUT: Hi Hi Hi Hi

EXPLANATION:

When the first fork() inside the conditional statement will execute, then two processes will be cloned, one is child process C1 with logic 0 and another is the parent process P with logic +1. Both the processes will run parallelly.

For C1, the logic of C1 is 0 , So, the logic of the conditional statement will be zero and the second fork() and third fork() will not be executed. The program will simply output Hi.

For P, the logic of P is 1, so the second fork() call will be executed and it will again create two processes, one is child process C2 and another is parent process P.

C2 will simply print Hi as C1.

In P, the logic of the conditional statement will be 1, so the third fork() will be called and it will again create one child process C3 and parent process P. Both of them will return Hi.

So, the output of the program will be

Hi Hi Hi Hi

C1 C2 C3 P

Write a program to create a class shape with functions to find the area of the shapes. Create derived classes circle, rectangle and trapezoid each having overridden functions area and display. Base class:

2021-06-07T13:01:00.004+05:30

Write a program to create a class shape with functions to find the area of the shapes. Create

derived classes circle, rectangle and trapezoid each having overridden functions area and display.

Base class:

● Member Variables: (protected) length, breadth

● Member Function: (public) virtual function area

CODE:


import java.util.*;

public abstract class Shape {

	public abstract double area();
	public abstract void display();
}

public class Circle extends Shape{
    public void display() {
        System.out.println(area());
    }
    public Circle(int radius) {
        this.radius = radius;
    }
    protected int radius;
    public double area() {
        return (3.14*radius*radius);
    }
}

public class Rectangle extends Shape {
	   
    public void display() {
        System.out.println(area());        
    }

    public Rectangle(double length, double width) {
        this.length = length;
        this.width = width;
    }
    protected double length;
    protected double width;
    
    public double area() {
        return length*width;
    }

}

public class Trapezoid extends Shape{
    public void display() {
        System.out.println(area());
    }
    public Trapezoid(double lenght, double breadth, double height) {
        this.lenght = lenght;
        this.breadth = breadth;
        this.height = height;
    }
    protected double lenght;
    protected double breadth;
    protected double height;
   
    public double area() {
        return ((lenght+breadth)/2)*height;
    }
}
public class Main2 {
	 public static void main (String[] args) {
	
	        Rectangle rectangle = new Rectangle(4,4);
	        rectangle.display();
	        Circle circle= new Circle(6);
	        circle.display();
	        Trapezoid trapezoid=new Trapezoid(5,7,4);
	        trapezoid.display();
	        System.out.println();
	    }
}

Create a class called 'Matrix' containing a constructor that initializes the number of rows and number of columns of a new Matrix object. The Matrix class has the following information:

2021-06-07T12:56:00.003+05:30

Create a class called 'Matrix' containing a constructor that initializes the number of rows and

number of columns of a new Matrix object. The Matrix class has the following information:

- number of rows of matrix

- number of columns of matrix

- elements of matrix in the form of 2D array

The Matrix class has the following methods:

- get the number of rows

- get the number of columns

- set the elements of the matrix at given position (i,j)

- adding two matrices. If the matrices are not addable, "Matrices cannot be added"

will be displayed.

- multiplying the two matrices

CODE:


import java.io.*;
import java.util.*;
class Matrix {
    int row;
    int col;
    int[][] a;
    public Matrix(int r, int c)
    {
        row = r;
        col = c;
        a = new int[row][col];
    }
    public int getrow()
    {
        return row;
    }
    public int getcol()
    {
        return col;
    }
    public int getelement(int r, int c)
    {
        return a[r][c];
    }
    public void setelement(int r, int c, int value)
    {
        a[r][c] = value;
    }
    public static Matrix add(Matrix a, Matrix b)
    {
        if((a.row != b.row) || (a.col != b.col))
        {
          System.out.println("Matrices can not be added");
          return new Matrix(0,0);
        }
        else
        {
          Matrix m = new Matrix(a.row,a.col);
          for(int i = 0;i < m.row;i++)
          {
            for(int j = 0;j < m.col;j++)
            {
              m.setelement(i,j,(a.getelement(i,j)+b.getelement(i,j)));
            }
          }
          return m;
        }
    }
    public static Matrix mult(Matrix a, Matrix b)
    {
      Matrix m = new Matrix(a.row,b.col);
      for(int i = 0;i < a.row;i++)
      {
        for(int j = 0;j < b.col;j++)
        {
          int s = 0;
          for(int k = 0;k < a.col;k++)
          {
            s = s+(a.getelement(i,k)*b.getelement(k,j));
          }
          m.setelement(i,j,s);
        }
      }
      return m;

    }

    public void printmat()
    {
        for(int i = 0;i < row;i++)
        {
            for(int j = 0;j < col;j++)
            {
              System.out.print(a[i][j]+"	");
            }
            System.out.println("");
         }
    }
}

GAME: Choice of area

2021-05-25T18:57:00.004+05:30

Consider a game where we are given two types of power A and B and three types of area X,Y,Z. Every second you have to switch between these areas, each area has specific properties by which your power A and power B increase or decrease. We need to keep choosing areas in such a way that our survival time is maximized. Survival time ends when any of the powers, A or B reaches less than 0.


#include < bits/stdc++.h >
using namespace std;

int Survival(int A,int B, int x1,int x2,int y1,int y2,int z1,int z2,int last,map < pair < int,int > ,int > mp)
{
          if(A <=0 || B <=0)
                    return 0;
          pair < int,int > current=make_pair(A,B);
          if(mp.find(current)!=mp.end())
                    return mp[current];

          int temp;
          if(last==1)
                    temp=1+max(Survival(A+y1,B+y2,x1,x2,y1,y2,z1,z2,2,mp),Survival(A+z1,B+z2,x1,x2,y1,y2,z1,z2,3,mp));
          else if(last==2)
                    temp=1+max(Survival(A+x1,B+x2,x1,x2,y1,y2,z1,z2,1,mp),Survival(A+z1,B+z2,x1,x2,y1,y2,z1,z2,3,mp));
          else
                    temp=1+max(Survival(A+y1,B+y2,x1,x2,y1,y2,z1,z2,2,mp),Survival(A+x1,B+x2,x1,x2,y1,y2,z1,z2,1,mp));
          mp[current]=temp;
          return temp;
}

int main()

{
       int A,B,x1,x2,y1,y2,z1,z2;
       cin >> A >> B;
       cin >> x1 >> x2;
       cin >> y1 >> y2;
       cin >> z1 >> z2;
       map < pair < int,int > ,int > mp;
       int ans=0;
       if(A <=0 || B <=0)
          ans=0;
       else
       {
                  ans=max(Survival(A+x1,B+x2,x1,x2,y1,y2,z1,z2,1,mp),max(Survival(A+y1,B+y2,x1,x2,y1,y2,z1,z2,2,mp),Survival(A+z1,B+z2,x1,x2,y1,y2,z1,z2,3,mp)));
       }
        cout << ans;
        return 0;
}

20 8

3 2

-5 -10

-20 5

How to print all subset of an array with given sum?

2021-05-25T17:57:00.004+05:30


#include < bits/stdc++.h >

using namespace std;
bool**dp;
void printvector(vector < int > &v)
{
          for(int i=0;i < v.size();i++)
                    cout << v[i] << " ";
          cout << endl;
}

void print(int a[], int i,int sum, vector < int > &v)
{
        if(i==1 && sum!=0 && dp[1][sum])
        {
                  v.push_back(a[i-1]);
                  if(a[i-1]==sum)
                    printvector(v);
                  return;
        }
        if(i==1 && sum==0)
        {
                  printvector(v);
                  return;
        }
        //excluding given element make a new path
        if(dp[i-1][sum])
        {
                  vector < int > p=v;
                  print(a,i-1,sum,p);
        }
        //including given element
         if (sum >= a[i-1] && dp[i-1][sum-a[i-1]])
          {
                    v.push_back(a[i-1]);
                    print(a, i-1, sum-a[i-1],v);
          }
}
int main()
{
        int n,sum;
        cin >> n >> sum;
        int a[n];
        for(int i=0;i < n;i++)
          cin >> a[i];
        dp = new bool*[n+1];
        for(int i=0;i <=n;i++)
          dp[i]=new bool[sum+1];
        //initialize if i==0 dp[i][j]=False if j==0 dp[i][j]=true but if i==0 && j==0 dp[i][j]=True
        dp[0][0]= 1;
        for(int i=1;i <=n;i++)
          dp[i][0]=1;
        for(int j=1;j <=sum;j++)
          dp[0][j]=0;
        for(int i=1;i <=n;i++)
        {
                  for(int j=1;j <=sum;j++)
                  {         //exclude a[i] for sum j
                            if(a[i-1] > j)
                              dp[i][j]=dp[i-1][j];
                            //else include if j-a[i] is true
                            else
                              dp[i][j]=dp[i-1][j-a[i-1]] || dp[i-1][j];
                  }
        }
        if(dp[n][sum])
          cout << "YES" << endl;
        else
          cout << "NO" << endl;
        vector < int > v;
        print(a,n,sum,v);

        return 0;
}

5 10

1 2 3 4 5

Find maximum sum i*arr[i] if only rotation of the array is allowed.

2021-05-24T19:37:00.004+05:30


#include < bits/stdc++.h >
using namespace std;

int main()
{
        int n;
        cin >> n;
        int a[n];
        for(int i=0;i < n;i++)
          cin >> a[i];

        int value=0,sum=0;
        for(int i=0;i < n;i++)
        {
                  sum=sum+a[i];
                  value=value+i*a[i];
        }
        int ans=value;
        for(int i=1;i < n;i++)
        {
                  value=value+sum-n*a[n-i];
                  ans=max(ans,value);
        }
        cout << ans;
        return 0;

}

How to rotate an array by d elements? (Efficient approach)

2021-05-24T18:46:00.004+05:30

How to rotate an array by d elements? (Efficient approach)

#include <bits/stdc++.h>

using namespace std;

void reverseArray(int a[],int i,int j)

{

while(i<j)

{

swap(a[i],a[j]);

i++;

j--;

}

int main()

{

int n,d;

cin>>n>>d;

int a[n];

for(int i=0;i<n;i++)

cin>>a[i];

reverseArray(a,0,d-1);

reverseArray(a,d,n-1);

reverseArray(a,0,n-1);

for(int i=0;i<n;i++)

cout<<a[i]<<" ";

return 0;

}

10 3

1 2 3 4 5 6 7 8 9 10

VARIOUS PROGRAMMING QUESTIONS AND DATA STRUCTURE AND ALGORITHM

2021-05-11T19:38:00.004+05:30

POSTS:

1. Compare and plot the graph of CPU time for different sorting algorithms- bubble sort, selection sort, insertion sort for different cases

2. Even array

3. HOW TO CONVERT DECIMAL TO ROMAN UPTO 10000000000 IN C and CPP

4. MOST EFFECTIVE WAY: COUNT AND PRINT THE NUMBERS WHICH ARE POWERS OF 2 || CHECK IF A NUMBER IS A POWER OF TWO

5. general case::::COUNT AND PRINT THE NUMBERS WHICH ARE POWERS OF ANOTHER GIVEN NUMBER || CHECK IF A NUMBER IS A POWER OF ANOTHER GIVEN NUMBER

6. TIME COMPLEXITY OF DIFFERENT ALGORITHMS

7. How to find Kth smallest element and Kth largest element of an array using heap

8. FIND K CLOSEST NUMBER OF A GIVEN NUMBER FROM A SET OF NUMBERS

10. FIND K MOST FREQUENT NUMBERS IN A SET OF NUMBERS

11. SORT A SET OF NUMBERS IN TERMS OF THE FREQUENCY OF THE NUMBERS

12. COMPARISON BETWEEN DIFFERENT SORTING TECHNIQUES

13. write a program to rotate a matrix in 90 degree angle using dynamic memory allocation and random number generator.

14. Find the number which occurs once in an array while all other elements occurs twice using XOR

15. Write a code to print large fibonacci numbers in minimum time

16. Write a program in c to insert and delete a middle element of a stack implemented using array

17. WHAT IS DYNAMIC PROGRAMMING? BASIC OF DP WITH EXAMPLE

18. DETERMINE IF THERE IS A SUBSET IN A GIVEN ARRAY WHOSE SUM IS EQUAL TO THE GIVEN SUM-( SUBSET SUM PROBLEM)

19. C PROGRAM TO IMPLEMENT QUEUE USING ARRAY AND PERFORM INSERTON, DELETION OPERATION

20. Write a program in c to insert an element in the middle and delete the middle element of queue

21. HOW TO IMPLEMENT STACK USING QUEUE IN C?

22. B TECH FIRST YEAR C PROGRAMMMING QUESTIONS

23. WRITE A PROGRAM TO COUNT THE MINIMUM NUMBER OF CHARACTERS TO BE REPLACED TO CONVERT A STRING TO A NEW STRING WITH NO PALINDROMIC SUBSTRING OF LENGTH MORE THAN ONE

24. EQUAL SUM PARTITION PROBLEM- YOU ARE GIVEN AN ARRAY, YOU NEED TO FIND IF IT IS POSSIBLE TO DIVIDE THE ARRAY IN TWO PARTS WHOSE SUMS ARE EQUAL.

25. COUNT THE NUMBER OF SUBSETS IN AN ARRAY WHOSE SUM IS EQUAL TO A GIVEN SUM

26. MINIMUM SUBSET SUM DIFFERENCE : HOW TO DIVIDE AN ARRAY IN TWO PARTS SUCH THAT THE DIFFERENCE BETWEEN THE SUM OF THE SUBSETS IS MINIMUM POSSIBLE

27. HOW TO COUNT THE NUMBER OF SUBSETS IN AN ARRAY WITH A GIVEN DIFFERECE OF THE SUM OF THE SUBSETS

28. TARGET SUM PROBLEM: DYNAMIC PROGRAMMING

29. ROD CUTTING PROBLEM-DYNAMIC PROGRAMMING

30. EGG DROPPING PROBLEM- DYNAMIC PROGRAMMING

31. SOME SIMPLE MATHEMATICS USED IN COMPETITIVE PROGRAMMING

32. COIN CHANGE PROBLEM- TOTAL NUMBER OF WAYS

33. COIN CHANGE PROBLEM II - MINIMUM NO OF COINS

34. WHAT IS BINARY SEARCH ? INTRODUCTION TO BINARY SEARCH

35. HOW TO APPLY BINARY SEARCH ON REVERSE SORTED ARRAY?

36. HOW TO FIND FIRST AND LAST OCCURRENCE OF AN ELEMENT IN A SORTED ARRAY?

37. HOW TO COUNT THE NUMBER OF OCCURRENCE OF AN ELEMENT IN A SORTED ARRAY USING BINARY SEARCH?

38. HOW MANY TIMES A SORTED ARRAY IS ROTATED TO GET ANOTHER GIVEN ARRAY?

39. HOW TO FIND AN ELEMENT IN A SORTED ROTATED ARRAY?

40. How To Perform Searching Operation In a Nearly Sorted Array?

41. How To Find The Next Nearest Letter Of A Character Present in a Sorted Character Array?

42. How To Find Position Of An Element in an Infinite Sorted Array?

43. PEAK ELEMENT-BINARY SEARCH

44. How To Find Maximum Element In a Bitonic Array?

45. Shortest path in a maze using BFS

46. WHAT IS GRAPH THEORY ALGORITHM?

47. Given a binary matrix of size 10x10, where 0 represents water and 1 represents land, and connected ones form an island, count the total number of islands (using queue data structure) in c.

48. Implement Huffman Encoding using java

49. Heap sort in c

50. Activity selection problem

51. How to rotate an array by d elements? (Efficient approach)

Activity selection problem

2021-05-10T18:50:00.001+05:30

You are given n activities with their start and finish times. Select the maximum number of activities that can be performed by a single person, assuming that a person can only work on a single activity at a time.

CODE:


#include <bits/stdc++.h>
using namespace std;

int main()
{
     int n;
     cin>>n;
     int start[n],ending[n];
     for(int i=0;i<n;i++)
      {
          cin>>start[i];
          cin>>ending[i];
      }
      //to sort the elements in terms of ending time
      priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>>pq;
	for(int i=0;i<n;i++)
	{
   		 pq.push(make_pair(ending[i],start[i]));
	}

	//to track the count of activities selected

	int c=0;

	//to store the activities selected

	vector<int>v1,v2;
	int p=pq.top().first;
	int q=pq.top().second;
	pq.pop();
	c++;
	v1.push_back(q);
	v2.push_back(p);
	while(!pq.empty())
	{
		if(pq.top().second>=p)
   		{

          c++;  
          p=pq.top().first;        
          q=pq.top().second;
          v1.push_back(q);
          v2.push_back(p);
    	}
   		pq.pop;
	}

	cout<<"Total no. of activities selected   "<<c<<endl;
	cout<<"The start time and end time of the activities are"<<endl;
	for(int i=0;i<v1.size();i++)
	{
    	cout<<"( "<<v1[i]<<" , "<<v2[i]<<" )"<<endl;

	}
}

OUTPUT:

Construct a DFA that accepts any strings over {a,b} that doesn't contain the string aabb in it.

2021-05-08T18:57:00.000+05:30

Construct a DFA that accepts any strings over {a,b} that doesn't contain the string aabb in it.

To solve this problem, first we will construct a DFA that accepts all strings over {a,b} that accepts the string aabb in it.

We can get the dfa that accepts all other strings except aabb by complementing above DFA. To do that, make all final states as non-final states and all non-final states as final states. The final DFA is shown below.