Student databse system (PHP project) free download

This is a simple php project (student database system) in php. The code is easy to understand and implement.
Please read the 'readme.txt' file in the folder. Download the file " student.zip " from the download box. 

PHP project free download

This is a simple php project (airline database system) project in php. The code is easy to understand and implement.
Actually this is a small tutorial which will help you understand how to create a simple front-end and its back-end in php-mysql.
Please read the 'readme.txt' file in the folder.
The folder for this project 'student.zip' and 'Airline.zip'  files can be downloaded from the Download Box in the blog.

Airline Database System (php project) free download

This is a simple airline database system project in php. The code is easy to understand and implement.
Actually this is a small tutorial which will help you understand how to create a simple front-end and its back-end in php-mysql.
Please read the 'readme.txt' file in the folder.
The folder for this project 'Airline.zip' file can be downloaded from the Download Box in the blog.

UDP protocol:Socket programming(Java Code)

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Sliding Window Protocol:Go Back N (Java code)

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Sliding Window Protocol:Stop & Wait(Java Code)

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Assembly level programming

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Socket programming for CN

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Sem 4 & Sem 7 (AOAD & RAI) subjects : Java Code / Program for Prim's algorithm for Minimum spanning tree

Given Below is the Java Code for Prim's algorithm for minimum spanning tree.

//Aim:To implement Prim's algorithm. //Program: import java.util.*; class Prim { static int v,e,cost; static int G[][]; static int T[][]; static int visited[]; static void cal_tree() { int temp1; int temp2=0,temp3=0; int flag=1; visited[1]=1; temp1=9999; //do for(int k=1;k<v;++k) { temp1=9999; for(int i=1;i<=v;++i) { for(int j=1;j<=v;++j) { if(G[i][j]<temp1 && ((visited[i]==1 && visited[j]!=1) ||(visited[i]!=1 && visited[j]==1)) ) { temp1=G[i][j]; temp2=i; temp3=j; } }// end of inner loop } System.out.println(temp2+"----"+temp3+" = "+G[temp2][temp3]); cost+=G[temp2][temp3]; visited[temp2]=1; visited[temp3]=1; T[temp2][temp3]=T[temp3][temp2]=1; flag=check(); } } static int check() { for(int i=1;i<=v;++i) if(visited[i]==-1) return 1; return 0; } public static void main(String args[]) { Scanner scr=new Scanner(System.in); System.out.println("Enter the number of vertices. "); v=scr.nextInt(); G=new int[v+1][v+1]; T=new int[v+1][v+1]; visited=new int[v+1]; /****Initialize to large value*****/ for(int i=1;i<=v;++i) { visited[i]=-1; for(int j=1;j<=v;++j) { G[i][j]=9999; T[i][j]=-1; } } System.out.println("Enter the number of edges. "); e=scr.nextInt(); System.out.println("\nPlease enter your graph as follows(v1 v2 e)where e is cost of edge between v1 & v2.\n"); int temp1,temp2; for(int i=0;i<e;++i) { System.out.print("\nEnter (v1 v2 e) "); temp1=scr.nextInt(); temp2=scr.nextInt(); G[temp1][temp2]=G[temp2][temp1]=scr.nextInt(); } System.out.println("\nThe minimum spanning tree is as follows.\n"); cal_tree(); System.out.println("\nThe cost of minimum spanning tree is "+cost+"."); } } /*OUTPUT Enter the number of vertices. 7 Enter the number of edges. 9 Please enter your graph as follows(v1 v2 e)where e is cost of edge between v1 & v2. Enter (v1 v2 e) 1 2 25 Enter (v1 v2 e) 2 3 14 Enter (v1 v2 e) 3 4 11 Enter (v1 v2 e) 4 5 20 Enter (v1 v2 e) 5 6 23 Enter (v1 v2 e) 6 1 10 Enter (v1 v2 e) 7 2 12 Enter (v1 v2 e) 7 5 22 Enter (v1 v2 e) 7 4 17 The minimum spanning tree is as follows. 1----6 = 10 5----6 = 23 4----5 = 20 3----4 = 11 2----3 = 14 2----7 = 12 The cost of minimum spanning tree is 90. */

Checksum Error detection (java program)

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CRC: Cyclic Redundancy Code (java code)

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Easy solution for SEM 5 pracs

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Hamming Code(Error Detection & correction)-Java Code

Given below is the JAVA program for Hamming code error detection and correction.
import java.util.*; class hamming_code { static int set_parity_bit(int a[]) { int count=0; //........Initialising count to zero which will count the number of 1. int l=a.length; for(int i=0;i<l;++i) if(a[i]==1) ++count; //............Incrementing count if value in array "a" is 1. if((count%2)==0) return 0;//........Returning 0 if even number of 1 else return 1;//........Returning 1 if odd number of 1 } public static void main(String args[]) { Scanner scr=new Scanner(System.in); System.out.println("This is hamming code error detection and correction using EVEN parity"); System.out.println(); System.out.println("Enter 4 data bits.D4 D3 D2 D1"); int n=4; //set number of data bits int d[]=new int[4]; //..........Array to accept data bits for(int i=n-1;i>=0;--i) { System.out.println("Enter the value of D"+(i+1)); d[i]=scr.nextInt(); } /*.............. Formula for calculating 2^k>=n+k+1 ...............*/ int k=0;// k stands for number of parity bits.Initializing it to zero. while(Math.pow(2,k)<(n+k+1)) // Calculating the value of k(number of parity bits). { ++k; } System.out.println(); System.out.println(k+" parity bits are required for the transmission of data bits."); int p[]=new int[k]; //..........Array to store parity bits int h[]=new int[n+k+1];//.........Array to hold the hamming code.(n+k+1) as we start from pos 1. /********** Initialising array h[] to -1 ************/ for(int i=0;i<7;++i) h[i]=-1; int count=0; int c=2; while(count<4) { ++c; if(c==4) continue; h[c]=d[count]; ++count; } int p1[]={h[1],h[3],h[5],h[7]}; int p2[]={h[2],h[3],h[6],h[7]}; int p3[]={h[4],h[5],h[6],h[7]}; int parity[]=new int[3]; /************Setting the value of parity bit*************/ parity[0]=set_parity_bit(p1); parity[1]=set_parity_bit(p2); parity[2]=set_parity_bit(p3); /************Inserting the parity bits in the hamming code**********/ h[1]=parity[0]; h[2]=parity[1]; h[4]=parity[2]; System.out.println("\nSENDER:"); System.out.print("\nThe data bits entered are: "); for(int i=3;i>=0;--i) System.out.print(d[i]+" "); System.out.println("\nThe Parity bits are: "); for(int i=2;i>=0;--i) System.out.println("Value of P"+(i+1)+" is "+parity[i]+" "); System.out.print("\nThe Hamming code is as follows :-\nD4 D3 D2 P3 D1 P2 P1 \n"); for(int i=(n+k);i>0;--i) System.out.print(h[i]+" "); System.out.println(); System.out.println("\nEnter the hamming code with error at any position of your choice.\nNOTE: ENTER A SPACE AFTER EVERY BIT POSITION.\nError should be present only at one bit position"); for(int i=7;i>0;--i) h[i]=scr.nextInt(); int p4[]={h[1],h[3],h[5],h[7]}; int p5[]={h[2],h[3],h[6],h[7]}; int p6[]={h[4],h[5],h[6],h[7]}; parity[0]=set_parity_bit(p4); parity[1]=set_parity_bit(p5); parity[2]=set_parity_bit(p6); int position=(int)(parity[2]*Math.pow(2,2)+parity[1]*Math.pow(2,1)+parity[0]*Math.pow(2,0)); System.out.println("\nRECEIVER:"); System.out.println("Error is detected at position "+position+" at the receiving end."); System.out.println("Correcting the error.... "); if(h[position]==1) h[position]=0; else h[position]=1; System.out.print("The correct code is "); for(int i=7;i>0;--i) System.out.print(h[i]+" "); }} /* Output C:\Users\Aditya\Desktop>java hamming_code This is hamming code error detection and correction using EVEN parity Enter 4 data bits.D4 D3 D2 D1 Enter the value of D4 1 Enter the value of D3 0 Enter the value of D2 1 Enter the value of D1 0 3 parity bits are required for the transmission of data bits. SENDER: The data bits entered are: 1 0 1 0 The Parity bits are: Value of P3 is 0 Value of P2 is 1 Value of P1 is 0 The Hamming code is as follows :- D4 D3 D2 P3 D1 P2 P1 1 0 1 0 0 1 0 Enter the hamming code with error at any position of your choice. NOTE: ENTER A SPACE AFTER EVERY BIT POSITION. Error should be present only at one bit position 1 0 1 0 1 1 0 RECEIVER: Error is detected at position 3 at the receiving end. Correcting the error.... The correct code is 1 0 1 0 0 1 0 */

CG project:Balloon Shooting Game & Power Point Presentation

To download the code and additional information(ppt) about it, use the download box !!
Hope you enjoy the game!!

YXN6EJRM78VY

Foreign key concept

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YXN6EJRM78VY

Two Phase Locking Protocol & Validation based protocol

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YXN6EJRM78VY

Linux programs(Shell Script)

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SQL Programs.(views,triggers,functions,procedures,nested,complex queries)

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AOAD programs for SEM 4(Java code free download)

friends use the download box for downloading the aoad programs (java code).

Keep visiting the blog as my next posts will be other important programs 

many sites and books provide a C++ code for these programs,but this is the only blog 
providing easy to understand JAVA CODE for AOAD

CG Programs Free Download(C,C++ code):Reflection about X,Y axis and Origin

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Polygon Clipping(Sutherland Hodgeman) C++ code free download

Given below is C code for Sutherland Hodgeman Polygon clipping algorithm.
//Sutherland Hodgeman Polygon clipping BY ADITYA MANDHARE #include<iostream.h> #include<conio.h> #include <graphics.h> #include <stdlib.h> #include <stdio.h> #include <conio.h> #include<dos.h> int px[]={0,0,0,0,0,0,0,0,0,0,0,0,0}; int py[]={0,0,0,0,0,0,0,0,0,0,0,0,0}; int lx[]={0,0,0,0,0,0,0,0,0,0,0,0,0}; int ly[]={0,0,0,0,0,0,0,0,0,0,0,0,0}; int rx[]={0,0,0,0,0,0,0,0,0,0,0,0,0}; int ry[]={0,0,0,0,0,0,0,0,0,0,0,0,0}; int bx[]={0,0,0,0,0,0,0,0,0,0,0,0,0}; int by[]={0,0,0,0,0,0,0,0,0,0,0,0,0}; int tx[]={0,0,0,0,0,0,0,0,0,0,0,0,0}; int ty[]={0,0,0,0,0,0,0,0,0,0,0,0,0}; float m[12]; int getx(int,float,int); int gety(int,float,int); int c=0; void main() { int gdriver = DETECT, gmode; initgraph(&gdriver, &gmode, "c:\\tc\\bgi"); cout<<"\t\tHello friends!! This is Polygon Clipping\n\n "; cout<<"\t\t\t\t***WINDOW***\n"; cout<<"\n\n\t\t (200,100)"; rectangle(200,100,400,250); cout<<"\n\n\n\n\n\n\n\n\n\n\n\t\t\t\t\t\t(400,250)"; cout<<"\n\n\n"; cout<<"\nEnter the number of vertices of polygon : "; int n=0; cin>>n; int i; cout<<"Enter the polygon in clockwise direction"; int x1,y1,x2,y2; for(i=0;i<n;++i) { cout<<"\nEnter the end points of edge (x1,y1) (x2,y2) "; cin>>x1>>y1; cin>>x2>>y2; if(x2==x1) m[i]=9999; else m[i]=( (0.1*(y2-y1)) / (0.1*(x2-x1)) ); px[i]=x1; py[i]=y1; } for(i=0;i<n;++i) { line(px[i],py[i],px[(i+1)%n],py[(i+1)%n]); } delay(3000); cleardevice(); cout<<"\n\t\tLeft clipping"; /*for(i=0;i<n;++i) if(px[i]<200) l=i;*/ int l=-1; for(i=0;i<n;++i) { ++l; float y; if(px[i]<200 && px[(i+1)%n]>200) { y=gety(200,m[i],i); lx[l]=200; ly[l]=y; lx[l+1]= px[(i+1)%n]; ly[l+1]= py[(i+1)%n]; ++l; } else if(px[i]<200 && px[(i+1)%n]<200) --l; else if(px[i]>200 && px[(i+1)%n]>200) { lx[l]=px[(i+1)%n]; ly[l]=py[(i+1)%n]; } else if(px[i]>200 && px[(i+1)%n]<200) { y=gety(200,m[i],i); lx[l]=200; ly[l]=y; } } /* for(i=0;i<=l;++i) cout<<" "<<lx[i]; for(i=0;i<=l;++i) cout<<" "<<ly[i]; */ setcolor(WHITE); rectangle(200,100,400,250); setcolor(RED); ++l; for(i=0;i<l-1;++i) { line(lx[i],ly[i],lx[(i+1)],ly[(i+1)]); } line(lx[l-1],ly[l-1],lx[0],ly[0]); ++c; /* cout<<"Value of l = "<<l; for(i=0;i<n;++i) cout<<"m= "<<m[i]; */ delay(6000); cleardevice(); cout<<"\n\t\t Right clip"; for(i=0;i<l;++i) { if(lx[(i+1)%l]==lx[i]) m[i]=999; else m[i]=((0.1*(ly[(i+1)%l]-ly[i]))/(0.1*(lx[(i+1)%l]-lx[i]))); } int r=-1; for(i=0;i<l;++i) { ++r; float y; if(lx[i]>400 && lx[(i+1)%l]<400) { y=gety(400,m[i],i); rx[r]=400; ry[r]=y; rx[r+1]= lx[(i+1)%l]; ry[r+1]= ly[(i+1)%l]; ++r; } else if(lx[i]>400 && lx[(i+1)%l]>400) --r; else if(lx[i]<400 && lx[(i+1)%l]<400) { rx[r]=lx[(i+1)%l]; ry[r]=ly[(i+1)%l]; } else if(lx[i]<400 && lx[(i+1)%l]>400) { y=gety(400,m[i],i); rx[r]=400; ry[r]=y; } //'''''./////////////////// } ++r; /* for(i=0;i<r;++i) cout<<" "<<rx[i]; for(i=0;i<r;++i) cout<<" "<<ry[i]; */ setcolor(WHITE); rectangle(200,100,400,250); setcolor(GREEN); for(i=0;i<r;++i) { line(rx[i],ry[i],rx[(i+1)%r],ry[(i+1)%r]); } ++c; // cout<<"Value of r = "<<r; delay(3000); cleardevice(); cout<<"\n\t\t Bottom clip"; for(i=0;i<r;++i) { if(rx[(i+1)%r]==rx[i]) m[i]=999; else m[i]=((0.1*(ry[(i+1)%r]-ry[i]))/(0.1*(rx[(i+1)%r]-rx[i]))); } int b=-1; for(i=0;i<r;++i) { ++b; float x; if(ry[i]>250 && ry[(i+1)%r]<250) { x=getx(250,m[i],i); //////////////////////////// by[b]=250; bx[b]=x; bx[b+1]= rx[(i+1)%r]; by[b+1]= ry[(i+1)%r]; ++b; } else if(ry[i]>250 && ry[(i+1)%r]>250) --b; else if(ry[i]<250 && ry[(i+1)%r]<250) { bx[b]=rx[(i+1)%r]; by[b]=ry[(i+1)%r]; } else if(ry[i]<250 && ry[(i+1)%r]>250) { x=getx(250,m[i],i); by[b]=250; bx[b]=x; } } ++b; /* for(i=0;i<b;++i) cout<<" "<<bx[i]; for(i=0;i<b;++i) cout<<" "<<by[i]; */ setcolor(WHITE); rectangle(200,100,400,250); setcolor(YELLOW); for(i=0;i<b;++i) { line(bx[i],by[i],bx[(i+1)%b],by[(i+1)%b]); } ++c; // cout<<"Value of b = "<<b ; delay(3000); cleardevice(); cout<<"\n\t\tTop clip"; cout<<"\n\t The final CLIPPED POLYGON is as shown above"; for(i=0;i<b;++i) { if(bx[(i+1)%b]==bx[i]) m[i]=999; else m[i]=((0.1*(by[(i+1)%b]-by[i]))/(0.1*(bx[(i+1)%b]-bx[i]))); } int t=-1; for(i=0;i<b;++i) { ++t; float x; if(by[i]<100 && by[(i+1)%b]>100) { x=getx(100,m[i],i); //////////////////////////// ty[t]=100; tx[t]=x; tx[t+1]=bx[(i+1)%b]; ty[t+1]=by[(i+1)%b]; ++t; } else if(by[i]<100 && by[(i+1)%b]<100) --t; else if(by[i]>100 && by[(i+1)%b]>100) { tx[t]=bx[(i+1)%b]; ty[t]=by[(i+1)%b]; } else if(by[i]>100 && by[(i+1)%b]<100) { x=getx(100,m[i],i); ty[t]=100; tx[t]=x; } } ++t; /* for(i=0;i<t;++i) cout<<" "<<tx[i]; for(i=0;i<t;++i) cout<<" "<<ty[i]; */ setcolor(WHITE); rectangle(200,100,400,250); setcolor(BLUE); for(i=0;i<t;++i) { line(tx[i],ty[i],tx[(i+1)%t],ty[(i+1)%t]); } ++c; // cout<<"Value of t = "<<t ; getch(); closegraph(); } int gety(int x,float m,int i) { if(c==0) return (py[i]+(m*(x-px[i]))); if(c==1) return (ly[i]+(m*(x-lx[i]))); if(c==2) return (ry[i]+(m*(x-rx[i]))); if(c==3) return (by[i]+(m*(x-bx[i]))); } int getx(int y,float m ,int i) { if(c==0) return (px[i]+((y-py[i])/m)); if(c==1) return (lx[i]+((y-ly[i])/m)); if(c==2) return (rx[i]+((y-ry[i])/m)); if(c==3) return (bx[i]+((y-by[i])/m)); }

Line Clipping(Cohen Sutherland) C++ code free download

Given below is the C code for Cohen Sutherland Line clipping algorithm.
//Cohensutherland line clip BY ADITYA MANDHARE #include<iostream.h> #include<conio.h> #include <graphics.h> #include <stdlib.h> #include <stdio.h> #include <conio.h> #include<dos.h> float getx(float); float gety(float); void allocate(float x,float y,int n); int checkbothzero(int a[],int b[]); int checknotzero(int a[],int b[]); float x1,y1,x2,y2; float m; int a[4]={0,0,0,0}; int b[4]={0,0,0,0}; void main() { int gdriver = DETECT, gmode; initgraph(&gdriver, &gmode, ""); cout<<"\t\t\t*****COHEN SUTHERLAND LINE CLIPPING*****"; cout<<"\nEnter the starting and end co-ordinates of line"; cout<<"\nstart co-ord(x,y)"; cin>>x1>>y1; cout<<"end co-ord(x,y)"; cin>>x2>>y2; m=((y2-y1)/(x2-x1)); line(x1,y1,x2,y2); rectangle(100,200,300,100); delay(2000); cleardevice(); delay(1000); int flag=1; int select; do { int l; /* cout<<"values1"; for( l=0;l<4;++l) cout<<" "<<a[l]; */ /* cout<<"values2"; for( l=0;l<4;++l) cout<<" "<<b[l];*/ int i; int val; select=0; allocate(x1,y1,1); allocate(x2,y2,2); if(checkbothzero(a,b)) break; if(checknotzero(a,b)) { flag=0; break; } for(i=3;i>=0;--i) { if(a[i]==0 && b[i]==0) continue; else { val=i; break; } } if(a[val]==1) select=1; else select=2; if(val==3) { if(select==1) { x1=100; y1=gety(x1); } else { x2=100; y2=gety(x2); } } if(val==2) { if(select==1) { x1=300; y1=gety(x1); } else { x2=300; y2=gety(x2); } } if(val==1) { if(select==1) { y1=200; x1=getx(y1); } else { y2=200; x2=getx(y2); } } if(val==0) { if(select==1) { y1=100; x1=getx(y1); } else { y2=100; x2=getx(y2); } } } while(1); rectangle(100,200,300,100); if(flag==0) cout<<"Line is completely outside"; else { cout<<"Clipped line is as follows"; setcolor(RED); line(x1,y1,x2,y2); } getch(); cleardevice(); } float getx(float y) { float x; x=(((y-y1)/m)+x1); return x; } float gety(float x) { float y; y=(((x-x1)*m)+y1); return y; } void allocate(float x,float y,int n) { for(int i=0;i<4;++i) { if(n==1) a[i]=0; else b[i]=0; } if(x<100) { if(n==1) a[3]=1; else b[3]=1; } if(x>300) { if(n==1) a[2]=1; else b[2]=1; } if(y<100) { if(n==1) a[0]=1; else b[0]=1; } if(y>200) { if(n==1) a[1]=1; else b[1]=1; } } int checkbothzero(int a[],int b[]) { for(int k=0;k<4;++k) { if((a[k]+b[k])>0) return 0; } return 1; } int checknotzero(int a[],int b[]) { for(int k=0;k<4;++k) { if(a[k]+b[k]==2) return 1; } return 0; }

JOB SEQUENCING WITH DEADLINES(Java Code)-Simple Easy To Understand(Free Download)

Given below is the java code for Job Sequencing with Deadlines.
/*********** JOB SEQUENCING WITH DEADLINES by:" ADITYA MANDHARE "*****************/ // PROGRAM: import java.util.*; class job { int p; //.............for profit of a job int d; //.............for deadline of a job int v; //.............for checking if that job has been selected /*************default constructor**************/ job() { p=0; d=0; v=0; } job(int x,int y,int z) // parameterised constructor { p=x; d=y; v=z; } } class js { static int n; static int out(job jb[],int x) { for(int i=0;i<n;++i) if(jb[i].p==x) return i; return 0; } public static void main(String args[]) { Scanner scr=new Scanner(System.in); System.out.println("Enter the number of jobs"); n=scr.nextInt(); int max=0; // this is to find the maximum deadline job jb[]=new job[n]; /***********************Accepting job from user*************************/ for(int i=0;i<n;++i) { System.out.println("Enter profit and deadline(p d)"); int p=scr.nextInt(); int d=scr.nextInt(); if(max<d) max=d; // assign maximum value of deadline to "max" variable jb[i]=new job(p,d,0); //zero as third parameter to mark that initially it is unvisited } //accepted jobs from user /*************************Sorting in increasing orser of deadlines*************************/ for(int i=0;i<=n-2;++i) { for(int j=i;j<=n-1;++j) { if(jb[i].d>jb[j].d) { job temp=jb[i]; jb[i]=jb[j]; jb[j]=temp; } } } // sorting process ends /******************************Displaying the jobs to the user***********************/ System.out.println("The jobs are as follows "); for(int i=0;i<n;++i) System.out.println("Job "+i+" Profit = "+jb[i].p+" Deadline = "+jb[i].d); // jobs displayed to the user int count; int hold[]=new int[max]; for(int i=0;i<max;++i) hold[i]=0; /*****************************Process of job sequencing begins*************************/ for(int i=0;i<n;++i) { count=0; for(int j=0;j<n;++j) { if(count<jb[j].d && jb[j].v==0 && count<max && jb[j].p>hold[count]) { int ch=0; if(hold[count]!=0) { ch=out(jb,hold[count]); jb[ch].v=0; } hold[count]=jb[j].p; jb[j].v=1; ++count; } // end of if } //end of inner for }// end of outer for /******************************job sequencing process ends******************************/ /******************************calculating max profit**********************************/ int profit=0; for(int i=0;i<max;++i) profit+=hold[i]; System.out.println("The maximum profit is "+profit); }//end main method }//end class /****************SAMPLE OUTPUT *********************/ /* Enter the number of jobs 4 Enter profit and deadline(p d) 70 2 Enter profit and deadline(p d) 12 1 Enter profit and deadline(p d) 18 2 Enter profit and deadline(p d) 35 1 The jobs are as follows Job 0 Profit = 12 Deadline = 1 Job 1 Profit = 35 Deadline = 1 Job 2 Profit = 18 Deadline = 2 Job 3 Profit = 70 Deadline = 2 The maximum profit is 105 */

Semester 4 Operating System : Pre Emptive SJF java program / code

Given below is Java program for Pre Emptive SJF.

//Aim:To implement preemptive SJF scheduling //Program: import java.util.*; class PR { int bt; int at; PR() { bt=0; at=0; } PR(int x,int y) { bt=x; at=y; } } class PESJF { static int sum(PR p[],int n) { int s=0; for(int i=0;i<n;++i) s+=p[i].bt; if(s==0) return 1; else return 0; } /* Assume that the none of the processes terminate before all the processes have arrived */ public static void main(String args[]) { PR p[]; Scanner scr=new Scanner(System.in); System.out.println("Enter the number of processes"); int n=scr.nextInt(); p=new PR[n]; int i=0,j=0; int bt,at; for(i=0;i<n;++i) { System.out.println("Enter the burst time and arrival time of a process"); bt=scr.nextInt(); at=scr.nextInt(); p[i]=new PR(bt,at); } PR temp; // Sorting processes in increasing order of arrival time for(i=0;i<=n-2;++i) { for(j=i;j<=n-1;++j) { if(p[i].at>p[j].at) { temp=p[i]; p[i]=p[j]; p[j]=temp; } } } int time=p[n-1].at; int tt[]=new int[n]; int wt[]=new int[n]; int b[]=new int[n]; for(i=0;i<n;++i) { System.out.println("Process "+(i+1)+" Burst time= "+p[i].bt+" Arrival time= "+p[i].at); b[i]=p[i].bt; } int mat=p[n-1].at; int pn=0; i=0; j=0; p1:while(j<mat) { while(j<p[i+1].at) { if(p[pn].bt>0) { p[pn].bt--; ++j; } else { System.out.println("Process "+(pn+1)+" executed.Turn around time= "+b[pn]+" Wait time=0"); ++i; for(int y=0;y<i;++y) { for(int z=y;z<i-1;++z) { if((p[y].bt<p[z].bt)&&p[y].bt>0) pn=y; } } continue p1; }} if(i<n-1) if(p[i+1].bt<p[pn].bt) pn=(i+1); ++i; } System.out.println("After arrival of all processes"); for(int g=0;g<n;++g) System.out.println("Burst time of process "+(g+1)+"left = "+ p[g].bt+" "); System.out.println("Time elapsed= "+i); time=i; int k=0; int min=0; do { min=0; if(p[0].bt==0) for(i=0;i<n;++i) { if(p[i].bt!=0) { min=i; break; } } for(i=0;i<n;++i) { if((p[i].bt<p[min].bt)&&p[i].bt>0) min=i; } while(p[min].bt>0) { ++time; p[min].bt--; } tt[min]=(time-p[min].at); wt[min]=(tt[min]-b[min]); System.out.println("Process "+(min+1)+" executed.Turn around time= "+tt[min]+" Wait time= "+wt[min]); ++k; } while(sum(p,n)!=1); double att=0,awt=0; double s1=0,s2=0; for(i=0;i<n;++i) { s1+=tt[i]; s2+=wt[i]; } att=s1/n; awt=s2/n; System.out.println("Average Turn around time= "+att+" Average wait time= "+awt); } } /* OUTPUT Enter the number of processes 4 Enter the burst time and arrival time of a process 8 0 Enter the burst time and arrival time of a process 4 1 Enter the burst time and arrival time of a process 9 2 Enter the burst time and arrival time of a process 5 3 Process 1 Burst time= 8 Arrival time= 0 Process 2 Burst time= 4 Arrival time= 1 Process 3 Burst time= 9 Arrival time= 2 Process 4 Burst time= 5 Arrival time= 3 After arrival of all processes Burst time of process 1left = 7 Burst time of process 2left = 2 Burst time of process 3left = 9 Burst time of process 4left = 5 Time elapsed= 3 Process 2 executed.Turn around time= 4 Wait time= 0 Process 4 executed.Turn around time= 7 Wait time= 2 Process 1 executed.Turn around time= 17 Wait time= 9 Process 3 executed.Turn around time= 24 Wait time= 15 Average Turn around time= 13.0 Average wait time= 6.5 */

Sem 4 AOAD : Java Code / Program for MAXMIN algorithm

Given below is the java code for MAXMIN concept.

import java.util.*; class Stack { int top; int s[]=new int[50]; Stack() { top=-1; } void push(int n) { s[++top]=n; } int pop() { return s[top--]; } void display() { for(int i=0;i<=top;++i) System.out.print(" "+s[i]); } } class MAXMIN2 { static int a[]; static int i,j,max,min; static int flag=0; Scanner scr=new Scanner(System.in); MAXMIN2() { System.out.println("Enter number of elements in array"); int n=scr.nextInt(); a=new int[n]; System.out.println("Enter the elements"); for(int i=0;i<n;++i) a[i]=scr.nextInt(); } Stack s1=new Stack(); Stack s2=new Stack(); void max_min(int i,int j) { if(i==j) { //if(flag==0) //{ max=a[j]; min=a[i]; s1.push(max); s2.push(min); //} /*else { nmax=a[j]; nmin=a[i]; } */ } else if(i==(j-1)) { //if(flag==0) //{ min=(a[i]<a[j])?a[i]:a[j]; max=(a[i]>a[j])?a[i]:a[j]; s1.push(max); s2.push(min); //} /*else { nmin=(a[i]<a[j])?a[i]:a[j]; nmax=(a[i]>a[j])?a[i]:a[j]; } */ } else { int m=(i+j)/2; max_min(i,m); max_min(m+1,j); int lmax,lmin,rmax,rmin; rmax=s1.pop(); rmin=s2.pop(); lmax=s1.pop(); lmin=s2.pop(); if(rmax>lmax) s1.push(rmax); else s1.push(lmax); if(rmin<lmin) s2.push(rmin); else s2.push(lmin); } } void display1() { int low=s2.pop(); int high=s1.pop(); System.out.println("max value in the array is= "+high+" min value in the array= "+low); } } class MainMAXMIN { public static void main(String args[]) { Scanner scr=new Scanner(System.in); System.out.println("Please enter the length of array"); int l=scr.nextInt(); MAXMIN2 m=new MAXMIN2(); m.max_min(0,(l-1)); m.display1(); /* System.out.println("max value in the array is= "++" min value in the array= "+min); System.out.println("max value in the array is= "+max+" min value in the array= "+min);*/ } }

Semester 4 Operating System (OS) : Optimal page replacement policy Java program / Code

Given below is the Java Code for Optimal page replacement. //Aim:To implement optimal page replacement policy //Program: import java.util.*; class PR3 { static int check2(int p[],int pf[],int count[],int j,int s) {int k; for(int i=0;i<s;++i) count[i]=0; for(k=0;k<s;++k) { loop:for(int i=j-1;;--i) { ++count[k]; if(p[i]==pf[k]) break loop; } //System.out.println("count = "+count[k]); } int max=count[0]; int pos=0; for(int l=0;l<s;++l) { if(count[l]>max) { max=count[l]; pos=l; } } return pos; } static int check(int p[],int pf[],int count[],int j,int s) { int k; for(int i=0;i<s;++i) count[i]=0;int found=0; for(k=0;k<s;++k) { loop:for(int i=j+1;i<p.length;++i) { ++count[k]; if(p[i]==pf[k]) { ++found; break loop; } } } int pos; if(found<(s-1)) { System.out.println(); System.out.println("More than 1 used pages not required for furthur operation.\nSwapping LRU page."); pos=check2(p,pf,count,j,s); } else { int max=count[0]; pos=0; for(int l=0;l<s;++l) { if(count[l]>max) { max=count[l]; pos=l; } } } return pos; } static boolean check1(int p[],int x,int n) { for(int i=0;i<n;++i) { if(p[i]==x) return true; } return false; } public static void main(String args[]) { Scanner scr=new Scanner(System.in); System.out.println("Enter the number of page entries"); int n=scr.nextInt(); int fault=0,hit=0; int p[]=new int[n]; System.out.println("Enter the pages "); for(int i=0;i<n;++i) p[i]=scr.nextInt(); System.out.print("The page table entries are "); for(int i=0;i<n;++i) System.out.print(" "+p[i]); System.out.println(); int s=3; System.out.println("Do you wish to set a the number of page frames?(1=yes/0=no)"); int ch=scr.nextInt(); if(ch==1) { System.out.println("Enter the number of page frames "); s=scr.nextInt(); System.out.println("The number of page frames set to "+s); } else System.out.println("The default number of page frames are "+s); int count[]=new int[s]; int pf[]=new int[s]; int t=0; int incr=0; for(int i=0;i<s || (t!=s);++i) { if(check1(pf,p[i],t)) { ++incr; System.out.println(); for(int y=0;y<t;++y) System.out.print(pf[y]+" "); System.out.print(" Hit"); ++hit; } else {++incr; ++t; pf[t-1]=p[i]; System.out.println(); for(int y=0;y<t;++y) System.out.print(pf[y]+" "); System.out.print(" F"); ++fault; } } for(int j=incr;j<n;++j) { if(check1(pf,p[j],s)) { ++hit; System.out.println(); for(int k=0;k<s;++k) System.out.print(pf[k]+" "); System.out.print(" H"); } else { int choice=check(p,pf,count,j,s); ++fault; pf[choice]=p[j]; System.out.println(); for(int k=0;k<s;++k) System.out.print(pf[k]+" "); System.out.print(" F"); } } System.out.println(); System.out.println("Page faults= "+fault+" Page hits= "+hit); } } /* OUTPUT Enter the number of page entries 20 Enter the pages 7 0 1 2 0 3 0 4 2 3 0 3 2 1 2 0 1 7 0 1 The page table entries are 7 0 1 2 0 3 0 4 2 3 0 3 2 1 2 0 1 7 0 1 Do you wish to set a the number of page frames?(1=yes/0=no) 0 The default number of page frames are 3 7 F 7 0 F 7 0 1 F 2 0 1 F 2 0 1 H 2 0 3 F 2 0 3 H 2 4 3 F 2 4 3 H 2 4 3 H 2 0 3 F 2 0 3 H 2 0 3 H 2 0 1 F 2 0 1 H 2 0 1 H 2 0 1 H 7 0 1 F 7 0 1 H 7 0 1 H Page faults= 9 Page hits= 11 Enter the number of page entries 20 Enter the pages 7 0 1 2 0 3 0 4 2 3 0 3 2 1 2 0 1 7 0 1 The page table entries are 7 0 1 2 0 3 0 4 2 3 0 3 2 1 2 0 1 7 0 1 Do you wish to set a the number of page frames?(1=yes/0=no) 1 Enter the number of page frames 4 The number of page frames set to 4 7 F 7 0 F 7 0 1 F 7 0 1 2 F 7 0 1 2 H 3 0 1 2 F 3 0 1 2 H 3 0 4 2 F 3 0 4 2 H 3 0 4 2 H 3 0 4 2 H 3 0 4 2 H 3 0 4 2 H 1 0 4 2 F 1 0 4 2 H 1 0 4 2 H 1 0 4 2 H 7 0 4 2 F 7 0 4 2 H 1 0 4 2 F Page faults= 9 Page hits= 11 */

Sem 4 Operating System (OS) : LRU page replacement Java Code / Program

Given below is the Java program for LRU page replacement policy. import java.util.*; class LRU { static int check1(int p[],int pf[],int n,int s) { int count[]=new int[s]; for(int i=0;i<s;++i) { int j=n; while(j>=0 && p[j]!=pf[i]) { ++count[i]; --j; } } int max=count[0]; int pos=0; for(int i=1;i<s;++i) { if(count[i]>max) { max=count[i]; pos=i; } } return pos; } static boolean check(int p[],int n,int x) { for(int i=0;i<n;++i) { if(p[i]==x) return true; } return false; } public static void main(String args[]) { Scanner scr=new Scanner(System.in); System.out.println("Enter the number of page entries"); int n=scr.nextInt(); int fault=0,hit=0; int p[]=new int[n]; System.out.println("Enter the pages "); for(int i=0;i<n;++i) p[i]=scr.nextInt(); System.out.print("The page table entries are "); for(int i=0;i<n;++i) System.out.print(" "+p[i]); System.out.println(); int s=3; System.out.println("Do you wish to set a the number of page frames?(1=yes/0=no)"); int ch=scr.nextInt(); if(ch==1) { System.out.println("Enter the number of page frames "); s=scr.nextInt(); System.out.println("The number of page frames set to "+s); } else System.out.println("The default number of page frames are "+s); int pf[]=new int[s]; for(int i=0;i<s;++i) pf[i]=-1; int d=0; int choice=0; for(int i=0;i<n;++i) { if(check(pf,s,p[i])) { ++hit; for(int j=0;j<s && pf[j]!=-1;++j) System.out.print(pf[j]+" "); System.out.print("H\n"); } else { ++fault; if(fault<=s) { pf[d]=p[i]; ++d; } else { choice=check1(p,pf,i-1,s); pf[choice]=p[i]; } for(int j=0;j<s && pf[j]!=-1;++j) System.out.print(pf[j]+" "); System.out.print("F\n"); } } System.out.println("Number of page Hits= "+hit+"\nNumber of page Faults= "+fault); } }

Sem 4 Operating System (OS) : FIFO Page replacement Java code / program

Given below is the Java program for FIFO page replacement algorithm.You can also use the DOWNLOAD BOX to download the '.java' file. import java.util.*; class FIFO { static boolean check(int p[],int n,int x) { for(int i=0;i<n;++i) { if(p[i]==x) return true; } return false; } public static void main(String args[]) { Scanner scr=new Scanner(System.in); System.out.println("Enter the number of page entries"); int n=scr.nextInt(); int fault=0,hit=0; int p[]=new int[n]; System.out.println("Enter the pages "); for(int i=0;i<n;++i) p[i]=scr.nextInt(); System.out.print("The page table entries are "); for(int i=0;i<n;++i) System.out.print(" "+p[i]); System.out.println(); int s=3; System.out.println("Do you wish to set a the number of page frames?(1=yes/0=no)"); int ch=scr.nextInt(); if(ch==1) { System.out.println("Enter the number of page frames "); s=scr.nextInt(); System.out.println("The number of page frames set to "+s); } else System.out.println("The default number of page frames are "+s); int pf[]=new int[s]; for(int i=0;i<s;++i) pf[i]=-1; int d=0; for(int i=0;i<n;++i) { if(check(pf,s,p[i])) { ++hit; for(int j=0;j<s && pf[j]!=-1;++j) System.out.print(pf[j]+" "); System.out.print("H\n"); } else { ++fault; if(d>s-1) d=0; pf[d]=p[i]; ++d; for(int j=0;j<s && pf[j]!=-1;++j) System.out.print(pf[j]+" "); System.out.print("F\n"); } } System.out.println("Number of page Hits= "+hit+"\nNumber of page Faults= "+fault); } }

Sem 4 Operating System (OS) : Round Robin process scheduling Java Code / Program

Given below is the Java code for Round Robin Process Scheduling algorithm.

import java.util.*; class Process { Scanner scr=new Scanner(System.in); int p[]; int bt,tt,wt; Process(int i) { System.out.println("Enter the burst time of process "+(i+1)); bt=scr.nextInt(); } } class RR1 { static boolean check(Process p[],int n) { int s=0; for(int i=0;i<n;++i) s+=p[i].bt; if(s<=0) return false; else return true; } public static void main(String args[]) { Scanner scr=new Scanner(System.in); System.out.println("Enter the number of processes"); int n =scr.nextInt(); Process p[]=new Process[n]; for(int i=0;i<n;++i) p[i]=new Process(i); System.out.println("The quantum size is 5 ms "); int burst[]=new int[n]; for(int i=0;i<n;++i) burst[i]=p[i].bt; for(int i=0;i<n;++i) System.out.println("The burst time of process "+(i+1)+" is "+burst[i]); int time=0; int temp; int flag[]=new int[n]; for(int i=0;i<n;++i) flag[i]=0; do { loop:for(int k=0;k<n;++k) { //System.out.println(" Flag value is "+flag[k]); if(flag[k]==0) { temp=p[k].bt; if(p[k].bt>5) { p[k].bt-=5; time+=5; //System.out.println("The time value "+time); } else { p[k].bt=0; time+=temp; p[k].tt=time; flag[k]=1; //System.out.println("The process "+(k+1)+" ends"); //System.out.println("flag value "+flag[k]); } } else continue loop; } } while(check(p,n)); for(int i=0;i<n;++i) p[i].wt=p[i].tt-burst[i]; double s1=0,s2=0; for(int i=0;i<n;++i) { System.out.println("The turn around time of process "+(i+1)+" is "+p[i].tt); System.out.println("The wait time of process "+(i+1)+" is "+p[i].wt); } for(int i=0;i<n;++i) s1+=p[i].tt; for(int i=0;i<n;++i) s2+=p[i].wt; double att=0,awt=0; att=s1/n; awt=s2/n; System.out.println("The average turn around time is "+att+"\nThe average wait time is "+awt); } } /* Output Enter the number of processes 4 Enter the burst time of process 1 12 Enter the burst time of process 2 3 Enter the burst time of process 3 7 Enter the burst time of process 4 15 The quantum size is 5 ms The burst time of process 1 is 12 The burst time of process 2 is 3 The burst time of process 3 is 7 The burst time of process 4 is 15 The turn around time of process 1 is 32 The wait time of process 1 is 20 The turn around time of process 2 is 8 The wait time of process 2 is 5 The turn around time of process 3 is 25 The wait time of process 3 is 18 The turn around time of process 4 is 37 The wait time of process 4 is 22 The average turn around time is 25.5 The average wait time is 16.25 */

Sem 4 Operating System : Process scheduling Algorithm - Pre-emptive Priority Algorithm Java program / code

Given below is Java program for Pre-emptive Priority algorithm.

//Aim:To implement preemptive priority scheduling //Program: import java.util.*; class PR { public int bt; public int at; public int pt; public int pr; PR() { bt=0; at=0; pt=0; } PR(int x,int y,int z) { bt=x; at=y; pt=z; } void t(PR p2) { pr=p2.at-at; } } class PEP { static int sum(PR p[],int n) { int s=0; for(int i=0;i<n;++i) s+=p[i].bt; if(s==0) return 1; else return 0; } public static void main(String args[]) { PR p[]; Scanner scr=new Scanner(System.in); System.out.println("Enter the number of processes"); int n=scr.nextInt(); p=new PR[n]; int i=0,j=0; int bt,at,pt; for(i=0;i<n;++i) { System.out.println("Enter the burst time and arrival time & priority of a process"); bt=scr.nextInt(); at=scr.nextInt(); pt=scr.nextInt(); p[i]=new PR(bt,at,pt); } PR temp; // Sorting processes in increasing order of arrival time for(i=0;i<=n-2;++i) { for(j=i;j<=n-1;++j) { if(p[i].at>p[j].at) { temp=p[i]; p[i]=p[j]; p[j]=temp; } } } int tt[]=new int[n]; int wt[]=new int[n]; int b[]=new int[n]; for(i=0;i<n;++i) b[i]=p[i].bt; for(i=0;i<n;++i) { System.out.println("Process "+(i+1)+" Burst time= "+p[i].bt+" priority= "+p[i].pt+" Arrival time= "+p[i].at); } int time=0; int av; for(int k=0;k<n-1;++k) { p[k].t(p[k+1]); } System.out.println(); int t1; do { t1=0; av=0; for(i=0;i<n;++i) if(p[i].at<=time) ++av; int hp; int max=0; if(p[0].bt==0) { for(i=1;i<n;++i) if(p[i].bt!=0) { max=i;break; } } for(i=0;i<av;++i) { if((p[max].pt>p[i].pt) && p[i].bt>0) max=i; } hp=max; p[hp].bt-=1; ++time; if(p[hp].bt==0) tt[hp]=time; ++t1; } while(sum(p,n)!=1); double att=0,awt=0; double s1=0,s2=0; for(i=0;i<n;++i) tt[i]=tt[i]-p[i].at; System.out.println(); for(int h=0;h<n;++h) System.out.println("Turn around time of process "+(h+1)+" = "+tt[h]+" wait time = "+(tt[h]-b[h])); for(i=0;i<n;++i) wt[i]=tt[i]-b[i]; for(i=0;i<n;++i) { s1+=tt[i]; s2+=wt[i]; } att=s1/n; awt=s2/n; System.out.println("Average Turn around time= "+att+" Average wait time= "+awt); } } /* output Enter the number of processes 5 Enter the burst time and arrival time & priority of a process 10 0 3 Enter the burst time and arrival time & priority of a process 1 1 1 Enter the burst time and arrival time & priority of a process 2 2 3 Enter the burst time and arrival time & priority of a process 1 3 4 Enter the burst time and arrival time & priority of a process 5 4 2 Process 1 Burst time= 10 priority= 3 Arrival time= 0 Process 2 Burst time= 1 priority= 1 Arrival time= 1 Process 3 Burst time= 2 priority= 3 Arrival time= 2 Process 4 Burst time= 1 priority= 4 Arrival time= 3 Process 5 Burst time= 5 priority= 2 Arrival time= 4 Turn around time of process 1 = 16 wait time = 6 Turn around time of process 2 = 1 wait time = 0 Turn around time of process 3 = 16 wait time = 14 Turn around time of process 4 = 16 wait time = 15 Turn around time of process 5 = 5 wait time = 0 Average Turn around time= 10.8 Average wait time= 7.0 */

Sem 4 (AOAD) : Graph traversal Techniques- Breadth First Search (BFS) & Depth First Search (DFS) Java program / code.

Given below are java codes for graph traversal techniques.

Breadth First Search Java Code

// Aim:To implement Breadth first search import java.util.*; class Graph { Scanner scr=new Scanner(System.in); static int a[][]; static int visit[]=new int[10]; void adjm(int v,int e) { a=new int[v+1][v+1]; for(int i=1;i<=v;++i) { for(int j=1;j<=v;++j) { a[i][j]=0; } } for(int i=1;i<=e;++i) { System.out.println("Enter vertices of edge"); int b=scr.nextInt(); int c=scr.nextInt(); a[b][c]=a[c][b]=1; } } void bft(int v,int e) { for(int i=1;i<=v;++i) { if(visit[i]==0) bfs(i,v); } } void bfs(int u,int v) { visit[u]=1; System.out.print(" "+u); for(int i=1;i<=v;++i) { if(a[u][i]==1) { if(visit[i]==0) { visit[i]=1; System.out.print(" "+i); } } } } void dft() { for(int i=1;i<=v;++i) { if(visit[i]==0) dfs(i); } } void dfs(int u) { System.out.print(" "+u); visit[u]=1; for(int i=1;i<=v;++i) { if(a[u][i]==1) { if(visit[i]==0) dfs(i); } } } } class GT { public static void main(String args[]) { Scanner scr=new Scanner(System.in); try { System.out.println("Enter v & e"); int v=scr.nextInt(); int e=scr.nextInt(); Graph g=new Graph(); g.adjm(v,e); g.bft(v,e); } catch(Exception e) {} } } /* output Enter v & e 4 4 Enter vertices of edge 1 2 Enter vertices of edge 2 3 Enter vertices of edge 3 4 Enter vertices of edge 1 4 1 2 4 3 */


Depth First Search Java code
// Aim:To implement Depth first search import java.util.*; class Graph { Scanner scr=new Scanner(System.in); static int a[][]; static int visit[]=new int[10]; static int v,e; void adjm(int v,int e) { a=new int[v+1][v+1]; this.v=v; this.e=e; for(int i=1;i<=v;++i) { for(int j=1;j<=v;++j) { a[i][j]=0; } } for(int i=1;i<=e;++i) { System.out.println("Enter vertices of edge"); int b=scr.nextInt(); int c=scr.nextInt(); a[b][c]=a[c][b]=1; } } void dft() { for(int i=1;i<=v;++i) { if(visit[i]==0) dfs(i); } } void dfs(int u) { System.out.print(" "+u); visit[u]=1; for(int i=1;i<=v;++i) { if(a[u][i]==1) { if(visit[i]==0) dfs(i); } } } } class GT1 { public static void main(String args[]) { Scanner scr=new Scanner(System.in); System.out.println("Enter v & e"); int v=scr.nextInt(); int e=scr.nextInt(); Graph g=new Graph(); g.adjm(v,e); g.dft(); } } /* output Enter v & e 7 10 Enter vertices of edge 1 4 Enter vertices of edge 4 7 Enter vertices of edge 4 6 Enter vertices of edge 1 2 Enter vertices of edge 2 3 Enter vertices of edge 2 5 Enter vertices of edge 3 5 Enter vertices of edge 1 5 Enter vertices of edge 1 6 Enter vertices of edge 5 6 1 2 3 5 6 4 7 */

Semester 3:Data Structures and Files (DSF) Java programs for searching, sorting, Stack, Linked List, Queue free download.

For Stack , Queue, Linked list, Searching, Sorting  programs, use the 'Download Box' in the blog to download the file 'dsf.rar'.