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 */