Robotics and AI : Wumpus world problem Java Code/Program.

Given below is the java program for "Wumpus world problem".
The program gives a perfect implementation of this problem.It accepts the environment from the user dynamically, and then the agent tries to explore the best possible path to reach the gold, preventing the pits and the wumpus.
This is a perfect code with sample outputs for two scenarios/environments.

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// Aim:To implement Wumpus world problem in Java //Program: import java.util.*; class Environment { Scanner scr=new Scanner(System.in); //char w[][]=new char[5][5]; int np; //number of pits int wp,gp; // wumpus position gold position int pos[]; // position of pits int b_pos[]=new int[20]; int s_pos[]=new int[20]; void accept(String w[][]) { for(int i=0;i<20;++i) { b_pos[i]=-1; s_pos[i]=-1; } for(int i=0;i<5;++i) for(int j=0;j<5;++j) w[i][j]=""; int count=1; System.out.println("\n\n********* Wumpus World Problem *********\n-by Aditya Mandhare.\n"); System.out.println("The positions are as follows."); for(int i=1;i<=4;++i) { System.out.println("\n-----------------------------------------------------------------"); System.out.print("|\t"); for(int j=1;j<=4;++j) System.out.print((count++)+"\t|\t"); } System.out.println("\n-----------------------------------------------------------------"); System.out.println("\nAgent start position: 13"); w[4][1]="A"; System.out.println("\nEnter the number of pits."); np=scr.nextInt(); pos=new int[np]; System.out.println("Positions of pit, gold and wumpus should not overlap."); System.out.println("Enter the position of pits."); for(int i=0;i<np;++i) { pos[i]=scr.nextInt(); show_sense(pos[i],1,w); } System.out.println("Enter the position of wumpus."); wp=scr.nextInt(); show_sense(wp,2,w); System.out.println("Enter the position of gold."); gp=scr.nextInt(); insert(w); } void insert(String w[][]) { int temp=0; int count=0; int flag1=0,flag2=0; for(int i=0;i<np;++i) { temp=pos[i]; count=0; for(int j=1;j<=4;++j) { for(int k=1;k<=4;++k) { ++count; if(count==temp) w[j][k]+="P"; else if(count==gp && flag1==0) { w[j][k]+="G"; flag1=1; } else if(count==wp && flag2==0) { w[j][k]+="W"; flag2=1; } } } } display(w); } void show_sense(int a,int b,String w[][]) { int t1,t2,t3,t4; t1=a-1; t2=a+1; t3=a+4; t4=a-4; if(a==5 || a==9) t1=0; if(a==8 || a==12) t2=0; if(a==4) t2=0; if(a==13) t1=0; if(t3>16) t3=0; if(t4<0) t4=0; //int temp[]=new int[4]; if(b==1) {b_pos[0]=t1;b_pos[1]=t2;b_pos[2]=t3;b_pos[3]=t4;} else if(b==2) {s_pos[0]=t1;s_pos[1]=t2;s_pos[2]=t3;s_pos[3]=t4;} int temp1,count; for(int i=0;i<4;++i) { if(b==1) temp1=b_pos[i]; else temp1=s_pos[i]; count=0; for(int j=1;j<=4;++j) { for(int k=1;k<=4;++k) { ++count; if(count==temp1 && b==1 && !w[j][k].contains("B")) { w[j][k]+="B"; } else if(count==temp1 && b==2 && !w[j][k].contains("S")) w[j][k]+="S"; } } } //display(w); } void display(String w[][]) { System.out.println("\nThe environment for problem is as follows.\n"); for(int i=1;i<=4;++i) { System.out.println("\n-----------------------------------------------------------------"); System.out.print("|\t"); for(int j=1;j<=4;++j) System.out.print(w[i][j]+"\t|\t"); } System.out.println("\n-----------------------------------------------------------------"); } } class tiles { int safe=0; int unsafe=0; int wump=0; int pit=0; int gold=0; int doubt_pit=0; int doubt_wump=0; String env; int num=0; int br=0; int bl=0; int bu=0; int bd=0; int visited=0; int l,r,u,d; String back=""; tiles(String s,int n) { env=s; num=n; l=r=u=d=0; if(n==9 || n==5) bl=1; if(n==8 || n==12) br=1; if(n==1) { bu=1;bl=1; } if(n==13) { bd=1;bl=1; } if(n==4) { bu=1;br=1; } if(n==16) { bd=1;br=1; } } int sense() { if(env.contains("B")) return 1; else if(env.contains("S")) return 2; else if(env.contains("G")) return 3; if(env.contains("W")) return 4; else return 0; } } class wumpus { static int scream=0; static int score=0; static int complete=0; static boolean check(tiles t) { int temp=t.sense(); if(temp==1 || temp==2) return false; return true; } public static void main(String args[]) { Scanner scr=new Scanner(System.in); Environment e=new Environment(); String w[][]=new String[5][5]; e.accept(w); System.out.println("\n\nFinding the solution..."); tiles t[]=new tiles[17]; int c=1; out:for(int i=1;i<5;++i) { for(int j=1;j<5;++j) { if(c>16) break out; t[c]=new tiles(w[i][j],c); ++c; } } t[13].safe=1; t[13].visited=1; int pos=13; int condition; int limit=0; String temp1,temp2; do { ++limit; condition=-1; if(t[pos].env.contains("G")) { complete=1; System.out.println("Gold Found!!"); break; } if(t[pos].br!=1 && t[pos].r!=1 && t[pos+1].doubt_pit<1 && t[pos+1].doubt_wump<1 && t[pos+1].pit!=1 && t[pos+1].wump!=1 && !(t[pos].back.contains("r") && (t[pos].l!=1 || t[pos].u!=1 || t[pos].d!=1) && check(t[pos]) )) { //////////// temp1="l"; /////////// t[pos].r=1; ++pos; System.out.println("\nfront pos="+pos); ++score; //t[pos].visited=1; //////////////// t[pos].back+=temp1; //////////////// condition=t[pos].sense(); if(condition==3) {complete=1;break;} else if(condition==1 && t[pos].visited==0) { if(t[pos].br!=1 && t[pos+1].safe!=1) t[pos+1].doubt_pit+=1; if(t[pos].bu!=1 && (pos-4)>=1 && t[pos-4].safe!=1) t[pos-4].doubt_pit+=1; if(t[pos].bl!=1 && t[pos-1].safe!=1 ) t[pos-1].doubt_pit+=1; if(t[pos].bd!=1 && (pos+4)<=16 && t[pos+4].safe!=1) t[pos+4].doubt_pit+=1; t[pos].safe=1; } else if(condition==2 && t[pos].visited==0) { if(t[pos].br!=1 && t[pos+1].safe!=1) t[pos+1].doubt_wump+=1; if(t[pos].bu!=1 && (pos-4)>=1 && t[pos-4].safe!=1) t[pos-4].doubt_wump+=1; if(t[pos].bl!=1 && t[pos-1].safe!=1) t[pos-1].doubt_wump+=1; if(t[pos].bd!=1 && (pos+4)<=16 && t[pos+4].safe!=1) t[pos+4].doubt_wump+=1; t[pos].safe=1; } /*else if(condition==4) { score=score+100; t[pos].safe=1; }*/ else if(condition==0) t[pos].safe=1; t[pos].visited=1; } else if(t[pos].bl!=1 && t[pos].l!=1 && t[pos-1].doubt_pit<1 && t[pos-1].doubt_wump<1 && t[pos-1].pit!=1 && t[pos-1].wump!=1 && !(t[pos].back.contains("l") && (t[pos].r!=1 || t[pos].u!=1 || t[pos].d!=1) && check(t[pos]) )) { //////////////////// temp1="r"; /////////////////// t[pos].l=1; pos=pos-1; System.out.println("\nback pos= "+pos); ++score; //t[pos].visited=1; ////////////////////// t[pos].back+=temp1; ///////////////////// condition=t[pos].sense(); if(condition==3) {complete=1;break;} else if(condition==1 && t[pos].visited==0) { if(t[pos].br!=1 && t[pos+1].safe!=1) t[pos+1].doubt_pit+=1; if(t[pos].bu!=1 && (pos-4)>=1 && t[pos-4].safe!=1) t[pos-4].doubt_pit+=1; if(t[pos].bl!=1 && t[pos-1].safe!=1) t[pos-1].doubt_pit+=1; if(t[pos].bd!=1 && (pos+4)<=16 && t[pos+4].safe!=1) t[pos+4].doubt_pit+=1; t[pos].safe=1; } else if(condition==2 && t[pos].visited==0) { if(t[pos].br!=1 && t[pos+1].safe!=1) t[pos+1].doubt_wump+=1; if(t[pos].bu!=1 && (pos-4)>=1 && t[pos-4].safe!=1) t[pos-4].doubt_wump+=1; if(t[pos].bl!=1 && t[pos-1].safe!=1) t[pos-1].doubt_wump+=1; if(t[pos].bd!=1 && (pos+4)<=16 && t[pos+4].safe!=1) t[pos+4].doubt_wump+=1; t[pos].safe=1; } else if(condition==0) t[pos].safe=1; t[pos].visited=1; } else if(t[pos].bu!=1 && t[pos].u!=1 && (pos-4)>=1 && t[pos-4].doubt_pit<1 && t[pos-4].doubt_wump<1 && t[pos-4].pit!=1 && t[pos-1].wump!=1 && !(t[pos].back.contains("u") && (t[pos].l!=1 || t[pos].r!=1 || t[pos].d!=1) && check(t[pos]) )) { ///////////////////// temp1="d"; ///////////////////// t[pos].u=1; pos=pos-4; System.out.println("\nUp pos= "+pos); ++score; //t[pos].visited=1; /////////////////////// t[pos].back+=temp1; ///////////////////// condition=t[pos].sense(); if(condition==3) {complete=1;break;} else if(condition==1 && t[pos].visited==0) { if(t[pos].br!=1 && t[pos+1].safe!=1) t[pos+1].doubt_pit+=1; if(t[pos].bu!=1 && (pos-4)>=1 && t[pos-4].safe!=1) t[pos-4].doubt_pit+=1; if(t[pos].bl!=1 && t[pos-1].safe!=1) t[pos-1].doubt_pit+=1; if(t[pos].bd!=1 && (pos+4)<=16 && t[pos+4].safe!=1) t[pos+4].doubt_pit+=1; t[pos].safe=1; } else if(condition==2 && t[pos].visited==0) { if(t[pos].br!=1 && t[pos+1].safe!=1) t[pos+1].doubt_wump+=1; if(t[pos].bu!=1 && (pos-4)>=1 && t[pos-4].safe!=1) t[pos-4].doubt_wump+=1; if(t[pos].bl!=1 && t[pos-1].safe!=1) t[pos-1].doubt_wump+=1; if(t[pos].bd!=1 && (pos+4)<=16 && t[pos+4].safe!=1) t[pos+4].doubt_wump+=1; t[pos].safe=1; } else if(condition==0) t[pos].safe=1; t[pos].visited=1; } else if(t[pos].bd!=1 && t[pos].d!=1 && (pos+4)<=16 && t[pos+4].doubt_pit<1 && t[pos+4].doubt_wump<1 && t[pos+4].pit!=1 && t[pos+4].wump!=1) { ///////////////// temp1="u"; //////////////// t[pos].d=1; pos=pos+4; System.out.println("\ndown pos= "+pos); ++score; //t[pos].visited=1; ////////////////// t[pos].back+=temp1; ////////////////// condition=t[pos].sense(); if(condition==3) {complete=1;break;} else if(condition==1 && t[pos].visited==0) { if(t[pos].br!=1 && t[pos+1].safe!=1) t[pos+1].doubt_pit+=1; if(t[pos].bu!=1 && (pos-4)>=1 && t[pos-4].safe!=1) t[pos-4].doubt_pit+=1; if(t[pos].bl!=1 && t[pos-1].safe!=1) t[pos-1].doubt_pit+=1; if(t[pos].bd!=1 && (pos+4)<=16 && t[pos+4].safe!=1) t[pos+4].doubt_pit+=1; t[pos].safe=1; } else if(condition==2 && t[pos].visited==0) { if(t[pos].br!=1 && t[pos+1].safe!=1) t[pos+1].doubt_wump+=1; if(t[pos].bu!=1 && (pos-4)>=1 && t[pos-4].safe!=1) t[pos-4].doubt_wump+=1; if(t[pos].bl!=1 && t[pos-1].safe!=1) t[pos-1].doubt_wump+=1; if(t[pos].bd!=1 && (pos+4)<=16 && t[pos+4].safe!=1) t[pos+4].doubt_wump+=1; t[pos].safe=1; } else if(condition==0) t[pos].safe=1; t[pos].visited=1; } else if(limit>50) { int temp3=pos; int flag_1=0,flag2=0,flag3=0,flag4=0; System.out.println("\nCurrently at position "+temp3+".\nThinking...."); //if(!(t[pos].back.contains("r") && (t[pos].l!=1 || t[pos].u!=1 || t[pos].d!=1) && check(t[pos]) )) while(t[pos].visited==1 && t[pos].br!=1) { ++pos; ++score; } if(t[pos].pit==1 || t[pos].wump==1 || (t[pos].br==1 && t[pos].visited==1 && t[pos].safe!=1)) { //System.out.println("\nUnsuccessful at pos "+pos); pos=temp3; //System.out.println("\nBack at pos "+pos); flag_1=1; } if(flag_1==0) t[pos].back+="l"; //if(!(t[pos].back.contains("u") && (t[pos].l!=1 || t[pos].r!=1 || t[pos].d!=1) && check(t[pos]) )) while(pos+4>=1 && t[pos].bu!=1 && t[pos].visited==1) { pos-=4; ++score; } if(t[pos].pit==1 || t[pos].wump==1 || (t[pos].bu==1 && t[pos].visited==1 && t[pos].safe!=1)) { //System.out.println("\nUnsuccessful at pos "+pos); pos=temp3; //System.out.println("\nBack at pos "+pos); flag3=1; } if(flag3==0) t[pos].back+="d"; //if(!(t[pos].back.contains("l") && (t[pos].r!=1 || t[pos].u!=1 || t[pos].d!=1) && check(t[pos]) )) while(t[pos].visited==1 && t[pos].bl!=1) { --pos; ++score; } if(t[pos].pit==1 || t[pos].wump==1 || (t[pos].bl==1 && t[pos].visited==1 && t[pos].safe!=1)) { //System.out.println("\nUnsuccessful at pos "+pos); pos=temp3; //System.out.println("\nBack at pos "+pos); flag2=1; } if(flag2==0) t[pos].back+="r"; //if(!(t[pos].back.contains("d") && (t[pos].l!=1 || t[pos].r!=1 || t[pos].u!=1) && check(t[pos]) )) while(pos+4<=16 && t[pos].bd!=1 && t[pos].visited==1) { pos+=4; ++score; } if(t[pos].pit==1 || t[pos].wump==1 || (t[pos].bd==1 && t[pos].visited==1 && t[pos].safe!=1)) { //System.out.println("\nUnsuccessful at pos "+pos); pos=temp3; //System.out.println("\nBack at pos "+pos); flag4=1; } //t[pos-1].r=0; //++pos; //if(!t[pos].env.contains("P") && !t[pos].env.contains("W")) if(flag4==0) t[pos].back+="u"; t[pos].safe=1; t[pos].visited=1; System.out.println("reached at position "+pos); limit=0; } if(t[pos].env.contains("W") && scream!=1) { score+=100; scream=1; t[pos].safe=1; System.out.println("\n\nWumpus killed >--0-->"); t[pos].env.replace("W"," "); for(int l=1;l<=16;++l) { t[l].doubt_wump=0; t[l].env.replace("S"," "); } } if(t[pos].env.contains("P")) { score+=50; t[pos].pit=1; System.out.println("\n\nFallen in pit of position "+pos+"."); } for(int k=1;k<=16;++k) { if(t[k].doubt_pit==1 && t[k].doubt_wump==1) { t[k].doubt_pit=0; t[k].doubt_wump=0; t[k].safe=1; } } for(int y=1;y<=16;++y) { if(t[y].doubt_wump>1) { t[y].wump=1; for(int h=1;h<=16;++h) { if(h!=y) { t[h].doubt_wump=0; t[h].env.replace("S"," "); } } } } /////////////////////////// for(int y=1;y<=16;++y) { if(t[y].doubt_pit>1) {t[y].pit=1; //System.out.println("\nPit confirmed at position "+y); } } /////////////////////////// try{Thread.sleep(200);}catch(Exception p){} } while(complete==0); if(complete==1) { //score=score*2; //if(scream==1) //score-=100; score*=-1; score+=1000; } System.out.println("The score of the agent till he reaches gold is "+score+".\nNow he will return back following the best explored path."); } } /* OUTPUT 1 C:\Users\Aditya\Desktop\aditya>javac wumpus.java C:\Users\Aditya\Desktop\aditya>java wumpus ********* Wumpus World Problem ********* The positions are as follows. ----------------------------------------------------------------- | 1 | 2 | 3 | 4 | ----------------------------------------------------------------- | 5 | 6 | 7 | 8 | ----------------------------------------------------------------- | 9 | 10 | 11 | 12 | ----------------------------------------------------------------- | 13 | 14 | 15 | 16 | ----------------------------------------------------------------- Agent start position: 13 Enter the number of pits. 3 Positions of pit, gold and wumpus should not overlap. Enter the position of pits. 4 7 15 Enter the position of wumpus. 5 Enter the position of gold. 6 The environment for problem is as follows. ----------------------------------------------------------------- | S | | B | P | ----------------------------------------------------------------- | W | BSG | P | B | ----------------------------------------------------------------- | S | | B | | ----------------------------------------------------------------- | A | B | P | B | ----------------------------------------------------------------- Finding the solution... front pos=14 back pos= 13 Up pos= 9 front pos=10 front pos=11 back pos= 10 Up pos= 6 Gold Found!! The score of the agent till he reaches gold is 993. Now he will return back following the best explored path. OUTPUT 2 C:\Users\Aditya\Desktop\aditya>javac wumpus.java C:\Users\Aditya\Desktop\aditya>java wumpus ********* Wumpus World Problem ********* The positions are as follows. ----------------------------------------------------------------- | 1 | 2 | 3 | 4 | ----------------------------------------------------------------- | 5 | 6 | 7 | 8 | ----------------------------------------------------------------- | 9 | 10 | 11 | 12 | ----------------------------------------------------------------- | 13 | 14 | 15 | 16 | ----------------------------------------------------------------- Agent start position: 13 Enter the number of pits. 2 Positions of pit, gold and wumpus should not overlap. Enter the position of pits. 5 11 Enter the position of wumpus. 6 Enter the position of gold. 4 The environment for problem is as follows. ----------------------------------------------------------------- | B | S | | G | ----------------------------------------------------------------- | SP | BW | BS | | ----------------------------------------------------------------- | B | BS | P | B | ----------------------------------------------------------------- | A | | B | | ----------------------------------------------------------------- Finding the solution... front pos=14 front pos=15 back pos= 14 Up pos= 10 down pos= 14 Currently at position 14. Thinking.... reached at position 16 Currently at position 16. Thinking.... reached at position 12 Up pos= 8 back pos= 7 front pos=8 Up pos= 4 The score of the agent till he reaches gold is 988. Now he will return back following the best explored path. */

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Sem 4 & Sem 7 (AOAD & RAI) : N queen Algorithm Java Code

import java.util.*; class nqueen { static int n=10; static int board[]=new int[n]; static boolean place(int row,int column) { for(int i=1;i<=row-1;++i) { if(column==board[i]) return false; else if(abs(column-board[i])==abs(row-i)) return false; } return true; } static int abs(int arg) { if(arg<0) return arg*-1; else if(arg==0) return 0; else return arg; } static void queen(int row,int n) { for(int column=1;column<=n;++column) { if(place(row,column)) { board[row]=column; if(row==n) display(); else queen(row+1,n); } } } static void display() { System.out.println(); System.out.print("Pattern "); for(int i=1;i<=n;++i) System.out.print(" "+board[i]); System.out.println(); int pos; System.out.print(" "); for(int i=1;i<=n;++i) System.out.print(" "+i); System.out.println(); for(int i=1;i<=n;++i) { pos=board[i]; System.out.print(" "+i); for(int j=1;j<=n;++j) { if(j==pos) System.out.print(" Q"); else System.out.print(" -"); } System.out.println(); } } public static void main(String args[]) { Scanner scr=new Scanner(System.in); System.out.println("**********N Queen problem*************"); System.out.println("Enter the number of queens"); n=scr.nextInt(); /* IF there are 'n' queens there are n rows and n columns */ queen(1,n); } }

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Water Jug Problem Java program

The java code for Water Jug Problem is given below.There are many solutions for it run the code to find all.

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//Aim:To implement Water jug problem in Java. //Program: import java.util.*; class wj2 { static int gallon1,gallon2; static int wastage; static int steps; static int rules[]=new int[15]; static int reset=0; static int ct=0; static int incr; static int tgallon1,tgallon2; static int check() { if(steps>10 || reset==1) return 1; else if(gallon1==2) return 2; else return 0; } static void empty1() { wastage=wastage+gallon1; gallon1=0; } static void empty2() { wastage=wastage+gallon2; gallon2=0; } static void fill1() { gallon1=gallon1+4; } static void fill2() { gallon2+=3; } static void gallon1_to_gallon2() { int temp=0; temp=3-gallon2; gallon2=3; gallon1=gallon1-temp; } static void gallon2_to_gallon1() { int temp=0; temp=4-gallon1; gallon1=4; gallon2=gallon2-temp; } static void empty1_fill2() { gallon2=gallon2+gallon1; gallon1=0; } static void empty2_fill1() { gallon1=gallon1+gallon2; gallon2=0; } static void call(int n) { tgallon1=gallon1; tgallon2=gallon2; try { //Thread.sleep(1000); } catch(Exception e) { System.out.println("Exception in thread"); } if(steps>10) { System.out.println("ERROR num of steps greater than 10"); } ++steps; if(n==1) fill1(); //empty1(); else if(n==2) fill2(); //empty2(); else if(n==3) empty1(); //fill1(); else if(n==4) empty2(); //fill2(); else if(n==5) gallon1_to_gallon2(); else if(n==6) gallon2_to_gallon1(); else if(n==7) empty1_fill2(); else if(n==8) empty2_fill1(); if(gallon1>4 || gallon1<0 || gallon2>3 || gallon2<0) reset=1; } static void repeat(int rules[]) { gallon1=0; gallon2=0; wastage=0; steps=0; System.out.println("\nGallon1\tGallon2\tRule"); System.out.println(gallon1+"\t"+gallon2+"\t"); call(rules[0]); if(rules[0]!=0) System.out.println(gallon1+"\t"+gallon2+"\t"+rules[0]); call(rules[1]); if(rules[1]!=0) System.out.println(gallon1+"\t"+gallon2+"\t"+rules[1]); call(rules[2]); if(rules[2]!=0) System.out.println(gallon1+"\t"+gallon2+"\t"+rules[2]); call(rules[3]); if(rules[3]!=0) System.out.println(gallon1+"\t"+gallon2+"\t"+rules[3]); call(rules[4]); if(rules[4]!=0) System.out.println(gallon1+"\t"+gallon2+"\t"+rules[4]); call(rules[5]); if(rules[5]!=0) System.out.println(gallon1+"\t"+gallon2+"\t"+rules[5]); call(rules[6]); if(rules[6]!=0) System.out.println(gallon1+"\t"+gallon2+"\t"+rules[6]); call(rules[7]); if(rules[7]!=0) System.out.println(gallon1+"\t"+gallon2+"\t"+rules[7]); call(rules[8]); if(rules[8]!=0) System.out.println(gallon1+"\t"+gallon2+"\t"+rules[8]); call(rules[9]); if(rules[9]!=0) System.out.println(gallon1+"\t"+gallon2+"\t"+rules[9]); } public static void main(String args[]) { Scanner scr=new Scanner(System.in); System.out.println("**********This is waterjug *************"); System.out.println("\nRules:\n1.Fill Gallon 1.\n2.Fill Gallon 2.\n3.Empty gallon 1.\n4.Empty Gallon 2.\n5.Gallon 1 to Gallon 2 till full.\n6.Gallon 2 to Gallon 1 till full.\n7.Empty Gallon 1 and add to Gallon 2.\n8.Empty Gallon 2 and add to Gallon 1."); System.out.println("\nThe Optimal solution is the one with minimum steps and minimum wastage.\n"); int flag=0; int count1=0; flag=0; count1=0; wastage=0; steps=0; for(int i=0;i<10;++i) rules[i]=0; reset=0; int temp1=0; incr=0; /************Data for testing the functionality****************/ /* rules[0]=1;rules[1]=5;rules[2]=3;rules[3]=8;rules[4]=2;rules[5]=6;rules[6]=3;rules[7]=8;rules[8]=0;rules[9]=0; repeat(rules); System.out.println("\nWastage= "+wastage); rules[0]=2;rules[1]=8;rules[2]=2;rules[3]=6;rules[4]=3;rules[5]=8;rules[6]=0;rules[7]=0;rules[8]=0;rules[9]=0; repeat(rules); System.out.println("\nWastage= "+wastage); rules[0]=1;rules[1]=5;rules[2]=4;rules[3]=7;rules[4]=1;rules[5]=5;rules[6]=0;rules[7]=0;rules[8]=0;rules[9]=0; repeat(rules); System.out.println("\nWastage= "+wastage); */ l1:for(int i=0;i<8;++i) { l2:for(int j=0;j<8;++j) { l3:for(int k=0;k<8;++k) { l4:for(int l=0;l<8;++l) { l5:for(int m=0;m<8;++m) { l6:for(int n=0;n<8;++n) { l7:for(int o=0;o<8;++o) { l8:for(int p=0;p<8;++p) { l9:for(int q=0;q<8;++q) { l10:for(int r=0;r<8;++r) { reset=0; call(i); call(j); call(k); call(l); call(m); call(n); call(o); call(p); call(q); call(r); if(gallon1==2 && reset==0) { try { Thread.sleep(10000); rules[0]=i;rules[1]=j;rules[2]=k;rules[3]=l;rules[4]=m;rules[5]=n;rules[6]=o;rules[7]=p;rules[8]=q;rules[9]=r; System.out.println("\nFound a pattern !!!!\n"); repeat(rules); int count2=0; for(int z=0;z<10;++z) { if(rules[z]!=0) { //System.out.print(" "+rules[z]); ++count2; } } System.out.println("\nNumber of Steps= "+(count2+1)); System.out.println("\nWastage= "+wastage); System.out.println("\n\n\nPlease wait for another solution....."); Thread.sleep(10000); } catch(Exception e) {} } gallon1=0; gallon2=0; steps=0; reset=0; wastage=0; } } } } } } } } } } } } /*OUTPUT C:\Users\Aditya\Desktop\aditya>javac wj2.java C:\Users\Aditya\Desktop\aditya>java wj2 **********This is waterjug ************* Rules: 1.Fill Gallon 1. 2.Fill Gallon 2. 3.Empty gallon 1. 4.Empty Gallon 2. 5.Gallon 1 to Gallon 2 till full. 6.Gallon 2 to Gallon 1 till full. 7.Empty Gallon 1 and add to Gallon 2. 8.Empty Gallon 2 and add to Gallon 1. The Optimal solution is the one with minimum steps and minimum wastage. Found a pattern !!!! Gallon1 Gallon2 Rule 0 0 4 0 1 1 3 5 0 3 3 3 0 8 3 3 2 4 2 6 0 2 3 2 0 8 Number of Steps= 9 Wastage= 5 Please wait for another solution..... Found a pattern !!!! Gallon1 Gallon2 Rule 0 0 0 3 2 3 0 8 3 3 2 4 2 6 0 2 3 2 0 8 Number of Steps= 7 Wastage= 4 Please wait for another solution..... Found a pattern !!!! Gallon1 Gallon2 Rule 0 0 4 0 1 1 3 5 1 0 4 0 1 7 4 1 1 2 3 5 Number of Steps= 7 Wastage= 3 Please wait for another solution..... */

System Security : C code for Playfair Cipher (Should be executed in Linux with gcc.)

The Playfair cipher or Playfair square is a manual symmetric encryption technique and was the first literal digraph substitution cipher. The scheme was invented in 1854 by Charles Wheatstone, but bears the name of Lord Playfair who promoted the use of the cipher.

The technique encrypts pairs of letters (digraphs), instead of single letters as in the simple substitution cipher and rather more complex Vigenère ciphersystems then in use. The Playfair is thus significantly harder to break since the frequency analysis used for simple substitution ciphers does not work with it. Frequency analysis can still be undertaken, but on the 600 possible digraphs rather than the 26 possible monographs. The frequency analysis of digraphs is possible, but considerably more difficult – and it generally requires a much larger ciphertext in order to be useful.

//Aim:To implement Playfair cipher in C. //Program: #include<stdio.h> int check(char table[5][5],char k) { int i,j; for(i=0;i<5;++i) for(j=0;j<5;++j) { if(table[i][j]==k) return 0; } return 1; } void main() { int i,j,key_len; char table[5][5]; for(i=0;i<5;++i) for(j=0;j<5;++j) table[i][j]='0'; printf("**********Playfair Cipher************\n\n"); printf("Enter the length of the Key. "); scanf("%d",&key_len); char key[key_len]; printf("Enter the Key. "); for(i=-1;i<key_len;++i) { scanf("%c",&key[i]); if(key[i]=='j') key[i]='i'; } int flag; int count=0; // inserting the key into the table for(i=0;i<5;++i) { for(j=0;j<5;++j) { flag=0; while(flag!=1) { if(count>key_len) goto l1; flag=check(table,key[count]); ++count; }// end of while table[i][j]=key[(count-1)]; }// end of inner for }// end of outer for l1:printf("\n"); int val=97; //inserting other alphabets for(i=0;i<5;++i) { for(j=0;j<5;++j) { if(table[i][j]>=97 && table[i][j]<=123) {} else { flag=0; while(flag!=1) { if('j'==(char)val) ++val; flag=check(table,(char)val); ++val; }// end of while table[i][j]=(char)(val-1); }//end of else }// end of inner for }// end of outer for printf("The table is as follows:\n"); for(i=0;i<5;++i) { for(j=0;j<5;++j) { printf("%c ",table[i][j]); } printf("\n"); } int l=0; printf("\nEnter the length length of plain text.(without spaces) "); scanf("%d",&l); printf("\nEnter the Plain text. "); char p[l]; for(i=-1;i<l;++i) { scanf("%c",&p[i]); } for(i=-1;i<l;++i) { if(p[i]=='j') p[i]='i'; } printf("\nThe replaced text(j with i)"); for(i=-1;i<l;++i) printf("%c ",p[i]); count=0; for(i=-1;i<l;++i) { if(p[i]==p[i+1]) count=count+1; } printf("\nThe cipher has to enter %d bogus char.It is either 'x' or 'z'\n",count); int length=0; if((l+count)%2!=0) length=(l+count+1); else length=(l+count); printf("\nValue of length is %d.\n",length); char p1[length]; //inserting bogus characters. char temp1; int count1=0; for(i=-1;i<l;++i) { p1[count1]=p[i]; if(p[i]==p[i+1]) { count1=count1+1; if(p[i]=='x') p1[count1]='z'; else p1[count1]='x'; } count1=count1+1; } //checking for length char bogus; if((l+count)%2!=0) { if(p1[length-1]=='x') p1[length]='z'; else p1[length]='x'; } printf("The final text is:"); for(i=0;i<=length;++i) printf("%c ",p1[i]); char cipher_text[length]; int r1,r2,c1,c2; int k1; for(k1=1;k1<=length;++k1) { for(i=0;i<5;++i) { for(j=0;j<5;++j) { if(table[i][j]==p1[k1]) { r1=i; c1=j; } else if(table[i][j]==p1[k1+1]) { r2=i; c2=j; } }//end of for with j }//end of for with i if(r1==r2) { cipher_text[k1]=table[r1][(c1+1)%5]; cipher_text[k1+1]=table[r1][(c2+1)%5]; } else if(c1==c2) { cipher_text[k1]=table[(r1+1)%5][c1]; cipher_text[k1+1]=table[(r2+1)%5][c1]; } else { cipher_text[k1]=table[r1][c2]; cipher_text[k1+1]=table[r2][c1]; } k1=k1+1; }//end of for with k1 printf("\n\nThe Cipher text is:\n "); for(i=1;i<=length;++i) printf("%c ",cipher_text[i]); } /*OUTPUT [aditya@localhost Desktop]$ gcc playfair.c [aditya@localhost Desktop]$ ./a.out **********Playfair Cipher************ Enter the length of the Key. 15 Enter the Key. playfairexample The table is as follows: p l a y f i r e x m b c d g h k n o q s t u v w z Enter the length length of plain text.(without spaces) 25 Enter the Plain text. hidethegoldinthetreestump The replaced text(j with i) h i d e t h e g o l d i n t h e t r e e s t u m p The cipher has to enter 1 bogus char.It is either 'x' or 'z' Value of length is 26. The final text is: h i d e t h e g o l d i n t h e t r e x e s t u m p The Cipher text is: b m o d z b x d n a b e k u d m u i x m m o u v i f */

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System Security : C code for Transposition Cipher (should be executed in linux with gcc)

In cryptography, a transposition cipher is a method of encryption by which the positions held by units of plaintext (which are commonly characters or groups of characters) are shifted according to a regular system, so that the ciphertext constitutes a permutation of the plaintext. That is, the order of the units is changed. Mathematically a bijective function is used on the characters' positions to encrypt and an inverse function to decrypt.
//Aim:To implement Transposition Cipher in C. //Program: #include<stdio.h> int check(int x,int y) { int a,b,c; if(x%y==0) return 0; a=x/y; b=y*(a+1); c=b-x; return c; } void main() { int l1,i,d,j; printf("\nEnter the length of the key. "); scanf("%d",&l1); int sequence[l1]; printf("\nEnter the sequence key. "); for(i=0;i<l1;++i) { scanf("%d",&sequence[i]); } int order[l1]; for(i=1;i<=l1;++i) { for(j=0;j<l1;++j) { if(sequence[j]==i) order[i-1]=j; } } printf("\nEnter the depth. "); scanf("%d",&d); int l2; printf("\nEnter the length of String without spaces . "); scanf("%d",&l2); int temp1=check(l2,l1); int r=(l2+temp1)/l1; char p[l2+temp1]; char p1[r][l1]; //char p2[r][l1]; if(temp1>0) printf("\nYou need to enter %d bogus characters.So enter total %d characters. ",temp1,(l2+temp1)); else printf("\nEnter the string. "); for(i=-1;i<(l2+temp1);++i) { scanf("%c",&p[i]); } int count=0; while(d>0) { count=0; for(i=0;i<r;++i) { for(j=0;j<l1;++j) { p1[i][j]=p[count]; count=count+1; } } printf("\n\n\n"); for(i=0;i<r;++i) { for(j=0;j<l1;++j) { printf("%c ",p1[i][j]); } printf("\n"); } count=0; for(i=0;i<l1;++i) { for(j=0;j<r;++j) { p[count]=p1[j][order[i]]; count=count+1; } } for(i=0;i<(l2+temp1);++i) printf("%c ",p[i]); d=d-1; } } /* OUTPUT [aditya@localhost Desktop]$ gcc transposition.c [aditya@localhost Desktop]$ ./a.out Enter the length of the key. 7 Enter the sequence key. 4 3 1 2 5 6 7 Enter the depth. 2 Enter the length of String without spaces . 23 You need to enter 5 bogus characters.So enter total 28 characters. attackpostponeduntiltwoamxyz a t t a c k p o s t p o n e d u n t i l t w o a m x y z t t n a a p t m t s u o a o d w c o i x k n l y p e t z t t n a a p t m t s u o a o d w c o i x k n l y p e t z n s c y a u o p t t w l t m d n a o i e p a x t t o k z */

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