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..... */