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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