Answered step by step
Verified Expert Solution
Link Copied!

Question

1 Approved Answer

If you cant answer this please dont waste my question. thank you. This cryptographic program run and produce text screen output. You are to create

If you cant answer this please dont waste my question. thank you.

This cryptographic program run and produce text screen output. You are to create a GUI that uses the program. Your program (GUI) must allow a user to input of a message to be coded. It must also have an area to show the plaintext, the ciphertext, and the decrypted text. If required by your choice of cryptographic method, the user should have an area to input a key. You should also include instructions to the user on what and how to use your program. Remember user-friendly.

You should not have to rewrite any of the example programs just modify them to use a GUI.

Please add comments to code

import java.util.Scanner;

public class OneTimePadCipher

{

public static String encryptionMessage(String s)

{

int i, j; int randomBitPattern[] = new int[8];

for (i = 0; i < 7; i++)

{

randomBitPattern[i] = (i % 2 == 0) ? 1 : 0;

}

char asc[] = new char[s.length()];

for (i = 0; i < s.length(); i++)

{

asc[i] = (char) ((int) s.charAt(i));

}

BasicOperation b1 = new BasicOperation();

String cipherText = new String("");

for (i = 0; i < asc.length; i++)

{

int temp = (int) (asc[i]);

int len = b1.decimalToBinary(temp); int bintemp[] = new int[7]; int xorlen;

if (len == 7)

{

for (j = 1; j <= len; j++)

{

bintemp[j - 1] = b1.binaryArrayAtPosition(j);

}

// XOR Operation

xorlen = b1.xorop(bintemp, randomBitPattern, len); }

else

{

// System.out.println(" less than 7 :"+len);

bintemp[0] = 0;

for (j = 1; j <= len; j++)

{

bintemp[j] = b1.binaryArrayAtPosition(j);

}

// XOR Operation

xorlen = b1.xorop(bintemp, randomBitPattern, len + 1);

}

int xor[] = new int[xorlen];

for (j = 0; j < xorlen; j++)

{

xor[j] = b1.xorinArrayAt(j);

cipherText = cipherText + xor[j];

}

cipherText += " ";

}

return (cipherText);

}

public static String decryptionMessage(String s)

{

int i, j;

// char cipherChar[]=new char[(s.length()/2)];

char cipherChar[] = new char[(s.length())];

int cnt = -1;

for (i = 0; i < s.length(); i++)

{

// we receive only Ascii of it is allow 0 and 1, do not accept white

// space

// int ascii=(int)s.charAt(i);

if ((int) s.charAt(i) == 48 || (int) s.charAt(i) == 49

|| (int) s.charAt(i) == 32)

{

cnt++;

cipherChar[cnt] = s.charAt(i);

}

}

String s1 = new String(cipherChar);

String s2[] = s1.split(" ");

int data[] = new int[s2.length];

for (i = 0; i < s2.length; i++)

{

data[i] = Integer.parseInt(s2[i]);

}

char randomBitPattern[] = new char[7];

for (i = 0; i < 7; i++)

{

randomBitPattern[i] = (i % 2 == 0) ? '1' : '0';

}

BasicOperation b1 = new BasicOperation();

String plain = new String("");

// do the XOR Operation

for (i = 0; i < s2.length; i++) { int xorlen = b1.xorop(s2[i], randomBitPattern); int xor[] = new int[xorlen]; for (j = 0; j < xorlen; j++)

{

xor[j] = b1.xorinArrayAt(j);

plain += xor[j];

}

plain += " "; }

String p[] = plain.split(" ");

BasicOperation ob = new BasicOperation();

int decryptedChar[] = new int[p.length];

char plainTextChar[] = new char[p.length];

for (i = 0; i < p.length; i++)

{

decryptedChar[i] = ob.binaryToDecimal(Integer.parseInt(p[i]));

plainTextChar[i] = (char) decryptedChar[i];

}

return (new String(plainTextChar));

}

public static void main(String[] args)

{

Scanner sc = new Scanner(System.in);

System.out.println("Enter the message: ");

String message = sc.next();

System.out.println("'" + message + "' in encrypted message : "

+ encryptionMessage(message));

System.out.println("'" + encryptionMessage(message)

+ "' in decrypted message : "

+ decryptionMessage(encryptionMessage(message)));

sc.close();

}

}

class BasicOperation

{

int bin[] = new int[100];

int xor[] = new int[100];

int temp1[] = new int[100];

int temp2[] = new int[100];

int len;

int xorlen;

// convert binary number to decimal number

public int binaryToDecimal(int myNum)

{

int dec = 0, no, i, n = 0;

no = myNum;

// Find total digit of no of inupted number

while (no > 0)

{

n++;

no = no / 10;

}

// Convert inputed number into decimal

no = myNum;

for (i = 0; i < n; i++) { int temp = no % 10; dec = dec + temp * ((int) Math.pow(2, i));

no = no / 10;

}

return dec;

}

public int decimalToBinary(int myNum)

{

int j, i = -1, no, temp = 0;

no = myNum;

int t[] = new int[100];

while (no > 0)

{

i++;

temp = no % 2;

t[i] = temp;

no = no / 2;

}

len = (i + 1);

j = -1;

for (i = len; i >= 0; i--)

{

j++;

bin[j] = t[i];

}

return len;

}

// find the specific bit value of binary number at given position

public int binaryArrayAtPosition(int pos)

{ return bin[pos];

}

public int xorinArrayAt(int pos)

{

return xor[pos];

}

// perform the binary X-OR operation

public int xorop(int a[], int b[], int arrlen)

{

int i;

for (i = 0; i < arrlen; i++)

{

xor[i] = (a[i] == b[i]) ? 0 : 1;

}

xorlen = i;

return xorlen;

}

// perform the binary X-OR operation

public int xorop(String s, char c[])

{

int i = -1; for (i = 0; i < s.length(); i++)

{

xor[i] = (s.charAt(i) == c[i]) ? 0 : 1;

}

xorlen = i; return xorlen;

}

public int getLen()

{

return len + 1;

}

// display binary bit pattern or the array

public void displayBinaryArray()

{

for (int i = 0; i <= len; i++)

{

System.out.println(" Binary Array :" + bin[i]);

}

}

}

Step by Step Solution

There are 3 Steps involved in it

Step: 1

blur-text-image

Get Instant Access to Expert-Tailored Solutions

See step-by-step solutions with expert insights and AI powered tools for academic success

Step: 2

blur-text-image

Step: 3

blur-text-image

Ace Your Homework with AI

Get the answers you need in no time with our AI-driven, step-by-step assistance

Get Started

Recommended Textbook for

Object Oriented Databases Prentice Hall International Series In Computer Science

Authors: John G. Hughes

1st Edition

0136298745, 978-0136298748

More Books

Students also viewed these Databases questions