XOR Encryption by Shifting Plaintext

Here is a cipher algorithm, based on hexadecimal strings that is implemented by XORing the given plaintext, N number of times where N is its length. But, the catch is that every next XOR operation is done after shifting the consecutive plain text entry to the right. A sample operation is shown below : Suppose the password is 'abcd' then the hexadecimal text is calculated as a1d0a1d by XORing the password with itself N times i.e. 4 times in this case. Similarly if the password is '636f646572', then 653cae8da8edb426052 is the hexadecimal text. So, the problem statement is to create a decryption algorithm (in any programming language) and deduce the plain text from the given hexadecimal string. Examples :

Input : a1d0a1d
Output : abcd
abcd once coded will return a1d0a1d

Input : 653cae8da8edb426052
Output : 636f646572

Approach : The key ingredient in encrypting and decrypting is in the properties of XOR. XOR is a bitwise operation where the result is 0 if the two possible inputs are same but 1 when the inputs are different. The XOR table is given below for reference :

An important and useful property of XOR that is widely popular in cryptography is that in case of multiple XORing of numbers (say M numbers), if we know only the M - 1 numbers (one is unknown) along with the XOR result then, we can easily calculate the missing number by XORing the known numbers and the XOR result. This property is discussed with the following hexadecimal numbers : We shall be using the above listed property the most in course of this problem. Now, if we look at the encryption diagram of 'abcd' at the base it is just the repeated XORing of the digits. The rightmost digit is d and the rightmost digit of the 'abcd' is d as well so the last digit of both plaintext and hexstring is the same. The next digit is 1 which is calculated by XORing the second right digit of abcd and the previous digit i.e. 1 = d ^ c using the property we know the plain text digit can be deduced as d ^ 1 = c. Similarly, the next digit is a which is found by d ^ c ^ b = a. We only need to do this till the half of the hex string as the rest is symmetrical so they are not required. Below is the implementation of above approach : 

// Implementation of the code in C++
#include <bits/stdc++.h>
using namespace std;

int main()
{

  // Hex String variable
  string hex_s = "653cae8da8edb426052";

  // Plain text variable
  string plain = "";

  // variable to store the XOR
  // of previous digits
  int x = 0;

  int l = hex_s.length();

  // Loop for loop from the end to
  // the mid section of the string
  for (int i = l - 1; i > (l / 2) - 1; i--) {
    string digit = "";
    digit += hex_s[i];

    // calculation of the plaintext digit
    unsigned int y = x ^ stoul(digit, nullptr, 16);

    // calculation of XOR chain
    x = x ^ y;
    stringstream sstream;
    sstream << hex << y;
    string z = sstream.str();
    plain = z[z.length() - 1] + plain;
  }

  cout << plain;
}

// This code is contributed by phasing17

// Implementation of the code in Java


class GFG {
    public static void main(String[] args) {
        // Hex String variable
        String hex_s = "653cae8da8edb426052";
        
        // Plain text variable
        String plain = "";
        
        // variable to store the XOR
        // of previous digits
        int x = 0;
        
        int l = hex_s.length();
        
        // Loop for loop from the end to
        // the mid section of the string
        for (int i = l - 1; i > (l / 2) - 1; i--)
        {
            // calculation of the plaintext digit
            int y = x ^ Integer.parseInt(Character.toString(hex_s.charAt(i)), 16);
            
            // calculation of XOR chain
            x = x ^ y; 
            String z = Integer.toString(y, 16);
            plain = z.charAt(z.length() - 1) + plain;
        }  
        
        System.out.println(plain);
 
    }
}

//This code is contributed by phasing17

# Implementation in Python 3

# Hex String variable
hex_s = '653cae8da8edb426052'

# Plain text variable
plain = ''

# variable to store the XOR
# of previous digits
x = 0

l = len(hex_s)

# Loop for loop from the end to
# the mid section of the string
for i in range(l - 1, int(l / 2) - 1, -1):
    
    # calculation of the plaintext digit
    y = x^int(hex_s[i], 16)
    
    # calculation of XOR chain
    x = x^y
    plain = hex(y)[-1] + plain
    
print(plain)

// Implementation of the code in C#
using System;

class GFG {

  public static void Main(string[] args)
  {

    // Hex String variable
    string hex_s = "653cae8da8edb426052";

    // Plain text variable
    string plain = "";

    // variable to store the XOR
    // of previous digits
    int x = 0;

    int l = hex_s.Length;

    // Loop for loop from the end to
    // the mid section of the string
    for (int i = l - 1; i > (l / 2) - 1; i--) {
      string digit = "";
      digit += hex_s[i];

      // calculation of the plaintext digit
      int y = x ^ Convert.ToInt32(digit, 16);

      // calculation of XOR chain
      x = x ^ y;
      string z = Convert.ToString(y, 16);
      plain = z[z.Length - 1] + plain;
    }

    Console.WriteLine(plain);
  }
}

// This code is contributed by phasing17

// Implementation of the code in JS

// Hex String variable
var hex_s = '653cae8da8edb426052';

// Plain text variable
var plain = '';

// variable to store the XOR
// of previous digits
var x = 0;

var l = hex_s.length;

// Loop for loop from the end to
// the mid section of the string
for (var i = l - 1; i > (l / 2) - 1; i--)
{
    // calculation of the plaintext digit
    var y = x ^ parseInt(hex_s[i], 16);
    
    // calculation of XOR chain
    x = x^y;l
    plain = y.toString(16).slice(-1) + plain;
}  

console.log(plain);

// this code is contributed by phasing17

636f646572

Time Complexity : O(l) ,where l is size of hex string

Auxiliary Space : O(1)