Program to find decode XORed permutation in Python

Suppose we have an array enc that represents an encoded permutation. There is an array perm that is a permutation of the first n (odd) positive integers. This list is encoded into array enc of length n-1, such that enc[i] = perm[i] XOR perm[i+1]. We need to find the original array perm.

So, if the input is like enc = [2,5,6,3], then the output will be [7, 5, 0, 6, 5]. Here [7 XOR 5, 5 XOR 0, 0 XOR 6, 6 XOR 5] = [2, 5, 6, 3].

Algorithm

To solve this, we will follow these steps ?

• Calculate n as the size of enc
• Create a result array of size (n+1) and fill with 0
• Calculate XOR of all numbers from 1 to n+1
• Find the first element by XORing with alternate elements of enc
• Generate remaining elements using the XOR relationship
• Return the result

How It Works

The key insight is that if we XOR all elements of the original permutation, we get the XOR of numbers 1 to n+1. Since enc[i] = perm[i] XOR perm[i+1], we can use this property to find the first element, then derive the rest.

Example

Let us see the following implementation to get better understanding ?

def solve(enc):
    n = len(enc)
    result = [0] * (n + 1)
    
    # Calculate XOR of all numbers from 1 to n+1
    x = 0
    for i in range(1, n + 2):
        x ^= i
    
    # Find the first element
    result[0] = x
    for i in range(1, n + 1, 2):
        result[0] ^= enc[i]
    
    # Generate remaining elements
    for i in range(1, n + 1):
        result[i] = result[i-1] ^ enc[i-1]
    
    return result

# Test the function
enc = [2, 5, 6, 3]
perm = solve(enc)
print("Original permutation:", perm)

# Verify the result
print("Verification:")
for i in range(len(enc)):
    print(f"perm[{i}] XOR perm[{i+1}] = {perm[i]} XOR {perm[i+1]} = {perm[i] ^ perm[i+1]}")

Original permutation: [7, 5, 0, 6, 5]
Verification:
perm[0] XOR perm[1] = 7 XOR 5 = 2
perm[1] XOR perm[2] = 5 XOR 0 = 5
perm[2] XOR perm[3] = 0 XOR 6 = 6
perm[3] XOR perm[4] = 6 XOR 5 = 3

Step-by-Step Breakdown

def solve_with_steps(enc):
    print(f"Input encoded array: {enc}")
    n = len(enc)
    print(f"Length n = {n}")
    
    # Step 1: Calculate XOR of 1 to n+1
    x = 0
    for i in range(1, n + 2):
        x ^= i
    print(f"XOR of numbers 1 to {n+1}: {x}")
    
    # Step 2: Find first element
    result = [0] * (n + 1)
    result[0] = x
    print(f"Initial result[0] = {x}")
    
    for i in range(1, n + 1, 2):
        result[0] ^= enc[i]
        print(f"After XOR with enc[{i}] = {enc[i]}: result[0] = {result[0]}")
    
    # Step 3: Generate remaining elements
    for i in range(1, n + 1):
        result[i] = result[i-1] ^ enc[i-1]
        print(f"result[{i}] = result[{i-1}] ^ enc[{i-1}] = {result[i-1]} ^ {enc[i-1]} = {result[i]}")
    
    return result

# Example with step-by-step output
enc = [2, 5, 6, 3]
result = solve_with_steps(enc)
print(f"\nFinal result: {result}")

Input encoded array: [2, 5, 6, 3]
Length n = 4
XOR of numbers 1 to 5: 1
Initial result[0] = 1
After XOR with enc[1] = 5: result[0] = 4
After XOR with enc[3] = 3: result[0] = 7
result[1] = result[0] ^ enc[0] = 7 ^ 2 = 5
result[2] = result[1] ^ enc[1] = 5 ^ 5 = 0
result[3] = result[2] ^ enc[2] = 0 ^ 6 = 6
result[4] = result[3] ^ enc[3] = 6 ^ 3 = 5

Final result: [7, 5, 0, 6, 5]

Conclusion

This algorithm efficiently decodes a XORed permutation by leveraging the mathematical properties of XOR operations. The key insight is using the XOR of all numbers 1 to n+1 and then applying the encoding relationship in reverse to reconstruct the original permutation.