XOR Cryptography

1606070;"># XOR Truth Table:
2606070;"># 0 ⊕ 0 = 0
3606070;"># 0 ⊕ 1 = 1
4606070;"># 1 ⊕ 0 = 1
5606070;"># 1 ⊕ 1 = 0
6 
7606070;"># Key Properties:
8606070;"># A ⊕ A = 0       (anything XOR itself = 0)
9606070;"># A ⊕ 0 = A       (anything XOR 0 = itself)
10606070;"># A ⊕ B = B ⊕ A   (commutative)
11606070;"># A ⊕ B ⊕ B = A   (self-inverse: XOR twice = original)
12 
13606070;"># This means:
14606070;"># plaintext ⊕ key = ciphertext
15606070;"># ciphertext ⊕ key = plaintext
16 
17606070;"># Example:
18606070;"># "A" = 0x41 = 01000001
19606070;"># key = 0x20 = 00100000
20606070;"># XOR:         01100001 = 0x61 = "a"
21 
22606070;"># XOR "A" with 0x20, get "a"
23606070;"># XOR "a" with 0x20, get "A"

1606070;"># If ciphertext is XORed with a single byte, brute force all 256 values
2 
3def single_byte_xor(ciphertext, key):
4    return bytes([b ^ key for b in ciphertext])
5 
6def brute_force_single_byte(ciphertext):
7    for key in range(256):
8        plaintext = single_byte_xor(ciphertext, key)
9        try:
10            decoded = plaintext.decode(606070;">#a5d6ff;">'ascii')
11            606070;"># Check if result looks like text
12            if decoded.isprintable():
13                print(f606070;">#a5d6ff;">"Key {key} (0x{key:02x}): {decoded}")
14        except:
15            pass
16 
17606070;"># Example usage:
18cipher = bytes.fromhex(606070;">#a5d6ff;">"1b37373331363f78151b7f2b783431333d")
19brute_force_single_byte(cipher)
20 
21606070;"># CyberChef: "XOR Brute Force" operation

1606070;"># If you know part of the plaintext, recover the key!
2606070;"># plaintext ⊕ key = ciphertext
3606070;"># Therefore: plaintext ⊕ ciphertext = key
4 
5def xor_bytes(a, b):
6    return bytes([x ^ y for x, y in zip(a, b)])
7 
8ciphertext = bytes.fromhex(606070;">#a5d6ff;">"...")
9known_plaintext = b606070;">#a5d6ff;">"flag{"  # We know flags start with this!
10 
11606070;"># Recover key bytes for known portion
12key_fragment = xor_bytes(ciphertext[:5], known_plaintext)
13print(f606070;">#a5d6ff;">"Key fragment: {key_fragment}")
14 
15606070;"># If key repeats, use this fragment to find key length
16606070;"># and decrypt more of the message

1606070;"># Encryption with repeating key
2def repeating_key_xor(plaintext, key):
3    result = []
4    for i, byte in enumerate(plaintext):
5        result.append(byte ^ key[i % len(key)])
6    return bytes(result)
7 
8606070;"># Breaking repeating key XOR:
9606070;"># 1. Find key length
10606070;"># 2. Split cipher into groups by key position
11606070;"># 3. Each group is single-byte XOR - brute force each
12 
13606070;"># Finding key length with Hamming distance:
14def hamming_distance(a, b):
15    return sum(bin(x ^ y).count(606070;">#a5d6ff;">'1') for x, y in zip(a, b))
16 
17def find_key_length(cipher, max_len=40):
18    scores = []
19    for keylen in range(2, max_len):
20        chunks = [cipher[i:i+keylen] for i in range(0, len(cipher), keylen)]
21        if len(chunks) < 4:
22            continue
23        606070;"># Compare first chunks and average Hamming distance
24        distances = []
25        for i in range(min(4, len(chunks)-1)):
26            d = hamming_distance(chunks[i], chunks[i+1])
27            distances.append(d / keylen)  606070;"># Normalize
28        scores.append((keylen, sum(distances)/len(distances)))
29 
30    606070;"># Lower score = more likely key length
31    scores.sort(key=lambda x: x[1])
32    return scores[:5]  606070;"># Return top 5 candidates

1606070;">#!/usr/bin/env python3
2606070;"># Complete attack on repeating-key XOR
3 
4from collections import Counter
5 
6def score_english(text):
7    606070;">#a5d6ff;">"""Score text based on English letter frequency"""
8    freq = 606070;">#a5d6ff;">'etaoinshrdlcumwfgypbvkjxqz'
9    score = 0
10    for c in text.lower():
11        if c in freq:
12            score += len(freq) - freq.index(c)
13    return score
14 
15def single_byte_xor_decrypt(cipher):
16    606070;">#a5d6ff;">"""Try all single-byte keys, return best result"""
17    best_score = 0
18    best_result = None
19    best_key = None
20 
21    for key in range(256):
22        plaintext = bytes([b ^ key for b in cipher])
23        try:
24            decoded = plaintext.decode(606070;">#a5d6ff;">'ascii')
25            score = score_english(decoded)
26            if score > best_score:
27                best_score = score
28                best_result = decoded
29                best_key = key
30        except:
31            pass
32 
33    return best_key, best_result
34 
35def break_repeating_xor(cipher, key_length):
36    606070;">#a5d6ff;">"""Break repeating XOR with known key length"""
37    606070;"># Split into columns
38    columns = [[] for _ in range(key_length)]
39    for i, byte in enumerate(cipher):
40        columns[i % key_length].append(byte)
41 
42    606070;"># Solve each column as single-byte XOR
43    key = []
44    for col in columns:
45        k, _ = single_byte_xor_decrypt(bytes(col))
46        key.append(k)
47 
48    606070;"># Decrypt with recovered key
49    key = bytes(key)
50    plaintext = bytes([c ^ key[i % len(key)] for i, c in enumerate(cipher)])
51 
52    return key, plaintext
53 
54606070;"># Usage:
55cipher = bytes.fromhex(606070;">#a5d6ff;">"...")
56key_lengths = find_key_length(cipher)
57for kl, _ in key_lengths:
58    key, plain = break_repeating_xor(cipher, kl)
59    print(f606070;">#a5d6ff;">"Key length {kl}: key={key}, plain={plain[:50]}...")

1606070;"># Trick 1: XOR with itself reveals nothing
2606070;"># But XOR two ciphertexts encrypted with same key:
3606070;"># c1 ⊕ c2 = (p1 ⊕ key) ⊕ (p2 ⊕ key) = p1 ⊕ p2
4606070;"># Key cancels out! Now you have XOR of plaintexts.
5 
6606070;"># Trick 2: Flag format known
7606070;"># If flag is flag{...}, XOR cipher with "flag{" to get key start
8 
9606070;"># Trick 3: Null bytes in key
10606070;"># If plaintext has patterns (like spaces), key reveals itself
11606070;"># plaintext: "HELLO WORLD"
12606070;"># If encrypted and has obvious pattern, XOR with common char
13 
14606070;"># Trick 4: CyberChef shortcuts
15606070;"># - XOR: Specify key in hex, UTF8, or decimal
16606070;"># - XOR Brute Force: Try all single bytes
17606070;"># - Guess the key from context clues

1606070;"># Common attack: Two plaintexts, same key (many-time pad)
2606070;"># c1 = p1 ⊕ key
3606070;"># c2 = p2 ⊕ key
4606070;"># c1 ⊕ c2 = p1 ⊕ p2
5 
6606070;"># If we guess p1[i] = space (0x20):
7606070;"># c1[i] ⊕ 0x20 = key[i]
8606070;"># Then: c2[i] ⊕ key[i] = p2[i]
9 
10606070;"># Tool: mtp (many-time pad attack)
11606070;"># https://github.com/CameronLonsworthy/mtp

1606070;"># CyberChef Operations:
2606070;"># - XOR (with known key)
3606070;"># - XOR Brute Force (single byte)
4606070;"># - Guess the key
5 
6606070;"># xortool - Automated analysis
7pip install xortool
8 
9xortool cipher.bin
10606070;"># Suggests key lengths and attempts decryption
11 
12xortool -l 5 cipher.bin  606070;"># Force key length 5
13xortool -c 20 cipher.bin  606070;"># Assume most common char is space (0x20)
14 
15606070;"># Python one-liner XOR
16python3 -c 606070;">#a5d6ff;">"print(bytes([a^b for a,b in zip(open('cipher','rb').read(), b'KEY'*1000)])[:100])"
17 
18606070;"># xxd for hex manipulation
19xxd -p cipher.bin | tr -d 606070;">#a5d6ff;">'\n'  # Hex dump

Why is XOR used for encryption?