SPIDER1CODE

اهلا و سهلا بك عزيز او عزيزتي في Spider1Ctf قدرت اعمل تشكيله جميله من تحديات صنعتها بنفسي منها الصعب جدا و منها سهل جدا و كل تحدي معا ملف .txt يشرح تحدي او حتا هنت 👀💥 مده ctf من وقت نزول اعلان دا ليوم الجمعه الساعه 12 بتوقيت مصر و اول ما مسابقه تخلص هنطلع لايف مع فائزين نشرح حل تحديات 🏆 كان الله معكم و في عونكم ❤️ Welcome, challengers, to

Spider1CTF!

I’ve prepared a diverse collection of challenges crafted entirely by myself — some extremely difficult and others very beginner-friendly. Each challenge comes with a .txt file containing a description or even a hint 👀💥 The CTF will run from the moment this announcement is posted until Friday at 12:00 PM (Egypt time). Once the competition ends, we’ll go live with the winners to walk through the solutions together 🏆 Wishing you all strength, focus, and a bit of luck ❤️ ارسل العلم لي : @Spider1Security Send flag to : @Spider1Security

طيب لاسبوع الجاي هيكون بدايه ctf عايز كل يكون جاهز 🏆✌🏻 خد حط دا في cmd او ترمنال عندك و قولي شوفت اي 😂

ssh -o StrictHostKeyChecking=no watch.ascii.theater

عايز اقول خبر مش جميل بخصوص ctf الي هعلمها بعد تفكير كتير مش لقيت منصه كويسه اقدر استعلمها كا سيرفر فا ctf هتكون كريبتو و ريفيرس و فرونزيكس طيب هل ليها جوائز ؟ لاشخاص الي قدرت تحل هنطلع كلنا لايف ب اذن الله يوم جمعه و نشارك معاكم طرق حل طيب مواعيد امتا ؟ هنزل كل حاجه قريب بحيث ان فكره تكون جاهزه اول ب اول طيب ازاي نقدر ندخل تحديات ؟ تحديات هتنزل هنا علي قناه تلجرام و مجتمع فقط !!! ولو حد عندو اي استفسار يقدر يسيب كومنت و هرد علي ❤️

Thanks to EC-Council for inviting me to HACKERVERSE. I hope this won’t be our last collaboration. Thank you for the wonderful CTF competition ❤️🏆

الحمدالله حمدا كثيره طيب به قدرت احل تحدي CyberTalents / cryptography/ giga-chad عدد الي حلو تحدي : 10 و دا رايتب : https://spider1sec.medium.com/cybertalents-cryptography-giga-chad-f1db97b3e8f4?postPublishedType=repub

لسه مخلص شرح Wireshark بفتح شباك لقيتها قدامي

الحمدالله قدرت احل تحدي cyber talent / genfei كريبتو و دا رايتب :

حاسس اني بغتت عليكم بس رايتب دول مهمين والله الحمدالله طبعا قدرت احل تحدي Red Stone Admin

الحمدالله رضا نعمه حليت تحدي : corrupt-key-1 و دا رايتب بتاعو

الحمدالله حمدا كثيرا طيبا مباركا فيه قدرت احل تحدي play nice picoctf و دا رايتب بتاعو :

#!/usr/bin/env python3
import socket
import time

SQUARE_SIZE = 6

def generate_square(alphabet):
    """Convert alphabet string into a 6x6 matrix."""
    matrix = []
    for i, letter in enumerate(alphabet):
        if i % SQUARE_SIZE == 0:
            row = []
        row.append(letter)
        if i % SQUARE_SIZE == (SQUARE_SIZE - 1):
            matrix.append(row)
    return matrix

def get_index(letter, matrix):
    """Find row and column index of a letter in the matrix."""
    for row in range(SQUARE_SIZE):
        for col in range(SQUARE_SIZE):
            if matrix[row][col] == letter:
                return (row, col)
    return None

def decrypt_pair(pair, matrix):
    """Decrypt a pair of characters using Playfair rules."""
    p1 = get_index(pair[0], matrix)
    p2 = get_index(pair[1], matrix)
    
    if p1[0] == p2[0]:  # Same row - shift left
        return matrix[p1[0]][(p1[1] - 1) % SQUARE_SIZE] + \
               matrix[p2[0]][(p2[1] - 1) % SQUARE_SIZE]
    
    elif p1[1] == p2[1]:  # Same column - shift up
        return matrix[(p1[0] - 1) % SQUARE_SIZE][p1[1]] + \
               matrix[(p2[0] - 1) % SQUARE_SIZE][p2[1]]
    
    else:  # Rectangle - swap columns
        return matrix[p1[0]][p2[1]] + matrix[p2[0]][p1[1]]

def decrypt_string(s, matrix):
    """Decrypt entire message by processing pairs."""
    result = ""
    for i in range(0, len(s), 2):
        result += decrypt_pair(s[i:i + 2], matrix)
    return result

# Connect to server
sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
sock.connect(("mercury.picoctf.net", 19354))
time.sleep(0.5)

# Receive challenge
response = sock.recv(1024).decode()

# Parse response
lines = response.strip().split('\n')
alphabet = lines[0].split("alphabet: ")[1]
encrypted = lines[1].split("encrypted message: ")[1]

# Decrypt
matrix = generate_square(alphabet)
plaintext = decrypt_string(encrypted, matrix)

# Remove padding if present
padding_char = alphabet[0]
plaintext = plaintext.rstrip(padding_char)

print(f"Plaintext: {plaintext}")

# Submit answer
sock.send((plaintext + "\n").encode())
time.sleep(1)

# Receive flag
flag_response = sock.recv(2048).decode()
print(flag_response)

sock.close()

--- ## Key Insights 1. 6×6 Matrix Variant: The challenge uses a 6×6 matrix instead of the traditional 5×5, expanding the keyspace to 36 characters. 2. Non-standard Flag Format: The plaintext doesn't look like a typical flag (picoCTF{...}) but is the actual answer. The description "The flag is not in standard format" hints at this. 3. Symmetric Decryption: To reverse Playfair encryption: - Same row pairs: Shift LEFT (not right) - Same column pairs: Shift UP (not down) - Rectangle pairs: Column swap remains the same 4. Padding Character: The first character of the alphabet is used for padding odd-length messages. 5. Modulo Arithmetic: Wraparound is handled using modulo to ensure indices stay within bounds. --- ## Flag

dbc8bf9bae7152d35d3c200c46a0fa30

--- ## References - [Playfair Cipher - Wikipedia](https://en.wikipedia.org/wiki/Playfair_cipher) - [Classical Cryptography](https://en.wikipedia.org/wiki/Classical_cipher) - [PicoCTF Challenges](https://picoctf.org/) --- ## Challenge Analysis | Aspect | Details | |--------|---------| | Cipher Type | Playfair (6×6 variant) | | Keyspace | 36! possible alphabets | | Message Length | 30 characters (15 pairs) | | Attack Method | Direct decryption (no brute force needed) | | Time to Solve | ~5 minutes | | Difficulty | high | The challenge demonstrates that while classical ciphers like Playfair are no longer secure by modern standards, understanding their mechanics is valuable for cryptographic education and CTF participation.

# Playfair Cipher CTF Writeup ## Challenge Overview Challenge Name: Playfair (Ancient Ciphers) Challenge Type: Cryptography Server: nc mercury.picoctf.net 19354 Difficulty: High Description: "Not all ancient ciphers were so bad... The flag is not in standard format." ## Challenge Description This is a Playfair cipher challenge that runs on a remote server. The server: 1. Provides a randomly generated 36-character alphabet 2. Provides an encrypted message using that alphabet 3. Expects the solver to decrypt the message and return the plaintext 4. Returns a flag upon successful decryption Key Twist: The challenge uses a 6×6 Playfair variant instead of the traditional 5×5, and the plaintext doesn't follow the standard picoCTF{...} format. --- ## Understanding the Playfair Cipher ### Traditional Playfair (5×5) The Playfair cipher is a classical symmetric encryption method that: - Uses a 5×5 matrix (25 characters) containing a mixed alphabet - Encrypts plaintext in pairs of characters (digraphs) - Applies different rules based on character positions: 1. Same Row: Shift each character right (with wraparound) 2. Same Column: Shift each character down (with wraparound) 3. Rectangle: Swap columns (each takes the other's column in the same row) ### This Challenge's Variant (6×6) This CTF uses a 6×6 matrix with 36 characters (letters, numbers, and special characters), making it a larger keyspace than traditional Playfair. #### Encryption Example (6×6)

Alphabet: n5vgru7ehz1klja8s9340m2wcxbd6pqfitoy
Matrix:
  n 5 v g r u
  7 e h z 1 k
  l j a 8 s 9
  3 4 0 m 2 w
  c x b d 6 p
  q f i t o y

Message: "hitherefriend"
Plaintext pairs: hi|th|er|ef|ri|en|dn (padded with 'n')
Encryption result: av|iz|15|j5|vo|75|cg

--- ## Solution Approach ### Step 1: Connect to Server

nc mercury.picoctf.net 19354

Server Output:

Here is the alphabet: n5vgru7ehz1klja8s9340m2wcxbd6pqfitoy
Here is the encrypted message: hnjm2e4t51v16gsg104i4oi9wmrqli
What is the plaintext message? 

### Step 2: Parse the Challenge Extract from the server response: - Alphabet: n5vgru7ehz1klja8s9340m2wcxbd6pqfitoy (36 characters) - Encrypted Message: hnjm2e4t51v16gsg104i4oi9wmrqli (30 characters) - Matrix Size: 6×6 ### Step 3: Implement Decryption The decryption process reverses encryption: - Same Row: Shift each character left (opposite of right) - Same Column: Shift each character up (opposite of down) - Rectangle: Swap columns (same as encryption)

def decrypt_pair(pair, matrix):
    p1 = get_index(pair[0], matrix)  # Get row, col of first char
    p2 = get_index(pair[1], matrix)  # Get row, col of second char
    
    if p1[0] == p2[0]:  # Same row
        # Shift left with wraparound
        return matrix[p1[0]][(p1[1] - 1) % SQUARE_SIZE] + \
               matrix[p2[0]][(p2[1] - 1) % SQUARE_SIZE]
    
    elif p1[1] == p2[1]:  # Same column
        # Shift up with wraparound
        return matrix[(p1[0] - 1) % SQUARE_SIZE][p1[1]] + \
               matrix[(p2[0] - 1) % SQUARE_SIZE][p2[1]]
    
    else:  # Rectangle - swap columns
        return matrix[p1[0]][p2[1]] + matrix[p2[0]][p1[1]]

### Step 4: Decrypt the Message

alphabet = "n5vgru7ehz1klja8s9340m2wcxbd6pqfitoy"
encrypted = "hnjm2e4t51v16gsg104i4oi9wmrqli"

matrix = generate_square(alphabet)
plaintext = decrypt_string(encrypted, matrix)

# Result: 7v8441mfrerhdr8rh20f2fya20noaq

### Step 5: Handle Padding and Submit The plaintext may be padded with the first character of the alphabet. Remove if necessary and submit:

padding_char = alphabet[0]  # 'n'
if plaintext.endswith(padding_char):
    plaintext = plaintext.rstrip(padding_char)

# Submit: 7v8441mfrerhdr8rh20f2fya20noaq

### Step 6: Receive Flag

Congratulations! Here's the flag: dbc8bf9bae7152d35d3c200c46a0fa30

--- ## Complete Solution Script