ECC Mini Challenge: Decrypting the Flag with Elliptic Curve Cryptography

The provided code snippet represents a classic example of Elliptic Curve Cryptography (ECC) used for encryption. Let's break down the code and then work through the process of decrypting the flag.

Code Breakdown:

1. Importing Libraries:

   • from Crypto.Util.number import *: Imports necessary functions for working with large numbers and cryptographic operations from the Crypto.Util.number module.
   • from secret import flag: Imports the flag variable from the secret module. This is likely where the hidden flag is stored.

2. Setting Up the Cryptographic Environment:

   • flag=bytes_to_long(flag): Converts the flag from bytes to a large integer, which is more suitable for ECC operations.
   • a = getPrime(256):
   • b = getPrime(256):
   • p = getPrime(256): Generates three large prime numbers (a, b, and p) of 256 bits each. These are likely used to define the parameters of the elliptic curve.

3. Elliptic Curve Definition:

   • m1=int(str(flag)[:5])-4585:
   • m2=int(str(flag)[5:]): Extracts portions of the flag to be used as part of the elliptic curve operations.
   • E = EllipticCurve(GF(p), [a, b]): Defines an elliptic curve over a finite field (GF(p)) using the previously generated prime numbers. The coefficients a and b determine the curve's equation.

4. Key Generation and Encryption:

   • X=E.lift_x(m1): Finds the x-coordinate of a point on the elliptic curve based on the first part of the flag (m1).
   • Y=7*X: Calculates the y-coordinate of a point on the curve (this might be a simplification and not necessarily related to the actual y-coordinate).
   • m = E.random_point(): Generates a random point (m) on the elliptic curve. This point likely plays a role in the message or secret data.
   • G = E.random_point(): Generates another random point (G) on the elliptic curve. This point acts as the generator for the ECC group.
   • k = getPrime(256): Generates a large prime number (k). This is used as the secret key for encryption.
   • K = k * G: Calculates the public key (K) by multiplying the secret key (k) with the generator point (G).
   • r = getPrime(256): Generates another prime number for a random value (r) used in the encryption process.
   • c1 = m + r * K: Computes the first ciphertext element (c1) by adding the message point (m) and the product of the random value (r) and the public key (K).
   • c2 = r * G: Calculates the second ciphertext element (c2) by multiplying the random value (r) with the generator point (G).

5. Additional Calculations:

   • w2=m[0]*m2: Performs a calculation using the x-coordinate of the message point (m) and the second part of the flag (m2). The purpose of this calculation is unclear without further context.

6. Output:

   • The code then prints the values of p, a, b, k, E, Y, c1, c2, and w2.

Decryption Steps

1. Understand the Encryption Scheme: The provided code suggests an ECC encryption scheme where the public key (K) is derived from a secret key (k) and the generator point (G). The message is encrypted using the public key and a random value (r).

2. Reconstruct the Elliptic Curve: Use the provided values for p, a, and b to recreate the Elliptic Curve (E) in a suitable library like SageMath or Python's cryptography module.

3. Calculate the Shared Secret: Using the provided public key (K) and the secret key (k), calculate the shared secret point: S = k * K. This shared secret is essential for decryption.

4. Decrypt the Message: Subtract the shared secret point (S) from the first ciphertext element (c1) to retrieve the original message point: m = c1 - S.

5. Extract the Flag: The m point likely contains the original flag encoded within its coordinates (either x or y). You'll need to decipher how the flag was encoded into the point. Since you know m1 and m2 are portions of the flag, you can likely use this information to reconstruct the original flag.

Python Script for Decryption (Using SageMath):

from sage.all import *

p = 71397796933602469825964946338224836258949974632540581233301840806613437378503
a = 106105288190268015217241182934677375171023341761047638573248022053052499733117
b = 76170541771321874396004434442157725545076211607587599314450304327736999807927
k = 58155941823118858940343657716409231510854647214870891375273032214774400828217
E = EllipticCurve(GF(p), [a, b])

c1 = E(37414446283406201193977113266234367761786780230360175925999700345196415953455, 17037724145039910971426670298726906655653040365428438334942732090559637519851)
c2 = E(60560423732267272277570046154733119097475794979191838027420415113112056962844, 54372226143125971429691267751299496959531971082475860532181772357190222938465)
G = E.random_point()
K = k * G
S = k * K
m = c1 - S

print(f'The message point is: {m}')  # Extract the flag from the x or y coordinate of the point

# You'll need to determine how the flag is encoded into the point's coordinates
# Based on the context, it seems the flag was broken into two parts (m1 and m2).
# You can try reconstructing the flag from these parts and the point's coordinates

Important Notes:

• Understanding the Encoding: The key to successfully decrypting the flag lies in understanding how the flag is encoded into the elliptic curve's points. The provided code hints at this encoding scheme.
• Working with ECC: Ensure you have a suitable library (like SageMath or cryptography) for working with Elliptic Curves.
• Complexity: Elliptic Curve Cryptography can be computationally intensive, especially for large prime numbers.

Let me know if you have any further questions or would like help with any specific step!