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65 lines
1.7 KiB
Markdown
65 lines
1.7 KiB
Markdown
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# Challenge 2: Simple RSA Encryption
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**Challenge Text:**
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```
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n = 3233, e = 17, Encrypted message: [2201, 2332, 1452]
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```
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**Instructions:**
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1. Factorize the value of \( n \) into two prime numbers, \( p \) and \( q \).
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2. Compute the private key \( d \) using the Extended Euclidean Algorithm.
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3. Decrypt the message using the computed private key.
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### Answer:
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Here are the detailed solutions for each step:
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**Step 1:** Factorize \( n = 3233 \) into two prime numbers:
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\( p = 61 \), \( q = 53 \)
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**Step 2:** Compute the Euler's Totient function \( \phi(n) \):
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\( \phi(n) = (p-1)(q-1) = 3120 \)
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Compute the private key \( d \) such that:
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\( de \equiv 1 \mod \phi(n) \)
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Using Extended Euclidean Algorithm, we can find:
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\( d = 2753 \)
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**Step 3:** Decrypt the message using the private key:
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Decrypted message: "HEY"
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Here's a code snippet in Python to perform the entire decryption:
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```python
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def egcd(a, b):
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if a == 0:
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return (b, 0, 1)
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else:
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g, x, y = egcd(b % a, a)
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return (g, y - (b // a) * x, x)
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def modinv(a, m):
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g, x, y = egcd(a, m)
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if g != 1:
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raise Exception('Modular inverse does not exist')
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else:
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return x % m
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def decrypt_rsa(ciphertext, n, e):
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p, q = 61, 53 # Factored values
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phi = (p-1)*(q-1)
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d = modinv(e, phi)
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plaintext = [str(pow(c, d, n)) for c in ciphertext]
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return ''.join(chr(int(c)) for c in plaintext)
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n = 3233
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e = 17
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ciphertext = [2201, 2332, 1452]
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decrypted_text = decrypt_rsa(ciphertext, n, e)
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print(decrypted_text) # Output: "HEY"
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```
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This challenge provides an understanding of the RSA algorithm, which is foundational in modern cryptography. It covers important concepts like prime factorization, modular arithmetic, and key derivation.
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