Homework Help: Questions and Answers: Assume a one-time-pad-like encryption with a short key of 128 bits. This key is then being used periodically to encrypt large volumes of data. Describe the security fault by exhibiting a possible attack scheme.
Answer:
In a one-time pad (OTP) encryption, a key is supposed to be as long as the plaintext and used only once. When properly implemented, it is theoretically unbreakable due to the lack of any statistical patterns in the ciphertext. However, in the scenario described, a short key of 128 bits is reused periodically to encrypt large volumes of data, which leads to significant vulnerabilities.
Here’s a step-by-step explanation of the security fault and a possible attack scheme:
Key Reuse Problem
The key is only 128 bits long, but it is being reused to encrypt large volumes of data. This means that the same key stream (derived from the key) is applied to different sections of the plaintext periodically. This is no longer a proper one-time pad, and key reuse opens up possibilities for an attacker to exploit the repeated patterns.
XOR Properties and Known Plaintext Attack
Encryption using a one-time pad or stream cipher typically involves XORing the plaintext with the key. That is: ciphertext = plaintext ⊕ key
If the key is reused and two different plaintexts (P1 and P2) are encrypted using the same key stream K, then an attacker can simply XOR the two ciphertexts: C1= P1⊕K C2= P1⊕K
Now, if the attacker XORs C1 and C2:
C1 ⊕ C2 = (P1⊕K) ⊕ (P2⊕K) = P1 ⊕ P2
Since the key stream cancels out, the attacker is left with the XOR of the two plaintexts. This operation can reveal significant information about the underlying plaintexts, especially if the attacker has knowledge or guesses about part of one plaintext. This is called a known plaintext attack or crib attack.
Frequency Analysis and Pattern Recognition
Once an attacker has P1⊕P2, they can further analyze patterns, especially if the data has predictable structure or redundancy (like headers, common phrases, or formatting). For example:
- If one of the plaintexts is in a known format (e.g., a file header, HTTP request), the attacker can deduce parts of the key stream.
- With part of the key stream recovered, the attacker can then decrypt other ciphertexts that used the same key, exposing more plaintext.
Ciphertext Manipulation and Forgery
If the key is reused and portions of the key stream are known or deduced, an attacker can modify ciphertexts in such a way that they control parts of the decrypted plaintext.
For example: C′=C1 ⊕ (P1⊕P1′) Here, C1 is the original ciphertext, and P1′ is the desired new plaintext. The attacker can craft a new ciphertext C′ that, when decrypted, produces P1′, leading to possible message forgery.
Brute-Forcing the Key (If Practical)
Given that the key is only 128 bits long, in principle, an attacker could try to brute-force the key by attempting every possible key until they find the one that decrypts the data correctly.
While 128-bit keys are generally considered secure against brute force for small amounts of data, the periodic reuse of the key significantly reduces the strength of the encryption. If the attacker can obtain enough ciphertext and partial knowledge of plaintext, they might reduce the effective key space.
Conclusion: Security Fault
The key vulnerability arises from the reuse of the same key for multiple encryptions. This breaks the core security assumption of the one-time pad, making the encryption scheme susceptible to:
- Known plaintext attacks, where XORing multiple ciphertexts cancels out the key and reveals relationships between plaintexts.
- Pattern recognition and statistical analysis.
- Ciphertext manipulation, enabling the attacker to alter encrypted messages in a controlled manner.
The security fault could be mitigated by using unique, non-repeating keys or key streams for each encryption (e.g., through proper use of cryptographic algorithms like AES with unique initialization vectors).
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