cryptography
Perform cryptanalysis to recover the plain text in a Caesar cipher system if the cipher text is “UPN HQJLQHHULQJ FROOHJH”. Use Bruteforce and frequency analysis and compare the complexity of the same.
1 answer
Plaintext: "RMK ENGINEERING COLLEGE" (Caesar key = 3).
Caesar cipher: \(C = (P + k) \bmod 26\), so decryption is \(P = (C - k) \bmod 26\).
1. Brute-force attack
Try every key \(k = 1, 2, \dots, 25\) and look for meaningful English. For the first word UPN:
- \(k = 1\): TOM; \(k = 2\): SNL; \(k = 3\): RMK; …
Applying each key to the whole message, only \(k = 3\) gives readable text:
- UPN → RMK
- HQJLQHHULQJ → ENGINEERING
- FROOHJH → COLLEGE
(\(k = 1\) gives "TOM GPIKPGGTKPI EQNNGIG", which is meaningless after the first word.)
2. Frequency analysis
Count letters in the ciphertext: H = 5, Q = 3, J = 3, U = 2, L = 2, O = 2, others 1. The most common letter in English is E, so assume H → E. The shift is \(H - E = 7 - 4 = 3\). Decrypting with \(k = 3\) gives RMK ENGINEERING COLLEGE immediately. (The doubled letters OO → LL and the pattern of HQJLQHHULQJ also fit "ENGINEERING".)
3. Comparison of complexity
| Point | Brute force | Frequency analysis |
|---|---|---|
| Work | Up to 25 trial decryptions of \(n\) letters: \(O(25n)\) | One counting pass \(O(n)\) plus 1–3 trial decryptions |
| Needs | Only the ciphertext and a way to recognise plaintext | Enough ciphertext for letter statistics to be reliable |
| Reliability | Always succeeds (key space only 25) | May guess wrong on very short texts; try the next most frequent letters (T, A, O) |
Both are easy because the Caesar key space is tiny, which is why the Caesar cipher is insecure.
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