Interactive Cryptanalysis Lab (Caesar, Vigenère, XOR & ECB Mode) (2026)

The Index of Coincidence measures the probability that two randomly selected letters from a text are identical: `IC = Σ n_i(n_i - 1) / (N(N - 1))`. Because a monoalphabetic Caesar cipher merely permutes the alphabet without flattening frequencies, its `IC` stays near standard English (`~0.0667`). A polyalphabetic Vigenère cipher distributes letters across multiple alphabets, flattening `IC` toward uniform random (`1/26 ≈ 0.0385`).

Once you determine the likely key length `L` (by finding which column stride `L` yields per-column `IC ≈ 0.066`), a Vigenère cipher of key length `L` reduces to `L` independent Caesar ciphers! Each column `0 .. L-1` can then be solved independently using standard Chi-Squared (`χ²`) monogram frequency matching.

AES is a secure 128-bit block permutation, but **Electronic Codebook (ECB)** mode encrypts every 16-byte block completely independently with the same key (`C_i = E_K(P_i)`). Whenever two plaintext blocks are identical (such as repeated background pixels in a bitmap, repeated JSON headers, or identical database values), their ciphertext blocks are also 100% identical, leaking structural patterns and enabling block-replay attacks.

AES-CBC provides confidentiality via ciphertext block chaining (`C_i = E_K(P_i ⊕ C_{i-1})`), but provides **zero integrity authentication** on its own. If a server leaks whether PKCS#7 padding (`0x01`, `0x02 0x02`, etc.) was valid after decryption (a Padding Oracle), an attacker can XOR-flip bits in `C_{i-1}` to decrypt every byte of `P_i` in at most `256 × 16` requests. **AES-GCM** is an Authenticated Encryption with Associated Data (AEAD) mode that verifies a 128-bit GHASH authentication tag *before* outputting any plaintext.

Stream ciphers (including ChaCha20, AES-CTR, and AES-GCM) encrypt by XORing plaintext with a pseudo-random keystream: `C_1 = P_1 ⊕ K` and `C_2 = P_2 ⊕ K`. If the same key and nonce are ever reused, an attacker simply XORs the two ciphertexts together: `C_1 ⊕ C_2 = (P_1 ⊕ K) ⊕ (P_2 ⊕ K) = P_1 ⊕ P_2`, completely canceling out the secret key `K`!