Feistel Cipher Structure and AES Key Expansion (First 4 Bytes)

Part A: Feistel Cipher Structure

Feistel Cipher is a symmetric block cipher structure that divides the plaintext into two equal halves and processes them through multiple rounds of substitution and permutation, using sub-keys derived from the main key.

Working Principle:

The plaintext block is split into two halves: Left (L) and Right (R). In each round, one half is modified using a round function F and a sub-key, then the halves are swapped.

Structure of Each Round:

For round $i$:

$$L_i = R_{i-1}$$

$$R_i = L_{i-1} \oplus F(R_{i-1}, K_i)$$

Where $K_i$ is the sub-key for round $i$ and $F$ is the round function.

Block Diagram (described in words):

• Plaintext is divided into L₀ (left half) and R₀ (right half)
• R₀ is passed through function F along with sub-key K₁
• Output of F is XORed with L₀ to produce R₁
• R₀ becomes L₁ (swap)
• This process repeats for n rounds
• After the final round, the two halves are combined to form ciphertext

Key Properties:

• Encryption and Decryption use the same structure (sub-keys applied in reverse order for decryption)
• Round function F need not be invertible
• Security increases with more rounds
• Examples: DES (16 rounds), Blowfish, Camellia

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Part B: AES Key Expansion — First 4 Bytes of Next Key After First Iteration

Given Key (16 bytes):

W0	W1	W2	W3
2B 7E 15 16	28 AE D2 A6	AB F7 97 66	01 02 03 04

AES Key Expansion Formula:

$$W_4 = W_0 \oplus g(W_3)$$

Where g() function involves:

• a) RotWord — Circular left shift of W3 by 1 byte
• b) SubBytes — Substitute each byte using S-Box
• c) XOR with Rcon — XOR first byte with round constant

Step-by-step Computation:

Step i: Take W3

$$W_3 = [01, 02, 03, 04]$$

Step ii: RotWord (circular left shift by 1 byte)

$$RotWord(01, 02, 03, 04) = [02, 03, 04, 01]$$

Step iii: SubBytes using S-Box

For each byte, the row = upper nibble and column = lower nibble:

• Byte 02 → Row 0, Col 2 → S-Box value = 77
• Byte 03 → Row 0, Col 3 → S-Box value = 7B
• Byte 04 → Row 0, Col 4 → S-Box value = F2
• Byte 01 → Row 0, Col 1 → S-Box value = 7C

$$SubBytes = [77, 7B, F2, 7C]$$

Step iv: XOR with Rcon(1)

Round constant for first round: $Rcon(1) = [01, 00, 00, 00]$

$$[77 \oplus 01, \ 7B \oplus 00, \ F2 \oplus 00, \ 7C \oplus 00] = [76, 7B, F2, 7C]$$

So, $g(W_3) = [76, 7B, F2, 7C]$

Step v: Compute W4 = W0 ⊕ g(W3)

$$W_0 = [2B, 7E, 15, 16]$$

$$W_4 = [2B \oplus 76, \ 7E \oplus 7B, \ 15 \oplus F2, \ 16 \oplus 7C]$$

$$W_4 = [5D, 05, E7, 6A]$$

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Final Answer:

The first 4 bytes of the next key after the first iteration are: 5D 05 E7 6A