AES Algorithm: State Block Transformation using S-Box - Detailed Explanation

The given state block is a 4x4 block, with each element being a byte (8 bits).

To apply the AES S-box transformation, we need to substitute each byte in the state block with a new byte, according to a predefined substitution table. This table is called the AES S-box, and it maps each possible byte value (0x00 to 0xFF) to a new byte value.

To find the new state after applying the AES S-box transformation, we need to substitute each byte in the state block with its corresponding value in the S-box. The S-box is a 16x16 table, where each row and column represents a hexadecimal value (0x00 to 0xFF), and each cell contains a new byte value. For example, the first row of the S-box is:

0x00 0x01 0x02 0x03 0x04 0x05 0x06 0x07 0x08 0x09 0x0A 0x0B 0x0C 0x0D 0x0E 0x0F 0x63 0x7C 0x77 0x7B 0xF2 0x6B 0x6F 0xC5 0x30 0x01 0x67 0x2B 0xFE 0xD7 0xAB 0x76

To apply the S-box transformation to the given state block, we substitute each byte in the block with its corresponding value in the S-box. For example, the first byte in the state block is 'D1', which has a hexadecimal value of 0xD1. We look up the row and column in the S-box corresponding to the value 0xD1, which is row 'D' and column '1'. The value in that cell is 0x8C, so we replace the byte 'D1' with '8C'. We repeat this process for each byte in the state block.

After applying the S-box transformation to the given state block, we get the following new state:

'8C' '32' '94' '2B' 'C3' '75' 'F1' '18' '1F' 'F1' '18' '38' 'C4' '08' '7D' '32'

Note that the size of the state block remains the same (4x4), but each byte in the block has been replaced with a new byte value according to the AES S-box.