Presentation on theme: "AES Advanced Encryption Standard"— Presentation transcript:
1AES Advanced Encryption Standard Prepared by:Divan Sir
2Requirements for AESAES had to be a private key algorithm. It had to use a shared secret key.It had to support the following key sizes:128 bits192 bits256 bitsDES uses only 56-bit keys.If you were able to search half the DES key space in 1 second, then on average, it would take 149 trillion years to crack a 128-bit AES key.
3Operation1) Expand 128 bits key or 16-byte ( 4 X 4 Arrays) key into 11 (4 X 4 Arrays) and each arrays containing 16 bytes (16 bytes – 176 bytes) (4 words – 44 words)(w0 to w43).2)16 – bytes plain text block is copied into 4x4 array is called state3)XOR state with key block.
5Key expansionAfter filling first array the remaining 10 arrays (w4 to w43) are filled one by one.If the word in W array is a multiple of four, some complex logic is used like w4,w8,w12…. But for others simple XOR is used.Like for w5, we would XOR w4 and w1 and store output as w5.
6Key expansionBut if word is multiple of four w4,w8,w12… then three functions are used.Rotate, Substitute and constantRotate :Suppose original 4 word key is:Byte123456789101112131415hex000102030405060708090A0B0C0D0E0F
7Key expansion For find w4: First Rotation will produce Rotate W3 ( 0C 0D 0E 0F) which is equals to ( 0D 0E 0F 0C ).
8Key expansion2) second substitute,we need to take one byte at a time and look up in S- box. For example first byte 0D is replace with 00,similarly, 0E is replace with 00..so on…
9Key expansion3) Finally at last stage substituted word XOR with constant ( with the help of constant table) as per round number.Like D7 AB 76 FE XOR = D6 AB 76 FEFinally this XOR with w [i-4] means w0. where I = word numberRound12345678910constant010204082040801B36
10Plain text First Plain text of 16 bytes arrange into 4 x 4 array. Apply s-box to each array of plain text bytes.
11Plain textSecond step rotate Row of plain text k bytes.