Presentation on theme: "AES Advanced Encryption Standard"— Presentation transcript:
1 AES Advanced Encryption Standard Prepared by:Divan Sir
2 Requirements 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.
3 Operation1) 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.
5 Key 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.
6 Key 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
7 Key expansion For find w4: First Rotation will produce Rotate W3 ( 0C 0D 0E 0F) which is equals to ( 0D 0E 0F 0C ).
8 Key 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…
9 Key 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
10 Plain text First Plain text of 16 bytes arrange into 4 x 4 array. Apply s-box to each array of plain text bytes.
11 Plain textSecond step rotate Row of plain text k bytes.
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