Where is the S-Box generated in Rijandel/AES? It's rather kind of lame questions, and I can't find good and clear explanation:


In which step of Rijandel is S-box generated?
Is the S-box reused in every round of cipher or is generated in every iteration?
In a course-ware movie where they explained AES (class.coursera.org/crypto-preview/lecture/16) the speaker says that the S-box could be precomputed. Does that mean that the S-box could be constant?
The S-Box was generated when Rijndael was designed, not in any step.
It's used in every round in the SubBytes step.
The S-box is constant. You could see it as a function taking a byte and returning a byte. It is used to reduce algebraic properties of Rijndael.


In fact, this is it:

   | 0  1  2  3  4  5  6  7  8  9  a  b  c  d  e  f
---|--|--|--|--|--|--|--|--|--|--|--|--|--|--|--|--|
00 |63 7c 77 7b f2 6b 6f c5 30 01 67 2b fe d7 ab 76 
10 |ca 82 c9 7d fa 59 47 f0 ad d4 a2 af 9c a4 72 c0 
20 |b7 fd 93 26 36 3f f7 cc 34 a5 e5 f1 71 d8 31 15 
30 |04 c7 23 c3 18 96 05 9a 07 12 80 e2 eb 27 b2 75 
40 |09 83 2c 1a 1b 6e 5a a0 52 3b d6 b3 29 e3 2f 84 
50 |53 d1 00 ed 20 fc b1 5b 6a cb be 39 4a 4c 58 cf 
60 |d0 ef aa fb 43 4d 33 85 45 f9 02 7f 50 3c 9f a8 
70 |51 a3 40 8f 92 9d 38 f5 bc b6 da 21 10 ff f3 d2 
80 |cd 0c 13 ec 5f 97 44 17 c4 a7 7e 3d 64 5d 19 73 
90 |60 81 4f dc 22 2a 90 88 46 ee b8 14 de 5e 0b db 
a0 |e0 32 3a 0a 49 06 24 5c c2 d3 ac 62 91 95 e4 79 
b0 |e7 c8 37 6d 8d d5 4e a9 6c 56 f4 ea 65 7a ae 08 
c0 |ba 78 25 2e 1c a6 b4 c6 e8 dd 74 1f 4b bd 8b 8a 
d0 |70 3e b5 66 48 03 f6 0e 61 35 57 b9 86 c1 1d 9e 
e0 |e1 f8 98 11 69 d9 8e 94 9b 1e 87 e9 ce 55 28 df 
f0 |8c a1 89 0d bf e6 42 68 41 99 2d 0f b0 54 bb 16 


You can read it by choosing your column by the low nibble of a byte, and the row by the high nibble of a byte.