AES S-box calculation

Question

I am trying to design the s-box for AES. Quoting from Wikipedia, the algorithm for applying affine transformation in AES s-box calculation is

Let s (an 8-bit unsigned variable) be the input number.
Let result be 0.
For 5 times: 
   XOR result with s.
   Rotate s one bit to the left.

My question is why the multiplication is done this way? Why only 5 steps?

look at the affine matrix, each column and row only has 5 bits set — Richie Frame, Mar 14, 2018 at 21:04

Filip Franik · Accepted Answer · 2018-03-16 14:48:12Z

2

This is first half of the calculation presented on the Wiki page.

Notice that every bit of the result is actually a XOR of just five bits:

Now we slightly modify the algorithm you wrote to create five numbers by rotating the original number to the left and only then XOR them together:

As you can see the result is exactly the same.

answered Mar 16, 2018 at 14:48

Filip Franik

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$\begingroup$ Thanks for the answer. So for any S-box , I'll just compute the no. of set bits and do that much shifting and xor-ing? $\endgroup$
– user56952
Mar 17, 2018 at 7:05

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