Question
b’i = b(i+2) XOR b(i+5) XOR b(i+7) XOR di
Here di is –
a.
di is the ith bit of a byte ‘d’ whose hex value is 0x15
b.
di is the ith bit of a byte ‘d’ whose hex value is 0x05
c.
di is the ith bit of a byte ‘d’ whose hex value is 0x25
d.
di is the ith bit of a byte ‘d’ whose hex value is 0x51
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Q. The inverse s-box permutation follows - b’i = b(i+2) XOR b(i+5) XOR b(i+7) XOR di Here di is –
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