The inverse s-box permutation follows, b _i = b_(i+2) XOR b(i+5) XOR b_(i+7) XOR d_i

Question

TuteeHUB · Accepted Answer

d_i is the ith bit of a byte  d  whose hex value is 0x05

TuteeHUB · Answer

nHere d_i is

TuteeHUB · Answer

d_i is the ith bit of a byte  d  whose hex value is 0x15

TuteeHUB · Answer

d_i is the ith bit of a byte  d  whose hex value is 0x05

TuteeHUB · Answer

d_i is the ith bit of a byte  d  whose hex value is 0x25

TuteeHUB · Answer

d_i is the ith bit of a byte  d  whose hex value is 0x51

1.	The inverse s-box permutation follows, b _i = b_(i+2) XOR b(i+5) XOR b_(i+7) XOR d_i
A.	nHere d_i is
B.	d_i is the ith bit of a byte d whose hex value is 0x15
C.	d_i is the ith bit of a byte d whose hex value is 0x05
D.	d_i is the ith bit of a byte d whose hex value is 0x25
E.	d_i is the ith bit of a byte d whose hex value is 0x51
Answer» C. d_i is the ith bit of a byte d whose hex value is 0x05

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