3.2. We want to verify that IP(·) and IP−1(·) are truly inverse operations. We consider a vector x = (x1,x2,...,x64) of 64 bit. Show that IP−1(IP(x)) = x for the first five bits of x, i. e. for xi, i = 1,2,3,4,5

IP−1(IP(x))=x

the initial permutation IP and the final permutation IP ⁻¹  are bitwise permutations.

Initial Permutation IP is as per below table

58 50 42 34 26 18 10 2

60 52 44 36 28 20 12 4

62 54 46 38 30 22 14 6

64 56 48 40 32 24 16 8

57 49 41 33 25 17 9 1

59 51 43 35 27 19 11 3

61 53 45 37 29 21 13 5

63 55 47 39 31 23 15 7

Final Permutation IP ⁻¹  is as per below table

40 8 48 16 56 24 64 32

39 7 47 15 55 23 63 31

38 6 46 14 54 22 62 30

37 5 45 13 53 21 61 29

36 4 44 12 52 20 60 28

35 3 43 11 51 19 59 27

34 2 42 10 50 18 58 26

33 1 41 9 49 17 57 25

that input bit 58 is mapped to output position 1, input bit 50 is mapped to the second output position, and so forth. The final permutation IP ⁻¹ performs the inverse operation of IP.

When x=1 then IP ⁻¹ (IP(1) ) is as below. and for all x to 2 to 5 are as below as well.

IP ⁻¹ (IP(1) ) = IP ⁻¹ (40) = 1

Similarly IP ⁻¹ (IP(2) ) = IP ⁻¹ (8) = 2

IP ⁻¹ (IP(3) ) = IP ⁻¹ (48) = 3

IP ⁻¹ (IP(4) ) = IP ⁻¹ (16) = 4

IP ⁻¹ (IP(5) ) = IP ⁻¹ (56) = 2

Explanation:

Explanation is given in the answer section.

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