1. - - Problems 1 through 5 will used hexadecimal notation to represent bytes. The following table can be used to perform an XOR operation on hexadecimal digits: 10 1 2 3 4 5 6 7 8 9 a b c d e f -1 0 0 1 2 3 4 5 6 7 8 9 a b c d e f 1 1 0 3 2 5 4 7 6 9 8 badcfe 212 3 0 16 7 4 5 a b 8 9 efcd 313 2 1 0 7 6 5 4 6 a 9 8 fedc 4/4 5 6 7 0 123 cdef 89 a b 515 4 7 6 1 0 3 2 dcfe98 b a 616 7 4 5 2 3 0 1 fcd a b 8 9 717 6 5 4 3 2 1 0fedcba 98 889 a b c d e f 0 1 2 3 4 5 6 7 998 b a dcfe 10 3 2 5 4 7 6 ala b89efcd 230 16 7 4 5 bb a 98fedc3 210 7 6 5 4 ccdef 89 a b 4 5 6 7 0 1 2 3 dldcfe98 b a 5 4 7 6 1 0 3 2 elefcd a b 8 9 6 7 4 5 2 3 0 1 f|fedcba 9 8 7 6 5 4 3 2 1 0 For example, 5 dwould give the entry in row 5 column d, which is 8 Given a plaintext block P and a round key K, with the following values: P=0001020304050607 08090A0B OCODOEOF Ko = 46369cf7 9345246d 450856fb 70c62e3e what would the internal state be after the initial AES AddRoundKey layer? 2. The AES byte substitution layer can be defined using the following table: 00 01 02 03 04 05 06 07 08 09 Oa ob oc Od Oe of -- 00 63 76 77 7bf2 6b6f c5 30 01 67 2b fe d7 ab 76 10 ca 82 c9 7d fa 59 47 fo ad d4 a 2 af 9c a4 72 co 20b7 fd 93 26 36 3f f7 cc 34 a5 e5 f1 71 d8 31 15 3004 c7 23 c3 18 96 05 9a 07 12 80 e2 eb 27 62 75 40 09 83 2c 1a 1b 6e 5a ao 52 3b d6 b3 29 e3 2f 84 50 53 d1 00 ed 20 fc b1 5b 6a cb be 39 40 4c 58 cf 60 do ef aa fb 43 4d 33 85 45 69 02 7f 50 30 9f a 8 70 51 a3 40 8f 92 9d 38 f5 bc b6 da 21 10 ff f3 d2 80 cd oc 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 ob db ao eo 32 3a Da 49 06 24 5C c2 d3 ac 62 91 95 e4 79 bo e7 C8 37 6d 8d d5 4e a 9 6 56 f4 ea 65 7a ae 08 co ba 78 25 2e 1c a6 b4 c6 e 8 dd 74 1f 4b bd 8b 8a de 70 3e b5 66 48 03 f6 Oe 61 35 57 69 86 c1 1d 9e ee1 f8 98 11 69 d9 8e 94 9b le 87 e 9 ce 55 28 df fo|8c a1 89 od bf e6 42 68 41 99 2d of be 54 bb 16 The row headings represent the first four bits of the input byte while the column headings represent the last four bits of the input byte. The entries in the table are the corresponding output bytes for an input byte whose first four bits match the row heading and whose last four bits match the column heading. For example, the input byte b4 would translate to the entry in row be and column 04, which is 8d. Given the resulting internal state that you computed in the previous problem, what would be the result after the SubBytes layer from the first round? 1. - - Problems 1 through 5 will used hexadecimal notation to represent bytes. The following table can be used to perform an XOR operation on hexadecimal digits: 10 1 2 3 4 5 6 7 8 9 a b c d e f -1 0 0 1 2 3 4 5 6 7 8 9 a b c d e f 1 1 0 3 2 5 4 7 6 9 8 badcfe 212 3 0 16 7 4 5 a b 8 9 efcd 313 2 1 0 7 6 5 4 6 a 9 8 fedc 4/4 5 6 7 0 123 cdef 89 a b 515 4 7 6 1 0 3 2 dcfe98 b a 616 7 4 5 2 3 0 1 fcd a b 8 9 717 6 5 4 3 2 1 0fedcba 98 889 a b c d e f 0 1 2 3 4 5 6 7 998 b a dcfe 10 3 2 5 4 7 6 ala b89efcd 230 16 7 4 5 bb a 98fedc3 210 7 6 5 4 ccdef 89 a b 4 5 6 7 0 1 2 3 dldcfe98 b a 5 4 7 6 1 0 3 2 elefcd a b 8 9 6 7 4 5 2 3 0 1 f|fedcba 9 8 7 6 5 4 3 2 1 0 For example, 5 dwould give the entry in row 5 column d, which is 8 Given a plaintext block P and a round key K, with the following values: P=0001020304050607 08090A0B OCODOEOF Ko = 46369cf7 9345246d 450856fb 70c62e3e what would the internal state be after the initial AES AddRoundKey layer? 2. The AES byte substitution layer can be defined using the following table: 00 01 02 03 04 05 06 07 08 09 Oa ob oc Od Oe of -- 00 63 76 77 7bf2 6b6f c5 30 01 67 2b fe d7 ab 76 10 ca 82 c9 7d fa 59 47 fo ad d4 a 2 af 9c a4 72 co 20b7 fd 93 26 36 3f f7 cc 34 a5 e5 f1 71 d8 31 15 3004 c7 23 c3 18 96 05 9a 07 12 80 e2 eb 27 62 75 40 09 83 2c 1a 1b 6e 5a ao 52 3b d6 b3 29 e3 2f 84 50 53 d1 00 ed 20 fc b1 5b 6a cb be 39 40 4c 58 cf 60 do ef aa fb 43 4d 33 85 45 69 02 7f 50 30 9f a 8 70 51 a3 40 8f 92 9d 38 f5 bc b6 da 21 10 ff f3 d2 80 cd oc 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 ob db ao eo 32 3a Da 49 06 24 5C c2 d3 ac 62 91 95 e4 79 bo e7 C8 37 6d 8d d5 4e a 9 6 56 f4 ea 65 7a ae 08 co ba 78 25 2e 1c a6 b4 c6 e 8 dd 74 1f 4b bd 8b 8a de 70 3e b5 66 48 03 f6 Oe 61 35 57 69 86 c1 1d 9e ee1 f8 98 11 69 d9 8e 94 9b le 87 e 9 ce 55 28 df fo|8c a1 89 od bf e6 42 68 41 99 2d of be 54 bb 16 The row headings represent the first four bits of the input byte while the column headings represent the last four bits of the input byte. The entries in the table are the corresponding output bytes for an input byte whose first four bits match the row heading and whose last four bits match the column heading. For example, the input byte b4 would translate to the entry in row be and column 04, which is 8d. Given the resulting internal state that you computed in the previous problem, what would be the result after the SubBytes layer from the first round