Step 3: The function f is actually completed by 3 sub-steps: (1) expanding 32-bits to 48-bits...
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Step 3: The function f is actually completed by 3 sub-steps: (1) expanding 32-bits to 48-bits using expansion permutation table E, and then exclusive OR the 48-bit key K1; (2) shrinking the 48-bits result from (1) back to 32 bits using the 8 S-boxes; (3) re-order the 32-bits from (2) using permutation function P. The 3 steps are illustrated as bellow: Step 3.1: show the result after the right 32-bit R is expanded to 48 bit using E table which is as follows: ban (10 points) 32 4 8 12 16 20 24 28 1 5 9 13 17 18 2 4 6 8 10 11 12 14 15 16 3 7 19 20 22 23 24 27 31 21 25 26 29 30 2 33 32 5 9 13 17 21 25 29 1 Step 3.2: show a new 48-bit result after doing exclusive OR the above 48-bit with the 48-bit round key K1. Step 3.3: Splitting the 48-bit into 8 boxes and each is 6-bit using the 8 Sboxes as illustrated below. Si 14 4 13 12 15 11 6 310 612 6007 015 7 414 213 1 10 0 12 11 9530 4 114 813 6 2 11 15 12 0 7 310 60 15 12 0 2 4 17 611 314 10 0013 S; 3 4 07 212 12 0 5 10 01100011 6 12 e 03215 6 7 12 0 514 9 15 1 014 011 313 4 7 15 2 814 12 014 7 11 10 413 1 5 13 810 1 315 4 2 11 Sy 10 0014 6 3 15 5 1 1312 711 4 20 13 70034 610 20 514 12 11 15 1 13640815 3 011 1 2 12 6 10 14 7 110 13 00087 415 14 3 11 5 2 12 S. (10 points) 713 14 30010 120 611 12 415 13 811 5015 0347 212 1 10 14 0 1001211 7 13 15 13 14 6 284 315 0010 113 004 611 12 7 2 14 S 2 12 4 1 7 10 11 085 3 15 13 014 0 1411 212 4 713 1 6 0 16 10 3 000 4 2 111 10 13 7 8 15 912 53014 11 012 7 114 213 015 0910 463 S 12 1 10 15 0 260013 3 414 7 511 10 15 4 2 7 12 066 1 13 14 011 38 01415 5 2 0 12 3 7 0 4 10 1 13 11 6 43 212 0 5 15 10 11 14 17 80813 S 411 2 14 15 0 013 312 07 510 61 13 011 7 4 0 110 14 3 512 215 06 1 411 13 12 3 7 14 10 16 880502 011 13 8 1 410 796 015 14 2 312 Sa 15 284 01511 110 0314 6 012 7 115 13 810 3 7 412 5 6 11 014 02 711 4 10 12 14 2 0 0 10 13 15 358 2 1 14 7 4 10 13 16 12 9035011 and then we can shrink each 6-bit block to 4-bit block. As a result we have the 32-bit result after the shrinking. Please show this 32-bit result after shrinking using Sboxes. (20 points) Step 3.4: based on the result from step 3.3, we re-order the 32-bit result from step 3.3 using the permutation function P which is i illustrated as follows, please show have the new 32-bit result after P permutation. (10 points) Step 3: The function f is actually completed by 3 sub-steps: (1) expanding 32-bits to 48-bits using expansion permutation table E, and then exclusive OR the 48-bit key K1; (2) shrinking the 48-bits result from (1) back to 32 bits using the 8 S-boxes; (3) re-order the 32-bits from (2) using permutation function P. The 3 steps are illustrated as bellow: Step 3.1: show the result after the right 32-bit R is expanded to 48 bit using E table which is as follows: ban (10 points) 32 4 8 12 16 20 24 28 1 5 9 13 17 18 2 4 6 8 10 11 12 14 15 16 3 7 19 20 22 23 24 27 31 21 25 26 29 30 2 33 32 5 9 13 17 21 25 29 1 Step 3.2: show a new 48-bit result after doing exclusive OR the above 48-bit with the 48-bit round key K1. Step 3.3: Splitting the 48-bit into 8 boxes and each is 6-bit using the 8 Sboxes as illustrated below. Si 14 4 13 12 15 11 6 310 612 6007 015 7 414 213 1 10 0 12 11 9530 4 114 813 6 2 11 15 12 0 7 310 60 15 12 0 2 4 17 611 314 10 0013 S; 3 4 07 212 12 0 5 10 01100011 6 12 e 03215 6 7 12 0 514 9 15 1 014 011 313 4 7 15 2 814 12 014 7 11 10 413 1 5 13 810 1 315 4 2 11 Sy 10 0014 6 3 15 5 1 1312 711 4 20 13 70034 610 20 514 12 11 15 1 13640815 3 011 1 2 12 6 10 14 7 110 13 00087 415 14 3 11 5 2 12 S. (10 points) 713 14 30010 120 611 12 415 13 811 5015 0347 212 1 10 14 0 1001211 7 13 15 13 14 6 284 315 0010 113 004 611 12 7 2 14 S 2 12 4 1 7 10 11 085 3 15 13 014 0 1411 212 4 713 1 6 0 16 10 3 000 4 2 111 10 13 7 8 15 912 53014 11 012 7 114 213 015 0910 463 S 12 1 10 15 0 260013 3 414 7 511 10 15 4 2 7 12 066 1 13 14 011 38 01415 5 2 0 12 3 7 0 4 10 1 13 11 6 43 212 0 5 15 10 11 14 17 80813 S 411 2 14 15 0 013 312 07 510 61 13 011 7 4 0 110 14 3 512 215 06 1 411 13 12 3 7 14 10 16 880502 011 13 8 1 410 796 015 14 2 312 Sa 15 284 01511 110 0314 6 012 7 115 13 810 3 7 412 5 6 11 014 02 711 4 10 12 14 2 0 0 10 13 15 358 2 1 14 7 4 10 13 16 12 9035011 and then we can shrink each 6-bit block to 4-bit block. As a result we have the 32-bit result after the shrinking. Please show this 32-bit result after shrinking using Sboxes. (20 points) Step 3.4: based on the result from step 3.3, we re-order the 32-bit result from step 3.3 using the permutation function P which is i illustrated as follows, please show have the new 32-bit result after P permutation. (10 points)
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To expand the 32bit right half R to 48 bits using the expansion permutation table E follow the given ... View the full answer
Related Book For
Fundamentals of Financial Management
ISBN: 978-0324664553
Concise 6th Edition
Authors: Eugene F. Brigham, Joel F. Houston
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