Since M = 3570 E2F1 BA46 82C7 HEX = (0011 0101 0111 0000 1110 0010 1111 0001 1011 1010 0100 0110 1000 0010 1100 0111), ● Initial Permutation ((IP) ● then IP(M) = [Lo Ro] where Lo= 1010 1110 0001 1011 1010 0001 1000 1001 AE1BA189 HEX R 1101 1100 0001 1111 0001 0000 1111 0100 = DC1F10F4 HEX Now we have Lo and R ready for iteration!
Since M = 3570 E2F1 BA46 82C7 HEX = (0011 0101 0111 0000 1110 0010 1111 0001 1011 1010 0100 0110 1000 0010 1100 0111), ● Initial Permutation ((IP) ● then IP(M) = [Lo Ro] where Lo= 1010 1110 0001 1011 1010 0001 1000 1001 AE1BA189 HEX R 1101 1100 0001 1111 0001 0000 1111 0100 = DC1F10F4 HEX Now we have Lo and R ready for iteration!
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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can you explain me this pic how di get all numbers Ik binary for M but coul not figure how did get keft and right numbers so explain me in step by step please
![Initial Permutation ((IP)
Since M = 3570 E2F1 BA46 82C7 HEX
(0011 0101 0111 0000 1110 0010 1111 0001
1011 1010 0100 0110 1000 0010 1100
0111),
then IP(M) = [L, R₂] where
L 1010 1110 0001 1011 1010 0001
1000 1001 = AE1BA189 HEX
R = 1101 1100 0001 1111 0001 0000 1111
0100 DC1F10F4 HEX
=
• Now we have L and R ready for iteration!](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5bad4e48-dad8-4710-a64b-b24b80d1efcf%2F3badf728-451d-4acd-8637-5cc1cbe49a5f%2Fv1ktly_processed.png&w=3840&q=75)
Transcribed Image Text:Initial Permutation ((IP)
Since M = 3570 E2F1 BA46 82C7 HEX
(0011 0101 0111 0000 1110 0010 1111 0001
1011 1010 0100 0110 1000 0010 1100
0111),
then IP(M) = [L, R₂] where
L 1010 1110 0001 1011 1010 0001
1000 1001 = AE1BA189 HEX
R = 1101 1100 0001 1111 0001 0000 1111
0100 DC1F10F4 HEX
=
• Now we have L and R ready for iteration!
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