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2.11 When discussing linear indexing in Section 2.4, we arrived at the lincar index in Eq. (2-14) by inspection. The same argument used there can be extended to a 3-D array with coordinates x, y, and z, and corresponding dimensions M, N, and P. The linear index for any (x.y, z) is s = x + M(y + Nz) Start with this expression and (a) Derive Eq. (2-15). (b) Derive Eq. (2-16). x = s mod M (2 – 15) - S - X y : (2 – 16) - M 5. *s = a in the lectures and book

2.11 When discussing linear indexing in Section 2.4, we arrived at the lincar index in Eq. (2-14) by inspection. The same argument used there can be extended to a 3-D array with coordinates x, y, and z, and corresponding dimensions M, N, and P. The linear index for any (x.y, z) is s = x + M(y + Nz) Start with this expression and (a) Derive Eq. (2-15). (b) Derive Eq. (2-16). x = s mod M (2 – 15) - S - X y : (2 – 16) - M 5. *s = a in the lectures and book

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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2.11 When discussing linear indexing in Section 2.4, we arrived at the lincar index in Eq. (2-14) by inspection. The same argument used there can be extended to a 3-D array with coordinates x, y, and z, and corresponding dimensions M, N, and P.The linear index for any (x.y,z) is s = x + M(y + Nz) Start with this expression and (a)* Derive Eq. (2-15). (b) Derive Eq. (2-16). x = s mod M (2 – 15) S - x y = (2 – 16) 5. *s = a in the lectures and book
Transcribed Image Text:2.11 When discussing linear indexing in Section 2.4, we arrived at the lincar index in Eq. (2-14) by inspection. The same argument used there can be extended to a 3-D array with coordinates x, y, and z, and corresponding dimensions M, N, and P.The linear index for any (x.y,z) is s = x + M(y + Nz) Start with this expression and (a)* Derive Eq. (2-15). (b) Derive Eq. (2-16). x = s mod M (2 – 15) S - x y = (2 – 16) 5. *s = a in the lectures and book
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