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5. Given an array of size n, what is the asymptotic runtime of the function Foo? Assume all elements of the array are 32-bits in size and the input size is a power of 8. 1: function Fo0(A[1..n]) 2: if n = 1 then return A[1] + 1 x = 2 · Foo(A[1..n/2) y = 3. Foo(A[n/2 + 1...n]) tmp = 0 for (i = 1 to n) do for (j = i to n) do // tmp will truncate to 32-bits if the sum is larger than 32-bits tmp = A[i] + A[j] – x + y + tmp 3: 4: 5: 6: // tmp is 32-bits long 7: 8: 9: 10: return tmp

5. Given an array of size n, what is the asymptotic runtime of the function Foo? Assume all elements of the array are 32-bits in size and the input size is a power of 8. 1: function Fo0(A[1..n]) 2: if n = 1 then return A[1] + 1 x = 2 · Foo(A[1..n/2) y = 3. Foo(A[n/2 + 1...n]) tmp = 0 for (i = 1 to n) do for (j = i to n) do // tmp will truncate to 32-bits if the sum is larger than 32-bits tmp = A[i] + A[j] – x + y + tmp 3: 4: 5: 6: // tmp is 32-bits long 7: 8: 9: 10: return tmp

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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5. Given an array of size n, what is the asymptotic runtime of the function Foo? Assume all elements of the array are 32-bits in size and the input size is a power of 8. 1: function Fo0(A[1..n]) 2: if n = 1 then return A[1] + 1 x = 2 · Foo(A[1..n/2]) y = 3· Foo(A[n/2+1...n]) tmp = 0 for (i = 1 to n) do for (j = i to n) do // tmp will truncate to 32-bits if the sum is larger than 32-bits tmp = A[i] + A[j] – x + y+ tmp 3: 4: 5: 6: // tmp is 32-bits long 7: 8: 9: 10: return tmp
Transcribed Image Text:5. Given an array of size n, what is the asymptotic runtime of the function Foo? Assume all elements of the array are 32-bits in size and the input size is a power of 8. 1: function Fo0(A[1..n]) 2: if n = 1 then return A[1] + 1 x = 2 · Foo(A[1..n/2]) y = 3· Foo(A[n/2+1...n]) tmp = 0 for (i = 1 to n) do for (j = i to n) do // tmp will truncate to 32-bits if the sum is larger than 32-bits tmp = A[i] + A[j] – x + y+ tmp 3: 4: 5: 6: // tmp is 32-bits long 7: 8: 9: 10: return tmp
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