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The objective of this assignment is to analyze a simple matrix transposition algorithm. Below is its pseudocode of the simplified Transposition algorithm of an nxn matrix A. Transpose (A) 1 for i = 1 to n-1 2 for j = i+1 to n 3 exchange A[i,j] with A[j,i] Running Time of Transpose (90 points) a. (2 points) What is the input size? b. (40 points) Count precisely the total number of comparisons C(n) performed by the for loops in Lines 1 and 2. Express C(n) as a function of n. (Use the same steps we used in the previous lectures and homework assignments to analyze nested for loops. Use a table like in Homework 1. Be as precise as Student 3 (See Appendix) C. (2 points) What is the asymptotic bound for C(n)? d. (40 points) Count precisely the total number of exchanges E(n) performed by Line 3. Express E(n) as a function of n. (Use the same steps we used in the previous lectures and homework assignments to analyze nested for loops. Use a table like in Homework 1. Be as precise as Student 3 (See Appendix) e. (2 points) What is the asymptotic bound for E(n)? f. (4 points) Which function (C(n)? or E(n)) yield the right asymptotic growth? Which one (C(n)? or E(n)) is more convenient to derive the asymptotic growth of the running time of the Transpose algorithm? for j INSERTION-SORT (A) 1 = 2 to A.length n comparisons 2 key A[j] 3 // Insert A[j] into the sorted sequence A[1.. j - 1]. 4 i = j - 1 5 while i > 0 and A[i] > key 2.j - I comparisons (t; = j all the time) 6 A[i + 1] = A[i] 7 i =i-1 T(n)=n+ 1 8 A[i + 1] = key = 11 23-1=n+2, 1= nin+1) = n+20 - 1)-(n-1) = n(n+1)-1= 2(n) 2 12