Q3/ What kind of optimization we can apply to the following code t16 +j+1 t17 ← 116-1 t8 ←-j-1 t9← 4*t8 temp

zay7 Jump to level 1 p q r-rv-q)A-p T TT TTF TFT T T LL F TT FTF FFT FFF Check Next Fill in (-rv-q)^-p.

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First Pass: ( 5 1 4 2 8 ) \to ( 1 5 4 2 8 ), Here, algorithm compares the first two elements, and swaps them. ( 1 5 4 2 8 ) \to ( 1 4 5 2 8 ), Swap since 5 > 4 ( 1 4 5 2 8 ) \to ( 1 4 2 5 8 ), Swap since 5 > 2 ( 1 4 2 5 8 ) \to ( 1 4 2 5 8 ), Now, since these elements are already in order (8 > 5), algorithm does not swap them. Second Pass: ( 1 4 2 5 8 ) \to ( 1 4 2 5 8 ) ( 1 4 2 5 8 ) \to ( 1 2 4 5 8 ), Swap since 4 > 2 ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) Now, the array is already sorted, but our algorithm does not know if it is completed. The algorithm needs one whole pass without any swap to know it is sorted. Third Pass: ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) ( 1 2 4 5 8 ) \to ( 1 2 4 5 8 ) Finally, the array is sorted, and the algorithm can terminate. Bubble sort is a simple sorting algorithm. It works by repeatedly stepping through the list to be sorted, comparing each pair of…

Below are the pseudocodes from the book for has_edge and in_edges. Use the pseudocode to show in a MS PowerPoint presentation why in_edges is slow and takes more than just O(n) time-complexity. You can do this work in pairs. If you do work in pairs, write in the online text and on the presentation who are the partners. has edge(i,j) for k in adj[i] do if k = j then in edges(i) ou ArrayStack() for j in 0,1,2,...,n – 1 do if has edge(j, i) then out.append(j) return true return false return out A T Fr B U S X2 x2

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This code deletes all nodes whose info field contains the value x. Please modify the code to delete the only first value of x : ( maybe we need to add one line )  ( with explanation for each line if possible) many thanks.

A = [1 1; 1 -1];b = [80; 20]; A_inverse = inv(A);x = A\b; disp('A inverse is');disp(A_inverse);disp(['x1 is ', num2str(x(1)), ' and x2 is ', num2str(x(2))]); if isequal(round(x), [50; 30])    disp('Success');else    disp('Incorrect: Please try again');end.  Output matlab. .

This code deletes all nodes whose info field contains the value x. Please modify the code to delete the only first value of x :  ( with explanation for each line if possible) many thanks. ( do not care about yellow )

debug the code for me % 1 gravity_si=9.81 gravity_eng=32.17   % 2 math_consts=[0.5772,1.1618,2.718,3.1416]   %3 r_vec1=50:-2:1   %4 r_vec2=linspace(0,pi,37)   %5 science_consts=[6.022*10^23;1.381*10^-23;6.626*10^-34]   %6 mat1=[10 4 7 9 11;12 8 2 3 4;7 0 12 6 5]'   %7 matw=zeros(6,8)   %8 chr1=char("discrete")   %9 str1="mathematics"   %10 str2=["Generation" "Transmission";"Distribution" "Supply"]