Question

l := 0; for i = 2 to n do for j = 1 to n \cdot i do l := l + 2 \cdot i + 3 \cdot j od od s := 0; for i = 1 to n do for j = 1 to \lfloor \log i \rfloor do s := s + i \cdot j od od

          l := 0;
for i = 2 to n do
for j = 1 to n \cdot i do
l := l + 2 \cdot i + 3 \cdot j
od
od
s := 0;
for i = 1 to n do
for j = 1 to \lfloor \log i \rfloor do
s := s + i \cdot j
od
od

l := 0;
for i = 2 to n do
for j = 1 to n ·i do
l := l + 2 ·i + 3 ·j
od
od
s := 0;
for i = 1 to n do
for j = 1 to ⌊logi ⌋do
s := s + i ·j
od
od

Added by Sarah M.

Computer Science and Information Technology

Trishna Knowledge Systems 2018 Edition

Instant Answer

Solved by Expert Supreeta N

09/05/2023

Step 1

Inside this loop, there is another loop that iterates from j=l to n.i. The inner loop performs a constant number of operations (2*i+3*j) for each iteration. Therefore, the total number of iterations of the inner loop is (n-2+1)(n-2+1) = (n-1)^2. Show more…

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Analyze the following pseudocode and give a tight (Θ) bound on the running time as a function of n. You can assume that all individual instructions are elementary. Show your work. l := 0; for i=2 to n do forj=l to n.i do l= l+2*i+3*j od od s :=0; for i=l to n do for j=l to [log i] do s := s+i*j od od