(d) algorithm if n ≥ 5 then PrintDs(n) print("D") print("D") if (x = 0 (mod 2)) then Print Ds([n/5]) Print Ds([n/51) x x +3 else Print Ds([n/5]) Print Ds([n/5]) c+52+3

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Need d)

Problem 1: Give an asymptotic estimate, using the O-notation, of the number of letters printed by the
algorithms given below. Give a complete justification for your answer, by providing an appropriate recurrence
equation and its solution.
(a) algorithm PrintAs(n)
if n ≤ 1 then
print("AAA")
else
3
for j1 to n³
do print ("A")
for i 1 to 5 do
Print As( [n/2])
(c) algorithm Print Cs(n)
if n ≤ 1 then
print("C")
else
for j1 to n
do print ("C")
Print Cs([n/2])
Print Cs([n/2])
Print Cs([n/2])
Print Cs([n/2])
(b) algorithm PrintBs(n)
if n ≥ 4 then
for j1 to n²
do print ("B")
for i 1 to 4 do
In part (d), variable x is a global variable initialized to 1.
PrintBs([n/4])
for i 1 to 12 do
Print Bs([n/4])
(d) algorithm PrintDs(n)
if n ≥ 5 then
print("D")
print("D")
if (x = 0 (mod 2)) then
Print Ds([n/5])
Print Ds([n/5])
x + 3
x
else
PrintDs([n/5])
Print Ds([n/5])
x+5 +3
Transcribed Image Text:Problem 1: Give an asymptotic estimate, using the O-notation, of the number of letters printed by the algorithms given below. Give a complete justification for your answer, by providing an appropriate recurrence equation and its solution. (a) algorithm PrintAs(n) if n ≤ 1 then print("AAA") else 3 for j1 to n³ do print ("A") for i 1 to 5 do Print As( [n/2]) (c) algorithm Print Cs(n) if n ≤ 1 then print("C") else for j1 to n do print ("C") Print Cs([n/2]) Print Cs([n/2]) Print Cs([n/2]) Print Cs([n/2]) (b) algorithm PrintBs(n) if n ≥ 4 then for j1 to n² do print ("B") for i 1 to 4 do In part (d), variable x is a global variable initialized to 1. PrintBs([n/4]) for i 1 to 12 do Print Bs([n/4]) (d) algorithm PrintDs(n) if n ≥ 5 then print("D") print("D") if (x = 0 (mod 2)) then Print Ds([n/5]) Print Ds([n/5]) x + 3 x else PrintDs([n/5]) Print Ds([n/5]) x+5 +3
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