Question

Q3. The number of operations executed by algorithms A and B are 12n3 + 40n log n and 5n4 -100n2 respectively.

Determine c and n0 such that B is greater than c*A for n ≥ n0.

Answer 1

12n B 5n-100 n2 Given Expressions A 40 n log n From the above expression, need to show that the number of operations executed by algorithms T(B)> T(A) Here T means the number of operations executed by the algorithm. That means 5n4-100 n2> 12n3 40 n logn Let us consider T(A) = 12n3 40 n log n (a) Divide the above equation with n2 then 40 log n T(A) 12n .. (1) Let us consider T(B) 5n4-100 n2 (b) Divide the above equation with n2 then 5n2 -100 T(B) T(A) then 40 5n2-100> 12n log n (c)

n value is growing faster than the logn value. So, it can be written as in the below format log n<n log n 1 from the above statement (c) can be written in the below manner 40 5n2-100 12 n +log n n log n 5n2 - 100 12 n +40 n log n 5n2 100 12 n 40 5n2- 100 12 n 40 5n2 12 n 40 100 From the calculation, the resultant statement is log n 1 Here substitutes the value in the above 5n2 -100 12n+ 40 (1) log n 2 5n -100 12n 40 multiply above expression with n'on both sides, then 5n4-100 n2>12n3 40 n logn A ..

5n4-100 n2>12n3 40 n logn from the above expression, using induction method 1 then suppose 5(1)-100 (1)212(1) + 40 (1 logl) 5-100 12 40 (1 log1) 5 12 40 (1 log 1)+100 5112+40(0)log1 0 5>112 which is false 5n4 -100 n2>12n3 40 n logn 2 then suppose n 5(2)-100 (2)212(2)40 (2 log2) 80- 400> 96+40 (0.6020) 96 40 (0.6020)+400 80 80 96+24.08 400 80 520.08 which is false

5n4 -100 n2>12n3 40 n logn suppose n 7 then 5(7)4-100 (7)2>12(7) 40 (7 log7) 5(2401) 100(49) 12(343)+ 40 (5.9156 12005 4900 4116 236.24 7105 4352.24 which is true from the above calculations , ifn 7 then it satisifies the condition T (B) > T(A) means the number of operations B is greater than A T(B) T (A) n 7 and n 7 then , the number of operations B will exceeds the value of A.