# Divide and Conquer, Sorting and Searching, and Randomized Algorithms Week 2 | Problem Set #2 Quiz Answer

## Divide and Conquer, Sorting and Searching, and Randomized Algorithms Week 2 | Problem Set #2 Quiz Answer

In this article i am gone to share Coursera Course Divide and Conquer, Sorting and Searching, and Randomized Algorithms Week 2 | Problem Set #2 Quiz Answer with you..

#### Problem Set #2

Question 1)

This question will give you further practice with the Master Method. Suppose the running time of an algorithm is governed by the recurrence T(n) = 7 ✳ T(n / 3) +n2. What's the overall asymptotic running time (i.e, the value of T(n))?

• Ө(n2 log n)
• Ө(n2)
• Ө(n2.81)
• Ө(n log n)

Question 2)

This question will give you further practice with the Master Method. Suppose the running time of an algorithm is governed by the recurrence T(n) = 9✳T(n / 3) +n2. What's the overall asymptotic running time (i.e, the value of T(n))?

• Ө(n2 log n)
• Ө(n2)
• Ө(n3.17)
• Ө(n log n)

Question 3)

This question will give you further practice with the Master Method. Suppose the running time of an algorithm is governed by the recurrence T(n) = 5✳T(n / 3) +4n. What's the overall asymptotic running time (i.e, the value of T(n))?

• Ө(n2.59)
• Ө(n2)
• Ө(nlog 3 / log 5)
• Ө(n5/3)
• Ө(n log (n))
• Ө(nlog 3 / (5))

Question 4)

Consider the following pseudocode for calculating a where a and bare positive integers)

FastPower(a,b) :

if b = 1

return a

else

c : = a*a

ans := Fast Power(c,[b/2])

if b is odd

return a*ans

else return ans

end

Here [x] denotes the floor function, that is, the largest integer less than or equal to x.

Now assuming that you use a calculator that supports multiplication and division (Le.. you can do multiplications and divisions in constant time), what would be the overall asymptotic running time of the above algorithm (as a function of b)?

• Ө(b log(b))
• Ө(log(b))
• Ө(b)
• Ө(√b)

Question 5)

Choose the smallest correct upper bound on the solution to the following recurrence T(1) = 1 and T (n) ≤ T( [√n]) +1 for n > 1. Here [x] denotes the "floor” function, which rounds down to the nearest integer. (Note that the Master Method does not apply.)

• 0(log n)
• 0(√n)
• 0(1)
• 0(log log n )