What is the expected time complexity of the function below?

import random

def foo(n):
    # draw a random number uniformly from 0, 1, 2, ..., 99
    # in constant time
    x = random.randrange(100)
    for i in range(x):
        for j in range(n):
            print("Here!")

What is the expected time complexity of the function below?

import random

def foo(n):
    # draw a number uniformly at random from 0, 1, 2, ..., n-1
    # in constant time
    x = random.randrange(n)
    if x < 100:
        for j in range(n**2):
            print(j)

What is the expected time complexity of the following function?

import random

def foo(n):
    for i in range(n):
        # draw a number uniformly at random from 0, 1, 2, ..., n-1 in Theta(1)
        x = random.randrange(n)
        for j in range(x):
            print(j)

What is the expected time complexity of the function below?

import random

def foo(n):
    # draw a number uniformly at random from 0, 1, 2, ..., n-1 in Theta(1)
    x = random.randrange(n)
    if x < math.sqrt(n):
        for j in range(n):
            print(j)

What is the expected time complexity of the following function? State your answer using asymptotic notation.

import random

def foo(n):
    x = random.randrange(n)

    if x < 8:
        for j in range(n**3):
            print(j)
    else:
        for j in range(n):
            print(j)

What is the expected time complexity of the function below? State your answer using asymptotic notation.

You may assume that math.sqrt and math.log take \(\Theta(1)\) time. math.log computes the natural log.

import random
import math

def foo(n):
    # draw a number uniformly at random from 0, 1, 2, ..., n-1 in Theta(1)
    x = random.randrange(n)

    if x < math.log(n):
        for j in range(n**2):
            print(j)
    elif x < math.sqrt(n):
        print('Ok!')
    else:
        for i in range(n):
            print(i)

What is the expected time complexity of the function below?

You may assume that math.sqrt takes \(\Theta(1)\) time.

import random
import math

def foo(n):
    # draw a number uniformly at random from 0, 1, 2, ..., n-1 in Theta(1)
    x = random.randrange(n)

    if x < math.sqrt(n):
        for j in range(n):
            print(j)
    elif x < n//2:
        print('Ok!')
    else:
        for i in range(n**2):
            print(i)

What is the expected time complexity of the function below?

import random

def foo(n):
    # draw a random number uniformly from 0, 1, 2, ..., n-1
    # in constant time
    x = random.randrange(n)
    for i in range(x):
        for j in range(n):
            print("Here!")

What is the expected time complexity of the function below? State your answer using asymptotic notation.

You may assume that math.sqrt and math.log2 take \(\Theta(1)\) time.

import random
import math

def boo(n):
    # draw a number uniformly at random from 0, 1, 2, ..., n-1 in Theta(1)
    x = random.randrange(n)

    if x < math.log2(n):
        for i in range(n**2):
            print("Very unlucky!")
    elif x < math.sqrt(n):
        for i in range(n):
            print("Unlucky!")
    else:
        print("Lucky!")

boo(100)

\(\Theta(n \log n)\) Show Answer

Solution

There are three cases: Case 1) \(x < \log_2 n\); Case 2) \(1 < x < \sqrt n\); and Case 3) \(x \geq\sqrt{n}\).

The probability of Case 1 is \(\log n / n\). The time taken in case 1 is \(\Theta(n^2)\).

The probability of Case 2 is

\[\frac{\sqrt n}{n} - \frac{\log_2 n}{n} = \Theta(\sqrt n / n) = \Theta(1/\sqrt n). \]

The time taken in this case is \(\Theta(n)\).

The probability of Case 3 is 1 minus the probability of the sum of the other two cases, which will simply be \(\Theta(1)\):

\[ 1 - \Theta(1 / n) - \Theta(1 / \sqrt{n}) = \Theta(1). \]

The time taken is \(\Theta(1)\).

Therefore, the expected time is: