Question-1

Python

What is the output of the following snippet of code?

def foo(n):
    if n % 2 == 0:
        return n // 2
    return 3 * n + 1

count = 0
x = 10
while x != 1:
    x = foo(x)
    count += 1
print(count)

\(6\)

For a given integer n, foo(n) returns n // 2 if n is even and 3 * n + 1 if n is odd. Starting with x initialized to 10, the while loop repeatedly applies foo(x) until x equals 1, incrementing count with each iteration.

First Iteration:

  • x = 10
  • Since 10 % 2 == 0, x becomes 10 // 2 = 5
  • Increment count to 1

Second Iteration:

  • x = 5
  • Since 5 % 2 != 0, x becomes 3 * 5 + 1 = 16
  • Increment count to 2
iteration previous value of x x % 2 == 0 current value of x count
first 10 True 10 // 2 = 5 1
second 5 False 3 * 5 + 1 = 16 2
third 16 True 16 // 2 = 8 3
fourth 8 True 8 // 2 = 4 4
fifth 4 True 4 // 2 = 2 5
sixth 2 True 2 // 2 = 1 6

As the value of x is 1, the loop terminates. So, the loop executes six times, making the final value of count 6.