An alternative strategy for the expo function uses the following recursive definition: expo(number, exponent) = 1, when exponent = 0 = number * expo(number, exponent – 1), when exponent is odd = (expo(number, exponent // 2)) ** 2, when exponent is even Define a recursive function expo that uses this strategy, and state its computational complexity using big O notation. Note: The program should output in the following format: 0 1 1 2 2 4 3 8 4 16 --------------------------------------------------------------------------------- def expo(base, exponent): expo.calls += 1 # Used to track recursive call count # Write you recursive expo function here expo.calls = 0 def main(): """Tests with powers of 2.""" for exponent in range (5): print(exponent, expo(2, exponent)) if __name__ == "__main__": main()

C++ Programming: From Problem Analysis to Program Design
8th Edition
ISBN:9781337102087
Author:D. S. Malik
Publisher:D. S. Malik
Chapter15: Recursion
Section: Chapter Questions
Problem 12PE
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An alternative strategy for the expo function uses the following recursive definition:

expo(number, exponent) = 1, when exponent = 0 = number * expo(number, exponent – 1), when exponent is odd = (expo(number, exponent // 2)) ** 2, when exponent is even

Define a recursive function expo that uses this strategy, and state its computational complexity using big O notation.

Note: The program should output in the following format:

0 1

1 2

2 4

3 8

4 16

 

---------------------------------------------------------------------------------

 

def expo(base, exponent):
    expo.calls += 1 # Used to track recursive call count
    # Write you recursive expo function here

expo.calls = 0

def main():
    """Tests with powers of 2."""
    for exponent in range (5):
        print(exponent, expo(2, exponent))

if __name__ == "__main__":
    main()
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