Select the correct answerObserve the following Python code?def x(num): if num == 0: return 0 else: return num*x(num - 1)def y(num, ct): if num == 0: return ct else: return y(num-2, ct-2)OptionsBoth x() and y() are tail recursiveBoth x() and y() aren’t tail recursivey() is tail recursive but x() isn’tx() is tail recursive but y() isn’t
Question
Select the correct answerObserve the following Python code?def x(num): if num == 0: return 0 else: return num*x(num - 1)def y(num, ct): if num == 0: return ct else: return y(num-2, ct-2)OptionsBoth x() and y() are tail recursiveBoth x() and y() aren’t tail recursivey() is tail recursive but x() isn’tx() is tail recursive but y() isn’t
Solution
The correct answer is "y() is tail recursive but x() isn’t".
Here's why:
Tail recursion is a concept in programming where the recursive call is the final operation in the function. In other words, the function will do nothing after the recursive call.
Looking at the function x(), we see that it performs a multiplication operation after the recursive call (num*x(num - 1)). This means that it is not tail recursive because it still has work to do (the multiplication) after making the recursive call.
On the other hand, the function y() makes its recursive call as the last operation (return y(num-2, ct-2)). There is no additional work to be done after this call, which means that y() is tail recursive.
Similar Questions
Select the correct answer Observe the following Python code?def a(n): if n == 0: return 0 else: return n*a(n - 1)def b(n, tot): if n == 0: return tot else: return b(n-2, tot-2)Optionsb() is tail recursive but a() isn’t Both a() and b() are tail recursive Both a() and b() aren’t tail recursivea() is tail recursive but b() isn’t
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