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What will be the output of the following code?
def power(x, n):
if n == 0:
return 1
elif n % 2 == 0:
return power(x, n // 2) ** 2
else:
return x * power(x, n - 1)
if n == 0:
return 1
elif n % 2 == 0:
return power(x, n // 2) ** 2
else:
return x * power(x, n - 1)
result = power(2, 4)
print(result)
print(result)
- a)8
- b)16
- c)4
- d)2
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
What will be the output of the following code?def power(x, n):if n == ...
The power function calculates the power of a number recursively. In this case, power(2, 4) calculates 2 to the power of 4, which is 16.
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Community Answer
What will be the output of the following code?def power(x, n):if n == ...
Explanation:
The given code defines a recursive function named "power" that takes two parameters: "x" and "n". This function calculates and returns the value of "x" raised to the power of "n" using recursion.
Here is a step-by-step explanation of how the code works:
1. The function "power" is defined with two parameters: "x" and "n".
2. The first condition checks if "n" is equal to 0. If it is, the function returns 1, as any number raised to the power of 0 is 1.
3. The second condition checks if "n" is even (i.e., if "n" is divisible by 2 without a remainder). If it is, the function recursively calls itself with the parameters "x" and "n // 2" (integer division), and then squares the result.
4. The third condition handles the case where "n" is odd. In this case, the function recursively calls itself with the parameters "x" and "n - 1", and then multiplies the result by "x".
5. Finally, outside the function, the variable "result" is assigned the value of calling the "power" function with the parameters 2 and 4. This means that it calculates 2 raised to the power of 4.
6. The value of "result" is then printed, which will be the final output of the code.
Explanation of the calculation:
When the function is called with the parameters 2 and 4, the following steps occur:
1. Since 4 is even, the function calls itself with the parameters 2 and 2 (4 // 2). This results in a recursive call to calculate 2 raised to the power of 2.
2. Again, since 2 is even, the function calls itself with the parameters 2 and 1 (2 // 2). This results in a recursive call to calculate 2 raised to the power of 1.
3. Now, the base case is reached as "n" is equal to 0. So, the function returns 1.
4. The previous recursive call receives the result 1 and squares it, resulting in 1^2 = 1.
5. The initial recursive call receives the result 1 and multiplies it by 2, resulting in 1 * 2 = 2.
6. Finally, the initial function call returns the value 2, which is stored in the variable "result" and printed as the output.
Therefore, the output of the code will be 2.
The given code defines a recursive function named "power" that takes two parameters: "x" and "n". This function calculates and returns the value of "x" raised to the power of "n" using recursion.
Here is a step-by-step explanation of how the code works:
1. The function "power" is defined with two parameters: "x" and "n".
2. The first condition checks if "n" is equal to 0. If it is, the function returns 1, as any number raised to the power of 0 is 1.
3. The second condition checks if "n" is even (i.e., if "n" is divisible by 2 without a remainder). If it is, the function recursively calls itself with the parameters "x" and "n // 2" (integer division), and then squares the result.
4. The third condition handles the case where "n" is odd. In this case, the function recursively calls itself with the parameters "x" and "n - 1", and then multiplies the result by "x".
5. Finally, outside the function, the variable "result" is assigned the value of calling the "power" function with the parameters 2 and 4. This means that it calculates 2 raised to the power of 4.
6. The value of "result" is then printed, which will be the final output of the code.
Explanation of the calculation:
When the function is called with the parameters 2 and 4, the following steps occur:
1. Since 4 is even, the function calls itself with the parameters 2 and 2 (4 // 2). This results in a recursive call to calculate 2 raised to the power of 2.
2. Again, since 2 is even, the function calls itself with the parameters 2 and 1 (2 // 2). This results in a recursive call to calculate 2 raised to the power of 1.
3. Now, the base case is reached as "n" is equal to 0. So, the function returns 1.
4. The previous recursive call receives the result 1 and squares it, resulting in 1^2 = 1.
5. The initial recursive call receives the result 1 and multiplies it by 2, resulting in 1 * 2 = 2.
6. Finally, the initial function call returns the value 2, which is stored in the variable "result" and printed as the output.
Therefore, the output of the code will be 2.
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