Answer:
The power function can be written as a recursive function (using Java) as follows:
- static int power(int x, int n)
- {
- if(n == 0){
- return 1;
- }
- else {
- return power(x, n-1 ) * x;
- }
- }
Explanation:
A recursive function is a function that call itself during run time.
Based on the question, we know x to the 0th power is 1. Hence, we can just create a condition if n = 0, return 1 (Line 3 - 5).
Next, we implement the logic "x to the nth power can be obtained by multiplying x to the n-1'th power with x " from the question with the code: return power(x, n-1 ) * x in the else block. (Line 6 -8)
In Line 7, power() function will call itself recursively by passing x and n-1 as arguments. Please note the value of n will be reduced by one for every round of recursive call. This recursive call will stop when n = 0.
Just imagine if we call the function as follows:
int result = power(2, 3);
What happen will be as follows:
- run Line 7 -> return power(2, 2) * 2
- run Line 7 -> return power(2, 1) * 2
- run Line 7 -> return power(1, 0) * 2
- run Line 4 -> return 1 (Recursive call stop here)
Next, the return value from the inner most recursive call will be return to the previous call stack:
- power(1, 0) * 2 -> 1 * 2
- power(2, 1) * 2 -> 1 * 2 * 2
- power(2, 2) * 2 -> 1 * 2 * 2 * 2 - > 8 (final output)
PYTHON CODE
# function to compare a and b
def compare(a,b):
# if a is greater than b , return 1
if a > b:
return 1
# if a and b are equal, return 0
elif a == b:
return 0
# if a less than b , return -1
elif a < b :
return -1
# testing
if __name__=="__main__":
# calling the compare function with a and b
print('compare(5,2) %d'%compare(5,2))
print('compare(2,5) %d'%compare(2,5))
print('compare(3,3) %d'%compare(3,3))
# getting values for a and b
a=int(input("Enter the value for a: "))
b=int(input("Enter the value for b: "))
print('compare(%d,%d) %d'%(a,b,compare(a,b)))
(see attachment for output)
The range of cells from zz1 to zz23
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