How would I go about justifying this algorithm is O(log n)?
public static long exponentiation(long x, int n){
if(n == 0){
return 1;
}
else if (n % 2 == 0){
x = exponentiation(x, n / 2);
return x * x;
}
else{
return x * exponentiation(x, n-1);
}
}
Each recursive call to method exponentiation
is a multiplication step . Hence you need to count the number of recursive calls. There are several ways to achieve this. I chose to add another parameter to the method.
public static long exponentiation(long x, int n, int count) {
if (n == 0) {
System.out.println("steps = " + count);
return 1;
}
else if (n % 2 == 0) {
x = exponentiation(x, n / 2, count + 1);
return x * x;
}
else {
return x * exponentiation(x, n - 1, count + 1);
}
}
Here is the initial call to method exponentiation
exponentiation(2, 63, 0);
When I run the above code, the following is printed
steps = 11
You can use a static
counter as well (without changing the prototype of the function):
public static long counter = 0;
public static long exponentiation(long x, int n){
if(n == 0){
return 1;
}
else if (n % 2 == 0){
x = exponentiation(x, n / 2);
counter++;
return x * x;
}
else{
counter++;
return x * exponentiation(x, n-1);
}
}
However, you need to reset the counter before calling the function each time, ie, set counter = 0
.
Note that you need to the counter to prove that it is in O(log(n))
. To prove the complexity, just you need to find the complexity term by looking at the flow of the code. Suppose T(n)
is the number of multiplications for computing x^n
. So, based on the written code, T(n) = T(n/2) + 1
, if n
is even, and T(n) = T(n-1) + 1
, if n
is odd. Now, at least in one of two consecutive recursions, input n
is even. Therefore, at most 2 log(n)
is required to reach to n = 0
. Because, for each even input, the next input will be halved. So, we can conclude that the algorithm is in O(log(n))
.
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