What is the time complexity of exponentiation by squaring? - c

Question

Here is a code to exponentiate a number to a given power:
#include <stdio.h>
int foo(int m, int k) {
if (k == 0) {
return 1;
} else if (k % 2 != 0) {
return m * foo(m, k - 1);
} else {
int p = foo(m, k / 2);
return p * p;
}
}
int main() {
int m, k;
while (scanf("%d %d", &m, &k) == 2) {
printf("%d\n", foo(m, k));
}
return 0;
}
How do I calculate the time complexity of the function foo?
I have been able to deduce that if k is a power of 2, the time complexity is O(log k).
But I am finding it difficult to calculate for other values of k. Any help would be much appreciated.

Answer 1

How do I calculate the time complexity of the function foo()?
I have been able to deduce that if k is a power of 2, the time complexity is O(logk).
First, I assume that the time needed for each function call is constant (this would for example not be the case if the time needed for a multiplication depends on the numbers being multiplied - which is the case on some computers).
We also assume that k>=1 (otherwise, the function will run endlessly unless there is an overflow).
Let's think the value k as a binary number:
If the rightmost bit is 0 (k%2!=0 is false), the number is shifted right by one bit (foo(m,k/2)) and the function is called recursively.
If the rightmost bit is 1 (k%2!=0 is true), the bit is changed to a 0 (foo(m,k-1)) and the function is called recursively. (We don't look at the case k=1, yet.)
This means that the function is called once for each bit and it is called once for each 1 bit. Or, in other words: It is called once for each 0 bit in the number and twice for each 1 bit.
If N is the number of function calls, n1 is the number of 1 bits and n0 is the number of 0 bits, we get the following formula:
N = n0 + 2*n1 + C
The constant C (C=(-1), if I didn't make a mistake) represents the case k=1 that we ignored up to now.
This means:
N = (n0 + n1) + n1 + C
And - because n0 + n1 = floor(log2(k)) + 1:
floor(log2(k)) + C <= N <= 2*floor(log2(k)) + C
As you can see, the time complexity is always O(log(k))

Answer 2

O(log(k))
Some modification added to output a statistics for spread sheet plot.
#include <stdio.h>
#include <math.h>
#ifndef TEST_NUM
#define TEST_NUM (100)
#endif
static size_t iter_count;
int foo(int m, int k) {
iter_count++;
if (k == 0) {
return 1;
} else if(k == 1) {
return m;
} else if (k % 2 != 0) {
return m * foo(m, k - 1);
} else {
int p = foo(m, k / 2);
return p * p;
}
}
int main() {
for (int i = 1; i < TEST_NUM; ++i) {
iter_count = 0;
int dummy_result = foo(1, i);
printf("%d, %zu, %f\n", i, iter_count, log2(i));
}
return 0;
}
Build it.
gcc t1.c -DTEST_NUM=10000
./a > output.csv
Now open the output file with a spread sheet program and plot the last two output columns.

Answer 3

For k positive, the function foo calls itself recursively p times if k is the p-th power of 2. If k is not a power of 2, the number of recursive calls is strictly inferior to 2 * p where p is the exponent of the largest power of 2 inferior to k.
Here is a demonstration:
let's expand the recursive call in the case k % 2 != 0:
int foo(int m, int k) {
if (k == 1) {
return m;
} else
if (k % 2 != 0) { /* 2 recursive calls */
// return m * foo(m, k - 1);
int p = foo(m, k / 2);
return m * p * p;
} else { /* 1 recursive call */
int p = foo(m, k / 2);
return p * p;
}
}
The total number of calls is floor(log2(k)) + bitcount(k), and bitcount(k) is by construction <= ceil(log2(k)).
There are no loops in the code and the time of each individual call is bounded by a constant, hence the overall time complexity of O(log k).

Answer 4

The number of times the function is called (recursively or not) per power call is proportional to the minimum number of bits in the exponent required to represent it in binary form.
Each time you enter in the function, it solves by reducing the number by one if the exponent is odd, OR reducing it to half if the exponent is even. This means that we will do n squares per significant bit in the number, and m more multiplications by the base for all the bits that are 1 in the exponent (which are, at most, n, so m < n) for a 32bit significant exponent (this is an exponent between 2^31 and 2^32 the routine will do between 32 and 64 products to get the result, and will reenter to itself a maximum of 64 times)
as in both cases the routine is tail-recursive, the code you post can be substituted with an iterative code in which a while loop is used to solve the problem.
int foo(int m, int k)
{
int prod = 1; /* last recursion foo(m, 0); */
int sq = m; /* squares */
while (k) {
if (k & 1) {
prod *= sq; /* foo(m, k); k odd */
}
k >>= 1;
sq *= sq;
}
return prod; /* return final product */
}
That's huge savings!!! (between 32 multiplications and 64 multiplications, to elevate something to 1,000,000,000 power)