Slightly faster is_prime function

[rrrlw]

Checks (on average) mod every third integer below sqrt(n) instead of every second. Benchmark test showed that current is_prime function is very slightly faster for faster test cases (about half the integers below 10,000) and the suggested is_prime function is faster (1-2.5x) for slower test cases. Benchmark code below.

#include <iostream>
#include <fstream>
#include <ctime>

using namespace std;

//ripser
bool is_prime1(const int16_t n)
{
	if (!(n & 1) || n < 2) return n == 2;

	for (int16_t p = 3, q = n / p, r = n % p; p <= q; p += 2, q = n / p, r = n % p)
		if (!r) return false;

	return true;
}

//checking only mod 6k+1, 6k-1
bool is_prime2(const int16_t n)
{
	if (!(n & 1) || n < 2)
		return (n == 2);

	if (n == 3)
		return true;
	else if (n % 3 == 0)
		return false;

	int16_t curr;
	for (int16_t i = 1; ; i++)
	{
		curr = 6 * i;
		if ((curr-1) * (curr-1) > n) break;
		if (n % (curr + 1) == 0 ||
			n % (curr - 1) == 0)
			return false;
	}
	return true;
}

int main()
{
	clock_t start, end;
	double duration;

	const int LEN = 10000,	//should be multiple of 10 for progress update to work properly
			NUM_RUNS = 25000;
	double time1[LEN];
	double time2[LEN];
	int progress = 0;

	for (int16_t currNum = 0; currNum < LEN; currNum++)
	{
		if (is_prime1(currNum) != is_prime2(currNum))
		{
			cout << "ERROR AT " << currNum << endl;
			return 1;
		}

		//record time for first prime method
		start = clock();

		for (int i = 0; i < NUM_RUNS; i++)
			is_prime1(currNum);

		end = clock();
		time1[currNum] = (end - start) / ((double) CLOCKS_PER_SEC);

		//record time for second prime method
		start = clock();

		for (int i = 0; i < NUM_RUNS; i++)
			is_prime2(currNum);

		end = clock();
		time2[currNum] = (end - start) / ((double) CLOCKS_PER_SEC);

		//update status
		if (currNum % (LEN / 10) == 0)
		{
			cout << progress << "% DONE" << endl;
			progress += 10;
		}
	}

	//update status
	cout << "DONE WITH TIMING" << endl;

	//output data to CSV
	ofstream fout("Times.csv");
	fout << "Method1,Method2\n";
	for (int i = 0; i < LEN; i++)
		fout << time1[i] << ',' << time2[i] << '\n';
	fout.close();

	//update status and exit
	cout << "DONE WITH DATA OUTPUT; PROGRAM TERMINATING." << endl;
	return 0;
}```