Finding the first number larger than N that is a relative prime to M

Question

Basically, the title says everything. The numbers are not too big (the maximum for N is ~2/3 * max(long) and max M is max(long)), so I think even a simple solution that I currently have is sufficient. M is always bigger than N.

What I currently have:

Most simple, just start from N + 1, perform plain Euclidean GCD, and if it returns 1 we are done, if not increment and try again.

I would like to know what is the worst case scenario with this solution. Performance is not a big issue, but still I feel like there must be a better way.

Thanks.

About the worst case, I made a small test:

Random r = new Random();
while (true)
            {
                long num = (long) r.Next();
                num *= r.Next();
                f((long)(num * 0.61), num);
            }

...

public static int max;

        public static int f(long N, long M)
        {
            int iter = 0;
            while (GCD(N++, M) != 1)
            {
                iter++;
            }

            if (iter > max)
            {
                max = iter;
                Console.WriteLine(max);
            }

            return 0;
        }

It is running for ~30 minutes and the worst case so far is 29 iterations. So I believe that there is a more precise answer then O(N).

swen · Accepted Answer · 2012-02-01T15:59:36.187

I don't know the worst scenario, but using the fact that M < 2⁶⁴, I can bound it above by 292 iterations and below by 53 (removing the restriction that the ratio N/M be approximately fixed).

Let p₁, …, p_k be the primes greater than or equal to 5 by which M is divisible. Let N' ≥ N be the least integer such that N' = 1 mod 6 or N' = 5 mod 6. For each i = 1, …, k, the prime p_i divides at most ceil(49/p_i) of the integers N', N' + 6, N' + 12, …, N' + 288. An upper bound on ∑_i=1,…,k ceil(49/p_i) is ∑_i=3,…,16 ceil(49/q_i) = 48, where q is the primes in order starting with q₁ = 2. (This follows because ∏_i=3,…,17 ≥ 2⁶⁴ implies that M is the product of at most 14 distinct primes other than 2 and 3.) We conclude that at least one of the integers mentioned is relatively prime to M.

For the lower bound, let M = 614889782588491410 (product of the first fifteen primes) and let N = 1. After 1, the first integer relatively prime to the first fifteen primes is the sixteenth prime, 53.

I expect both bounds could be improved without too much work, though it's not clear to me for what purpose. For the upper bound, handle separately the case where 2 and 3 are both divisors of M, as then M can be the product of at most thirteen other primes. For the lower bound, one could try to find a good M by running the sieve of Eratosthenes to compute, for a range of integers, the list of primes dividing those integers. Then sweep a window across the range; if the product of the distinct primes in the window is too large, advance the trailing end of the window; otherwise, advance the leading end.

Saeed Amiri · Answer 2 · 2012-02-01T18:01:28.057

Sure it's not O(n), By knowing that prime number gaps is log_en we can simply say your algorithm has at most log_en iterations,(because after passing at most log_en number you will see new prime number which is prime respect to your given number n) for more detail about this gap, you can see prime numbers gap.

So for your bounded case it is smaller than log_en = log_e2⁶⁴ <= 44 and it will be smaller than 44 iterations.

Finding the first number larger than N that is a relative prime to M

2 Answers2