Find a number which generates all the integers mod q

9

1

Consider the integers modulo q where q is prime, a generator is any integer 1 < x < q so that x^1, x^2, ..., x^(q-1) covers all q-1 of the integers between 1 and q-1. For example, consider the integers modulo 7 (which we write as Z_7). Then 3, 3^2 mod 7 = 2, 3^3 = 27 mod 7 = 6, 3^4 = 81 mod 7 = 4, 3^5 = 243 mod 7 = 5, 3^6 = 729 mod 7 = 1 covers all the values 3, 2, 6, 4, 5, 1 covers all the integers 1..6 as required.

The task is to write code that takes an input n and outputs a generator for Z_n. You cannot use any builtin or library that does this for you of course.

The only restriction on the performance of your code is that you must have tested it to completion with n = 4257452468389.

Note that 2^n means 2 to the power of n. That is ^ represents exponentiation.

user9206

Posted 2017-07-02T11:36:07.003

Reputation:

Hmm...1 < x < q makes the challenge a lot easier imo. – Erik the Outgolfer – 2017-07-02T11:49:39.107

@EriktheOutgolfer I am not sure I know what you mean? Those are just all the distinct integers that are not 0 or 1. – None – 2017-07-02T11:52:41.657

I mean that it's easier than what many probably think...or maybe some inactive moment on PPCG. – Erik the Outgolfer – 2017-07-02T11:53:53.183

3But I think that requiring people to test it to completion to a large number is unnecessary...basically tio would just memory-error. – Erik the Outgolfer – 2017-07-02T11:56:38.337

@Lembik Is there any case where there is no generator for a certain number? Some test cases would be good. – Mr. Xcoder – 2017-07-02T11:58:59.673

@Mr.Xcoder No. This is because q is prime. – None – 2017-07-02T11:59:17.727

@EriktheOutgolfer The point of that test is to exclude really naive solutions without specify exactly how efficient the solution should be. – None – 2017-07-02T12:00:41.587

@Lembik you need a tag for that probably, that's not normal code-golf – Stephen – 2017-07-02T12:02:56.557

@StepHen Agree, [tag:restricted-complexity]. – Mr. Xcoder – 2017-07-02T12:03:23.810

@Mr.Xcoder It was suggested to me in chat that this is an alternative way of doing this. It has some advantages we can discuss in chat. – None – 2017-07-02T12:04:30.050

This is just finding a generator of the group of units of Z_p, right? Or do I need to refresh my abstract algebra notes... – Giuseppe – 2017-07-02T12:28:35.757

@Giuseppe Yes although notice that p is prime in this case. – None – 2017-07-02T12:30:49.597

Would the curator of SCP-155 care to participate? – Joshua – 2017-07-02T16:16:03.947

Answers

13

Pyth, 16 15 bytes

f-1m.^T/tQdQPtQ

Test suite

If p is the input, we know that g^(p-1) = 1 mod p, so we just need to check that g^a != 1 mod p for any smaller a. But a must be a factor of p-1 for that to be possible, and any multiple of an a with that property will also have that property, so we only need to check that g^((p-1)/q) != 1 mod p for all prime factors q of p-1. So, we check all integers g in increasing order until we find one that works.

Explanation:

f-1m.^T/tQdQPtQ
f                  Return the first value T such that the following is truthy:
            PtQ    Take the prime factorization of the input - 1.
   m               Map those prime factors to
       /tQd        Take the input - 1 divided by the factor
    .^T    Q       Raise T to that exponent mod input,
                   performed as modular exponentiation, for performance.
 -1                Check that 1 is not found among the results.

isaacg

Posted 2017-07-02T11:36:07.003

Reputation: 39 268

Pretty awesome! – None – 2017-07-02T12:22:11.853

Does your code perform the factorization? – None – 2017-07-02T12:23:59.443

@Lembik It does (the PtQ part). – Erik the Outgolfer – 2017-07-02T12:24:15.493

5

Mathematica, 52 bytes

Inspired by isaacg's answer.

1//.i_/;PowerMod[i,Divisors[#-1],#]~Count~1!=1:>i+1&

alephalpha

Posted 2017-07-02T11:36:07.003

Reputation: 23 988

-3

%MATLAB CODE
%Program to generate Z_n for an integer n
n = input('Enter a number to find modulo')
q = input ('Enter a prime number greater than the number you wished to find modulo')
if n>=q 
   fprintf('Error')
   exit(1)
end
for R=1:q-1
    fprintf(rem(n.^R, q))
    fprintf('\n')
end

John Brookfields

Posted 2017-07-02T11:36:07.003

Reputation: 1

1This isn't solving the right problem. Your code should, for example, take one input and give one output. – None – 2017-07-02T16:29:17.447

1Also, this code isn't golfed at all. Golfed code needs to be as short as possible, so you can remove the input text and spaces around equals signs and such. – Comrade SparklePony – 2017-07-02T16:59:50.187

3@ComradeSparklePony I think the first problem that it isn't solving the right problem should be tackled first :) – None – 2017-07-02T17:08:36.760