minimal prime generator C - c

Question

I have to create a minimal prime generator in C (we need the minimal primes that have at least two digits) and i cannot use tables.So my thought was first find all the primes,second with using masks to find all the subsequences of each number and finally check if every subsequence is not a prime number.i cannot find the minimal primes because i didnt put a condition to check if a subsequence is prime or no?(my code is not ready so it may have some mistakes but it runs)
my code
#include <stdio.h>
#define MAXNUMB 100
int main ()
{
int i,j,x,mask,max=1,mult,sub;
for (i = 11 ; i < MAXNUMB; i += 2 ) {
for (j = 3; j * j <= i; j += 2) {
if (i % j == 0) {
break;
}
}
if (j * j > i) {
int length = 0;
int tmp=i;
while (tmp != 0) {
tmp /= 10;
length++;
}
for (x=1;x<length*2;x++) {
mask=x;
mult=1;
sub=0;
int num=i;
while ( num != 0 ) {
if ( mask % 2 == 1 ) {
sub += num % 10 * mult;
mult *= 10;
}
num /= 10;
mask /= 2;
}
printf ("%d \n",sub);
}
printf ("%d is minimal prime \n",i);
}
}
}

Answer 1

If MAXNUMB is not too large then you can very quickly find all the primes until MAXNUMB using the Sieve_of_Eratosthenes. After you have that you can get every subsequence of an n-digit number by counting from 1 to 2^n-1 and use the bit-pattern of the current count to specify the current subsequence. Check each subsequence in the sieve.
If MAXNUMB is too big then you can build the sieve up to sqrt(MAXNUMB), so that when you test a number whether it's prime or not you just have to check whether it's divisible with any prime, instead of checking whether it's divisible any odd number.

Answer 2

Wikipedia has a nice pseudocode algorithm for primality. It's naive, but it works.
function is_prime(n : integer)
if n ≤ 1
return false
else if n ≤ 3
return true
else if n mod 2 = 0 or n mod 3 = 0
return false
let i ← 5
while i×i ≤ n
if n mod i = 0 or n mod (i + 2) = 0
return false
i ← i + 6
return true
I'll leave translating this into C for you. My one note is that i should be of a type larger than n's type. For the reason, consider what happens to i×i ≤ n when n is the largest integer.
When you're ready, you can look at my solution to this problem.

Answer 3

The concept of a minimal prime is moderately interesting. I wrote the following code to do the checking.
/* SO 33838621: Minimal Primes */
/*
** Find the minimal primes less than 100,000.
**
** A minimal prime is a prime number for which no subsequence of the digits
** that make up the number is itself prime.
** The question gives two examples:
** = 881 is prime and is a minimal prime because none of { 8, 8, 1, 88,
** 81, 81 } are prime.
** = 109 is prime but is not a minimal prime because { 1, 0, 9, 10, 9,
** 19 } includes the prime 19.
** Clearly, the single digit primes are all trivially minimal.
**
** Additional wrinkle: the code may not build up a table of primes.
**
** NB: All primes except 2 and 3 have the form 6N±1
*/
/*
** There are two problems to solve:
** (1) Check for primality without using a table of primes.
** (2) Generate all subsequences of a number.
** The latter problem is somewhat harder than the former.
** The surviving solution uses recursive string manipulation.
**
** NB: Command subsequences is derived from this, and helps check
** the validity of the minimal primes.
*/
#include <assert.h>
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
static bool is_prime(int n)
{
if (n < 2)
return false;
if (n == 2 || n == 3 || n == 5 || n == 7)
return true;
if (n % 2 == 0 || n % 3 == 0 || n % 5 == 0 || n % 7 == 0)
return false;
for (int c = 12; (c - 1) * (c - 1) <= n; c += 6)
{
if (n % (c - 1) == 0 || n % (c + 1) == 0)
return false;
}
return true;
}
static bool has_prime_n_digit_subset(int p0, int n_digits, const char *buffer, int buflen)
{
//printf("-->> p0 = %5d, n = %d (%s)\n", p0, n_digits, buffer);
assert(buflen >= 0 && strlen(buffer) == (size_t)buflen);
for (int i = 0; i < buflen; i++)
{
int p1 = 10 * p0 + buffer[i] - '0';
if (n_digits > 1)
{
if (has_prime_n_digit_subset(p1, n_digits - 1, &buffer[i+1], buflen - i - 1))
{
//printf("<<-- true\n");
return true;
}
}
else
{
//printf("C %d\n", p1);
if (is_prime(p1))
{
//printf("<<-- p1 = %d: true\n", p1);
return true;
}
}
}
//printf("<<-- false\n");
return false;
}
static void check_minimal_prime(int n)
{
assert(n > 0);
if (is_prime(n))
{
//printf("P %d\n", n);
char buffer[20];
sprintf(buffer, "%d", n);
char n_digits = strlen(buffer);
for (int i = 1; i < n_digits; i++)
{
if (has_prime_n_digit_subset(0, i, buffer, n_digits))
return;
}
printf("%d\n", n); /* It's a minimal prime */
}
}
int main(int argc, char **argv)
{
int max = 100000;
if (argc > 2)
{
fprintf(stderr, "Usage: %s [maximum]\n", argv[0]);
exit(1);
}
else if (argc == 2)
{
max = atoi(argv[1]);
if (max <= 0)
{
fprintf(stderr, "Invalid number (%d from %s)\n", max, argv[1]);
exit(1);
}
}
max /= 6;
check_minimal_prime(2);
check_minimal_prime(3);
for (int c = 1; c < max; c++)
{
check_minimal_prime(6 * c - 1);
check_minimal_prime(6 * c + 1);
}
return 0;
}
The list of numbers generated was:
2
3
5
7
11
19
41
61
89
409
449
499
881
991
6469
6949
9001
9049
9649
9949
60649
666649
946669
60000049
66000049
66600049
I didn't find any more minimal primes when checking up to 1,000,000,000. Timing was:
100 0m0.006s
1000 0m0.006s
10000 0m0.006s
100000 0m0.012s
1000000 0m0.129s
10000000 0m2.617s
100000000 1m8.200s
1000000000 32m34.561s