for loop unexpectedly jumping down in value - c

Question

Goldbach's conjecture states that every even integer over 4 is the sum of two primes, I am writing a program in C to find these pairs. To do this it first finds all the primes less than a user given number. I have a for loop to iterate from 4 to the user given number and find the pairs within the loop body. When that loop gets to about around 40, suddenly jumps back down by about 30 and then continues to iterate up (with user input 50 it jumped from 38 to 9, with input 60 it jumped from 42 to 7). I can't figure out why this is happening. Here is my code:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <sys/types.h>
#include <unistd.h>
struct pair{
int a;
int b;
}pair_t;
int main(){
int N;
int numPrimes = 1;
int *primes = malloc(100*sizeof(int));
int isPrime = 1;
primes[0] = 2;
int timesRealloc = 0;
int availableSlots = 100;
printf("Please enter the largest even number you want to find the Goldbach pair for: \n");
scanf("%d", &N);
struct pair pairs[N/2 + 4];
int j = 0;
int i;
for (i = 3; i <= N; i+=2){
j = 0;
isPrime = 1;
while (primes[j] <= sqrt(i)) {
if (i%primes[j] == 0) {
isPrime = 0;
break;
}
j++;
}
if (isPrime == 1){
primes[numPrimes] = i;
numPrimes++;
}
if (availableSlots == numPrimes){
timesRealloc++;
availableSlots += 100;
primes = realloc(primes, availableSlots*sizeof(int));
}
}
printf("The largest prime I found was %d\n", primes[(numPrimes-1)]);
int k;
for (i=4; i<=N; i+=2){
printf("i is %d, N is %d\n", i, N);
if (i > N){ break; }
for (j=0; j<numPrimes; j++){
for (k=0; k<numPrimes; k++){
int sum = primes[j] + primes[k];
if(sum == i){
pairs[i].a = primes[j];
pairs[i].b = primes[k];
}
}
}
}
for (i=4; i<=N; i+=2){
printf("%d is the sum of %d and %d\n", i, pairs[i].a, pairs[i].b);
}
return 0;
}

Answer 1

You attempt to be space efficient by compressing the pairs array to just hold every other (even) number and start from 4 instead of zero. However, you miscalculate its size and then when you go to use it, you treat it like it hasn't been compressed and that there's a slot for every natural number.
The code suffers from having the prime array calculation in main() along with the other code, this is best separated out. And when it looks for pairs, it doesn't quit when it finds one, nor when it starts getting sums greater than the target. My rework below attempts to address all of these issues:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <stdbool.h>
#define INITIAL_SLOTS (100)
struct pair {
int a;
int b;
} pair_t;
int compute_primes(int limit, unsigned **primes, int size) {
int numPrimes = 0;
(*primes)[numPrimes++] = 2;
for (int i = 3; i <= limit; i += 2) {
bool isPrime = true;
for (int j = 0; (*primes)[j] <= i / (*primes)[j]; j++) {
if (i % (*primes)[j] == 0) {
isPrime = false;
break;
}
}
if (isPrime) {
(*primes)[numPrimes++] = i;
}
if (numPrimes == size) {
size *= 2;
*primes = realloc(*primes, size * sizeof(unsigned));
}
}
return numPrimes;
}
int main() {
int N;
printf("Please enter the largest even number you want to find the Goldbach pair for: \n");
scanf("%d", &N);
unsigned *primes = calloc(INITIAL_SLOTS, sizeof(unsigned));
int numPrimes = compute_primes(N, &primes, INITIAL_SLOTS);
printf("The largest prime I found was %d\n", primes[numPrimes - 1]);
struct pair pairs[(N - 4) / 2 + 1]; // compressed data structure
for (int i = 4; i <= N; i += 2) {
int offset = (i - 4) / 2; // compressed index
bool found = false;
for (int j = 0; ! found && j < numPrimes; j++) {
for (int k = 0; ! found && k < numPrimes; k++) {
int sum = primes[j] + primes[k];
if (sum == i) {
pairs[offset].a = primes[j];
pairs[offset].b = primes[k];
found = true;
} else if (sum > i) {
break;
}
}
}
}
for (int i = 4; i <= N; i += 2) {
int offset = (i - 4) / 2; // compressed index
printf("%d is the sum of %d and %d\n", i, pairs[offset].a, pairs[offset].b);
}
free(primes);
return 0;
}
OUTPUT
> ./a.out
Please enter the largest even number you want to find the Goldbach pair for:
10000
The largest prime I found was 9973
4 is the sum of 2 and 2
6 is the sum of 3 and 3
8 is the sum of 3 and 5
10 is the sum of 3 and 7
12 is the sum of 5 and 7
14 is the sum of 3 and 11
...
9990 is the sum of 17 and 9973
9992 is the sum of 19 and 9973
9994 is the sum of 53 and 9941
9996 is the sum of 23 and 9973
9998 is the sum of 31 and 9967
10000 is the sum of 59 and 9941
>