The Sieve of Eratosthenes and Goldbach's Conjecture

Question

Implement the Sieve of Eratosthenes and use it to find all prime numbers less than or equal to one million. Use the result to prove Goldbach's Conjecture for all even integers between four and one million, inclusive.

Implement a function with the following declaration:

void sieve(int array[], int num); This function takes an integer array as its argument. The array should be initialized to the values 1 through 1000000. The function modifies the array so that only the prime numbers remain; all other values are zeroed out. This function must be written to accept an integer array of any size. You must should output for all primes numbers between 1 and 1000000, but when I test your function it may be on an array of a different size.

Implement a function with the following declaration:

void goldbach(int array[], int num); This function takes the same argument as the previous function and displays each even integer between 4 and 1000000 with two prime numbers that add to it. The goal here is to provide an efficient implementation. This means no multiplication, division, or modulus when determining if a number is prime. It also means that the second function must find two primes efficiently.

Output for your program: All prime numbers between 1 and 1000000 and all even numbers between 4 and 1000000 and the two prime numbers that sum up to it.

DO NOT provide output or a session record for this project!

This is the code that I have so far, my problem is that it displays numbers higher than 1,000 as 1s, how can I go about this, thank you!

#include <iostream>
#include <stdio.h>
#include <math.h>

using namespace std;

void sieve(int array[], int num);
void goldbach(int array[], int num);

const int arraySize = 1000000;
int nums[arraySize];

int main(){

    for (int i = 0; i <= arraySize; ++i)
        nums[i] = 1;

    nums[0] = nums[1] = 0;

    sieve(nums, arraySize);

    for(int i = 0; i < 10000; ++i){
        if (nums[i] > 0){
            cout << nums[i] << " "; 
        }
    }

    goldbach(nums, arraySize);

    return 0;
}


void sieve(int array[], int num) {

    int squareR = (int)sqrt(num);

    for(int i = 2; i <= squareR; ++i){

        int k;
        if(array[i]){
            for(k = i*i; k <= num; k += i)
                array[k] = 0;
        }

        if (array[i] == 1){
            array[i] = i;   
        }     
    }
}


void goldbach(int array[], int num){

    int i, r = 0;
    for (i = 4; i <= num; i += 2){

        for (int j = 2; j <= i/2; j++)

            if (array[j] && array[i-j]) r ++;
    }
}

I strongly suggest changing `arraySize` from 1000000 to (say) 20 for the purposes of debugging, and then step through `sieve` in your debugger. — Martin Bonner supports Monica, Mar 19 '17 at 18:50

score 0 · Answer 1 · answered Mar 19 '17 at 20:02

my problem is that it displays numbers higher than 1,000 as 1s, how can I go about this

That's because you're not updating the values in the array above 1000, here:

for(int i = 2; i <= squareR; ++i){
   ...
    if (array[i] == 1){
        array[i] = i;

clearly the array's entries above squareR are not updated and remain at the value you initialized them, which is 1.

However I you don't need this update at all. You can drop it and simplify your code, keeping the array's entries as either 1 (for primes) or 0 (for non-primes). with this, and display your result like this (in main):

for(int i = 0; i < arraySize; ++i){
    if (nums[i] != 0){
        // cout << nums[i] << " "; // <-- drop this
        cout << i << " ";          // <-- use this
    }
}

The Sieve of Eratosthenes and Goldbach's Conjecture

1 Answers1