Sieve of Eratosthenes C++ using single linked list - c++

Question

Hi I have a problem with Sieve of Eratosthenes in C++. I have to do this using single linked list. My program is running and showing first declaration of list but I don't know how to delete non prime numbers properly. My function just isn't working for me. How should I change my delete function?
#include <iostream>
#include <cmath>
using namespace std;
struct List
{
int number;
List* next;
};
List* head = new List;
void l_add(int n)
{
List* temp = head;
for (int i = 2; i <= n; i++)
{
temp->next = new List();
temp->number = i;
temp = temp->next;
}
}
void l_print()
{
List* temp = head;
while (temp->next != 0)
{
cout << temp->number << " ";
temp = temp->next;
}
cout << endl;
}
void l_delete(int n)
{
List* temp = head;
for (int i = 2; i < sqrt(n); i++)
{
if (temp->number % i == 0)
{
head = temp->next;
delete temp;
temp = head;
}
while (temp->next != 0)
{
if (temp->next->number % i == 0)
{
temp->next = temp->next->next;
delete temp->next;
}
temp = temp->next;
}
}
}
int main()
{
int n;
cout << "Enter up to which number to find prime numbers using Sieve of Eratosthenes: " << endl;
cin >> n;
l_add(n);
l_print();
l_delete(n);
l_print();
return 0;
}

Answer 1

This would be a working version of the l_delete method:
void l_delete(int n)
{
List* temp = head;
for (int i = 2; i < sqrt(n); i++)
{
while (temp->next != 0)
{
if (temp->next->number % i == 0 && temp->next->number != i)
{
List* temp2 = temp->next->next;
delete temp->next;
temp->next = temp2;
}
if(temp->next == 0) break;
temp = temp->next;
}
temp = head;
if (temp->number % i == 0 && temp->number != i)
{
head = temp->next;
delete temp;
temp = head;
}
}
}
There were several problems with your deletion method.
Problems with algorithm logic: With your algorithm head should be checked last because otherwise if its deleted, the new head is not checked for primality, you immediately check new next, which is old ->next->next. Also you didn't check if number is equal to divider in which case it should not be deleted.
Problems with programming logic:
When you're deleting next node in the while loop, same as when deleting head, you need another temporary variable to store temp->next->next and then after deleting assign it to temp->next.
But the biggest problem here is that this is not Eratosthenes sieve at all, you are
just checking all numbers for divisibility with all others smaller than sqrt(n). It
is suboptimal compared to the Eratosthenes sieve. If you Google Eratosthenes sieve, you’ll find a lot of detailed tutorials and explanations.

Answer 2

I like what Bran is pitching, and I'm going to add a few tricks.
Comment: Global variables suck. They make things harder to trace when projects get bigger. I'd move head into main and pass it around as a parameter. I'm also ditching the sentry node because I find them more trouble than they are worth most of the time.
int main()
{
int n;
cout << "Enter up to which number to find prime numbers using Sieve of Eratosthenes: " << endl;
if (cin >> n)
{
// should be a bit more logic here to automatically handle ns of 1 or 2
List* head = nullptr; // not using a sentinel node
l_add(head, n); // passing in head rather than global variable free-for-all
l_delete(head);
l_print(head);
return 0;
}
else
{
cout << "invalid input." << endl;
return 1;
}
}
When adding to the linked list, you don't need any even numbers other than 2. So don't add them. Less time spent iterating the list. After that it's a matter of making sure nodes go in in the right order.
void l_add(List*& head, // passing in head. Easier to track
int n)
{
List** temp = &head; // head is a next pointer with a different name
// hiding it behind another pointer allows us to treat
// it like a next
// temp is now a pointer to next pointers. We can add directly to the
// last nodes's next pointer and also use it to access the current
// pointer if we need to
(*temp) = new List {2, nullptr}; // 2 is only even prime
temp = &(*temp)->next;
for (int i = 3; i <= n; i+=2) // start at 3 and only add odd numbers
{
(*temp) = new List {i, nullptr};
temp = &(*temp)->next; // Advance to next node
}
}
When we're iterating through the list looking to eliminate multiples we need two loops. One to keep track of the node we're eliminating multiples and another loop to do the hunting and eliminating. Note that I'm using the pointer-to-pointer trick again. This trick annihilates about half of the code needed to traverse and remove nodes.
void l_delete(List * head)
{
List* last = head->next; // track the last known prime node. skip node 2.
// note this will blow up if there is no node 2.
while (last) // for every node still in the list
{
List** current = &last->next; // similar to trick above.
// if we have a pointer to the next to be
// updated, we don't need to track the previous node
while ((*current)) // look at every node after the last prime
{
if ((*current)->number % last->number == 0) // if it's a multiple, remove it.
{
List * to_del = (*current); //save node to delete
(*current) = (*current)->next; // link previous node to next node.
// effectively automatically advances the node
delete to_del;
}
else // proceed to next node
{
current = &(*current)->next;
}
}
last = last->next; // advance to next prime number
}
}
Probably plenty of room in there for optimization, but I'm aiming more for readability because if I drop ten lines of cryptic gibberish nobody learns nuthin'.