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# Prime numbers. Sieve of Eratosthenes

The calculator finds prime numbers using "Sieve of Eratosthenes" method.

This calculator finds prime numbers using a method known from ancient times as the Sieve of Eratosthenes. Let us recall that prime numbers have no other divisors except themselves and 1.
As a result of the calculator's work, a matrix containing composite numbers (gray) and prime numbers (black) will be displayed.

#### Sieve of Eratosthenes

Sieve

It was a demo calculator having a naive algorithm. The range of numbers is limited to 1000. The calculator and its source code would rather be useful for those who want to understand the logic of the ancient Greek scientist who invented the method in the 3rd century BC.
The following calculator evolves the Eratosthenes idea; it has a memory-optimized implementation and fewer excessive operations. Using this calculator (if your computer allows it), you can find prime numbers up to several billion. However, be careful - with a large boundary, your device's memory, and processor will be mercilessly used.

#### Sieve of Eratosthenes, optimised

Prime numbers count

The file is very large. Browser slowdown may occur during loading and creation.
Execution time, ms

Output time, ms

All found prime numbers can be downloaded as a CSV file, but here I warn you again - everything happens in your computer's memory. When downloading, five times the amount of memory will be grabbed, compared to that necessary for storing numbers. My old laptop with 4GB of RAM coped quite easily with finding 26+ million primes from the half-billion range, but I could not download it as CSV.

## Sieve of Eratosthenes algorithm

The algorithm in a pseudocode:

//Boundary
n;
//fill an array with ones (upper bound = n)
a ⟵ [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1...];
//first loop
for i=2,3,4..≤n:
if  a[i] = 1:
//second loop
for j = 2i,3i,4i .. ≤n:
a[i]⟵0
output ⟵ all i in the range 2 ≤ i ≤ n, for which the condition a[i]=1 is met

### Algorithm optimization

• as it is easy to see the original algorithm is passed several times over the same array elements, to avoid this, we change the following
• first loop для i=2,3,4..while i2≤n
• second loop: j=i2,i2+i,i2+2i... ≤n
• instead boolean type of 1 byte, we can use 1 bit - to 8 times reduce the memory required
• as it is easy to see all even numbers except 2 are not prime, taking this fact into account, we do the following:
• reduce the amount of memory required by another half
• change the algorithm in this way:
//first loop
for i=3,5,7..while  i²≤n
if  a[(i-1)/2] = 1:
//second loop
for j = i²,i²+2i,i²+4i .. ≤n:
a[(i-1)/2]⟵0
output ⟵ 2, all odd i in the range: 3 ≤ i ≤ n, for which the condition a[(i-1)/2]=1 is met
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