# Sieve of Atkin

#### Algorithms List of Mathematical Algorithms prime number

The Sieve of Atkin is an efficient algorithm used to find all prime numbers upto a given number (say N) and does so in O(N) time complexity. With a modified version with enumerating lattice points variation, the time complexity goes to O(N / log log N). Sieve of Atkin is computationally efficient than Sieve of Eratosthenes as it marks multiple of square of prime numbers.

Other similar ways to calculate prime numbers using sieve techniques are:

1. Sieve of Eratosthenes
2. Sieve of Sundaram

### Algorithm

• Create a list named final list and fill it with:
• 2, 3 and 5
• Create a list named sieve list with an entry for each positive integer and intially mark them as non prime
• For each entry number N in the sieve list, find the modulo-sixty remainder r (remainder when N is divided by 60) for N:
• If r is 1, 13, 17, 29, 37, 41, 49, or 53:
• flip the entry for each possible solution to 4x^2 + y^2 = N.
• All numbers with modulo-sixty remainder 1, 13, 17, 29, 37, 41, 49, or 53 have a modulo-twelve remainder of 1 or 5. These numbers are prime if and only if the number of solutions to 4×^2+y^2=n is odd and the number is square free. A square free integer is one which is not divisible by any perfect square other than 1.
• If r is 7, 19, 31, or 43:
• flip the entry for each possible solution to 3x^2 + y^2 = N.
• All numbers with modulo-sixty remainder 7, 19, 31, or 43 have a modulo-six remainder of 1. These numbers are prime if and only if the number of solutions to 3x^2 + y^2 = n is odd and the number is square free.
• If r is 11, 23, 47, or 59:
• flip the entry for each possible solution to 3x^2 – y^2 = N when x > y.
• All numbers with modulo-sixty remainder 11, 23, 47, or 59 have a modulo-twelve remainder of 11. These numbers are prime if and only if the number of solutions to 3x^2 – y^2 = n is odd and the number is square free.
• If r is something else, ignore it completely.
• Take the next number in the sieve list still marked prime
• Include the number in the results list
• Square the number and mark all multiples of that square as non-prime. Note that the multiples that can be factored by 2, 3, or 5 need not be marked, as these will be ignored in the final enumeration of primes

### Complexity

Sieve of Atkin:

Time complexity: Θ(N)
Space complexity: Θ(N)

The algorithm described above can compute primes up to N using O(N) operations with only O(N) bits of memory.

Sieve of Eratosthenes which uses O(N log log N) operations and the same O(N^1/2/log N) bits of memory plus a minimal page buffer.

Sieve of Atkin (modified enumerating lattice points variation)

Time complexity: Θ(N / (log log N))
Space complexity: Θ(N^0.5)

A special modified "enumerating lattice points" variation which is not the above version of the Sieve of Atkin can theoretically compute primes up to N using O( N / log log N) operations with N^1/2 + o(1) bits of memory but this variation is rarely implemented.

That is a little better in performance at a very high cost in memory as compared to both the ordinary page segmented version and to an equivalent but rarely-implemented version of the sieve of Eratosthenes which uses O(N) operations and O(N^1/2(log log N)/log N) bits of memory.

### Example

Input- limit to be 20

Input numbers are:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10
11, 12, 13, 14, 15, 16, 17, 18, 19, 20.


Printing prime numbers and ignoring non-prime.
Output:

2, 3, 5, 7, 11, 13, 17, 19


Explanation

• Status for all the numbers in the start is False. Special number is 2, 3 and 5 which are known to be prime.
• Generate Values for the conditions.(take help of the table given below).

• Flipping the status according to condition.The above values of n in the table generated in x, y loop will be tested for modulo condition.
* if (colum1 value) % 12 == 1 or 5:
* then flip the sieve status for that number.
*
We are taking mod with 12 in place of 60, this is because if we take mod 60 then we have to consider many r as 1, 13, 17, 29, 37, 41, 49, or 53 and for all these r, mod 12 is 1 or 5. (done only to reduce the expression size).
status of 5, 13, 17 became true

• if (column 2 value) % 12 == 7
• then flip the sieve status for that number.
status of 7, 19 becomes true

• if (column 3 value) % 12 == 11:
• then flip the sieve status for that number.
status of 11 becomes true)</li>

• Checking for Square free Condition: If any number in our list in square of any number then remove it. // If square them make them non-prime and not added in the result.

• Creating array of prime numbers for which status is true and printing it as the output

2 3 5 7 11 13 17 19


### Implementation

Following is the implementation of Sieve of Atkin in Python:

   def SieveOfAtkin(limit):
#for 2 and 3
if (limit > 2):
print(2 , end = " ")
if (limit > 3):
print(3 , end = " ")
#intialize sieve
sieve = [False] * limit
for i in range( 0 , limit ):
sieve[i] = False # here we are intializing each number as non-prime.
#for step 3
x = 1
while(x * x < limit ) :
y = 1
while(y * y < limit ) :
#main part
n = (4 * x * x) + (y * y)
if (n <= limit and (n % 12 == 1 or
n % 12 == 5)):
sieve[n] ^= True # changing status to prime.
n = (3 * x * x) + (y * y)
if (n <= limit and n % 12 == 7):
sieve[n] ^= True # changing status to prime.
n = (3 * x * x) - (y * y)
if (x > y and n <= limit and
n % 12 == 11):
sieve[n] ^= True # changing status to prime.
y += 1
x += 1
#mark all multiples of square as non-prime
r = 5
while(r * r < limit) :
if (sieve[r]) :
for i in range(r * r, limit, r * r):
sieve[i] = False # here we are changing status to non-prime and ignoring it.
# Print primes
for a in range(5 , limit ):
if (sieve[a]):
print(a , end = " ")

limit = 20
SieveOfAtkin(limit)


### Applications

The Sieve of Atkin is a method considered as an optimized version of the Sieve of Eratosthenes to find all prime numbers up to a limit fixed N.

This algorithm can be used in the following cases:

• Determine whether a number N is a prime number or not.
• Factorize a number N.
• Find all prime numbers within a range N to M.
• Prove prime number theorems for a range like Goldbach’sConjecture.

## Question

#### How many prime numbers are there upto 100?

25
20
27
24
Run the algorithm and find it out