How to create the most compact mapping n → isprime(n) up to a limit N? - algorithm

Question

Naturally, for bool isprime(number) there would be a data structure I could query.
I define the best algorithm, to be the algorithm that produces a data structure with lowest memory consumption for the range (1, N], where N is a constant.
Just an example of what I am looking for: I could represent every odd number with one bit e.g. for the given range of numbers (1, 10], starts at 3: 1110
The following dictionary can be squeezed more, right? I could eliminate multiples of five with some work, but numbers that end with 1, 3, 7 or 9 must be there in the array of bits.
How do I solve the problem?

Answer 1

The fastest algorithm for general prime testing is AKS. The Wikipedia article describes it at lengths and links to the original paper.
If you want to find big numbers, look into primes that have special forms like Mersenne primes.
The algorithm I usually implement (easy to understand and code) is as follows (in Python):
def isprime(n):
"""Returns True if n is prime."""
if n == 2:
return True
if n == 3:
return True
if n % 2 == 0:
return False
if n % 3 == 0:
return False
i = 5
w = 2
while i * i <= n:
if n % i == 0:
return False
i += w
w = 6 - w
return True
It's a variant of the classic O(sqrt(N)) algorithm. It uses the fact that a prime (except 2 and 3) is of form 6k - 1 or 6k + 1 and looks only at divisors of this form.
Sometimes, If I really want speed and the range is limited, I implement a pseudo-prime test based on Fermat's little theorem. If I really want more speed (i.e. avoid O(sqrt(N)) algorithm altogether), I precompute the false positives (see Carmichael numbers) and do a binary search. This is by far the fastest test I've ever implemented, the only drawback is that the range is limited.

Answer 2

A pretty simple and concise brute-force solution to check whether a number N is prime: simply check if there is any divisor of N from 2 up to the square root of N (see why here if interested).
The following code is compatible with both Python 2 and Python 3:
from math import sqrt
from itertools import count, islice
def is_prime(n):
return n > 1 and all(n % i for i in islice(count(2), int(sqrt(n) - 1)))
And here's a simpler Python 3 only implementation:
def is_prime(n):
return n > 1 and all(n % i for i in range(2, int(n ** 0.5) + 1))
Here are the extended versions of the above for clarity:
from math import sqrt
from itertools import count, islice
def is_prime(n):
if n < 2:
return False
for divisor in islice(count(2), int(sqrt(n) - 1)):
if n % divisor == 0:
return False
return True
def is_prime(n):
if n < 2:
return False
for divisor in range(2, int(n ** 0.5) + 1):
if n % divisor == 0:
return False
return True
This is not meant to be anything near the fastest nor the most optimal primality check algorithm, it only accomplishes the goal of being simple and concise, which also reduces implementation errors. It has a time complexity of O(sqrt(n)).
If you are looking for faster algorithms to check whether a number is prime, you might be interested in the following:
Finding primes & proving primality: brief overview and explanation of the most famous primality tests and their history.
Probabilistic primality tests (Wikipedia): these can be incorporated in the above code rather easily to skip the brute force if they do not pass, as an example there is this excellent answer to the duplicate of this question.
Fast deterministic primaliry tests (Wikipedia)
This Q&A Fastest way to list all primes below N along with the pyprimesieve library.
Implementation notes
You might have noticed that in the Python 2 compatible implementation I am using itertools.count() in combination with itertools.islice() instead of a simple range() or xrange() (the old Python 2 generator range, which in Python 3 is the default). This is because in CPython 2 xrange(N) for some N such that N > 263 ‒ 1 (or N > 231 ‒ 1 depending on the implementation) raises an OverflowError. This is an unfortunate CPython implementation detail.
We can use itertools to overcome this issue. Since we are counting up from 2 to infinity using itertools.count(2), we'll reach sqrt(n) after sqrt(n) - 1 steps, and we can limit the generator using itertools.islice().

Answer 3

There are many efficient ways to test primality (and this isn't one of them), but the loop you wrote can be concisely rewritten in Python:
def is_prime(a):
return all(a % i for i in xrange(2, a))
That is, a is prime if all numbers between 2 and a (not inclusive) give non-zero remainder when divided into a.

Answer 4

This is the most efficient way to see if a number is prime, if you only have a few queries. If you ask a lot of numbers if they are prime, try Sieve of Eratosthenes.
import math
def is_prime(n):
if n == 2:
return True
if n % 2 == 0 or n <= 1:
return False
sqr = int(math.sqrt(n)) + 1
for divisor in range(3, sqr, 2):
if n % divisor == 0:
return False
return True

Answer 5

If a is a prime then the while x: in your code will run forever, since x will remain True.
So why is that while there?
I think you wanted to end the for loop when you found a factor, but didn't know how, so you added that while since it has a condition. So here is how you do it:
def is_prime(a):
x = True
for i in range(2, a):
if a%i == 0:
x = False
break # ends the for loop
# no else block because it does nothing ...
if x:
print "prime"
else:
print "not prime"

Answer 6

I compared the efficiency of the most popular suggestions to determine if a number is prime. I used python 3.6 on ubuntu 17.10; I tested with numbers up to 100.000 (you can test with bigger numbers using my code below).
This first plot compares the functions (which are explained further down in my answer), showing that the last functions do not grow as fast as the first one when increasing the numbers.
And in the second plot we can see that in case of prime numbers the time grows steadily, but non-prime numbers do not grow so fast in time (because most of them can be eliminated early on).
Here are the functions I used:
this answer and this answer suggested a construct using all():
def is_prime_1(n):
return n > 1 and all(n % i for i in range(2, int(math.sqrt(n)) + 1))
This answer used some kind of while loop:
def is_prime_2(n):
if n <= 1:
return False
if n == 2:
return True
if n == 3:
return True
if n % 2 == 0:
return False
if n % 3 == 0:
return False
i = 5
w = 2
while i * i <= n:
if n % i == 0:
return False
i += w
w = 6 - w
return True
This answer included a version with a for loop:
def is_prime_3(n):
if n <= 1:
return False
if n % 2 == 0 and n > 2:
return False
for i in range(3, int(math.sqrt(n)) + 1, 2):
if n % i == 0:
return False
return True
And I mixed a few ideas from the other answers into a new one:
def is_prime_4(n):
if n <= 1: # negative numbers, 0 or 1
return False
if n <= 3: # 2 and 3
return True
if n % 2 == 0 or n % 3 == 0:
return False
for i in range(5, int(math.sqrt(n)) + 1, 2):
if n % i == 0:
return False
return True
Here is my script to compare the variants:
import math
import pandas as pd
import seaborn as sns
import time
from matplotlib import pyplot as plt
def is_prime_1(n):
...
def is_prime_2(n):
...
def is_prime_3(n):
...
def is_prime_4(n):
...
default_func_list = (is_prime_1, is_prime_2, is_prime_3, is_prime_4)
def assert_equal_results(func_list=default_func_list, n):
for i in range(-2, n):
r_list = [f(i) for f in func_list]
if not all(r == r_list[0] for r in r_list):
print(i, r_list)
raise ValueError
print('all functions return the same results for integers up to {}'.format(n))
def compare_functions(func_list=default_func_list, n):
result_list = []
n_measurements = 3
for f in func_list:
for i in range(1, n + 1):
ret_list = []
t_sum = 0
for _ in range(n_measurements):
t_start = time.perf_counter()
is_prime = f(i)
t_end = time.perf_counter()
ret_list.append(is_prime)
t_sum += (t_end - t_start)
is_prime = ret_list[0]
assert all(ret == is_prime for ret in ret_list)
result_list.append((f.__name__, i, is_prime, t_sum / n_measurements))
df = pd.DataFrame(
data=result_list,
columns=['f', 'number', 'is_prime', 't_seconds'])
df['t_micro_seconds'] = df['t_seconds'].map(lambda x: round(x * 10**6, 2))
print('df.shape:', df.shape)
print()
print('', '-' * 41)
print('| {:11s} | {:11s} | {:11s} |'.format(
'is_prime', 'count', 'percent'))
df_sub1 = df[df['f'] == 'is_prime_1']
print('| {:11s} | {:11,d} | {:9.1f} % |'.format(
'all', df_sub1.shape[0], 100))
for (is_prime, count) in df_sub1['is_prime'].value_counts().iteritems():
print('| {:11s} | {:11,d} | {:9.1f} % |'.format(
str(is_prime), count, count * 100 / df_sub1.shape[0]))
print('', '-' * 41)
print()
print('', '-' * 69)
print('| {:11s} | {:11s} | {:11s} | {:11s} | {:11s} |'.format(
'f', 'is_prime', 't min (us)', 't mean (us)', 't max (us)'))
for f, df_sub1 in df.groupby(['f', ]):
col = df_sub1['t_micro_seconds']
print('|{0}|{0}|{0}|{0}|{0}|'.format('-' * 13))
print('| {:11s} | {:11s} | {:11.2f} | {:11.2f} | {:11.2f} |'.format(
f, 'all', col.min(), col.mean(), col.max()))
for is_prime, df_sub2 in df_sub1.groupby(['is_prime', ]):
col = df_sub2['t_micro_seconds']
print('| {:11s} | {:11s} | {:11.2f} | {:11.2f} | {:11.2f} |'.format(
f, str(is_prime), col.min(), col.mean(), col.max()))
print('', '-' * 69)
return df
Running the function compare_functions(n=10**5) (numbers up to 100.000) I get this output:
df.shape: (400000, 5)
-----------------------------------------
| is_prime | count | percent |
| all | 100,000 | 100.0 % |
| False | 90,408 | 90.4 % |
| True | 9,592 | 9.6 % |
-----------------------------------------
---------------------------------------------------------------------
| f | is_prime | t min (us) | t mean (us) | t max (us) |
|-------------|-------------|-------------|-------------|-------------|
| is_prime_1 | all | 0.57 | 2.50 | 154.35 |
| is_prime_1 | False | 0.57 | 1.52 | 154.35 |
| is_prime_1 | True | 0.89 | 11.66 | 55.54 |
|-------------|-------------|-------------|-------------|-------------|
| is_prime_2 | all | 0.24 | 1.14 | 304.82 |
| is_prime_2 | False | 0.24 | 0.56 | 304.82 |
| is_prime_2 | True | 0.25 | 6.67 | 48.49 |
|-------------|-------------|-------------|-------------|-------------|
| is_prime_3 | all | 0.20 | 0.95 | 50.99 |
| is_prime_3 | False | 0.20 | 0.60 | 40.62 |
| is_prime_3 | True | 0.58 | 4.22 | 50.99 |
|-------------|-------------|-------------|-------------|-------------|
| is_prime_4 | all | 0.20 | 0.89 | 20.09 |
| is_prime_4 | False | 0.21 | 0.53 | 14.63 |
| is_prime_4 | True | 0.20 | 4.27 | 20.09 |
---------------------------------------------------------------------
Then, running the function compare_functions(n=10**6) (numbers up to 1.000.000) I get this output:
df.shape: (4000000, 5)
-----------------------------------------
| is_prime | count | percent |
| all | 1,000,000 | 100.0 % |
| False | 921,502 | 92.2 % |
| True | 78,498 | 7.8 % |
-----------------------------------------
---------------------------------------------------------------------
| f | is_prime | t min (us) | t mean (us) | t max (us) |
|-------------|-------------|-------------|-------------|-------------|
| is_prime_1 | all | 0.51 | 5.39 | 1414.87 |
| is_prime_1 | False | 0.51 | 2.19 | 413.42 |
| is_prime_1 | True | 0.87 | 42.98 | 1414.87 |
|-------------|-------------|-------------|-------------|-------------|
| is_prime_2 | all | 0.24 | 2.65 | 612.69 |
| is_prime_2 | False | 0.24 | 0.89 | 322.81 |
| is_prime_2 | True | 0.24 | 23.27 | 612.69 |
|-------------|-------------|-------------|-------------|-------------|
| is_prime_3 | all | 0.20 | 1.93 | 67.40 |
| is_prime_3 | False | 0.20 | 0.82 | 61.39 |
| is_prime_3 | True | 0.59 | 14.97 | 67.40 |
|-------------|-------------|-------------|-------------|-------------|
| is_prime_4 | all | 0.18 | 1.88 | 332.13 |
| is_prime_4 | False | 0.20 | 0.74 | 311.94 |
| is_prime_4 | True | 0.18 | 15.23 | 332.13 |
---------------------------------------------------------------------
I used the following script to plot the results:
def plot_1(func_list=default_func_list, n):
df_orig = compare_functions(func_list=func_list, n=n)
df_filtered = df_orig[df_orig['t_micro_seconds'] <= 20]
sns.lmplot(
data=df_filtered, x='number', y='t_micro_seconds',
col='f',
# row='is_prime',
markers='.',
ci=None)
plt.ticklabel_format(style='sci', axis='x', scilimits=(3, 3))
plt.show()

Answer 7

One can use sympy.
import sympy
sympy.ntheory.primetest.isprime(33393939393929292929292911111111)
True
From sympy docs. The first step is looking for trivial factors, which if found enables a quick return. Next, if the sieve is large enough, use bisection search on the sieve. For small numbers, a set of deterministic Miller-Rabin tests are performed with bases that are known to have no counterexamples in their range. Finally if the number is larger than 2^64, a strong BPSW test is performed. While this is a probable prime test and we believe counterexamples exist, there are no known counterexamples

Answer 8

According to wikipedia, the Sieve of Eratosthenes has complexity O(n * (log n) * (log log n)) and requires O(n) memory - so it's a pretty good place to start if you aren't testing for especially large numbers.

Answer 9

bool isPrime(int n)
{
// Corner cases
if (n <= 1) return false;
if (n <= 3) return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n%2 == 0 || n%3 == 0) return false;
for (int i=5; i*i<=n; i=i+6)
if (n%i == 0 || n%(i+2) == 0)
return false;
return true;
}
This is just c++ implementation of above

Answer 10

For large numbers you cannot simply naively check whether the candidate number N is divisible by none of the numbers less than sqrt(N). There are much more scalable tests available, such as the Miller-Rabin primality test. Below you have implementation in python:
def is_prime(x):
"""Fast implementation fo Miller-Rabin primality test, guaranteed to be correct."""
import math
def get_sd(x):
"""Returns (s: int, d: int) for which x = d*2^s """
if not x: return 0, 0
s = 0
while 1:
if x % 2 == 0:
x /= 2
s += 1
else:
return s, x
if x <= 2:
return x == 2
# x - 1 = d*2^s
s, d = get_sd(x - 1)
if not s:
return False # divisible by 2!
log2x = int(math.log(x) / math.log(2)) + 1
# As long as Riemann hypothesis holds true, it is impossible
# that all the numbers below this threshold are strong liars.
# Hence the number is guaranteed to be a prime if no contradiction is found.
threshold = min(x, 2*log2x*log2x+1)
for a in range(2, threshold):
# From Fermat's little theorem if x is a prime then a^(x-1) % x == 1
# Hence the below must hold true if x is indeed a prime:
if pow(a, d, x) != 1:
for r in range(0, s):
if -pow(a, d*2**r, x) % x == 1:
break
else:
# Contradicts Fermat's little theorem, hence not a prime.
return False
# No contradiction found, hence x must be a prime.
return True
You can use it to find huge prime numbers:
x = 10000000000000000000000000000000000000000000000000000000000000000000000000000
for e in range(1000):
if is_prime(x + e):
print('%d is a prime!' % (x + e))
break
# 10000000000000000000000000000000000000000000000000000000000000000000000000133 is a prime!
If you are testing random integers probably you want to first test whether the candidate number is divisible by any of the primes smaller than, say 1000, before you call Miller-Rabin. This will help you filter out obvious non-primes such as 10444344345.

Answer 11

Python 3:
def is_prime(a):
return a > 1 and all(a % i for i in range(2, int(a**0.5) + 1))

Answer 12

Way too late to the party, but hope this helps. This is relevant if you are looking for big primes:
To test large odd numbers you need to use the Fermat-test and/or Miller-Rabin test.
These tests use modular exponentiation which is quite expensive, for n bits exponentiation you need at least n big int multiplication and n big int divison. Which means the complexity of modular exponentiation is O(n³).
So before using the big guns, you need to do quite a few trial divisions. But don't do it naively, there is a way to do them fast.
First multiply as many primes together as many fits into a the words you use for the big integers. If you use 32 bit words, multiply 3*5*7*11*13*17*19*23*29=3234846615 and compute the greatest common divisor with the number you test using the Euclidean algorithm. After the first step the number is reduced below the word size and continue the algorithm without performing complete big integer divisions. If the GCD != 1, that means one of the primes you multiplied together divides the number, so you have a proof it's not prime. Then continue with 31*37*41*43*47 = 95041567, and so on.
Once you tested several hundred (or thousand) primes this way, you can do 40 rounds of Miller-Rabin test to confirm the number is prime, after 40 rounds you can be certain the number is prime there is only 2^-80 chance it's not (it's more likely your hardware malfunctions...).

Answer 13

I have got a prime function which works until (2^61)-1 Here:
from math import sqrt
def isprime(num): num > 1 and return all(num % x for x in range(2, int(sqrt(num)+1)))
Explanation:
The all() function can be redefined to this:
def all(variables):
for element in variables:
if not element: return False
return True
The all() function just goes through a series of bools / numbers and returns False if it sees 0 or False.
The sqrt() function is just doing the square root of a number.
For example:
>>> from math import sqrt
>>> sqrt(9)
>>> 3
>>> sqrt(100)
>>> 10
The num % x part returns the remainder of num / x.
Finally, range(2, int(sqrt(num))) means that it will create a list that starts at 2 and ends at int(sqrt(num)+1)
For more information about range, have a look at this website!
The num > 1 part is just checking if the variable num is larger than 1, becuase 1 and 0 are not considered prime numbers.
I hope this helped :)

Answer 14

In Python:
def is_prime(n):
return not any(n % p == 0 for p in range(2, int(math.sqrt(n)) + 1))
A more direct conversion from mathematical formalism to Python would use all(n % p != 0... ), but that requires strict evaluation of all values of p. The not any version can terminate early if a True value is found.

Answer 15

best algorithm for Primes number javascript
function isPrime(num) {
if (num <= 1) return false;
else if (num <= 3) return true;
else if (num % 2 == 0 || num % 3 == 0) return false;
var i = 5;
while (i * i <= num) {
if (num % i == 0 || num % (i + 2) == 0) return false;
i += 6;
}
return true
}

Answer 16

A prime number is any number that is only divisible by 1 and itself. All other numbers are called composite.
The simplest way, of finding a prime number, is to check if the input number is a composite number:
function isPrime(number) {
// Check if a number is composite
for (let i = 2; i < number; i++) {
if (number % i === 0) {
return false;
}
}
// Return true for prime numbers
return true;
}
The program has to divide the value of number by all the whole numbers from 1 and up to the its value. If this number can be divided evenly not only by one and itself it is a composite number.
The initial value of the variable i has to be 2 because both prime and composite numbers can be evenly divided by 1.
for (let i = 2; i < number; i++)
Then i is less than number for the same reason. Both prime and composite numbers can be evenly divided by themselves. Therefore there is no reason to check it.
Then we check whether the variable can be divided evenly by using the remainder operator.
if (number % i === 0) {
return false;
}
If the remainder is zero it means that number can be divided evenly, hence being a composite number and returning false.
If the entered number didn't meet the condition, it means it's a prime number and the function returns true.

Answer 17

Smallest memory? This isn't smallest, but is a step in the right direction.
class PrimeDictionary {
BitArray bits;
public PrimeDictionary(int n) {
bits = new BitArray(n + 1);
for (int i = 0; 2 * i + 3 <= n; i++) {
bits.Set(i, CheckPrimality(2 * i + 3));
}
}
public PrimeDictionary(IEnumerable<int> primes) {
bits = new BitArray(primes.Max());
foreach(var prime in primes.Where(p => p != 2)) {
bits.Set((prime - 3) / 2, true);
}
}
public bool IsPrime(int k) {
if (k == 2) {
return true;
}
if (k % 2 == 0) {
return false;
}
return bits[(k - 3) / 2];
}
}
Of course, you have to specify the definition of CheckPrimality.

Answer 18

To find if the number or numbers in a range is/are prime.
#!usr/bin/python3
def prime_check(*args):
for arg in args:
if arg > 1: # prime numbers are greater than 1
for i in range(2,arg): # check for factors
if(arg % i) == 0:
print(arg,"is not Prime")
print(i,"times",arg//i,"is",arg)
break
else:
print(arg,"is Prime")
# if input number is less than
# or equal to 1, it is not prime
else:
print(arg,"is not Prime")
return
# Calling Now
prime_check(*list(range(101))) # This will check all the numbers in range 0 to 100
prime_check(#anynumber) # Put any number while calling it will check.

Answer 19

myInp=int(input("Enter a number: "))
if myInp==1:
print("The number {} is neither a prime not composite no".format(myInp))
elif myInp>1:
for i in range(2,myInp//2+1):
if myInp%i==0:
print("The Number {} is not a prime no".format(myInp))
print("Because",i,"times",myInp//i,"is",myInp)
break
else:
print("The Number {} is a prime no".format(myInp))
else:
print("Alas the no {} is a not a prime no".format(myInp))

Answer 20

public static boolean isPrime(int number) {
if(number < 2)
return false;
else if(number == 2 || number == 3)
return true;
else {
for(int i=2;i<=number/2;i++)
if(number%i == 0)
return false;
else if(i==number/2)
return true;
}
return false;
}

Answer 21

Most of previous answers are correct but here is one more way to test to see a number is prime number. As refresher, prime numbers are whole number greater than 1 whose only factors are 1 and itself.(source)
Solution:
Typically you can build a loop and start testing your number to see if it's divisible by 1,2,3 ...up to the number you are testing ...etc but to reduce the time to check, you can divide your number by half of the value of your number because a number cannot be exactly divisible by anything above half of it's value.
Example if you want to see 100 is a prime number you can loop through up to 50.
Actual code:
def find_prime(number):
if(number ==1):
return False
# we are dividiing and rounding and then adding the remainder to increment !
# to cover not fully divisible value to go up forexample 23 becomes 11
stop=number//2+number%2
#loop through up to the half of the values
for item in range(2,stop):
if number%item==0:
return False
print(number)
return True
if(find_prime(3)):
print("it's a prime number !!")
else:
print("it's not a prime")

Answer 22

We can use java streams to implement this in O(sqrt(n)); Consider that noneMatch is a shortCircuiting method that stops the operation when finds it unnecessary for determining the result:
Scanner in = new Scanner(System.in);
int n = in.nextInt();
System.out.println(n == 2 ? "Prime" : IntStream.rangeClosed(2, ((int)(Math.sqrt(n)) + 1)).noneMatch(a -> n % a == 0) ? "Prime" : "Not Prime");

Answer 23

With help of Java-8 streams and lambdas, it can be implemented like this in just few lines:
public static boolean isPrime(int candidate){
int candidateRoot = (int) Math.sqrt( (double) candidate);
return IntStream.range(2,candidateRoot)
.boxed().noneMatch(x -> candidate % x == 0);
}
Performance should be close to O(sqrt(N)). Maybe someone find it useful.

Answer 24

### is_prime(number) =
### if number % p1 !=0 for all p1(prime numbers) < (sqrt(number) + 1),
### filter numbers that are not prime from divisors
import math
def check_prime(N, prime_numbers_found = [2]):
if N == 2:
return True
if int(math.sqrt(N)) + 1 > prime_numbers_found[-1]:
divisor_range = prime_numbers_found + list(range(prime_numbers_found[-1] + 1, int(math.sqrt(N)) + 1+ 1))
else:
divisor_range = prime_numbers_found
#print(divisor_range, N)
for number in divisor_range:
if number not in prime_numbers_found:
if check_prime(number, prime_numbers_found):
prime_numbers_found.append(number)
if N % number == 0:
return False
else:
if N % number == 0:
return False
return True

Answer 25

bool isPrime(int n) {
if(n <= 3)
return (n > 1)==0? false: true;
else if(n%2 == 0 || n%3 == 0)
return false;
int i = 5;
while(i * i <= n){
if(n%i == 0 || (n%(i+2) == 0))
return false;
i = i + 6;
}
return true;
}
For any number, the minimum iterations to check if the number is prime or not can be from 2 to square root of the number. To reduce the iterations, even more, we can check if the number is divisible by 2 or 3 as maximum numbers can be eliminated by checking if the number is divisible by 2 or 3. Further any prime number greater than 3 can be expressed as 6k+1 or 6k-1. So the iteration can go from 6k+1 to the square root of the number.

Answer 26

Let me suggest you the perfect solution for 64 bit integers. Sorry to use C#. You have not already specified it as python in your first post. I hope you can find a simple modPow function and analyze it easily.
public static bool IsPrime(ulong number)
{
return number == 2
? true
: (BigInterger.ModPow(2, number, number) == 2
? ((number & 1) != 0 && BinarySearchInA001567(number) == false)
: false)
}
public static bool BinarySearchInA001567(ulong number)
{
// Is number in list?
// todo: Binary Search in A001567 (https://oeis.org/A001567) below 2 ^ 64
// Only 2.35 Gigabytes as a text file http://www.cecm.sfu.ca/Pseudoprimes/index-2-to-64.html
}

Answer 27

BEST SOLUTION
I an unsure if I understand the concept of Time complexity: O(sqrt(n)) and Space complexity: O(1) in this context but the
function prime(n) is probably the fastest way (least iterations)
to calculate if a number is prime number of any size.
https://github.com/ganeshkbhat/fastprimenumbers
This probably is the BEST solution in the internet as of today 11th
March 2022. Feedback and usage is welcome.
This same code can be applied in any languages like C, C++, Go Lang,
Java, .NET, Python, Rust, etc with the same logic and have performance
benefits. It is pretty fast. I have not seen this implemented before
and has been indigenously done.
If you are looking at the speed and performance here is the """BEST""" hopeful solution I can give:
Max iterations 16666 for n == 100000 instead of 100000 of conventional
way
The codes can also be found here: https://github.com/ganeshkbhat/fastprimecalculations
If you use it for your project please spend 2 minutes of your time crediting me by letting me know by either sending me an email, or logging an Github issue with subject heading [User], or star my Github project. But let me know here https://github.com/ganeshkbhat/fastprimecalculations. I would love to know the fans and users of the code logic
def prime(n):
if ((n == 2 or n == 3 or n == 5 or n == 7)):
return True
if (n == 1 or ((n > 7) and (n % 5 == 0 or n % 7 == 0 or n % 2 == 0 or n % 3 == 0))):
return False
if ( type((n - 1) / 6) == int or type((n + 1) / 6) == int):
for i in range(1, n):
factorsix = (i * 6)
five = n / (5 + factorsix)
seven = n / (7 + factorsix)
if ( ((five > 1) and type(five) == int) or ((seven > 1) and type(five) == int) ):
return False;
if (factorsix > n):
break;
return True
return False
Here is an analysis of all the ways of calculation:
Conventional way of checking for prime:
def isPrimeConventionalWay(n):
count = 0
if (n <= 1):
return False;
# Check from 2 to n-1
# Max iterations 99998 for n == 100000
for i in range(2,n):
# Counting Iterations
count += 1
if (n % i == 0):
print("count: Prime Conventional way", count)
return False;
print("count: Prime Conventional way", count)
return True;
SQUAREROOT way of checking for prime:
def isPrimeSquarerootWay(num):
count = 0
# if not is_number num return False
if (num < 2):
print("count: Prime Squareroot way", count)
return False
s = math.sqrt(num)
for i in range(2, num):
# Counting Iterations
count += 1
if (num % i == 0):
print("count: Prime Squareroot way", count)
return False
print("count: Prime Squareroot way", count)
return True
OTHER WAYS:
def isprimeAKSWay(n):
"""Returns True if n is prime."""
count = 0
if n == 2:
return True
if n == 3:
return True
if n % 2 == 0:
return False
if n % 3 == 0:
return False
i = 5
w = 2
while i * i <= n:
count += 1
if n % i == 0:
print("count: Prime AKS - Mersenne primes - Fermat's little theorem or whatever way", count)
return False
i += w
w = 6 - w
print("count: Prime AKS - Mersenne primes - Fermat's little theorem or whatever way", count)
return True
SUGGESTED way of checking for prime:
def prime(n):
count = 0
if ((n == 2 or n == 3 or n == 5 or n == 7)):
print("count: Prime Unconventional way", count)
return True
if (n == 1 or ((n > 7) and (n % 5 == 0 or n % 7 == 0 or n % 2 == 0 or n % 3 == 0))):
print("count: Prime Unconventional way", count)
return False
if (((n - 1) / 6).is_integer()) or (((n + 1) / 6).is_integer()):
for i in range(1, n):
# Counting Iterations
count += 1
five = 5 + (i * 6)
seven = 7 + (i * 6)
if ((((n / five) > 1) and (n / five).is_integer()) or (((n / seven) > 1) and ((n / seven).is_integer()))):
print("count: Prime Unconventional way", count)
return False;
if ((i * 6) > n):
# Max iterations 16666 for n == 100000 instead of 100000
break;
print("count: Prime Unconventional way", count)
return True
print("count: Prime Unconventional way", count)
return False
Tests to compare with the traditional way of checking for prime numbers.
def test_primecalculations():
count = 0
iterations = 100000
arr = []
for i in range(1, iterations):
traditional = isPrimeConventionalWay(i)
newer = prime(i)
if (traditional == newer):
count = count + 1
else:
arr.push([traditional, newer, i])
print("[count, iterations, arr] list: ", count, iterations, arr)
if (count == iterations):
return True
return False
# print("Tests Passed: ", test_primecalculations())
You will see the results of count of number of iterations as below for check of prime number: 100007:
print("Is Prime 100007: ", isPrimeConventionalWay(100007))
print("Is Prime 100007: ", isPrimeSquarerootWay(100007))
print("Is Prime 100007: ", prime(100007))
print("Is Prime 100007: ", isprimeAKSWay(100007))
count: Prime Conventional way 96
Is Prime 100007: False
count: Prime Squareroot way 96
Is Prime 100007: False
count: Prime Unconventional way 15
Is Prime 100007: False
count: Prime AKS - Mersenne primes - Fermat's little theorem or whatever way 32
Is Prime 100007: False
Here are some performance tests and results below:
import time
isPrimeConventionalWayArr = []
isPrimeSquarerootWayArr = []
primeArr = []
isprimeAKSWayArr = []
def tests_performance_isPrimeConventionalWayArr():
global isPrimeConventionalWayArr
for i in range(1, 1000000):
start = time.perf_counter_ns()
isPrimeConventionalWay(30000239)
end = time.perf_counter_ns()
isPrimeConventionalWayArr.append(end - start)
tests_performance_isPrimeConventionalWayArr()
def tests_performance_isPrimeSquarerootWayArr():
global isPrimeSquarerootWayArr
for i in range(1, 1000000):
start = time.perf_counter_ns()
isPrimeSquarerootWay(30000239)
end = time.perf_counter_ns()
isPrimeSquarerootWayArr.append(end - start)
tests_performance_isPrimeSquarerootWayArr()
def tests_performance_primeArr():
global primeArr
for i in range(1, 1000000):
start = time.perf_counter_ns()
prime(30000239)
end = time.perf_counter_ns()
primeArr.append(end - start)
tests_performance_primeArr()
def tests_performance_isprimeAKSWayArr():
global isprimeAKSWayArr
for i in range(1, 1000000):
start = time.perf_counter_ns()
isprimeAKSWay(30000239)
end = time.perf_counter_ns()
isprimeAKSWayArr.append(end - start)
tests_performance_isprimeAKSWayArr()
print("isPrimeConventionalWayArr: ", sum(isPrimeConventionalWayArr)/len(isPrimeConventionalWayArr))
print("isPrimeSquarerootWayArr: ", sum(isPrimeSquarerootWayArr)/len(isPrimeSquarerootWayArr))
print("primeArr: ", sum(primeArr)/len(primeArr))
print("isprimeAKSWayArr: ", sum(isprimeAKSWayArr)/len(isprimeAKSWayArr))
Sample 1 Million Iterations
Iteration 1:
isPrimeConventionalWayArr: 1749.97224997225
isPrimeSquarerootWayArr: 1835.6258356258356
primeArr (suggested): 475.2365752365752
isprimeAKSWayArr: 1177.982377982378
Iteration 2:
isPrimeConventionalWayArr: 1803.141403141403
isPrimeSquarerootWayArr: 2184.222484222484
primeArr (suggested): 572.6434726434726
isprimeAKSWayArr: 1403.3838033838033
Iteration 3:
isPrimeConventionalWayArr: 1876.941976941977
isPrimeSquarerootWayArr: 2190.43299043299
primeArr (suggested): 569.7365697365698
isprimeAKSWayArr: 1449.4147494147494
Iteration 4:
isPrimeConventionalWayArr: 1873.2779732779734
isPrimeSquarerootWayArr: 2177.154777154777
primeArr (suggested): 590.4243904243905
isprimeAKSWayArr: 1401.9143019143019
Iteration 5:
isPrimeConventionalWayArr: 1891.1986911986912
isPrimeSquarerootWayArr: 2218.093218093218
primeArr (suggested): 571.6938716938716
isprimeAKSWayArr: 1397.6471976471976
Iteration 6:
isPrimeConventionalWayArr: 1868.8454688454688
isPrimeSquarerootWayArr: 2168.034368034368
primeArr (suggested): 566.3278663278663
isprimeAKSWayArr: 1393.090193090193
Iteration 7:
isPrimeConventionalWayArr: 1879.4764794764794
isPrimeSquarerootWayArr: 2199.030199030199
primeArr (suggested): 574.055874055874
isprimeAKSWayArr: 1397.7587977587978
Iteration 8:
isPrimeConventionalWayArr: 1789.2868892868894
isPrimeSquarerootWayArr: 2182.3258823258825
primeArr (suggested): 569.3206693206694
isprimeAKSWayArr: 1407.1486071486072

Answer 28

Similar idea to the algorithm which has been mentioned
public static boolean isPrime(int n) {
if(n == 2 || n == 3) return true;
if((n & 1 ) == 0 || n % 3 == 0) return false;
int limit = (int)Math.sqrt(n) + 1;
for(int i = 5, w = 2; i <= limit; i += w, w = 6 - w) {
if(n % i == 0) return false;
numChecks++;
}
return true;
}

Answer 29

When I have to do a fast verification, I write this simple code based on the basic division between numbers lower than square root of input.
def isprime(n):
if n%2==0:
return n==2
else:
cota = int(n**0.5)+1
for ind in range(3,2,cota):
if n%ind==0:
print(ind)
return False
is_one = n==1
return True != is_one
isprime(22783)
The last True != n==1 is to avoid the case n=1.