The task in Problem 35 of Project Euler is:

The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime. There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97. How many circular primes are there below one million?

## How to solve

If you have heard of the sieve of Eratosthenes, this one sounds quite easy:

- Find all primes below one million
- For each prime, do:
- Generate all rotations
- Check for every rotation if it is a prime

- Count the number of circular primes

## The implementation

### Sieve of Eratosthenes

The finds all primes below \(n \in \mathbb{N}\). But you can make a lot of mistakes in the implementation.

First, this is the way the sieve of Eratosthenes works:

For example, this implementation is not good:

```
def getPrimesBelowN(n=1000000):
""" Sieve of Eratosthenes """
from math import ceil
roundUp = lambda n, prime: int(ceil(float(n) / prime))
primes = range(2, n)
for currentPrime in primes:
for multiplicant in range(2, roundUp(n, currentPrime)):
noPrime = multiplicant * currentPrime
if noPrime in primes:
primes.remove(noPrime)
return primes
```

Whats bad with this code?
Well, just think about what it does: For every `noPrime`

Python has to go through the whole list. I couldn't find how `in`

is implemented, but I guess it is linear. So Python has to go through the whole list for `in`

. Additionally, `remove`

could also be expensive.

How could this get improved? Here is a better solution:

```
def getPrimesBelowN(n=1000000):
""" Sieve of Eratosthenes """
from math import ceil
roundUp = lambda n, prime: int(ceil(float(n) / prime))
primes = [True] * n
primes[0] = False
primes[1] = False
primeList = []
for currentPrime in range(2, n):
if not primes[currentPrime]:
continue
primeList.append(currentPrime)
for multiplicant in range(2, roundUp(n, currentPrime)):
primes[multiplicant * currentPrime] = False
return primeList
```

This solution does not need to search for `noPrime`

, it simply jumps there in the list.

A generator version of the sieve of Erasthostenes can be found on code.activestate.com.

### isCircularPrime

Rotation the digits of a number is the same as cutting the number into two pieces and switching the position of the pieces:

```
def isCircularPrime(primes, number):
number = str(number)
for i in range(0, len(number)):
rotatedNumber = number[i : len(number)] + number[0:i]
if int(rotatedNumber) not in primes:
return False
return True
```

Here is the same problem as above, in the sieving algorithm: Searching through the list takes much more time than jumping to a position in the list. So this one is better:

```
def isCircularPrime(primes, number):
number = str(number)
for i in range(0, len(number)):
rotatedNumber = number[i : len(number)] + number[0:i]
if not primes[int(rotatedNumber)]:
return False
return True
```

### Some more speedups

Every prime that contains one of the digits 0, 2, 4, 6 or 8 can't be a circular prime, because one rotation exist where that digit is at the end. This rotation would be divisible by 2 and thus not be a prime (except for 2, of course). You can use the same thought for the digit 5. So you can skip those digits### The final snippet

```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
def getPrimesBelowN(n=1000000):
"""Get all primes below n with the sieve of Eratosthenes.
@return: a list 0..n with boolean values that indicate if
i in 0..n is a prime.
"""
from math import ceil
primes = [True] * n
primes[0] = False
primes[1] = False
primeList = []
roundUp = lambda n, prime: int(ceil(float(n) / prime))
for currentPrime in range(2, n):
if not primes[currentPrime]:
continue
primeList.append(currentPrime)
for multiplicant in range(2, roundUp(n, currentPrime)):
primes[multiplicant * currentPrime] = False
return primes
def isCircularPrime(primes, number):
"""Check if number is a circular prime.
Keyword arguments:
primes -- a list from 0..n with boolean values that indicate if
i in 0..n is a prime
number -- the integer you want to check
"""
number = str(number)
for i in range(0, len(number)):
rotatedNumber = number[i : len(number)] + number[0:i]
if not primes[int(rotatedNumber)]:
return False
return True
if __name__ == "__main__":
print("Start sieving.")
primes = getPrimesBelowN(1000000)
print("End sieving.")
numberOfPrimes = 2
print(2) # I print them now, because I want to skip all primes
print(5) # that contain one of those digits: 0,2,4,5,6,8
for prime, isPrime in enumerate(primes):
if (
(not isPrime)
or ("2" in str(prime))
or ("4" in str(prime))
or ("6" in str(prime))
or ("8" in str(prime))
or ("0" in str(prime))
or ("5" in str(prime))
):
continue
if isCircularPrime(primes, prime):
print(prime)
numberOfPrimes += 1
print("Number of circular primes: %i" % numberOfPrimes)
```

It takes about 1.096 seconds (in comparison: having the version of `isCircularPrime`

that searches through the list of primes took over 5 minutes!)