示例#1
0
def remove_digits(number):
    from maths.misc import get_digits, digit_list_to_num
    digits = get_digits(number)
    num_digits = len(digits)
    
    for i in xrange(1, num_digits):
        # Remove digits from the right
        yield digit_list_to_num(digits[:i])
        
        # Remove digits from the left
        yield digit_list_to_num(digits[i:])
示例#2
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def list_to_num(digits):
    length = len(digits)
    num = 0
    for i in xrange(length):
        power = length - 1 - i
        num += digits[i] * 10**power
        
    return num

if __name__ == "__main__":
    from maths.misc import get_digits, is_prime
    
    circular_primes = []
    
    for i in xrange(2, 1000000):
        if not is_prime(i):
            continue
        prime = True
        digits = get_digits(i)
        for offset in xrange(1,len(digits)):
            shifted = shift_digits(digits, offset)

            num = list_to_num(shifted)
            
            if not is_prime(num):
                prime = False
                break
        if prime:
            circular_primes.append(i)
    print('The number of circular primes below 1,000,000 is %d'%len(circular_primes))
示例#3
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        # We don't want to "over-reduce"
        if num % i == 0 and denom % i == 0:
            num = num / i
            denom = denom / i 
    return num, denom
    

if __name__ == "__main__":
    from maths.misc import get_digits
    numerators = []
    denominators = []
    
    for i in xrange(10,100):
        for j in xrange(i + 1,100):
            # Get digits of i,j
            digits_i, digits_j = get_digits(i), get_digits(j)
            
            # Check for overlap
            overlap = filter(set(digits_i).__contains__, digits_j)
            
            if len(overlap) > 0:
                # Pull first char from overlap
                char_to_remove = overlap[0]
                
                # Check for trivial case where digits to be removed are in the same power of 10s position
                index_i = digits_i.index(char_to_remove)
                index_j = digits_j.index(char_to_remove)
                if index_i == index_j:
                    # Trivial case
                    continue