def odd(n):
    return n % 2 == 1

def is_square(n):
    """ From the internet. I should implement my own version of this. """
    ## Trivial checks
    if type(n) != int:  ## integer
        return False
    if n < 0:      ## positivity
        return False
    if n == 0:      ## 0 pass
        return True

    ## Reduction by powers of 4 with bit-logic
    while n&3 == 0:
        n=n>>2

    ## Simple bit-logic test. All perfect squares, in binary,
    ## end in 001, when powers of 4 are factored out.
    if n&7 != 1:
        return False

    if n==1:
        return True  ## is power of 4, or even power of 2

    ## Simple modulo equivalency test
    c = n%10
    if c in {3, 7}:
        return False  ## Not 1,4,5,6,9 in mod 10
    if n % 7 in {3, 5, 6}:
        return False  ## Not 1,2,4 mod 7
    if n % 9 in {2,3,5,6,8}:
        return False
    if n % 13 in {2,5,6,7,8,11}:
        return False

    ## Other patterns
    if c == 5:  ## if it ends in a 5
        if (n//10)%10 != 2:
            return False    ## then it must end in 25
        if (n//100)%10 not in {0,2,6}:
            return False    ## and in 025, 225, or 625
        if (n//100)%10 == 6:
            if (n//1000)%10 not in {0,5}:
                return False    ## that is, 0625 or 5625
    else:
        if (n//10)%4 != 0:
            return False    ## (4k)*10 + (1,9)

    ## Babylonian Algorithm. Finding the integer square root.
    ## Root extraction.
    s = (len(str(n))-1) // 2
    x = (10**s) * 4
    A = {x, n}
    while x * x != n:
        x = (x + (n // x)) >> 1
        if x in A:
            return False
        A.add(x)
    return True


def euler_094():
    n_max = 10 ** 9
    n, total_peri = 2, 0
    while n < n_max:
        n += 2
        for b2, c in [(n, n + 1), (n, n - 1)]:
            b = b2 // 2
            a_squared = c * c - b * b

            if not is_square(a_squared):
                continue

            peri = 2 * c + b2
            # print("b2={} c={} peri={}".format(b2, c, peri))

            if peri > n_max:
                return total_peri

            total_peri += peri

            if b2 > 10000:
                n = int(n * 3.725) - 5
                if odd(n):
                    n -= 1


if __name__ == "__main__":
    solution = euler_094()
    print("e094.py: " + str(solution))
    assert(solution == 518408346)