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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def estimate(self):
 """ Returns the estimate of the cardinality """

        E = self.alpha * float(self.m ** 2) / np.power(2.0, - self.M).sum()
        if E <= 2.5 * self.m:             # Small range correction
            V = self.m - np.count_nonzero(self.M)
            return int(self.m * np.log(self.m / float(V))) if V > 0 else int(E)
        # intermidiate range correction -> No correction
        elif E <= float(long(1) << self.precision) / 30.0:
            return int(E)
        else:
            return int(-(long(1) << self.precision) *
                       np.log(1.0 - E / (long(1) << self.precision)))