That's what I thought , but I tried it to no avail. Plus the syntax is wrong. Thanks anyway Colin
2009/1/28 John Fouhy <j...@fouhy.net> > 2009/1/28 col speed <ajarnco...@gmail.com>: > > Hello there, > > I got the following function while googling: > > > > def totient(n): > > """calculate Euler's totient function. > > > > If [[p_0,m_0], [p_1,m_1], ... ] is a prime factorization of 'n', > > then the totient function phi(n) is given by: > > > > (p_0 - 1)*p_0**(m_0-1) * (p_1 - 1)*p_1**(m_1-1) * ... > [...] > > return reduce(mult, [(p-1) * p**(m-1) for p,m in > prime_factors_mult(n)]) > > > > I already have the "prime_factors" function. The problem is that I cannot > > find "mult". I tried using "mul" which is in "operator" but that is > > obviously not the same thing. > > It seems pretty obvious that operator.mul is what they mean (based on > what the reduce function does). Perhaps it's a typo? > > -- > John. >
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