[Tutor] operator, mult
ajarncolin at gmail.com
Thu Jan 29 07:59:35 CET 2009
That sorted me out, sometimes I just can't get things worked out in my head,
then get a sense of "instant enlightenment", which your comments did for me.
I am ashamed to say I was using the wrong prime factors function, then
changing the mult to mul all started to make sense.
2009/1/29 John Fouhy <john at fouhy.net>
> 2009/1/29 col speed <ajarncolin at gmail.com>:
> > What I expected "mult" to do was (somehow)to work out what the powers
> > the prime factors would be. Another reason I didn't think it was "mul" is
> > the part that says " prime_factors_mult(n)" as the prime_factors
> > is just "prime_factors(n)" - without the "_mult".
> Well, it's been a while since my number theory course, so I was just
> going from the code comments:
> 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) * ...
> >>> phi(1)
> >>> phi(10)
> from operator import mult
> if n == 1: return 1
> return reduce(mult, [(p-1) * p**(m-1) for p,m in prime_factors_mult(n)])
> If we imagine for a moment that we have:
> prime_facs = [(p_0, m_0), (p_1, m_1), (p_2, m_2), (p_3, m_3)]
> reduce(operator.mul, [(p-1) * p**(m-1) for p,m in prime_facs])
> translates exactly to
> (p_0-1)*p_0**(m_0-1) * (p_1-1)*p_1**(m_1-1) * (p_2-1)*p_2**(m_2-1)
> * (p_3-1)*p_3**(m_3-1)
> which seems to match the description in the comment.
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