Python Solutions to [[http://projecteuler.net/index.php?section=problems|Project Euler]].

'''Index'''
<<TableOfContents(2)>>

= Solutions to the first 40 problems in functional Python =

Just found this site which is apparently devoted to solutions for the Euler problem set, in python, with a functional flavor.

http://pyeuler.wikidot.com/

= Problem 1: Add all the natural numbers below 1000 that are multiples of 3 or 5. =

http://projecteuler.net/index.php?section=problems&id=1

{{{
    def multiples_of_3_or_5():
        for number in xrange(1000):
            if not number % 3 or not number % 5:
                yield number

    print sum(multiples_of_3_or_5())
}}}

In one line:
{{{
    print sum( number for number in xrange(1000) if not (number % 3 and number % 5) )
}}}

= Problem 2: Find the sum of all the even-valued terms in the Fibonacci sequence which do not exceed one million. =

http://projecteuler.net/index.php?section=problems&id=2

{{{
    def fib():
        x,y = 0,1
        while True:
            yield x
            x,y = y, x+y

    def even(seq):
        for number in seq:
            if not number % 2:
                yield number

    def under_a_million(seq):
        for number in seq:
            if number > 1000000:
                break
            yield number   

    print sum(even(under_a_million(fib())))
}}}

Alternately:
{{{
    import itertools

    def fib():
        x,y = 0,1
        while True:
            yield x
            x,y = y, x+y

    print sum(x for x in itertools.takewhile(lambda x: x <= 1000000, fib()) if x % 2 == 0)
}}}

= Problem 3: Find the largest prime factor of 317584931803. =

http://projecteuler.net/index.php?section=problems&id=3

{{{
    plist = [2]

    def primes(min, max):
        if 2 >= min: yield 2
        for i in range(3, max, 2):
            for p in plist:
                if i%p == 0 or p*p > i: break
            if i%p:
                plist.append(i)
                if i >= min: yield i
        
    def factors(number):
        for prime in primes(2, number):
            if number % prime == 0:
                number /= prime
                yield prime
            if number == 1:
                break

    print max(factors(317584931803))
}}}