This notebook has a minor addition to euler-001-multiples-of-3-and-5-20160319.ipynb.

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.



In [1]:

    
from __future__ import print_function

...

I will have another go at the union thing, this time as a function.



In [29]:

    
def union(sets):
    u = set([])
    for s in sets:
        u |= s
    return u



In [30]:

    
def foo(n, divisors):
    return sum(union(set(range(0, n, d)) for d in divisors))



In [31]:

    
n=1000
divisors = (3, 5)
%timeit foo(n, divisors)
foo(n, divisors)









    



10000 loops, best of 3: 70.6 µs per loop






    Out[31]:





233168

It works, but is not as readable, so forget it.

2016-06-29: I have another way of writing the union function from above, this time using reduce() instead of iterating through a loop. Why did I not think of reduce() earlier?

The code and output above is preserved as it ran on 2016-03-19 in Python 2. The code and output below ran on 2016-06-29 in Python 3.



In [1]:

    
from functools import reduce
from operator import or_



In [2]:

    
def union(sets):
    return reduce(or_, sets)



In [3]:

    
def foo(n, divisors):
    return sum(union(set(range(0, n, d)) for d in divisors))



In [4]:

    
n=1000
divisors = (3, 5)
%timeit foo(n, divisors)
foo(n, divisors)









    



10000 loops, best of 3: 59.1 µs per loop






    Out[4]:





233168