On Tue, Jun 15, 2010 at 9:26 AM, Ian Kelly <ian.g.ke...@gmail.com> wrote:
> On Tue, Jun 15, 2010 at 6:21 AM, Alain Ketterlin
> <al...@dpt-info.u-strasbg.fr> wrote:
> > You compute i**2 too many times (7/5 times more than necessary) and
> > twice too many modulos. I suggest:
> >
> > c = { 0:1, 1:1, 2:1, 3:-1, 4:-1 }
> > #or, why not: c = lambda i : +1 if (i%5) < 3 else -1
> >
> > s = 0
> > for i in range(1,2011):
> > s += c[(i-1)%5]*(i**2)
> > print s
>
> In fact, most of them are unnecessary:
>
> from itertools import izip, cycle
>
> def squares(start, stop):
> square = start * start
> step = start * 2 + 1
> for root in xrange(start, stop):
> yield square
> square += step
> step += 2
>
> print sum(sign * square for sign, square in izip(cycle([1,1,1,-1,-1]),
> squares(1, 2011)))
>
> Now, anybody know how to make that version a one-liner without making
> it go quadratic in run-time?
>
> Cheers,
> Ian
> --
> http://mail.python.org/mailman/listinfo/python-list
>
I don't really consider long one liners that pythonic - they're often not
that clear.
Here's something I do consider pythonic, which uses the algorithm Ian
mentions. It's far from the shortest, but it's quite clear, IMO:
#!/usr/bin/python
import itertools
def signs():
while True:
yield 1
yield 1
yield 1
yield -1
yield -1
# take advantage of the property that a perfect square is the sum of odd
integers to get a radical speed boost
# 1, 4, 9, 16, 25
# 4 - 1 is 3
# 9 - 4 is 5
# 16 - 9 is 7
# and so on. IOW, if you build a square and then build onto two of the
sides to increase the dimensions by one unit, you're always
# adding the next odd integer - you can do it nicely with a child's blocks.
def squares():
oddno = 1
square = oddno
while True:
yield square
oddno += 2
square += oddno
for squareno, sign, square in itertools.izip(xrange(20), signs(),
squares()):
# what you want to do with these is your business :)
print squareno, sign, square
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