This is not a homework, I am just curious.
INFINITE is the key word here.
I wish to use it as for p in primes()
. I believe that this is a built-in function in Haskell.
So, the answer cannot be as naive as "Just do a Sieve".
First of all, you do not know how many consecutive primes will be consumed. Well, suppose you could concoct 100 of them at a time. Would you use the same Sieve approach as well as the frequency of prime numbers formula?
I prefer non-concurrent approach.
Thank you for reading (and writing ;) )!
The erat2
function from the cookbook can be further sped up (by about 20-25%):
import itertools as it
def erat2a( ):
D = { }
yield 2
for q in it.islice(it.count(3), 0, None, 2):
p = D.pop(q, None)
if p is None:
D[q*q] = q
yield q
else:
# old code here:
# x = p + q
# while x in D or not (x&1):
# x += p
# changed into:
x = q + 2*p
while x in D:
x += 2*p
D[x] = p
The not (x&1)
check verifies that x
is odd. However, since both q
and p
are odd, by adding 2*p
half of the steps are avoided along with the test for oddity.
If one doesn't mind a little extra fanciness, erat2
can be sped up by 35-40% with the following changes (NB: needs Python 2.7+ or Python 3+ because of the itertools.compress
function):
import itertools as it
def erat3( ):
D = { 9: 3, 25: 5 }
yield 2
yield 3
yield 5
MASK= 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0,
MODULOS= frozenset( (1, 7, 11, 13, 17, 19, 23, 29) )
for q in it.compress(
it.islice(it.count(7), 0, None, 2),
it.cycle(MASK)):
p = D.pop(q, None)
if p is None:
D[q*q] = q
yield q
else:
x = q + 2*p
while x in D or (x%30) not in MODULOS:
x += 2*p
D[x] = p
The erat3
function takes advantage of the fact that all primes (except for 2, 3, 5) modulo 30 result to only eight numbers: the ones included in the MODULOS
frozenset. Thus, after yielding the initial three primes, we start from 7 and work only with the candidates.
The candidate filtering uses the itertools.compress
function; the “magic” is in the MASK
sequence; MASK
has 15 elements (there are 15 odd numbers in every 30 numbers, as chosen by the itertools.islice
function) with a 1
for every possible candidate, starting from 7. The cycle repeats as specified by the itertools.cycle
function.
The introduction of the candidate filtering needs another modification: the or (x%30) not in MODULOS
check. The erat2
algorithm processed all odd numbers; now that the erat3
algorithm processes only r30 candidates, we need to make sure that all D.keys()
can only be such —false— candidates.
On an Atom 330 Ubuntu 9.10 server, versions 2.6.4 and 3.1.1+:
$ testit
up to 8192
==== python2 erat2 ====
100 loops, best of 3: 18.6 msec per loop
==== python2 erat2a ====
100 loops, best of 3: 14.5 msec per loop
==== python2 erat3 ====
Traceback (most recent call last):
…
AttributeError: 'module' object has no attribute 'compress'
==== python3 erat2 ====
100 loops, best of 3: 19.2 msec per loop
==== python3 erat2a ====
100 loops, best of 3: 14.1 msec per loop
==== python3 erat3 ====
100 loops, best of 3: 11.7 msec per loop
On an AMD Geode LX Gentoo home server, Python 2.6.5 and 3.1.2:
$ testit
up to 8192
==== python2 erat2 ====
10 loops, best of 3: 104 msec per loop
==== python2 erat2a ====
10 loops, best of 3: 81 msec per loop
==== python2 erat3 ====
Traceback (most recent call last):
…
AttributeError: 'module' object has no attribute 'compress'
==== python3 erat2 ====
10 loops, best of 3: 116 msec per loop
==== python3 erat2a ====
10 loops, best of 3: 82 msec per loop
==== python3 erat3 ====
10 loops, best of 3: 66 msec per loop
A primegen.py
module contains the erat2
, erat2a
and erat3
functions. Here follows the testing script:
#!/bin/sh
max_num=${1:-8192}
echo up to $max_num
for python_version in python2 python3
do
for function in erat2 erat2a erat3
do
echo "==== $python_version $function ===="
$python_version -O -m timeit -c \
-s "import itertools as it, functools as ft, operator as op, primegen; cmp= ft.partial(op.ge, $max_num)" \
"next(it.dropwhile(cmp, primegen.$function()))"
done
done