Remarkable results with psyco and sieve of Eratosthenes

[Steve Bergman]

Just wanted to report a delightful little surprise while experimenting
with psyco.
The program below performs astonoshingly well with psyco.

It finds all the prime numbers < 10,000,000

Processor is AMD64 4000+ running 32 bit.

Non psyco'd python version takes 94 seconds.

psyco'd version takes 9.6 seconds.

But here is the kicker. The very same algorithm written up in C and
compiled with gcc -O3, takes 4.5 seconds. Python is runng half as fast
as optimized C in this test!

Made my day, and I wanted to share my discovery.

BTW, can this code be made any more efficient?


============

#!/usr/bin/python -OO
import math
import sys
import psyco

psyco.full()

def primes():
primes=[3]
for x in xrange(5,10000000,2):
maxfact = int(math.sqrt(x))
flag=True
for y in primes:
if y > maxfact:
break
if x%y == 0:
flag=False
break
if flag == True:
primes.append(x)
primes()

 

[Will McGugan]

#!/usr/bin/python -OO
import math
import sys
import psyco

psyco.full()

def primes():
primes=[3]
for x in xrange(5,10000000,2):
maxfact = int(math.sqrt(x))
flag=True
for y in primes:
if y > maxfact:
break
if x%y == 0:
flag=False
break
if flag == True:
primes.append(x)
primes()

Some trivial optimizations. Give this a whirl.

def primes():
sqrt=math.sqrt
primes=[3]
for x in xrange(5,10000000,2):
maxfact = int(sqrt(x))
for y in primes:
if y > maxfact:
primes.append(x)
break
if not x%y:
break
return primes

 

[Will McGugan]

Just wanted to report a delightful little surprise while experimenting
with psyco.
The program below performs astonoshingly well with psyco.

It finds all the prime numbers < 10,000,000

Actualy, it doesn't. You forgot 1 and 2.


Will McGugan

 

[Beliavsky]

Actualy, it doesn't. You forgot 1 and 2.

The number 1 is not generally considered to be a prime number -- see
http://mathworld.wolfram.com/PrimeNumber.html .

 

[Will McGugan]

The number 1 is not generally considered to be a prime number -- see
http://mathworld.wolfram.com/PrimeNumber.html .

I stand corrected.

 

[dickinsm]

BTW, can this code be made any more efficient?

I'm not sure, but the following code takes around 6 seconds on my
1.2Ghz iBook. How does it run on your machine?

def smallPrimes(n):
"""Given an integer n, compute a list of the primes < n"""

if n <= 2:
return []

sieve = range(3, n, 2)
top = len(sieve)
for si in sieve:
if si:
bottom = (si*si - 3)//2
if bottom >= top:
break
sieve[bottom::si] = [0] * -((bottom-top)//si)

return [2]+filter(None, sieve)

smallPrimes(10**7)

 

[Steve Bergman]

Some trivial optimizations. Give this a whirl.

I retimed and got 9.7 average for 3 runs on my version.

Yours got it down to 9.2.

5% improvement. Not bad.

(Inserting '2' at the beginning doesn't seem to impact performance
much.;-) )

BTW, strictly speaking, shouldn't I be adding something to the floating
point sqrt result, before converting to int, to allow for rounding
error? If it is supposed to be 367 but comes in at 366.99999999, don't
I potentially classify a composite as a prime?

How much needs to be added?

 

[Steve Bergman]

I'm not sure, but the following code takes around 6 seconds on my
1.2Ghz iBook. How does it run on your machine?

Hmm. Come to think of it, my algorithm isn't the sieve.

Anyway, this is indeed fast as long as you have enough memory to handle
it for the range supplied.

It comes in at 1.185 seconds average on this box.

Come to think of it, there is a supposedly highly optimized version of
the sieve in The Python Cookbook that I've never bothered to actually
try out. Hmmm...

 

[dickinsm]

Hmm. Come to think of it, my algorithm isn't the sieve.

Right. I guess the point of the sieve is that you don't have to spend
any time
finding that a given odd integer is not divisible by a given prime;
all the
multiplies are done up front, so you save all the operations
corresponding to
the case when x % y != 0 in your code. Or something.

Anyway, this is indeed fast as long as you have enough memory to handle
it for the range supplied.

The sieve can be segmented, so that the intermediate space requirement
for
computing the primes up to n is O(sqrt(n)). (Of course you'll still
need
O(n/log n) space to store the eventual list of primes.) Then there
are all sorts
of bells and whistles (not to mention wheels) that you can add to
improve the
running time, most of which would considerably complicate the code.

The book by Crandall and Pomerance (Primes: A Computational
Perspective)
goes into plenty of detail on all of this.

Mark Dickinson

 

[Steven D'Aprano]

BTW, strictly speaking, shouldn't I be adding something to the floating
point sqrt result, before converting to int, to allow for rounding
error?

If you don't mind doing no more than one unnecessary test per candidate,
you can just add one to maxfact to allow for that. Or use round()
rather than int(). Or don't convert it at all, just say:

maxfact = math.sqrt(x)

and compare directly to that.

If it is supposed to be 367 but comes in at 366.99999999, don't
I potentially classify a composite as a prime?

Do you fear the math.sqrt() function is buggy? If so, all bets are off

How much needs to be added?

No more than 1, and even that might lead you to sometimes performing an
unnecessary test.

 

[Gabriel Genellina]

BTW, strictly speaking, shouldn't I be adding something to the floating
point sqrt result, before converting to int, to allow for rounding
error? If it is supposed to be 367 but comes in at 366.99999999, don't
I potentially classify a composite as a prime?

You could avoid sqrt using divmod (which gets the % result too); stop
when quotient<=divisor.
But this approach creates a tuple and then unpacks it, so you should
time it to see if there is an actual speed improvement.


--
Gabriel Genellina
Softlab SRL

__________________________________________________
Correo Yahoo!
Espacio para todos tus mensajes, antivirus y antispam ¡gratis!
¡Abrí tu cuenta ya! - http://correo.yahoo.com.ar

 

[George Sakkis]

#!/usr/bin/python -OO
import math
import sys
import psyco

psyco.full()

def primes():
primes=[3]
for x in xrange(5,10000000,2):
maxfact = int(math.sqrt(x))
flag=True
for y in primes:
if y > maxfact:
break
if x%y == 0:
flag=False
break
if flag == True:
primes.append(x)
primes()

Some trivial optimizations. Give this a whirl.

def primes():
sqrt=math.sqrt
primes=[3]
for x in xrange(5,10000000,2):
maxfact = int(sqrt(x))
for y in primes:
if y > maxfact:
primes.append(x)
break
if not x%y:
break
return primes

You can also save an attribute lookup for append; just add
append = primes.append
outside of the loop and replace primes.append(x) with append(x)
That should cut down a few fractions of second.

George

 

[bearophileHUGS]

George Sakkis:

You can also save an attribute lookup for append; just add
append = primes.append
outside of the loop and replace primes.append(x) with append(x)
That should cut down a few fractions of second.

We were talking about Psyco, and I think with Psyco (just released for
Py 2.5, BTW) such tricks are less useful.

Bye,
bearophile

 

[Pekka Karjalainen]

BTW, can this code be made any more efficient?

def primes():
primes=[3]
for x in xrange(5,10000000,2):
maxfact = int(math.sqrt(x))
flag=True
for y in primes:
if y > maxfact:
break

[...]

You can omit the call to math.sqrt if you test this instead.

y*y > x

in place of if y > maxfact: .

Pka

 

[Klaus Alexander Seistrup]

You can omit the call to math.sqrt if you test this instead.

y*y > x

in place of if y > maxfact: .

Or use

sqrt = lambda x: x ** .5

Cheers,

 

[Klaas]

Or use

sqrt = lambda x: x ** .5

Test it:

$ python -m timeit -s "from math import sqrt" "sqrt(5.6)"
1000000 loops, best of 3: 0.445 usec per loop
$ python -m timeit -s "sqrt = lambda x: x**.5" "sqrt(5.6)"
1000000 loops, best of 3: 0.782 usec per loop

Note that this overhead is almost entirely in function calls; calling
an empty lambda is more expensive than a c-level sqrt:

$ python -m timeit -s "sqrt = lambda x: x" "sqrt(5.6)"
1000000 loops, best of 3: 0.601 usec per loop

Just math ops:
$ python -m timeit -s "x = 5.6" "x*x"
10000000 loops, best of 3: 0.215 usec per loop
$ python -m timeit -s "x = 5.6" "x**.5"
1000000 loops, best of 3: 0.438 usec per loop

Of course, who knows that psyco does with this under the hood.

-Mike