However my attempts to use this in Python keep returning none
(obviously no match), however I don't see why, I was under the
impression Python used the same RE system as Perl/Ruby and I know the
convert is producing the correct display of 1's...Any thoughts?
def re_prime(n):
import re
convert = "".join("1" for i in xrange(n))
return re.match("^1?$|^(11+?)\1+$", convert)
print re_prime(2)
Also on a side note the quickest method I have come across as yet is
the following
def prime_numbers(n):
if n < 3: return [2] if n == 2 else []
nroot = int(n ** 0.5) + 1
sieve = range(n + 1)
sieve[1] = 0
for i in xrange(2, nroot):
if sieve[i]:
sieve[i * i: n + 1: i] = [0] * ((n / i - i) + 1)
return [x for x in sieve if x]
Damn clever whoever built this (note sieve will produce a list the
size of your 'n' which is unfortunate)
That needs to be either
return re.match(r"^1?$|^(11+?)\1+$", convert)
or
return re.match("^1?$|^(11+?)\\1+$", convert)
in order to prevent "\1" from being read as "\x01".
--
Carsten Haese
http://informixdb.sourceforge.net
return re.match(r"^1?$|^(11+?)\1+$", convert)
but it match the non-prime numbers. So re_prime(2) will return null
and re_prime(4) will return a match
2008/2/13, Carsten Haese <car...@uniqsys.com>:
> On Wed, 2008-02-13 at 07:31 -0800, cokof...@gmail.com wrote:
> > return re.match("^1?$|^(11+?)\1+$", convert)
>
> That needs to be either
>
> return re.match(r"^1?$|^(11+?)\1+$", convert)
>
> or
>
> return re.match("^1?$|^(11+?)\\1+$", convert)
>
> in order to prevent "\1" from being read as "\x01".
>
> --
> Carsten Haese
> http://informixdb.sourceforge.net
>
>
> --
> http://mail.python.org/mailman/listinfo/python-list
>
--
Rafael Sachetto Oliveira
Sir - Simple Image Resizer
http://rsachetto.googlepages.com
But why doesn't it work when you make that change?
>
> --
> Carsten Haesehttp://informixdb.sourceforge.net
It does work. Read the referenced website.
If there is a match then
the number isn't prime
else # no match
the number is prime.
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I can't answer that question, because it *does* work when you make that
change.
Well, the OP said the function was returning None which meant
no match which implies None means composite for the given example 2.
If None was supposed to mean prime, then why would returing None
for 2 be a problem?
But isn't this kind of silly?
## 3 None
## 4 <_sre.SRE_Match object at 0x011761A0>
## 5 None
## 6 <_sre.SRE_Match object at 0x011761A0>
## 7 None
## 8 <_sre.SRE_Match object at 0x011761A0>
## 9 <_sre.SRE_Match object at 0x011761A0>
## 10 <_sre.SRE_Match object at 0x011761A0>
## 11 None
## 12 <_sre.SRE_Match object at 0x011761A0>
## 13 None
## 14 <_sre.SRE_Match object at 0x011761A0>
## 15 <_sre.SRE_Match object at 0x011761A0>
## 16 <_sre.SRE_Match object at 0x011761A0>
## 17 None
## 18 <_sre.SRE_Match object at 0x011761A0>
## 19 None
>
> --
> Carsten Haesehttp://informixdb.sourceforge.net
here is another way you can find prime numbers
http://love-python.blogspot.com/2008/02/find-prime-number-upto-100-nums-range2.html
re.match(r"^(oo+?)\1+$", 'o'*i ) drops i in [0,1]
and
re.match(r"^(ooo+?)\1+$", 'o'*i ), which only drops i in [0,1,4].
Isn't the finite state machine "regular expression 'object'" really
large?
There's no finite state machine involved here, since this isn't a
regular expression in the strictest sense of the term---it doesn't
translate to a finite state machine, since backreferences are
involved.
Mark
What is it?
Sadly that is pretty slow though...
If you don't mind readability you can make the example I gave into
five lines.
def p(_):
if _<3:return[2]if _==2 else[]
a,b,b[1]=int(_**0.5)+1,range(_+1),0
for c in xrange(2,a):
if b[c]:b[c*c:_+1:c]=[0]*((_/c-c)+1)
return[_ for _ in b if _]
But then, I would have to kill you...
It's quadratic, and it's not even short, you can do (quadratic still):
print [x for x in range(2, 100) if all(x%i for i in range(2, x))]
In D you can write similar code.
Bye,
bearophile
all(x%i ha