Regular Expression for Prime Numbers (or How I came to fail at them,and love the bomb)

C

cokofreedom

I was reading up on this site [http://www.noulakaz.net/weblog/
2007/03/18/a-regular-expression-to-check-for-prime-numbers/] of an
interesting way to work out prime numbers using Regular Expression.

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:
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)
 
C

Carsten Haese

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".
 
R

Rafael Sachetto

with this works:

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
 
M

mensanator

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".

But why doesn't it work when you make that change?
 
R

Reedick, Andrew

-----Original Message-----
From: [email protected] [mailto:python-
[email protected]] On Behalf Of (e-mail address removed)
Sent: Wednesday, February 13, 2008 1:41 PM
To: (e-mail address removed)
Subject: Re: Regular Expression for Prime Numbers (or How I came to
fail at them, and love the bomb)

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".

But why doesn't it work when you make that change?


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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M

mensanator

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
 
C

castironpi

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".

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?
 
M

Mark Dickinson

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
 
C

castironpi

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?
 
C

cokofreedom

hmm... interesting
here is another way you can find prime numbershttp://love-python.blogspot.com/2008/02/find-prime-number-upto-100-nu...

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...
 
B

bearophileHUGS

cokofree:
Sadly that is pretty slow though...

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
 
C

castironpi

cokofree:
Sadly that is pretty slow though...

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
 

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