# [Edu-sig] Hello from a CS Teacher

**Kirby Urner
**
pdx4d@teleport.com

*Fri, 11 Feb 2000 17:31:41 -0800*

At 07:07 PM 02/11/2000 -0600, Dustin James Mitchell wrote:
>*On Fri, 11 Feb 2000, Kirby Urner wrote:
*>*
*>>* while len(primes)<nbprimes:
*>>* if addprime(i): primes.append(i)
*>>* # skip even numbers (only 2 is prime)
*>>* if i==2: i=i+1
*>>* else: i=i+1
*>*
*>*Shouldn't the latter be i=i+2, to go to the next odd number?
*>*
*
Dustin --
Yes, exactly, good eye! The algorithm still worked, but
less efficiently.
Below is a variant which computes all prime numbers _up to
a maximum_ i.e. get2max().
With this one in place, I'm able to do getfactors(), which
lists all the prime factors of a number.
I also threw in isprime(), which answers whether any number
is a prime number (1 means yes, 0 means no).
Action:
>>>* seive.getfactors(10981)
*[79, 139]
>>>* seive.isprime(99901)
*1
>>>* seive.getfactors(30012)
*[2, 2, 3, 41, 61]
>>>* 2*2*3*41*61
*30012
I'm not claiming this is the most efficient or super-elegant
implementation of getfactor(). If others have pointers as
to how to improve these examples (to put Python in a good
light), I'm happy to hear them. But I'm not looking for a
lot of error checking on the passed parameters. Yes you
can crash these algorithms if you pass out-of-context
values, but no, I'm not interested in cluttering up the
code based on the "naive user from hell" model.
Kirby
primes = [2] # dynamic list of primes
def get2max(maxnb):
# return list of primes
# maxnb = number not to exceed on list
global primes # keep at module level
while primes[-1]>maxnb:
primes = primes[:-1]
i=primes[-1]
while i<=maxnb:
if addprime(i): primes.append(i)
if i==2: i=i+1
else: i=i+2
return primes
def isprime(n):
global primes # keep at module level
retval = 0
primes = get2max(n)
if n in primes: retval = 1
return retval
def getfactors(n):
global primes # keep at module level
# return list containing prime factors of a number
primefactors = []
primes = get2max(n)
while not (n in primes):
for i in primes:
if n%i == 0: # no remainder
primefactors.append(i)
n = n/i
break
primefactors.append(n)
return primefactors