My backwards logic
ckaynor at zindagigames.com
Sat Sep 6 00:14:41 CEST 2014
On Fri, Sep 5, 2014 at 2:49 PM, Seymore4Head <Seymore4Head at hotmail.invalid>
> On Fri, 05 Sep 2014 12:48:56 -0400, Seymore4Head
> <Seymore4Head at Hotmail.invalid> wrote:
> >I'm still doing practice problems. I haven't heard from the library
> >on any of the books I have requested.
> >This is not a hard problem, but it got me to thinking a little. A
> >prime number will divide by one and itself. When setting up this
> >loop, if I start at 2 instead of 1, that automatically excludes one of
> >the factors. Then, by default, Python goes "to" the chosen count and
> >not "through" the count, so just the syntax causes Python to rule out
> >the other factor (the number itself).
> >So this works:
> >while True:
> > a=random.randrange(1,8)
> > print (a)
> > for x in range(2,a):
> > if a%x==0:
> > print ("Number is not prime")
> > break
> > wait = input (" "*40 + "Wait")
> >But, what this instructions want printed is "This is a prime number"
> >So how to I use this code logic NOT print (not prime) and have the
> >logic print "This number is prime"
> I am sure this has already been done, but after it was pointed out
> that you don't need to test for any number that multiplies by 2 it
> made me think again.
> If you start with the list [3,5,7] and step through the list of all
> remaining odd numbers (step 2), and start appending numbers that won't
> divide by numbers already appended in the list, that would seem like a
> pretty efficient way to find all prime numbers.
To be correct, you only need to check for n being divisible by primes less
than sqrt(n). For example, the following code will produce a list of primes
from 2 to 1000 (the result will be in the "primes" list):
primes = 
for i in range(3, 1000):
end = math.sqrt(i)
for x in primes:
if x > end: # Once x is larger than the sqrt(i), we know it must be
prime, so we can early exit.
#print(i, "is a prime number")
if (i % x) == 0:
#print(i, "is a composite number")
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