On Fri, Sep 5, 2014 at 2:49 PM, Seymore4Head <Seymore4Head@hotmail.invalid> wrote:
> On Fri, 05 Sep 2014 12:48:56 -0400, Seymore4Head > <Seymore4Head@Hotmail.invalid> wrote: > > >I'm still doing practice problems. I haven't heard from the library > >on any of the books I have requested. > > > > > http://www.practicepython.org/exercise/2014/04/16/11-check-primality-functions.html > > > >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): import math primes = [2] 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") primes.append(i) break if (i % x) == 0: #print(i, "is a composite number") break
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