Is_prime and range(2,x-1)


#1



is_prime

I am wondering about the logic in the more common solution for this problem using for n in range(2,(x-1)):
Am I correct in thinking that the logic in this is to save from the other if statements of catching when x is ==2? And that it does not check 1 number away from the max of input through then on out. Such as if the input is 10 it will not check against 10%9 but it doesn't need to because we know that after 2 a number ,say y%(y-1), will never equal 0?


def is_prime_again(x):
    if x>2:
        for n in range(2,x):
            print ""
            print x
            print n
            print x%n
            print range(2,x)
            if x%n ==0:
                print "not prime"
                return False
                break
        return True
    elif x==2:
        print "2 is prime"
        return True
    else:
        print "1 and less are not prime"
        return False

print is_prime_again(10)


#2

don't be fooled by the forum (even i made that mistake :stuck_out_tongue: ) this are the instructions:

For each number n from 2 to x - 1, test if x is evenly divisible by n

however, if you use range, the stop (x) is not included so range(2,x) is what you need to use, but many people get this wrong.

if you use range(2,x) you don't need a if statement to check if x equals two, you only need one for the lower numbers (smaller then 2)


#3

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