Question

Help on p1?

Answer 1

So I have this code so far...


count=1
number=3
divisor=2

while(count<1000):
(while (divisor<number):
if (number%divisor==0):
print number, "is not a prime"
divisor=2
number=number+2
else:
divisor=divisor+1
if(divisor==number):
count=count+1
print number, "is a prime"
number=number+2
divisor=2
print number

From what I can tell (by the print statements in the if/else), it is correctly finding the prime numbers, but I can't get the count to work. i.e. the loop won't stop. Any suggestions?

Answer 2

Nevermind! Got it!

Answer 3

what was it ? share the code please !

Answer 4

I just inserted a break statement at the end of the inner while loop. I also had to change the output to print number-2 (the end of the loop increases number, even in the last loop). The new code looks like


#File: ps1_1.py
# To print the 1000th prime number
count=1
number=3
divisor=2

while(count<1000):
while (divisor<number):
# Number is not prime
if (number%divisor==0):
divisor=2
number=number+2
# Number is prime
else:
divisor=divisor+1
if(divisor==number):
count=count+1
number=number+2
divisor=2
break
print number-2


However, this seems to take awhile to calculate so I'm trying to see if there's a better way to write this

Answer 5

https://gist.github.com/944461

Try reading through this, it should help you out no end.

Answer 6

here's a good example:
#Determine the 1000th prime number
candidate=1
#Already know that 2 is prime
primeCount=1
while (primeCount<=1000):
isPrime="true"
i=2
halfCand=(candidate/2)
while (isPrime=="true") and (i<=halfCand):
if ((candidate%i)==0):
isPrime="false"
else:
i+=1
if isPrime=="true":
print(candidate, "is a prime.")
primeCount+=1
#else:
#print(candidate, " is not a prime.")
candidate+=2
print("The 1000th prime number is ",(candidate-2))

Answer 7

How about
= = = = = = = = = = =
def is_prime(number):
if number < 2:
return False
for each in range(2, (number+2) / 2):
if number % each == 0:
return False
return True

tally, n = 0, 0
while tally < 1000:
n += 1
if is_prime(n):
tally += 1
print n
= = = = = = = = = = = =

Answer 8

looking good, :D

Answer 9

would explain this when u have a chance:
for each in range(2, (number+2) / 2):
I see that "(number + 2) " is divided by 2, but what is that first 2 in the expression doing?

Answer 10

Ah, that ~ it just makes sure the value that range() will run up to is always just more than half of number. We could just do

for each in range(2, number):

but it's more efficient to do the range from 2, up to not-less-than-half-of-number (because, more than half of number, times by 2 (or more than 2) can never produce number, it'll always be too big). This way, we only try about half as many values for the variable `each` and so we get our result about twice as fast, but still get the right result.

It's a bit of a sketchy way to do it, but it works. Welcome to the world of 'quick hacks'. Well spotted though :)

Answer 11

I wrote this earlier to help explain it to some one else, it is the above, but with a ton of comments that walk you through it. If anyone fancies it

https://gist.github.com/944461

You'll have to copy and paste it, I think because it's secure http, but that's just the way github does their URLs.

Answer 12

well, i'll understand it better, by and by... :D