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- anonymous

Help on p1?

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- anonymous

Help on p1?

- jamiebookeater

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- anonymous

So I have this code so far...
count=1
number=3
divisor=2
while(count<1000):
(while (divisor

- anonymous

Nevermind! Got it!

- anonymous

what was it ? share the code please !

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- anonymous

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

- carlsmith

https://gist.github.com/944461
Try reading through this, it should help you out no end.

- anonymous

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))

- carlsmith

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
= = = = = = = = = = = =

- anonymous

looking good, :D

- anonymous

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?

- carlsmith

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 :)

- carlsmith

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.

- anonymous

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

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