state =0 candidate=1 n=1 while (state <10): if candidate%2!=0: while(n<(candidate**(1/2)+1)): count =0 if candidate%n==0: count+=1 if count<2==True: candidate +=1 state+=1 else: candidate+=1 print 'iteration count',state print 'prime number at this position',candidate
1000th prime number generator problem!
it'se a beginner's prog, no fancy keywords or functions used
22 Replies  31239 Views  Last Post: 22 October 2012  08:27 PM
#1
1000th prime number generator problem!
Posted 04 June 2010  06:02 AM
Replies To: 1000th prime number generator problem!
#2 Guest_konstanz*
Re: 1000th prime number generator problem!
Posted 04 June 2010  12:15 PM
x=input('enter number=') pcntr=0 i=0 for i in range(1,x): for num in range(1,i): if i%num==0: pcntr+=1 if pcntr<=1: num=i print 'counter',pcntr print x,'is a prime number'
#3
Re: 1000th prime number generator problem!
Posted 05 June 2010  06:11 PM
But first, I have to ask that this is python right?
I'm also not an expert when it comes to python so feel free to question all my inputs.
In anycase you guys should stop and go back to the beginning. Your trying to find prime numbers. So start there and find out what makes certain numbers prime. Start with what you know and move on from there. Step by step, do some pseudocode if that helps you.
I guess I can start you off. You can choose to ignore 1 since 1 isn't prime. And we know that 2 is prime so we can incorporate that somehow.
The way I did mine was by making a prime function that focuses on analyzing 1 number. Then I would ask the user what number or how many numbers, range of numbers etc. And call the function from there.
Hope that helps
#4
Re: 1000th prime number generator problem!
Posted 05 June 2010  06:17 PM
I've got a Java snippet to do this here: http://www.dreaminco...snippet4601.htm
Same concept, but a different language.
#5
Re: 1000th prime number generator problem!
Posted 05 June 2010  06:46 PM
Make a function called isPrime that simply returns true or false. If you write it correctly, then the following code will work.
def nthPrime(n): count = 0 value = 1 while count<n: # repeat until we get up to n value += 1 # we start at 2 if isPrime(value): count += 1 return value
Hope this helps.
#6 Guest_Anonymous*
Re: 1000th prime number generator problem!
Posted 10 June 2010  04:41 PM
import math testnum = 3 maxprime = input("nth prime number:") if maxprime >= 1: print 2 primecount = 1 while primecount < maxprime: for num in range(2,int(math.sqrt(testnum)+2)): if testnum%num ==0: break if num == int(math.sqrt(testnum)+1): print testnum primecount = primecount + 1 testnum = testnum + 2
I did it like this, and it looks similar to yours, and I think mine works. However, it would be nice if somebody could tell me how to clean it up some more but still use the same method.
This post has been edited by atraub: 10 May 2012  10:07 AM
Reason for edit:: no code tags here either.
#7 Guest_Anonymous*
Re: 1000th prime number generator problem!
Posted 10 June 2010  04:43 PM
#8
Re: 1000th prime number generator problem!
Posted 12 June 2010  09:01 AM
Anonymous, on 10 June 2010  03:41 PM, said:
testnum = 3
maxprime = input("nth prime number:")
if maxprime >= 1:
print 2
primecount = 1
while primecount < maxprime:
for num in range(2,int(math.sqrt(testnum)+2)):
if testnum%num ==0:
break
if num == int(math.sqrt(testnum)+1):
print testnum
primecount = primecount + 1
testnum = testnum + 2
I did it like this, and it looks similar to yours, and I think mine works. However, it would be nice if somebody could tell me how to clean it up some more but still use the same method.
hey man this doesnt run above n=10
#9
Re: 1000th prime number generator problem!
Posted 12 June 2010  11:33 AM
I see that you tried to use a method of finding multiples between 2 and sqrt(n).
But like I said before your making things way to complicated and confusing. The way baavgai explained before is probably the easiest and simplest way to solve finding primes.
Quote
def nthPrime(n): count = 0 value = 1 while count<n: # repeat until we get up to n value += 1 # we start at 2 if isPrime(value): count += 1 return value
Primes are numbers divisible by 1 and itself. In other words, if a number is divisible by anything besides 1 and itself, it wouldn't be prime. But trying to find if a number is divisible by 2, 3, 4, 5 ,6 etc... would be annoying wouldn't it? Maybe theres a simpler way to know?
#10
Re: 1000th prime number generator problem!
Posted 14 June 2010  05:29 AM
cnampheonix, on 12 June 2010  10:33 AM, said:
I see that you tried to use a method of finding multiples between 2 and sqrt(n).
But like I said before your making things way to complicated and confusing. The way baavgai explained before is probably the easiest and simplest way to solve finding primes.
Quote
def nthPrime(n): count = 0 value = 1 while count<n: # repeat until we get up to n value += 1 # we start at 2 if isPrime(value): count += 1 return value
Primes are numbers divisible by 1 and itself. In other words, if a number is divisible by anything besides 1 and itself, it wouldn't be prime. But trying to find if a number is divisible by 2, 3, 4, 5 ,6 etc... would be annoying wouldn't it? Maybe theres a simpler way to know?
but thats the problem. i apparently cant calculate primes! my 'prime number check' doesnt work like its supose to! actually it just checks for the odd numbers which is very odd! i kind of tries to exclude all the even numbers as they are not primes! n then tried that 3 to sqrt(n) range check method! its cuts the checking to less than half other then that i cant think of anyother way of checking
ramu, on 14 June 2010  01:20 AM, said:
ramu, on 14 June 2010  01:11 AM, said:
konstanz, on 04 June 2010  05:02 AM, said:
state =0 candidate=1 n=1 while (state <10): if candidate%2!=0: while(n<(candidate**(1/2)+1)): count =0 if candidate%n==0: count+=1 if count<2==True: candidate +=1 state+=1 else: candidate+=1 print 'iteration count',state print 'prime number at this position',candidate
ummmm is this suppose to mean anything?
#11
Re: 1000th prime number generator problem!
Posted 14 June 2010  09:52 AM
We'll use the process baavgai pointed out earlier.
We start off with working on finding if a number is prime. We know that the number 1 cannot be prime so in our isPrime(value): we will first state that conditional:
if value == 1: return False
Next we know that number 2 is prime and being that it is a even number can account for all other even numbers:
if value == 2: return True if value % 2 == 0: return False
This will check for all other even numbers and return that they are not primes.
To check for all other numbers
we'll use an "else" so all other numbers are funneled here. And we should probably start with 3 since we already checked for 1 and 2. So starting with 3 we'll see if the number given lets say for example number 11 is prime. Well we could first start off with checking if the number % 3 == 0. That would allow us to know if there are any numbers divisible by 3. And if there are then we can say False. But what about all the other numbers in between 3 and 11. We could increment 3 and keep modding so we know that no number between 3 and 11 is divisible by 11. After finding that all checks did not return False. Then we can finally say that 11 is True and thus Prime.
Knowing this we would need:
To initialize 3.
Probably have a while loop.
also increment the variable with 3
Out of the while loop have a return True.
This is just the way I did mine for isPrime. After your able to get the function you can mess around and call it in another function and print, or do what baavgai did. Whatever you like. Hope this helps
This post has been edited by cnampheonix: 14 June 2010  09:55 AM
#12
Re: 1000th prime number generator problem!
Posted 07 May 2011  01:48 PM
number = input('enter prime you wish to find') x = 5 y = 3 while (y<(number+1)): if ((x/2)*2)<x: z = x1 while(x%z>0): z = z1 if z == 1: y = y+1 x = x+1 else: x = x+1 else: x = x+1 x = x1 print x
It's slightly different from the one you need in that you input any number (n) and it gives you the nth prime, but it can be changed to only find the 1000th prime with a few small modifications.
This post has been edited by RussianScience: 07 May 2011  01:50 PM
#13
Re: 1000th prime number generator problem!
Posted 24 May 2011  02:36 PM
for n in range(4, 20): for x in range(2, n): if n % x == 0: break else: # loop fell through without finding a factor print n, 'is a prime number'
you can change the range if you want more or less primes
hope it helps!
This post has been edited by atraub: 13 July 2011  07:29 PM
Reason for edit:: added code tags
#14
Re: 1000th prime number generator problem!
Posted 13 July 2011  06:13 AM
It seems most people miss failing primes after the 10th or 20th due to not checking for % of 3, 7, 11 etc...
Also, not sure how to make the code appear indented here, you will have to figure out indents.
#ensure number entered is int and > 1(this is copied from someone elses code to ensure number is int, sorry cannot #remember who it was though) isnum = False while isnum == False: tarnum = raw_input('Enter which nth prime you would like to find: ') try : tarnum = int(tarnum) if tarnum < 1: isnum = False print 'You entered a negative number, please enter which prime to calculate' else: isnum = True except: isnum = False print 'This is not a number, please enter which prime to calculate' #Set up variables lisprime = [2] testpass = True curprime = 3 listindex = 1 testindex = 1 rootprime = 0 lisprime.append(curprime) #loop to get nth prime while listindex <= tarnum: rootprime = curprime**0.5 #check if divisible by any primes less than the root while lisprime[testindex]<=rootprime: if curprime%lisprime[testindex] > 0: testindex += 1 testpass = True else: testpass = False break #If it is a prime, add it to lisprime if testpass == True: lisprime[listindex]=curprime lisprime.append(curprime) listindex += 1 #only test odd numbers(inc by 2) curprime += 2 testindex = 1 print 'This is the requested prime number:', lisprime[listindex2]
This post has been edited by macosxnerd101: 13 July 2011  06:15 AM
Reason for edit:: Fixed code tags
#15
Re: 1000th prime number generator problem!
Posted 17 July 2011  01:50 PM
def prime(x): if x == 1: return 0 else: limit = x/2 div = 2 check = 0 while div <= limit: if x % div == 0: check = 1 div = div + 1 if check == 0: return 1 else: return 0 num = 1 count = 0 while count < 999: if prime(num) == 1: count = count + 1 num = num + 2 num = num  2 print(num)
