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Yubing_Wang
 2 years ago
Hello Guys,
I'm a very beginner in computer science.I have trouble solving this problemfind the 1000th prime. i have tried a lot of time,but can not write working code.
the following is one of the code i tried, still not working, can anyone help me with the finding prime problem? thanks
candidate=3
a=[]
prime=[]
for i in range(2,candidate):
b=candidate%i
c=a.append(b)
if 0 in c:
candidate+=2
else:
final_prime=prime.append(candidate)
candidate+=2
while len(final_prime)=999:
Final_prime=[2]+final_prime
print Final_prime
Yubing_Wang
 2 years ago
Hello Guys, I'm a very beginner in computer science.I have trouble solving this problemfind the 1000th prime. i have tried a lot of time,but can not write working code. the following is one of the code i tried, still not working, can anyone help me with the finding prime problem? thanks candidate=3 a=[] prime=[] for i in range(2,candidate): b=candidate%i c=a.append(b) if 0 in c: candidate+=2 else: final_prime=prime.append(candidate) candidate+=2 while len(final_prime)=999: Final_prime=[2]+final_prime print Final_prime

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JonnyTruelove
 2 years ago
Best ResponseYou've already chosen the best response.0Im just as lost as you bro. But even tho its wrong your code has inspired me to think in a different direction so thank you. I think somewhere you need to multiply n**0.5 + 1. I dunno where yet lol but maybe it will help you.

Yubing_Wang
 2 years ago
Best ResponseYou've already chosen the best response.0i worked out the code about the prime number, as following: candidate=3 final_prime=[2] while len(final_prime)<1000: result=True for i in range(2,candidate): if(candidate%i==0): result=False if result==True: prime=candidate final_prime.append(prime) candidate+=2 print final_prime,final_prime[999] even though it's not perfectly efficient, it worked. if you have more effient way to solve this problem,please sent me the code, thanks.

slotema
 2 years ago
Best ResponseYou've already chosen the best response.1Nice job getting your code working :) One thing you could try to get a bit more efficient code is by breaking (using the `break` keyword) from the loop after you set result to False. That way, the number that are higher won't be checked. Antoher thing you can have a look at is the Sieve of Eratosthenes (see: http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes ) and try to implement that in python.

Yubing_Wang
 2 years ago
Best ResponseYou've already chosen the best response.0yes,you are right, thank you

bwCA
 2 years ago
Best ResponseYou've already chosen the best response.0here is a prime test that is actually pretty fast: http://dpaste.com/1004316/

rsmith6559
 2 years ago
Best ResponseYou've already chosen the best response.0One thing that everybody seems to forget is that any nonprime number can be factored into prime numbers. If you implement that fact, your program will run fast.

JonnyTruelove
 2 years ago
Best ResponseYou've already chosen the best response.0what does that even mean rsmith? any non prime number can be factored into prime numbers? how will that make the program run fast?

bwCA
 2 years ago
Best ResponseYou've already chosen the best response.0don't you still have to perform primality tests to generate prime numbers? those tests consume the time. http://dpaste.com/1010237/

bwCA
 2 years ago
Best ResponseYou've already chosen the best response.0@rsmith6559  so you are talking about an algorithm that uses previously found primes in the primality tests. do you have an example?

rsmith6559
 2 years ago
Best ResponseYou've already chosen the best response.0primes = [ 2, 3 ] testNum = 5 numberOfPrimes = 10000 while( len( primes ) < numberOfPrimes ): for i in xrange( 0, len( primes ) ): if( testNum % primes[i] ): if( i == ( len( primes )  1 )): primes.append( testNum ) else: break testNum += 2

bwCA
 2 years ago
Best ResponseYou've already chosen the best response.0@rsmith6559  the algorithm you posted takes 70 times longer to find the 10,000th prime than one using the test i posted. http://dpaste.com/1011884/

rsmith6559
 2 years ago
Best ResponseYou've already chosen the best response.0I tuned the code a bit: primes = [ 2, 3 ] testNum = 5 numberOfPrimes = 10000 while( len( primes ) < numberOfPrimes ): for i in xrange( 0, len( primes ) ): if( testNum % primes[i] ): if( primes[i] * primes[i] > testNum ): primes.append( testNum ) break else: break testNum += 2 I'm getting 0.67 second execution time.

bwCA
 2 years ago
Best ResponseYou've already chosen the best response.0still 20% slower http://dpaste.com/1012984/

rsmith6559
 2 years ago
Best ResponseYou've already chosen the best response.0I don't see how you can be getting 20 second times: [zeus@stamRSMITH:~/Desktop]$ time ./ps1.py The 10000th prime is: 104729 real 0m0.688s user 0m0.662s sys 0m0.017s [zeus@stamRSMITH:~/Desktop]$ time ./ps1.py The 10000th prime is: 104729 real 0m0.674s user 0m0.656s sys 0m0.016s [zeus@stamRSMITH:~/Desktop]$ time ./ps1.py The 10000th prime is: 104729 real 0m0.676s user 0m0.657s sys 0m0.016s [zeus@stamRSMITH:~/Desktop]$ time ./ps1.py The 10000th prime is: 104729 real 0m0.678s user 0m0.657s sys 0m0.016s [zeus@stamRSMITH:~/Desktop]$ time ./ps1.py The 10000th prime is: 104729 real 0m0.679s user 0m0.659s sys 0m0.017s [zeus@stamRSMITH:~/Desktop]$ time ./ps1.py The 10000th prime is: 104729 real 0m0.674s user 0m0.656s sys 0m0.016s

bwCA
 2 years ago
Best ResponseYou've already chosen the best response.0http://docs.python.org/2.7/library/timeit.html those times are for 100 executions  finding the 10000th prime 100 times on my machine your algorithm takes 25. seconds for 100 executions  which is still 20% slower than the algorithm from the wikipedia article. i made a couple of readability changes to your code at lines 7, 8, 9  http://dpaste.com/1013998/ now yours is 25% faster  cool

rsmith6559
 2 years ago
Best ResponseYou've already chosen the best response.0I'm not saying that my code is the fastest possible. Originally, I was just commenting on a mathmatical fact that nobody was taking advantage of. My purpose for mentioning it was that the brute force algorithms, with thought, could be replaced with more efficient algorithms. Certainly a lesson to be taken from ps1 (Problem Set 1).
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