Question

Fastest Algorithm to calculate 10000 primes in python ?is it

Answer 1

import time
st=time.time()
mdu=0
check =3
noofprime=2
primeLs=[2,3]
while noofprime!=10000:
check=check+2
a=len(primeLs)
b=primeLs[a-1]
for div in primeLs:
#print check
mdu=check%div
if mdu==0:
break
if div==b:
noofprime= noofprime+1
primeLs=primeLs+[check]



et=time.time()
#print primeLs
print st
print et
print et-st
print primeLs[9999]

Answer 2

time python test.py
1296951327.14
1296951364.78
37.6408441067
104729
real 0m37.667s
user 0m37.242s
sys 0m0.321s

Answer 3

is it fastest in python as i could not find fastest time in google but found this on c++ http://www.troubleshooters.com/codecorn/primenumbers/primenumbers.htm
below is the code avoid printing list it will take too much time but to verify i have put it there but commented and printed the last value ( 9999th value) verified from http://primes.utm.edu/lists/small/10000.txt

Answer 4

Mine isn't bad
http://www.youtube.com/watch?v=p6fHllFsy7k

Answer 5

But it would have to be recoded a bit to stop after 1000 primes

Answer 6

time ./primgen.py 7179 # I said explicitly to stop at 7919 since it doesn't count primes
real 0m1.302s
user 0m0.356s
sys 0m0.008s
Not bad I suppose

Answer 7

Small point. You don't need to check for divisibility by 2 since you are generating only odd numbers to test.

Big point. You can stop testing divisors when they exceed sqrt(x). At 7919, this is the difference between 999 test divisions (up to 7907) and 24 divisions (up to 83).

You can also get a small speedup by generalizing the method of generating only odd numbers (not div by 2) to generating numbers not divisible by 2 or 3.
Odd numbers not divisible by 3:
5,7, 11,13, 17,19, etc
So instead of always counting by 2s, alternate counting by 2 and 4. This eliminates 1/3 of the numbers you are testing (and you could carry this on to 5,7, etc if it was worth doing, but diminishing returns hit fast, especially since the loops you skip would have been quick ones that failed on the first division tests).

Answer 8

On my computer, I got bored waiting for your program to finish generating 10k primes, but 5k primes took about 11 sec.

Changing the loop to recognize numbers are prime when div>sqrt(x) took just 0.5 sec for 5000 primes, and about 1.4 sec for 10k primes.

Answer 9

@zifmia thanks for the tips i made changes as you suggested and it works, if you please verify the time as well posting the code below , will try for dict method as well and by the way do you think this code compromises space over time

Answer 10

import time
from math import sqrt
st=time.time()
mdu=0
check =3
noofprime=2
primeLs=[2,3]#,5,7,11,13]
prime=True
while noofprime!=10000:
check=check+2
if check%3==0:check=check+2
sqrcheck=int (sqrt(check))+1
a=len(primeLs)
b=primeLs[a-1]
for div in range (3,sqrcheck):
prime=True
mdu=check%div
if mdu==0 :
prime=False
break

if prime==True:
noofprime= noofprime+1
primeLs=primeLs+[check]

et=time.time()
print primeLs[noofprime-1]
print st
print et
print (et-st)/60

Answer 11

i get this time when checked without writing to list
prime 104729
st:1298174950.11
et:1298174952.94
0.0471333344777